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[ "# Which is the largest? Algebra Level pending $\\large \\frac{ 2^2}{1} , \\ \\frac{ 3^2 }{ 2}, \\ \\frac{ 4^2 } { 3}, \\ \\frac{ 5^2 } { 4}$ Which of the above is the largest? ×", "# If the sum of three consecutive odd integers is p, then in terms of p, what is the greatest of the three integers? Mar 9, 2016 $x = \\frac{p - 6}{3}$ #### Explanation: If the largest integer is $x$ then the other two integers are $x - 2$ and $x - 4$. For example if the largest integer was $11$ then the other two numbers would be $9$ and $7$. $\\left(x - 4\\right) + \\left(x - 2\\right) + x = p$ $3 x - 6 = p$ $x = \\frac{p + 6}{3}$", "## Problem of the Day #388: Find the Biggest PrimeApril 10, 2012 Posted by Saketh in : potd , trackback For certain values of $a$ and $b$, the expression $$\\frac{b}{4}\\sqrt{\\frac{2a-b}{2a+b}}$$ produces an integer. If we consider the set of all such integers, what is the largest prime member?", "# The data items in a list are 75,86,87,91, and 93. What is the largest integer you can add to the list so that the mean of six items is less than their median? Oct 1, 2016 Largest integer is $101$ #### Explanation: There are 5 numbers in the list, but a sixth one is to be added. (as large as possible) $75 \\text{ \"86\" \"87\" \"91\" \"93\" } x$ $\\textcolor{w h i t e}{\\times \\times \\times \\times \\times} \\uparrow$ The median will be $\\frac{87 + 91}{2} = 89$ Mean will be: $\\frac{75 + 86 + 87 + 91 + 93 + x}{6} < 89$ $432 + x < 6 \\times 89$ $x < 534 - 432$ $x < 102$ The largest integer can be 101. Check; If $x = 101$ Mean $= \\frac{533}{6} = 88.83$ $88.83 < 89$", "\\frac{1}{100}$$ With numerators of both sides same, for LHS > RHS, denominator of LHS should be less than that of RHS $$2^6 = 64$$ $$2^7 = 128$$ _________________ Kindly press \"+1 Kudos\" to appreciate Intern Joined: 22 Feb 2014 Posts: 27 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Jul 2014, 10:40 Bunuel wrote: SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 [m]\\frac{1}{2^n}> 0.01[/m]; $$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. Hi I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice Thanks Komal Intern Joined: 22 Feb 2014 Posts: 27 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Jul 2014, 10:40 Bunuel wrote: SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 [m]\\frac{1}{2^n}> 0.01[/m]; [color=#ff0000]$$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; [/color] $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. Hi I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know", "i may be missing some basic concept here..Please advice Thanks Komal Math Expert Joined: 02 Sep 2009 Posts: 50627 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Jul 2014, 11:01 GGMAT760 wrote: Bunuel wrote: SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 [m]\\frac{1}{2^n}> 0.01[/m]; $$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. Hi I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice Thanks Komal Cross-multiply $$\\frac{1}{2^n}> \\frac{1}{100}$$ to get $$2^n<100$$. _________________ Manager Joined: 29 May 2013 Posts: 67 GMAT 1: 710 Q49 V38 GPA: 4 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Sep 2014, 12:56 Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND Edition What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 Problem Solving Question: 86 Category: Arithmetic Exponents; Operations with rational numbers Page: 73 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative", "$$\\big[ x\\big]$$ is the largest integer $$y$$ that $$y\\leq x$$ Hence, the maximum possible value of $$x$$ is $$\\Big[ \\frac{14}{3}\\Big]+\\Big[ \\frac{14}{3^2}\\Big]=4+1=5$$ _________________ Manager Joined: 08 Oct 2016 Posts: 205 Location: United States Concentration: General Management, Finance GPA: 2.9 WE: Engineering (Telecommunications) Re: If 14!/3^x is an integer, what is the maximum possible value of x? [#permalink] ### Show Tags 24 Mar 2017, 07:59 nguyendinhtuong wrote: Bunuel wrote: If 14!/3^x is an integer, what is the maximum possible value of x? A. 2 B. 4 C. 5 D. 6 E. 9 The highest power of a prime $$p$$ dividing $$n!$$ is given by: $$\\Big[ \\frac{n}{p}\\Big]+\\bigg[ \\frac{n}{p^2}\\bigg]+\\bigg[ \\frac{n}{p^3}\\bigg]+...$$ with $$\\big[ x\\big]$$ is the largest integer $$y$$ that $$y\\leq x$$ Hence, the maximum possible value of $$x$$ is $$\\Big[ \\frac{14}{3}\\Big]+\\Big[ \\frac{14}{3^2}\\Big]=4+1=5$$ Hi Thanks for explanation _________________ Got Q42,V17 Target#01 Q45,V20--April End Intern Joined: 18 Feb 2017 Posts: 27 Location: Myanmar GMAT 1: 700 Q46 V40 GPA: 3.5 WE: Law (Law) Re: If 14!/3^x is an integer, what is the maximum possible value of x? [#permalink] ### Show Tags 25 Mar 2017, 08:25 Count how many factors of 3 in 14!: 3 - 1 6 - 1 9 - 2 12 - 1 Re: If 14!/3^x is an integer, what is the maximum possible value of x? &nbs [#permalink] 25 Mar 2017,", "Technology (Consulting) What is the largest integer n such that 3^n < 0.001? [#permalink] ### Show Tags 26 Feb 2017, 01:05 6 Bunuel wrote: What is the largest integer n such that 3^n < 0.001? A. −13 B. −9 C. −7 D. 0 E. 3 Consider this pattern: $$3^-1 = \\frac{1}{3} = 0.3..$$ $$3^-2 = \\frac{1}{9} = 0.1..$$ $$3^-3 = \\frac{1}{27} = 0.03..$$ $$3^-4 = \\frac{1}{81} = 0.01..$$ With this pattern: $$3^-5 ~ 0.003$$ $$3^-6 ~ 0.001$$ $$3^-7 < 0.001$$ ##### General Discussion Retired Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1333 Location: Viet Nam Re: What is the largest integer n such that 3^n < 0.001? [#permalink] ### Show Tags 30 Nov 2016, 04:02 2 2 Bunuel wrote: What is the largest integer n such that 3^n < 0.001? A. −13 B. −9 C. −7 D. 0 E. 3 It's clearly that $$n<0$$ $$3^n<10^{-3} \\implies \\frac{1}{3^{|n|}}<\\frac{1}{10^3} \\implies 3^{|n|}>10^3 >9^3 = 3^6 \\implies |n| \\geq 7 \\implies n \\leq -7$$ _________________ Manager Joined: 02 Nov 2013 Posts: 71 Location: India Re: What is the largest integer n such that 3^n < 0.001? [#permalink] ### Show Tags 30 Nov 2016, 04:45 Multiplying both side by 1000. 3^n *1000 < 1. By looking at answer choice only we can eliminate two options i.e. 0 and 3.", "# Factorials! Let $$f(x) = 1! + 2! + \\ldots+ x!$$. What is the largest integer value of $$x$$ such that $$f(x)$$ is a perfect square? ×", "+0 # Logs 0 105 1 What is the largest integer less than log_2(2/1) + log_2(3/2) + ... + log_2(64/63)? May 9, 2022 #1 +9461 0 $$\\log_2 \\left(\\dfrac{2}1\\right) + \\log_2\\left(\\dfrac32\\right)+\\cdots + \\log_2 \\left(\\dfrac{64}{63}\\right) \\\\ = \\log_2 \\left(\\dfrac21 \\times \\dfrac32 \\times \\cdots \\times \\dfrac{64}{63}\\right)\\\\ = \\log_2 64\\\\ = 6$$ The answer is obvious at this point. May 9, 2022", "integer. The closest such fraction to $r$ is $\\frac 27.$ What is the number of possible values for $r$?", "+0 # Let $$f(x) = \\frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \\cdot f(3) \\cdot f(4) \\cdots f(n-1) \\cdot f(n) < 1.98.$ +1 945 1 +601 Let $$f(x) = \\frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \\cdot f(3) \\cdot f(4) \\cdots f(n-1) \\cdot f(n) < 1.98.$ Oct 5, 2017", "# Find the largest integer such that 8 more than six times the integer is less than -42.? Jul 1, 2017 The largest integer that satisfies this is $- 9$. #### Explanation: Translating this into an equation would be: \"Find the largest integer $x$ such that $6 x + 8 < - 42$.\" Simplifying the inequality: $6 x + 8 < - 42$ $6 x < - 50$ $x < - \\setminus \\frac{50}{6}$ $x < - 8.333$ So, the largest integer less than $- 8.333$ would be $- 9$.", "- 15 => 3x = 15 Divide each side by 3 => $\\frac{3x}{3} = \\frac{15}{3}$ => x = 5, first integer. Step 3: Largest or third integer = x + 4 = 5 + 4 = 9 Hence the largest integer is 9. answer", "# Math Help - [SOLVED] Sequence involving largest integer function. 1. ## [SOLVED] Sequence involving largest integer function. Let $(a_n)$be the sequence defined by $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? 2. Hello wizard654zzz Originally Posted by wizard654zzz Let $(a_n)$be the sequence defined $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? Divide and get quotient and remainder: $\\frac{n^2 + 8n +10}{n+9}=(n-1) + \\frac{19}{n+9}$ Now for $1\\le n\\le 10, \\;2>\\frac{19}{n+9}\\ge 1$. Hence $a_n = n$. But for $n > 10, \\;\\frac{19}{n+9}<1$. Hence $a_n = n-1$. $1+2+3+...+10+10+11+12+...+29 = 445$", "1. ## Arithmetic problem What is the largest positive integer n for which there is a unique integer k such that 8/15 < n/(n+K) < 7/13 $\\frac{8}{15} < \\frac{n}{n+K} < \\frac{7}{13}$ $\\frac{104}{195} < \\frac{n}{n+K} < \\frac{105}{195}$ $104 < \\frac{195n}{n+K} < 105$ 3. $\\frac{8}{15} < \\frac{n}{n+K} < \\frac{7}{13}$ Multiply by a common divisor of $15$ and $13$ so as to get everything in integers. $\\frac{104}{195} < \\frac{195n}{195(n+K)} < \\frac{105}{195}$ Multiply everything by $195$ so we get integers ... $104 < \\frac{195n}{n+K} < 105$ It might help ... EDIT : oh, no ! owned by Pickslides ! (nearly the same contents though)", "Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1373 Location: Viet Nam Re: If 14!/3^x is an integer, what is the maximum possible value of x? [#permalink] ### Show Tags 06 Dec 2016, 07:30 1 3 Bunuel wrote: If 14!/3^x is an integer, what is the maximum possible value of x? A. 2 B. 4 C. 5 D. 6 E. 9 The highest power of a prime $$p$$ dividing $$n!$$ is given by: $$\\Big[ \\frac{n}{p}\\Big]+\\bigg[ \\frac{n}{p^2}\\bigg]+\\bigg[ \\frac{n}{p^3}\\bigg]+...$$ with $$\\big[ x\\big]$$ is the largest integer $$y$$ that $$y\\leq x$$ Hence, the maximum possible value of $$x$$ is $$\\Big[ \\frac{14}{3}\\Big]+\\Big[ \\frac{14}{3^2}\\Big]=4+1=5$$ _________________ Manager Joined: 08 Oct 2016 Posts: 204 Location: United States Concentration: General Management, Finance GPA: 2.9 WE: Engineering (Telecommunications) Re: If 14!/3^x is an integer, what is the maximum possible value of x? [#permalink] ### Show Tags 24 Mar 2017, 06:59 nguyendinhtuong wrote: Bunuel wrote: If 14!/3^x is an integer, what is the maximum possible value of x? A. 2 B. 4 C. 5 D. 6 E. 9 The highest power of a prime $$p$$ dividing $$n!$$ is given by: $$\\Big[ \\frac{n}{p}\\Big]+\\bigg[ \\frac{n}{p^2}\\bigg]+\\bigg[ \\frac{n}{p^3}\\bigg]+...$$ with $$\\big[ x\\big]$$ is the largest integer $$y$$ that $$y\\leq x$$ Hence, the maximum possible value of $$x$$ is $$\\Big[ \\frac{14}{3}\\Big]+\\Big[ \\frac{14}{3^2}\\Big]=4+1=5$$ Hi Thanks for explanation _________________ Got Q42,V17 Target#01 Q45,V20--April", "Schools: LBS '17 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 11 Feb 2014, 13:24 1 Took me a while but i think it should be 1/2^n > 1/100 so we need to compare really the maximum value of 2^n such that it is less than 100, because if it is more than 100 then 1/2^n becomes less than 1/100 2^5 = 32 2^6 = 64 2^7 = 128 So n cannot be 7 and the largest value it can have while still keeping 1/2^n > 1/100 will be 6 B Math Expert Joined: 02 Sep 2009 Posts: 50627 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 15 Feb 2014, 05:51 1 SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 $$\\frac{1}{2^n}> 0.01$$; $$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. _________________ SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1826 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 26 Jun 2014, 01:24 1 $$\\frac{1}{2^n} >", "40. The largest integer is 3 [#permalink] ### Show Tags 29 Jan 2014, 02:28 1 a+b+c=40 (1) Let a<b<c Therefore,c=3b and a=c-23. Replacing, a and b in (1) we get c=27 a=4 and b=9 a*b*c=4*9*27=972 Ans. B Intern Joined: 06 Jan 2014 Posts: 35 Re: The sum of three integers is 40. The largest integer is 3 [#permalink] ### Show Tags 30 Jan 2014, 10:37 1 When you guys post questions from the Official Review...How do you know the difficulty level? The difficulty is crazy low, but its a 600 Level Question? and yah, I got 972 as well A+B+C=40 A+B+3B=40 3B-23+B+3B=40 7B=63 $$\\frac{7B}{7}$$ $$=$$ $$\\frac{63}{7}$$ $$B=9$$ The rest is history I represent my three numbers as three different variables because it shows how they can be non or interrelated when I started the problem, I had no idea about the relationship(s) of the numbers and never assume they are \"n\" related, besides being integers. Manager Joined: 04 Oct 2013 Posts: 150 Location: India GMAT Date: 05-23-2015 GPA: 3.45 Re: The sum of three integers is 40. The largest integer is 3 [#permalink] ### Show Tags 02 Feb 2014, 06:09 The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What", "we have most precise evaluation. Thank you! _________________ Math Expert Joined: 02 Sep 2009 Posts: 50627 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Feb 2014, 23:45 8 10 SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 $$\\frac{1}{2^n}> 0.01$$; $$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. _________________ ##### General Discussion Intern Joined: 30 Jan 2014 Posts: 17 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 11 Feb 2014, 00:38 1 1 1 0.01=1/100 So the question is $$2^n<100$$ We should know that: 2^5=32 2^6=64 2^7=128 so the ans is 6. B _________________ Please +1 KUDO if my post helps. Thank you. Manager Joined: 04 Jan 2014 Posts: 118 GMAT 1: 660 Q48 V32 GMAT 2: 630 Q48 V28 GMAT 3: 680 Q48 V35 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 11 Feb 2014, 00:45 2 1/(2^n) > 1/100 Since, 2^6 = 64, 1/64 > 1/100. Answer (B). It cannot be 7 because 2^7 = 128 and 1/128 < 1/100. Intern Joined: 20 Nov 2013 Posts: 25" ]

[ "4$ $\\implies$ $2 \\cdot \\left(- 10\\right) > 4 \\cdot \\left(- 10\\right)$ $2 < 4$ $\\implies$ $2 \\cdot \\left(- \\frac{1}{4}\\right) > 4 \\cdot \\left(- \\frac{1}{4}\\right)$ $- 8 < 4$ $\\implies$ $- 8 \\cdot \\left(- 10\\right) > 4 \\cdot \\left(- 10\\right)$ $- 8 < 4$ $\\implies$ $- 8 \\cdot \\left(- \\frac{1}{4}\\right) > 4 \\cdot \\left(- \\frac{1}{4}\\right)$ $- 8 < - 4$ $\\implies$ $- 8 \\cdot \\left(- 10\\right) > - 4 \\cdot \\left(- 10\\right)$ $- 8 < - 4$ $\\implies$ $- 8 \\cdot \\left(- \\frac{1}{4}\\right) > - 4 \\cdot \\left(- \\frac{1}{4}\\right)$ We conclude with the following rule 2 about transformations of inequalities: both sides of inequality can be multiplied by any negative real number, but a sign of inequality should be changed to opposite. In the problem at hand 90 < −6w we can multiply both sides of the inequality by a negative number $- \\frac{1}{6}$. As a result, the right side will be equal to $- 6 w \\cdot \\left(- \\frac{1}{6}\\right) = w$, while the left side will be equal to $90 \\cdot \\left(- \\frac{1}{6}\\right) = - 15$. Since we multiplied by a negative number $- \\frac{1}{6}$, the sign of inequality should be changed to opposite, that is the resulting inequality will be $- 15 > w$. Since it is customary to place the unknown variable ($w$ in this case)", "number by which you multiply or divide. Multiplication and Division of Positive Numbers Property of Inequality If you multiply both sides of an inequality by a positive number or you divide both sides by a positive number you can keep the inequality symbol. Let's have a look at the antelopes example: • $80<132$ • Now we multiply both sides by $\\frac78$. That's the same as multiplying by $7$ and dividing by $8$, both positive numbers. $\\begin{array}{rcl} \\frac78\\cdot 80&<&\\frac78\\cdot 132\\\\ \\frac{7\\cdot 10 \\cdot 8}{8}&<&\\frac{7\\cdot 33\\cdot 4}{2\\cdot 4}\\\\ 7\\cdot 10&<&115.5 \\end{array}$ Multiplication and Division of Negative Numbers Property of Inequality Make sure to pay attention to the sign. If you multiply both sides of an inequality by a negative number or divide both sides by a negative number, you have to change the inequality symbol. To see this in action, let's have a look at the following example: • $-4<6$ • Now we divide both sides by $-2$ and get $2$ ??? $-3$. • What inequality symbol should we use? • $2$ is greater than $-3$. This gives us $2>-3$. • You see the sign changed from $<$ to $>$. • #### Determine the corresponding inequality. ##### Hints You only have to flip the inequality symbol if you multiply or divide both sides by negative numbers. Pay attention to the", "# How to prove that $7^{31} > 8^{29}$ How can I prove that $7^{31}$ is bigger than $8^{29}$? I tried to write exponents as multiplication, $2\\cdot 15 + 1$, and $2\\cdot 14+1$, then to write this inequality as $7^{2\\cdot 15}\\cdot 7 > 8^{2\\cdot 14}\\cdot 8$. I also tried to write the right hand side as $\\frac{8^{31}}{8^2}$. • This inequality is equivalent to $31<29 \\log_7 8$ or $\\frac{31}{29}>\\log_7 8$, so if you can find the logarithm and then multiply it by $29$ then you've got it. How to find the logarithm numerically is a substantial question in its own right. ${}\\qquad{}$ – Michael Hardy Jul 7 '15 at 14:25 • This fails:$$\\left( 1 + \\frac{1}{7} \\right)^{7\\cdot 4 \\frac{1}{7}} \\leq e^{4 \\frac{1}{7}} \\geq 7^2$$ Any better ideas? – abnry Jul 7 '15 at 14:41 • I think maybe this problem is from my old problem when Christian Blatter solution a step math.stackexchange.com/questions/431461/… – math110 Jul 7 '15 at 16:42 • Maybe you could try to use the binomial theorem. Note that $7=8-1$ and $8=7+1$. – Surb Jul 7 '15 at 16:52 • I'd love to see a proof using the binomial theorem, but it seems that the two values are two close (relatively) for simple bounds to work. – lhf Jul 7 '15 at 16:52 The following (not particularly elegant)", "# How do you solve x/3=-10? Jun 14, 2018 $x = - 30$ #### Explanation: When dealing with equalities, you can multiply both sides by a same number, and the inequality will preserve its truth value. By this, I mean that if you start with a true equality, it will still be true, for example $3 = 3 \\setminus \\to 3 \\cdot 4 = 3 \\cdot 4 \\setminus \\to 12 = 12$ So, we started with a true equality ($3 = 3$), and we multiplied both sides by $4$, obtaining another true identity ($12 = 12$). On the other hand, if the two sides are different, they will still differ after being multiplied by the same number: $3 \\setminus \\ne 2 \\setminus \\to 3 \\cdot 6 \\setminus \\ne 2 \\cdot 6 \\setminus \\to 18 \\setminus \\ne 12$ So, we started with a false equality ($3 \\setminus \\ne 2$), and we multiplied both sides by $6$, obtaining another true identity ($18 \\setminus \\ne 12$). In your case, you only need to multiply both sides by $3$: the expression becomes $3 \\cdot \\setminus \\frac{x}{3} = - 10 \\cdot 3$ Why did we choose $3$? Because our goals is to isolate the $x$ on the left side, obtaining an expression like $x = \\ldots$. And the $3$ we multiplied by simplifies", "# How do you solve the inequality 3/4+3d < 6 3/4? Sep 18, 2015 $d < 2$ #### Explanation: So from $\\frac{3}{4} + 3 d < 6 + \\frac{3}{4}$ we start by taking the LDC $\\frac{3 + 4 \\cdot 3 d}{4} < \\frac{6 \\cdot 4 + 3}{4}$ Then, we multiply both sides by 4 $4 \\cdot \\frac{3 + 4 \\cdot 3 d}{4} < 4 \\cdot \\frac{6 \\cdot 4 + 3}{4} \\rightarrow 3 + 4 \\cdot 3 d < 6 \\cdot 4 + 3$ The 3s cancel each other out $3 + 4 \\cdot 3 d < 6 \\cdot 4 + 3 \\rightarrow 4 \\cdot 3 d < 6 \\cdot 4$ We divide both sides by 4 $\\frac{4 \\cdot 3 d}{4} < \\frac{6 \\cdot 4}{4} \\rightarrow 3 d < 6$ Finally, divide both sides by 3 $\\frac{3 d}{3} < \\frac{6}{3} \\rightarrow d < 2$", "# How do you solve - 3/5y + 7 = - 8? Nov 5, 2015 $y = \\frac{75}{3}$ or $y = 25$, simplified. #### Explanation: In this equation, your end goal is to get $y$ by itself. Your first step is to get $y$ isolated on one side. To do this, you would use the $\\text{subtraction property of equality}$. You will subtract $7$ from both sides to cancel out the left side $7$ and leave $y$ by itself. $- \\frac{3}{5} y + 7 = - 8 \\implies - \\frac{3}{5} y + 0 = - 15$ Now you have $- \\frac{3}{5} y = - 15$ and you want to get y completely by itself. To do this you would use the $\\text{multiplication property of equality}$, and multiply by the $\\text{reciprocal}$ to make the $- \\frac{3}{5}$ to 1. $- \\frac{3}{5} y \\cdot \\left(- \\frac{5}{3}\\right) = - \\frac{15}{1} \\cdot \\left(- \\frac{5}{3}\\right)$ Doing this will give you $y = \\frac{75}{3}$ which reduces down to $y = 25$.", "since $$\\sum_{cyc} (-\\frac{1}{36} a^2 + \\frac{25}{36} b^2) = \\frac{24}{36}\\sum_{cyc} a^2 = \\frac{2}{3} (a^2+b^2+c^2)$$ ) Indeed, by AP-GP, we have $$41 (a^3 + b^3+b^3) \\ge 41 \\cdot 3 ab^2$$ and $$b^3 + a^2b \\ge 2 ab^2$$ Summing up, we have $$41a^3 + 83b^3 + a^2 b \\ge 125 ab^2$$ which is equivalent to: $$36(a^3 + 3b^3) \\ge (5a + b)(-a^2 + 25b^2)$$ Dividing $36(5a+b)$ from both sides, the desired inequality is proved. --Lightest 15:31, 7 May 2012 (EDT)", "gives us $$150+(x-50)>3x-600$$. When we group like terms we get $$x+100>3x-600$$. Then, we subtract $$100$$ from both sides, giving us $$x>3x-700$$. Subtracting $$3x$$ from each side gives us $$-2x>-700$$. When we multiply each side by $$-1$$, this also reverses the inequality so, we have $$2x<700$$. Dividing each side by $$2$$, gives us $$x<350$$. Since we know $$x\\ge200$$ but $$x<350$$, we can write the interval $$[200, 350)$$. Dec 11, 2021", "# How do you solve the inequality -5/6d+8>13? Jan 31, 2016 $d < - 6$ We start off with $\\frac{- 5}{6} d + 8 > 13$. My goal is to isolate $d$ on one side of the expression. My first step would be to subtract $8$ on both sides, making the expression now $\\frac{- 5}{6} d > 5$. Now, I want to get $d$ alone, so I would divide by $\\frac{- 5}{6}$ on both sides, making the expression now show $d < \\frac{5}{- \\frac{5}{6}}$, which can be rewritten as $d < 5 \\cdot \\left(- \\frac{6}{5}\\right)$. The $- 5 \\cdot \\left(- \\frac{6}{5}\\right)$will divide out (like so:$\\cancel{5} \\cdot \\left(- \\frac{6}{\\cancel{5}}\\right)$), leaving $d < - 6$. Please note that the sign changed when I divided $\\frac{- 5}{6}$ on both sides. This is because when I multiply or divide by a negative number, the sign of the expression changes, from $<$ to $>$ or vice versa.", "# Tag Info ## Hot answers tagged inequality 293 The four numbered areas are congruent. [Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version ... 178 It is probably easier to note that $2^{40} = 4^{20}$. The only ones of the 20 factors in $20!$ that are smaller than $4$ are $1$, $2$ and $3$. But, on the other hand, $18$, $19$ and $20$ are all larger than $4^2$, so we can see $$20! = 1\\cdot 2\\cdot3\\cdots 18\\cdot19\\cdot 20 > \\underbrace{1\\cdot1\\cdot1}_{3\\text{ ones}}\\cdot\\underbrace{4\\cdot4\\cdots 4\\cdot ... 165 You have to be careful when multiplying by x since x might be negative and hence flip the inequality. Suppose x>0. Then$$\\frac{1}{x}<4\\iff4x>1\\iff x>1/4.$$If x>0 and x>1/4, then x>1/4. Now suppose x<0. Then$$\\frac{1}{x}<4\\iff4x<1\\iff x<1/4.$$If x<0 and x<1/4, then x<0. So the solution set is ... 141 I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily. 123 Here is the solution$$\\frac { 1 }{ x } <4 \\frac { 1-4x }{ x } <0 \\frac { x\\left(", "# How do you solve -6x + 3< -18? Jul 14, 2015 $x > \\frac{7}{2}$ #### Explanation: $- 6 x + 3 < - 18$ Solve for $x$ as if the expression is an equality. Subtract $3$ from both sides. $- 6 x < - 18 - 3$ = $- 6 x < - 21$ Divide both sides by $- 6$. This will reverse the inequality. $x > \\frac{- 21}{- 6}$ = $x > \\frac{21}{6}$ Factor $\\frac{21}{6}$. $x > \\frac{3 \\cdot 7}{3 \\cdot 2}$ Cancel $3$. $x > \\frac{\\cancel{3} \\cdot 7}{\\cancel{3} \\cdot 2}$ $x > \\frac{7}{2}$", "# Thread: help with proving a few identities 1. ## help with proving a few identities hi guys, I'm almost certain that this inequality holds, however for s , n > 0 , n being an integer, s being a real value (s + n) ! > s^(s + n) as s --> infinity (fixed n) (s + n)! > s^(s + n) as n --> infinity (fixed s) Dave 2. Originally Posted by Dave82 hi guys, I'm almost certain that this inequality holds, however for s , n > 0 , n being an integer, s being a real value (s + n) ! > s^(s + n) as s --> infinity (fixed n) (s + n)! > s^(s + n) as n --> infinity (fixed s) Dave $\\frac{(s+n)!}{s^{s+n}}=\\frac{1\\cdot 2\\cdot\\ldots\\cdot n\\cdot (n+1)\\cdot\\ldots\\cdot (n+s)}{s\\cdot s\\cdot\\ldots\\cdot s}$ , with $n+s$ factors in the denominator. It seems pretty obvious that the denominator is way bigger than the numerator when $s\\to\\infty$ , isn't it? So the inequality is exactly the other way than what you wrote! Tonio 3. If you look at Tonio's ratio for n and n+1, $\\frac{(s+n+1)!}{s^{s+n+1}}$ divided by $\\frac{(s+n)!}{s^{s+n}}=\\frac{s+n+1}{s}$ is always greater than 1, and much greater as n increases, so the ratio goes to infinity. That proves your second inequality. But if you look at the ratio", "remove the denominator from the middle expression: $-3\\cdot (5) < \\dfrac{-2x-7}{5} \\cdot (5) < 7 \\cdot (5)$ $-15 < -2x-7 < 35$ Isolate $x$ in the middle of the inequality: $- 15 + 7 < -2x -7 + 7 < 35 + 7$ $- 8 < -2x < 42$ Now divide each part by -2 (and remember to change the direction of the inequality symbol!): $\\displaystyle \\frac{-8}{-2} > \\frac{-2x}{-2} > \\frac{42}{-2}$ $4 > x > -21$ Finally, it is customary (though not necessary) to write the inequality so that the inequality arrows point to the left (i.e., so that the numbers proceed from smallest to largest): $-21 < x < 4$ ## Inequalities with Absolute Value Inequalities with absolute values can be solved by thinking about absolute value as a number’s distance from 0 on the number line. ### Learning Objectives Solve inequalities with absolute value ### Key Takeaways #### Key Points • Problems involving absolute values and inequalities can be approached in at least two ways: through trial and error, or by thinking of absolute value as representing distance from 0 and then finding the values that satisfy that condition. • When solving inequalities that involve an an absolute value within a larger expression (for example, $\\left|2x\\right|+ 3>8$), it is necessary to algebraically isolate the absolute value", "Multiply both sides of the inequality by $$\\left( {{2^n} + 1} \\right)$$ $$\\left( {{2^{n + 1}} + 1} \\right):$$ $\\left( {{2^n} + 3} \\right)\\left( {{2^{n + 1}} + 1} \\right) \\ge \\left( {{2^{n + 1}} + 3} \\right)\\left( {{2^n} + 1} \\right),\\;\\; \\Rightarrow {2^n}{2^{n + 1}} + 3 \\cdot {2^{n + 1}} + {2^n} + 3 \\ge {2^n}{2^{n + 1}} + 3 \\cdot {2^n} + {2^{n + 1}} + 3,\\;\\; \\Rightarrow 2 \\cdot {2^{n + 1}} \\ge 2 \\cdot {2^n},\\;\\; \\Rightarrow {2^{n + 1}} \\ge {2^n},\\;\\;\\Rightarrow 2 \\ge 1.$ Since the last inequality is true, we can conclude that the sequence is decreasing.", "the fractions in the inequality is $10\\text{,}$ so we'll multiply both sides of the inequality by $10\\text{.}$ Since we are multiplying by a positive number, the direction of the inequality sign is not effected. \\begin{align*} \\frac{2}{5}t+\\frac{1}{2} \\amp\\gt \\frac{3}{10}\\\\ \\multiplyleft{10}\\left(\\frac{2}{5}t+\\frac{1}{2}\\right) \\amp\\gt \\multiplyleft{10}\\frac{3}{10}\\\\ \\frac{10}{1} \\cdot \\frac{2}{5}t+\\frac{10}{1} \\cdot \\frac{1}{2} \\amp\\gt \\frac{10}{1} \\cdot \\frac{3}{10}\\\\ 4t+5 \\amp\\gt 3\\\\ 4t+5\\subtractright{5} \\amp\\gt 3\\subtractright{5}\\\\ 4t \\amp\\gt -2\\\\ \\divideunder{4t}{4} \\amp\\gt \\divideunder{-2}{4}\\\\ t \\amp\\gt -\\frac{1}{2} \\end{align*} Let's check that the two sides are equal at $-\\frac{1}{2}\\text{.}$ \\begin{align*} \\frac{2}{5}t+\\frac{1}{2} \\amp= \\frac{3}{10}\\\\ \\frac{2}{5} \\cdot \\highlight{-\\frac{1}{2}}+\\frac{1}{2} \\amp\\stackrel{?}{=} \\frac{3}{10}\\\\ -\\frac{1}{5}+\\frac{1}{2} \\amp\\stackrel{?}{=} \\frac{3}{10}\\\\ -\\frac{2}{10}+\\frac{5}{10} \\amp\\stackrel{?}{=} \\frac{3}{10}\\\\ \\frac{3}{10} \\amp\\stackrel{\\checkmark}{=} \\frac{3}{10} \\end{align*} Let's test $0$ in the original inequality. \\begin{align*} \\frac{2}{5}t+\\frac{1}{2} \\amp\\gt \\frac{3}{10}\\\\ \\frac{2}{5} \\cdot \\highlight{0}+\\frac{1}{2} \\amp\\stackrel{?}{\\gt} \\frac{3}{10}\\\\ 0+\\frac{1}{2} \\amp\\stackrel{?}{\\gt} \\frac{3}{10}\\\\ \\frac{1}{2} \\amp\\stackrel{?}{\\gt} \\frac{3}{10}\\\\ \\frac{5}{10} \\amp\\stackrel{\\checkmark}{\\gt} \\frac{3}{10} \\end{align*} The solution set is $\\left\\{t|t \\gt -\\frac{1}{2}\\right\\}$ which is equivalent to $\\left(-\\frac{1}{2},\\infty\\right)\\text{.}$ ###### Example8.4.5. Determine the solution set for the inequality $3(x-2) \\lt 3x+2\\text{.}$ State the solution set using both set builder notation and interval notation. Solution \\begin{align*} 3(x-2) \\amp\\lt 3x+2\\\\ 3x-6 \\amp\\lt 3x+1\\\\ 3x-6\\addright{6}\\subtractright{3x} \\amp\\lt 3x+1\\addright{6}\\subtractright{3x}\\\\ 0 \\amp\\lt 6 \\end{align*} The final inequality is an identity - a true mathematical statement. Since the original inequality is equivalent to an identity, it is true no matter the value of $t\\text{.}$ The solution set is $\\left\\{x|x \\in \\mathbb{R}\\right\\}$ which is equivalent to $(-\\infty,\\infty)\\text{.}$ ###### Example8.4.6. Determine the solution", "1, \\text{square both sides} \\\\ \\sqrt{11} (12 - 2\\sqrt{11}) & \\leq & 11^{1/6} + 2 \\cdot 11^{1/12} + 1 < 2 + 2 \\cdot 2 + 1 = 7 \\\\ 12 \\sqrt{11} - 22 < 7 \\\\ 12\\sqrt{11} < 29 \\\\ \\end{eqnarray}$ Square both sides yields a contradiction, thus $$11 < 11^{1/2} + 11^{1/3} + 11^{1/4}$$ Alternatively, it's trivial to show that $$11^{1/2} < 4, 11^{1/3} < 3, 11^{1/4} < 2$$ Add them up gives $$11^{1/2} + 11^{1/3} + 11^{1/4} < 9$$ Now consider the negation for the given inequality, that is $$11 \\leq 11^{1/2} + 11^{1/3} + 11^{1/4} < 9$$ which is absurd. Another method is to apply the Power Mean Inequality. Suppose that $$11 \\leq 11^{1/2} + 11^{1/3} + 11^{1/4}$$, then $$\\LARGE \\Rightarrow \\frac {11}{3} \\leq \\frac {11^{1/2} + 11^{1/3} + 11^{1/4}}{3} < \\sqrt[12] { \\frac { 11^6 + 11^4 + 11^3}{3} } < \\sqrt[12] { \\frac {2 \\cdot 11^6}{3} }$$ $$\\large \\Rightarrow \\left (\\frac {11}{3} \\right )^{12} < \\frac {2 \\cdot 11^6}{3}$$ $$\\large \\Rightarrow \\left (\\frac {11}{3} \\right )^{12} < \\frac {2 \\cdot 11^6}{3} < \\frac {3 \\cdot 11^6}{3} = 11^6$$ $$\\large \\Rightarrow 11^6 < 3^{12} \\Rightarrow 11 < 3^2$$ which is invalid. - 3 years, 1 month ago NICE SOL. .... - 3 years, 1 month ago $$(2.2)^{2}=4.84<5$$ or$$\\sqrt{5}>2.2$$. $$5^{1/4}>\\sqrt{2.2}>1.4$$ since $$(1.4)^2=1.96<2.2$$ Hence,$$5^{1/3}>5^{1/4}>1.4$$. So,$$5^{1/2}+5^{1/3}+5^{1/4}>2.2+1.4+1.4=5$$.", "# Proving an inequality involving binomial coefficients I have $3$ positive integers $i, k, q$, satisfying $1 \\le k \\le i \\le q$. I am trying to prove the following inequality: $$\\frac{{\\left( {\\begin{array}{*{20}{c}} {q - k}\\\\ {i - k} \\end{array}} \\right)}}{{\\left( {\\begin{array}{*{20}{c}} q\\\\ i \\end{array}} \\right)}} = \\frac{{\\left( {i - k + 1} \\right) \\cdot \\left( {i - k + 2} \\right) \\cdot ... \\cdot \\left( i \\right)}}{{\\left( {q - k + 1} \\right) \\cdot \\left( {q - k + 2} \\right) \\cdot ... \\cdot \\left( q \\right)}} \\le \\frac{{{i^k}}}{{{q^k}}}$$ Note that ${\\left( {i - k + 1} \\right) \\cdot \\left( {i - k + 2} \\right) \\cdot ... \\cdot \\left( i \\right) \\le {i^k}}$ and ${\\left( {q - k + 1} \\right) \\cdot \\left( {q - k + 2} \\right) \\cdot ... \\cdot \\left( q \\right) \\le {q^k}}$, making the inequality non-trivial. I considered the following cases: 1. $k=1$. The inequality becomes $1 \\le 1$ which is of course true. 2. $k=2$. The inequality becomes $i \\le q$ which is true. 3. $k=3$. The inequality becomes $\\frac{{q + i}}{{qi}} \\le \\frac{3}{2}$ which seems true, but proof is needed. For $k=4$ and on the expressions become complicated (unless they can be simplified somehow). Any help would be highly appreciated. Hint. Show that $\\frac{i-k+1}{q-k+1} \\leq \\frac i q$ •", "+0 # Linear inequality problem.. Help? +1 40 1 +476 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? gueesstt Apr 13, 2018 Sort: #1 +19207 +2 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? $$\\begin{array}{|rcll|} \\hline 12 > \\dfrac{x}{2} +1 \\quad & | \\quad -1 \\\\\\\\ 11 > \\dfrac{x}{2} \\quad & | \\quad \\cdot 2 \\\\\\\\ 22 > x \\quad & | \\quad \\cdot 2 \\\\\\\\ x < 22 \\\\ \\hline \\end{array}$$ The greatest integer is 21 heureka Apr 13, 2018 ### 32 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "## Algebra 1 $n\\lt5\\frac{1}{3}$ Use the multiplication property of inequality to multiply both sides by $\\frac{4}{3}$. $\\frac{3}{4}n\\lt4$ $\\frac{3}{4}n\\times\\frac{4}{3}\\lt4\\times\\frac{4}{3}$ $n\\lt\\frac{16}{3}\\longrightarrow$ reduce $n\\lt5\\frac{1}{3}$", "{y}^{2} + 12 y$ Therefore, $1 - 16 {y}^{2} + 12 y \\ge 0$ $\\implies$, $16 {y}^{2} - 12 y - 1 \\le 0$ The solutions of this inequality are $y \\in \\left[\\frac{12 - \\sqrt{{\\left(- 12\\right)}^{2} - 4 \\cdot \\left(- 1\\right) \\cdot 16}}{32} , \\frac{\\left(- 12\\right) + \\sqrt{{\\left(- 12\\right)}^{2} - 4 \\cdot \\left(- 1\\right) \\cdot 16}}{32}\\right]$ $y \\in \\left[\\frac{12 - \\sqrt{208}}{32} , \\frac{12 + \\sqrt{208}}{32}\\right]$ $y \\in \\left[- 0.076 , 0.826\\right]$ graph{(x+3)/(x^2+4) [-6.774, 3.09, -1.912, 3.016]}" ]

[ "• # question_answer Let $n(>1)$ be a positive integer, then the largest integer $m$ such that $({{n}^{m}}+1)$ divides $(1+n+{{n}^{2}}+.......+{{n}^{127}})$, is [IIT 1995] A) 32 B) 63 C) 64 D) 127 Since ${{n}^{m}}+1$ divides $1+n+{{n}^{2}}+.......+{{n}^{127}}$ Therefore $\\frac{1+n+{{n}^{2}}+......+{{n}^{127}}}{{{n}^{m}}+1}$ is an integer $\\Rightarrow$ $\\frac{1-{{n}^{128}}}{1-n}\\times \\frac{1}{{{n}^{m}}+1}$ is an integer $\\Rightarrow$ $\\frac{(1-{{n}^{64}})(1+{{n}^{64}})}{(1-n)({{n}^{m}}+1)}$ is an integer when largest $m=64$.", "+0 # Linear inequality problem.. Help? +1 40 1 +476 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? gueesstt Apr 13, 2018 Sort: #1 +19207 +2 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? $$\\begin{array}{|rcll|} \\hline 12 > \\dfrac{x}{2} +1 \\quad & | \\quad -1 \\\\\\\\ 11 > \\dfrac{x}{2} \\quad & | \\quad \\cdot 2 \\\\\\\\ 22 > x \\quad & | \\quad \\cdot 2 \\\\\\\\ x < 22 \\\\ \\hline \\end{array}$$ The greatest integer is 21 heureka Apr 13, 2018 ### 32 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "1. ## Arithmetic problem What is the largest positive integer n for which there is a unique integer k such that 8/15 < n/(n+K) < 7/13 $\\frac{8}{15} < \\frac{n}{n+K} < \\frac{7}{13}$ $\\frac{104}{195} < \\frac{n}{n+K} < \\frac{105}{195}$ $104 < \\frac{195n}{n+K} < 105$ 3. $\\frac{8}{15} < \\frac{n}{n+K} < \\frac{7}{13}$ Multiply by a common divisor of $15$ and $13$ so as to get everything in integers. $\\frac{104}{195} < \\frac{195n}{195(n+K)} < \\frac{105}{195}$ Multiply everything by $195$ so we get integers ... $104 < \\frac{195n}{n+K} < 105$ It might help ... EDIT : oh, no ! owned by Pickslides ! (nearly the same contents though)", "+0 # Help!!!!! If $x^2+y^2=1$, what is the largest possible value of $|x|+|y|$? 0 82 1 If $x^2+y^2=1$, what is the largest possible value of $|x|+|y|$? Thanks! Feb 21, 2021 #1 +939 -1 Equality is achieved when $$x^2 = y^2 = \\frac{1}{2}$$. $$\\max({|x|+|y|}) = \\frac{\\sqrt{2}}{2} \\cdot 2 = \\boxed{\\sqrt{2}}$$ Feb 21, 2021", "+0 # Let $$f(x) = \\frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \\cdot f(3) \\cdot f(4) \\cdots f(n-1) \\cdot f(n) < 1.98.$ +1 945 1 +601 Let $$f(x) = \\frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \\cdot f(3) \\cdot f(4) \\cdots f(n-1) \\cdot f(n) < 1.98.$ Oct 5, 2017", "# identity for greatest integer function by Primavera Last Updated August 14, 2019 05:20 AM How to show that the following relation? : for $$n\\in\\mathbb{N}$$, $$\\sum_{k=1}^{\\infty}\\left[\\frac{n}{5^{k}}\\right]=15\\iff\\left[\\frac{n}{5}\\right]=13.$$ It's not obvious to me. Can anyone help me? Thank you! Tags : There are three parts to the claim: 1. If $$\\frac{n}{5}<13$$ then $$\\sum_{k=1}^{\\infty}\\left[\\frac{n}{5^{k}}\\right]<15$$ 2. If $$13\\leq\\frac{n}{5}<14$$ then $$\\sum_{k=1}^{\\infty}\\left[\\frac{n}{5^{k}}\\right]=15$$ 3. If $$\\frac{n}{5}\\geq14$$ then $$\\sum_{k=1}^{\\infty}\\left[\\frac{n}{5^{k}}\\right]>15$$ So choose whichever part strikes your fancy and tackle it in isolation. Repeat until all three are done. Chris Culter August 14, 2019 05:01 AM Observe that $$\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{5^k}\\right\\rfloor=\\nu_5(n!)$$, where $$\\nu_5(n!)$$ is the exponent of the largest power of $$5$$ that divides $$n!$$. Since this sum is $$15$$, thus the highest power of $$5$$ that divides $$n!$$ is $$15$$. Consider $$n!=(1\\cdot 2 \\cdot 3 \\cdot 4 \\cdot \\color{red}{5})(6\\cdot 7 \\cdot 8 \\cdot 9 \\cdot \\color{red}{10})\\dotsb$$ Each group of $$((a+1)\\cdot (a+2) \\cdot (a+3) \\cdot (a+4) \\cdot (a+5))$$ contributes a single power of $$5$$, until we reach $$25$$, where we get a contribution of $$2$$ towards the exponent of $$5$$. Likewise until we reach $$50$$ each such group of five numbers contribute only a single power of $$5$$ and so on. Thus the number $$n$$ must be such that $$n!=\\underbrace{(1\\cdot 2 \\cdot 3 \\cdot 4 \\cdot \\color{red}{5})}_{5^1}\\underbrace{(6\\cdot 7 \\cdot 8 \\cdot 9 \\cdot \\color{red}{10})}_{5^1}\\dotsb \\underbrace{(21\\cdot 22 \\cdot 23 \\cdot 24", "Question 104 # The smallest integer x for which the inequality $$\\frac{x-7}{x^2 + 5x-36}$$ > 0 is given by Solution $$\\frac{x-7}{x^2 + 5x-36}$$ > 0 $$\\frac{x-7}{(x+9)(x-4)}$$ > 0 So $$x \\epsilon (-9,4) U (7, \\infty$$) So smallest integer in this range is $$x = -8$$.", "$\\large \\frac{1}{\\sqrt{5x+1}}+\\frac{1}{\\sqrt{x^2-3x+4}}\\geq\\frac{x+3}{2(x+1)}$ How many integer $$x$$ satisfy the inequality above", "# Find the largest integer $k$ for which $3^k$ divides $400 \\choose 200$. I was working through a problem set for number theory and needed some help with this problem: Find the largest integer $$k$$ for which $$3^k$$ divides $$400\\choose 200$$. I know this will reduce to $$\\frac{400!}{(200!)^2}$$ and I found that the largest power of $$3$$ that divides $$400!$$ is $$196$$ and the largest power that divides $$200!$$ is $$97$$ but I don't know how to put them together. • $$196-2\\cdot97=?$$ – lab bhattacharjee May 7 at 15:02 You have found that $$400!=3^{196}m$$ and $$200!=3^{97}p$$ where $$m,p$$ are integers not divisible by $$3$$. Then $$\\frac {400!}{200!^2}=\\frac {3^{196}m}{(3^{97}p)^2}=\\frac {3^{196}m}{3^{2\\cdot97}p^2}=3^2\\frac m{p^2}$$", "+0 # Linear inequality problem.. Help? +1 324 1 +603 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? Apr 13, 2018 #1 +22489 +2 What is the greatest integer that is a solution to the inequality $12 > \\frac{x}{2} +1$? $$\\begin{array}{|rcll|} \\hline 12 > \\dfrac{x}{2} +1 \\quad & | \\quad -1 \\\\\\\\ 11 > \\dfrac{x}{2} \\quad & | \\quad \\cdot 2 \\\\\\\\ 22 > x \\quad & | \\quad \\cdot 2 \\\\\\\\ x < 22 \\\\ \\hline \\end{array}$$ The greatest integer is 21 Apr 13, 2018", "# Greatest Integer Value of a Side Using Angle Bisector Theorem In triangle XYZ, the bisector of $\\angle XYZ$ intersects $\\overline{XZ}$ at E if $\\frac{XY}{YZ}=\\frac{3}{4}$ an $XZ=42$, find the greatest integer value of XY. Thus far, I have determined that $XE=18$ and $ZE=24$ by the angle bisector theorem, but I am unsure how to find XY. The bisector theorem gives $\\frac{XE}{EZ}=\\frac{XY}{YZ}=\\frac{3}{4}$, hence $XZ=42$ implies $XE=18$ and $EZ=24$. By the triangle inequality, $XY+YZ$ has to be greater than $XZ=42$, hence $XY=\\frac{3}{7}(XY+YZ)$ has to be greater than $18$. On the other hand, $YZ-YX$ has to be smaller than $XZ=42$, hence the length of $XY$ is at most $3\\cdot 42=\\color{red}{126}$. • But what if we let $XY=3x$ and $YZ=4x$. Using your inequalities, we get $6 \\lt x \\lt 42$, which would make the greatest integer value $41*3=123$. Is this incorrect? – Db3010 Sep 18 '16 at 1:30", "- 15 => 3x = 15 Divide each side by 3 => $\\frac{3x}{3} = \\frac{15}{3}$ => x = 5, first integer. Step 3: Largest or third integer = x + 4 = 5 + 4 = 9 Hence the largest integer is 9. answer", "# Largest integer less than $2013$ obtained by repeatedly doubling an integer $x$. I am working on an old math contest and came across this problem: What is the largest integer less than $2013$ that can be obtained by repeatedly doubling a positive integer less than $100$? Here's what I've done: $$x<100$$ $$2^nx<2013$$ $$x<\\frac{2013}{2^n}$$ The first time $\\frac{2013}{2^n}$ is less than $100$ occurs when $n=5$. Therefore $x < 62.90625$. The largest integer that satisfies the inequality is $x=62$. As it turns out $62$ is the correct answer, but I feel like my solution is missing something. All I have done is narrow down the possibilities, going from $x<100$ to $x\\leq 62$. Can anyone point me in the direction of what I am missing? I don't want the solution just a hint/nudge in the right direction. Any tips would be appreciated! • You got a multiple of $64$. Since $2012\\lt 1984+32$, we can't do better by dividing by $32$. Aug 7 '16 at 16:43 If you start with a number less than $100$, you can double it at least $4$ times without exceeding $2012$, so the answer must be a multiple of $2^4=16$. The largest multiple of $16$ less than $2013$ is $2000=16\\cdot 125$, which doesn't work: $125\\not<100$. The next largest is $1984=31\\cdot 2^6$, which works and must therefore", "\\frac{1}{100}$$ With numerators of both sides same, for LHS > RHS, denominator of LHS should be less than that of RHS $$2^6 = 64$$ $$2^7 = 128$$ _________________ Kindly press \"+1 Kudos\" to appreciate Intern Joined: 22 Feb 2014 Posts: 27 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Jul 2014, 10:40 Bunuel wrote: SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 [m]\\frac{1}{2^n}> 0.01[/m]; $$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. Hi I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice Thanks Komal Intern Joined: 22 Feb 2014 Posts: 27 Re: What is the largest integer n such that 1/2^n > 0 ? [#permalink] ### Show Tags 10 Jul 2014, 10:40 Bunuel wrote: SOLUTION What is the largest integer n such that $$\\frac{1}{2^n}> 0.01$$ ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 [m]\\frac{1}{2^n}> 0.01[/m]; [color=#ff0000]$$\\frac{1}{2^n}> \\frac{1}{100}$$; $$2^n<100$$; [/color] $$2^6=64<100$$ and $$2^7=128>100$$, therefore, 6 is the largest integer such that $$\\frac{1}{2^n}> 0.01$$. Hi I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know", "# How do you determine the largest integer value of x in the solution for 5x-9<=7? Jan 4, 2017 Re-write to get $x$´s and numbers on opposite sides #### Explanation: Add 9 to both sides: $5 x - \\cancel{9} + \\cancel{9} \\le 7 + 9 \\to 5 x \\le 16 \\to$ Divide by 5: $x \\le \\frac{16}{5} \\mathmr{and} x \\le 3 \\frac{1}{5}$ The largest integer value would be $x = 3$", "# The data items in a list are 75,86,87,91, and 93. What is the largest integer you can add to the list so that the mean of six items is less than their median? Oct 1, 2016 Largest integer is $101$ #### Explanation: There are 5 numbers in the list, but a sixth one is to be added. (as large as possible) $75 \\text{ \"86\" \"87\" \"91\" \"93\" } x$ $\\textcolor{w h i t e}{\\times \\times \\times \\times \\times} \\uparrow$ The median will be $\\frac{87 + 91}{2} = 89$ Mean will be: $\\frac{75 + 86 + 87 + 91 + 93 + x}{6} < 89$ $432 + x < 6 \\times 89$ $x < 534 - 432$ $x < 102$ The largest integer can be 101. Check; If $x = 101$ Mean $= \\frac{533}{6} = 88.83$ $88.83 < 89$", "the smallest integer greater than $\\frac{p-1}{4} - \\frac{4}{4} = \\frac{p - 5}{4}$. • Furthermore, -(a - 1) is the largest integer less than $-\\frac{p - 5}{4}$. Hence it follows that: • Hence if $p \\equiv 1 \\pmod 8$ or $p \\equiv 7 \\pmod 8$, then g is even and $(a/p) = (-1)^g = 1$. If $p \\equiv 3 \\pmod 8$ or $p \\equiv 5 \\pmod 8$, then g is odd, and $(a/p) = (-1)^g = -1$. Hence, the proof is complete, and:", "# Math Help - [SOLVED] Sequence involving largest integer function. 1. ## [SOLVED] Sequence involving largest integer function. Let $(a_n)$be the sequence defined by $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? 2. Hello wizard654zzz Originally Posted by wizard654zzz Let $(a_n)$be the sequence defined $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? Divide and get quotient and remainder: $\\frac{n^2 + 8n +10}{n+9}=(n-1) + \\frac{19}{n+9}$ Now for $1\\le n\\le 10, \\;2>\\frac{19}{n+9}\\ge 1$. Hence $a_n = n$. But for $n > 10, \\;\\frac{19}{n+9}<1$. Hence $a_n = n-1$. $1+2+3+...+10+10+11+12+...+29 = 445$", "the largest integer number $For \\ x \\in R$ , let [x] be the largest integer less than or equal to x . prove that : $[x] +[x+ \\frac{1}{m}]+[x+\\frac{2}{m}…+[x+\\frac{m-1}{m}] =[mx], \\forall x \\in R , \\ \\forall m \\in N$", "can assert that it is say between $$\\frac{1}{3}$$ and $$\\frac{2}{3}$$ so that the whole thing will be largest when the denominator is smallest, namely when it is $$\\frac{1}{3}$$ giving the inequality $|\\frac{2-x}{x}|<3|x-2|$ 11. Study23 Group Title How would I find $$\\ \\delta ?$$" ]

[ "# 144...44 \\begin{aligned} 144 &= 12^2 \\\\ 1444 &= 38^2 \\end{aligned} Are there any other numbers $144 \\ldots 44$ (starting with 1 and followed by all 4s) that are perfect squares? ×", "well choose them in order $1, 2, 3, \\ldots$.", "24 x + 48 {x}^{2} + 80 {x}^{3} + 120 {x}^{4} + 168 {x}^{5} + 224 {x}^{6} + \\ldots \\ldots .$", "10404, 10816, \\ldots$ for $x=49,\\ldots,53$ and beyond. But that doesn't make for a very satisfying puzzle, so I suspect that's not the answer you're looking for. Next number could be 530 (Which is square of 23 plus 1) As, the first 5 numbers were obtained by adding 4 to the squares of even numbers- 2,4,6,8 and 10. Then, 1 is getting added to the squares of odd numbers - 11, 13, 15, 17, 19, 21. So the next number could be square of 23 that is 529 and 1 added to it. • Not bad, but there is a more elegant way, which doesn't discriminate between the first five numbers and the remaining numbers. Jun 12, 2018 at 0:12 It looks like the answer is 530 Because the difference increases by 8 each time although that breaks down for 104, 122, 170 - so unless that's a mistake my answer is wrong", "-44 & 15 & -47 \\ldots \\\\ \\hline -41 & 200 & -92 & -72 & 17 & -48 & 38 & -35 & 0 & 0 & 0 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1 \\\\ \\hline \\ldots 10 & -144 & 136 & -80 & -126 & 189 & -14 & 27 & 60 & 3 & 0 & 0 \\end{array}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$1, 10, 3, 30, 9, -7, 27, -21, -16, 34, -48, 5, -47, 15, -44, 45, -35, 38, -8, 17, -24, -46, 25, -41.$$ For instance, in your example, you can put the $24$-digit number $182\\ldots100$ above and these $24$ numbers (in reversed order) below to get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 & -44 & 15 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1\\end{array}$$ Calculating the product of those pairs of numbers above and below, you get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 &", "45 and 135 (of course there are an infinite number of them.) Hiope this helps.", "# Solutions to Problems 1064-1071 Q1064. The numbers $1, 2, \\ldots , 16$ are placed in the cells of a $4 \\times 4$ table as shown in the left hand diagram below.", "3 are there from 100 to 500? Again, let's start by listing some of them out: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots$ At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots 492,\\, 495,\\, 498$ Now let's divide each number in the list by 3: $34,\\, 35,\\, 36,\\, 37,\\, \\ldots 164,\\, 165,\\, 166$ See what happened? The numbers are now consecutive. Now the question becomes, \"How many numbers are there from 34 to 166?\" If you're not sure, ask yourself how many numbers there are from 5 to 10. There are 6. Now how did you get that? Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1: $10 - 5 + 1 = 6$ So to get the number of numbers from 34 to 166, we do the same thing: $166 - 34 + 1 = 133$ Finally, we have the answer, 133. Although this problem may seem long and complicated, the method we used to solve it applies to many other questions like this (namely, multiple questions). Once you understand it, solving similar questions is a breeze. Hopefully, you now have a good grasp", "{1, 4, 9, 16, \\ldots}$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "# Sequence of Powers of 16 The sequence of powers of $16$ begins: $1, 16, 256, 4096, 65 \\, 536, 1 \\, 048 \\, 576, 16 \\, 777 \\, 216, 268 \\, 435 \\, 456, \\ldots$", "# 144...44 \\begin{align} 144 &= 12^2 \\\\ 1444 &= 38^2 \\end{align} Are there any other numbers $$144 \\ldots 44$$ (starting with 1 and followed by all 4s) that are perfect squares? ×", "number is $47.$", "8 are: $\\ldots , 136 , 144 , 152 , 160 , \\textcolor{red}{168} , \\ldots$ The (additional) multiples of 7 are: $\\ldots , 119 , 126 , 133 , 140 , 147 , 154 , 161 , \\textcolor{red}{168} , \\ldots$ The least common factor is therefore 168.", "the list becomes $$1,2,3,5,7,11$$, which square to $$1,4,9,25,49,121$$", "# 126 number Template:Numbers 120s Cardinal One hundred [and]twenty-six Ordinal 126th Factorization ${\\displaystyle 2\\cdot 3^{2}\\cdot 7}$ Divisors 2, 3, 6, 7, 9, 14, 18, 21, 42, 63 Roman numeral CXXVI Binary 1111110 Hexadecimal 78", "42 = 168$", "3 are there from 100 to 500? Again, let’s start by listing some of them out: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots$ At this point, it’s not hard to see that this will take forever. Here’s the trick—instead of listing out all the numbers in the middle, let’s list out a few at the end: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots 492,\\, 495,\\, 498$ Now let’s divide each number in the list by 3: $34,\\, 35,\\, 36,\\, 37,\\, \\ldots 164,\\, 165,\\, 166$ See what happened? The numbers are now consecutive. Now the question becomes, “How many numbers are there from 34 to 166?” If you’re not sure, ask yourself how many numbers there are from 5 to 10. There are 6. Now how did you get that? Perhaps you didn’t realize it, but you subtracted 5 from 10 and then added 1: $10 – 5 + 1 = 6$ So to get the number of numbers from 34 to 166, we do the same thing: $166 – 34 + 1 = 133$ Finally, we have the answer, 133. Although this problem may seem long and complicated, the method we used to solve it applies to many other questions like this (namely, multiple questions). Once you understand it, solving similar questions is a breeze. Hopefully, you now have a good grasp", "4, 16, 37, 58........ You can find a lot of 'happy numbers' such as: 310, 70, 86, 130, 7, 5555, 1212, 13, 1000 and other numbers which are variations of these, e.g. swapping the numbers the other way around (13, 31) and adding zeros onto the end of numbers (31, 310). Also you get numbers that add up to other happy numbers e.g. 5555 - four fives squared = 100. The following sequences are the patterns of fixed points and cycles in base eight that I have found. These are fixed points so they are happy numbers and any numbers which start sequences ending like this are happy numbers. 64, 64, 62, 64, 64, $\\ldots$ 24, 24, 24, 24, 24, $\\ldots$ 1, 1, 1, 1, ,1, $\\ldots$ These are 2-cycles: 32, 15, 32, 15, 32, 15, $\\ldots$ 20, 4, 20, 4, 20, 4, $\\ldots$ This is a 3-cycle: 31, 12, 5, 31, 12, 5, 31, 12, 5, $\\ldots$ Pen Areecharoenlert, also from The Mount School, York found some more happy numbers and cycles in base eight. Pen says \"There does not appear to be any pattern in which ones are happy, but it looks as though there are 3 fixed points: 1, 24 and 64.\" This is Pen's list. 1, 1, 1, $\\ldots$ so 1 is happy. 2," ]

[ "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "# Another Think Out Of The Boxes(The Id-\"1\"-- ) What is the next number in the sequence $1, 11, 21, 1211, \\ldots?$ ×", "-44 & 15 & -47 \\ldots \\\\ \\hline -41 & 200 & -92 & -72 & 17 & -48 & 38 & -35 & 0 & 0 & 0 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1 \\\\ \\hline \\ldots 10 & -144 & 136 & -80 & -126 & 189 & -14 & 27 & 60 & 3 & 0 & 0 \\end{array}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "us to find all the values taking $\\pm 1$ in the three following recurrent linear sequences satisfying $a_n = 5a_{n-1} - 7a_{n-2}$, • $1,2,3,1,-16,-87,-323,\\ldots$ • $0,1,5,18,55,149,360,\\ldots$ • $1,3,8,19,39,62,37,\\ldots$ Looking at the recurrence relation mod $18$, we have $a_n \\equiv 5a_{n-1}+11a_{n-2} \\equiv 5(5a_{n-2}+11a_{n-3}) + 11a_{n-2} \\equiv a_{n-3}$. Looking at the first three terms of each sequence modulo $18$, we can already discard $2/3$ of all the terms and also remove the possibility of a $-1$ appearing, which reduces the problem to finding the $1$ values in the linear recurrent sequences satisfying $a_n = 20a_{n-1}-343_{n-2}$ • $1,1,-323,-6803,\\ldots$ • $1,55,757,-3725,\\ldots$ • $1,19,37,-5777,\\ldots$ Now each sequence is a linear combination of the coefficient sequences of $(2-\\omega)^{3n} = (1-18\\omega)^n = (10 - 9 \\sqrt{-3})^n$, that is, the two sequences • $1,10,-143,-6290,\\ldots$ • $0,-9,-180,-513, \\ldots$ (with integer coefficents in all three cases) With the binomial theorem, we can expand $(1+9(1-\\sqrt{-3}))^n = 1 + 9(1-\\sqrt{-3})n + 81(1-\\sqrt{-3})^2n(n-1)/2 + \\ldots$. If we place ourselves in $\\Bbb Z_3[\\sqrt{-3}]$ and we notice that $v_3(9^k/k!) >= 3k/2 \\to \\infty$, we can reorder the summation and get $1 + [9(1-\\sqrt{-3}) + O(81)]n + O(81)n^2 + \\ldots$ i.e. each coefficient sequence can be extended as a function $\\Bbb Z_3 \\mapsto \\Bbb Z_3$ which is a power series in $n$ with coefficients in $\\Bbb Z_3$ Now, if a power series is of", "opposite direction. So we need to create similar imbalances in the Subsets $A_1$ to $A_{117}$, so that we could add the above $117$ numbers to those imbalances to keep the total sum unchanged. Now start swapping the $1st$ number of each pair that we added to subsets $A_1$ to $A_{117}$ to create the above imbalances: Swap $1$ from $A_1$ with $59$ from $A_{59}$ - this creates an imbalance of $+58$ and $-58$ Swap $2$ from $A_2$ with $58$ from $A_{58}$ - this creates an imbalance of $+56$ and $-56$ $\\ldots$ Swap $29$ from $A_{29}$ with $31$ from $A_{31}$ - creates an imbalance of $+2$ and $-2$ Leave $30$ in $A_{30}$ as it is - $0$ imbalance. Similarly, Swap $60$ from $A_{60}$ with $117$ from $A_{117}$ - this creates an imbalance of $+57$ and $-57$ Swap $61$ from $A_{61}$ with $116$ from $A_{116}$ - this creates an imbalance of $+55$ and $-55$ $\\ldots$ Swap $88$ from $A_{88}$ with $89$ from $A_{89}$ - this creates an imbalance of $+1$ and $-1$ Now start adding the $117$ numbers from $(1)$ to the subsets $A_1$ to $A_{117}$ to counter the imbalances $-58$ to $+58$ so that the sum remains unchanged. Notice that each subset now has $17$ elements with total sum = $8*1990 + 995 = 16915$. $- Kris17$", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "10404, 10816, \\ldots$ for $x=49,\\ldots,53$ and beyond. But that doesn't make for a very satisfying puzzle, so I suspect that's not the answer you're looking for. Next number could be 530 (Which is square of 23 plus 1) As, the first 5 numbers were obtained by adding 4 to the squares of even numbers- 2,4,6,8 and 10. Then, 1 is getting added to the squares of odd numbers - 11, 13, 15, 17, 19, 21. So the next number could be square of 23 that is 529 and 1 added to it. • Not bad, but there is a more elegant way, which doesn't discriminate between the first five numbers and the remaining numbers. Jun 12, 2018 at 0:12 It looks like the answer is 530 Because the difference increases by 8 each time although that breaks down for 104, 122, 170 - so unless that's a mistake my answer is wrong", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$1, 10, 3, 30, 9, -7, 27, -21, -16, 34, -48, 5, -47, 15, -44, 45, -35, 38, -8, 17, -24, -46, 25, -41.$$ For instance, in your example, you can put the $24$-digit number $182\\ldots100$ above and these $24$ numbers (in reversed order) below to get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 & -44 & 15 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1\\end{array}$$ Calculating the product of those pairs of numbers above and below, you get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 &", "we are ignoring representations that contain infinite sequences of zeroes. Now instead of interleaving digits, we interleave chunks. To interleave $0.004999\\ldots$ and $0.01003430901111\\ldots$, we get $0.004\\ 01\\ 9\\ 003\\ 9\\ 4\\ 9\\ldots$. This is obviously reversible. It can never produce a result that ends with an infinite sequence of zeroes, and similarly the reverse mapping can never produce a number with an infinite sequence of trailing zeroes, so we win. A problem example similar to the one from a few paragraphs ago is resolved as follows: $\\frac12 = 0.4999\\ldots$ is the unique image of $\\langle 0.4999\\ldots, 0.999\\ldots\\rangle$ and $\\frac9{22} = 0.40909\\ldots$ is the unique image of $\\langle 0.40909\\ldots, 0.0909\\ldots\\rangle$.", "value to \\$10000? Month Interestearned Total amountof moneyin the account $0$ $\\dollar0.00$ $\\dollar5000.00$ $1$ $\\dollar33.33$ $\\dollar5033.33$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ # Subsection8.1.1Sequences As our discussion in the introduction and Preview Activity 8.1.1 illustrate, many discrete phenomena can be represented as lists of numbers (like the amount of money in an account over a period of months). We call these any such list a sequence. In other words, a sequence is nothing more than list of terms in some order. To be able to refer to a sequence in a general sense, we often list the entries of the sequence with subscripts, \\begin{equation*} s_1, s_2, \\ldots, s_n \\ldots \\text{,} \\end{equation*} where the subscript denotes the position of the entry in the sequence. More formally, ##### Definition8.1.3 A sequence is a list of terms $s_1, s_2, s_3, \\ldots$ in a specified order. As an alternative to Definition 8.1.3, we can also consider a sequence to be a function $f$ whose domain is the set of positive integers. In this context, the sequence $s_1\\text{,}$ $s_2\\text{,}$ $s_3\\text{,}$ $\\ldots$ would correspond to the function $f$ satisfying $f(n) = s_n$ for each positive integer $n\\text{.}$ This alternative view will be be useful in many situations. We will often write the sequence \\begin{equation*} s_1, s_2, s_3, \\ldots \\end{equation*}", "same digits but faster n/10 + 0,9 = 7n + ε n + 9 = 70n + ε n = (9-ε)/69 = (3-ε)/23 = 3/23 - ε n = 0,(Period)1304347826086956521739 - ε n = 0,1304347826086956521739 - Yes, this is a fine way to do it. – Ross Millikan Apr 2 '14 at 13:48 Let us call the original number $N$, and the modified number $M$. Thus $N$ ends with a 9 and $M$ begins with a 9. Since $N = M\\div 7$, we can set up a short division problem that looks like this: $$\\require{enclose} \\begin{array}{rl} & \\ \\,1\\ldots\\\\ 7&\\enclose{longdiv}{9\\ldots} \\end{array}$$ Here $N$ is the quotient and $M$ the dividend. We see that $N$ begins with the digit 1, so this is the second digit of $M$: $$\\begin{array}{rl} & \\ \\,1\\ldots\\\\ 7&\\enclose{longdiv}{91\\ldots} \\end{array}$$ Now that we know the second digit of $M$, we can divide one more step, computing the next digit of $N$: $$\\begin{array}{rl} & \\ \\,13\\ldots\\\\ 7&\\enclose{longdiv}{913\\ldots} \\\\ \\\\ & \\ \\,130\\ldots\\\\ 7& \\enclose{longdiv}{9130\\ldots} \\\\ \\\\ & \\ \\,1304\\ldots\\\\ 7& \\enclose{longdiv}{91304\\ldots} \\\\ \\\\ \\end{array}$$ We must continue in this fashion until we reach a point where there the remainder is 0 and the quotient ends with 9. This first occurs at: $$\\begin{array}{rl} & \\ \\,1304347826086956521739\\\\ 7& \\enclose{longdiv}{9130434782608695652173} \\\\ \\\\ \\end{array}$$ This is the smallest answer. By", "# How do you convert 0.bar(45) (meaning the 45 is being repeated) to a fraction? Jun 27, 2018 $0.45454545 \\ldots \\ldots \\ldots \\ldots . = \\frac{5}{11}$ #### Explanation: Let $x = 0.45454545 \\ldots \\ldots \\ldots \\ldots .$ ............................(1) then $100 x = 45.45454545 \\ldots \\ldots \\ldots \\ldots$ ............................(2) subtracting (1) from (2) we get $99 x = 45$ or $x = \\frac{45}{99} = \\frac{5}{11}$ Hence $0.45454545 \\ldots \\ldots \\ldots \\ldots . = \\frac{5}{11}$ Jun 28, 2018 $\\frac{5}{11}$ #### Explanation: When two numbers in a decimal sequence repeat forever, the denominator will be $11$. To think about the numerator, let's just take the number $45$. This number would round to $50$ or $5$ tens. Whatever number of tens the numerator rounds to, this will be our numerator. Thus, we have $\\frac{5}{11}$ Hope this helps!", "iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,\\!000$th number in the sequence? When counting from $3$ to $201$, $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$, $53$ is the $n^{th}$ number counted. What is $n$?", "# How many $6$-lentgh increasing sequence are there from $1$ to $49$? I have to figure out a way to count how many sequences are there s.t: 1. My alphabet is $1$ up to $49$. 2. Each number that is chosen, is chosen only once. 3. The sequence is $6$ digits long. 4. The sequence is in increasing order. e.g that is a valid sequence: $(1, 2, 40, 42, 43, 49)$ and this is not: $(1, 2, 40, 25, 25, 12)$ I know that for the first 3 rules the answer is: $\\binom{49}{6}\\cdot 6!$ How do I deal with the fourth? • Just remove $6!$. It's the same as the combinations of 6 elements out of 49. Oct 28, 2014 at 19:54 • Thanks, I just figured it out. Oct 28, 2014 at 19:55 I think I got it, I would say it's $\\binom{49}{6}$, because once you chose your 6 numbers, there is only 1 way to order them. You’ve found the nicest and easiest solution. Just for fun, here’s another (much less efficient!) approach showing how to convert this problem into a more familiar one. Let the sequence be $\\langle a_1,a_2,a_3,a_4,a_5,a_6\\rangle$. Let $x_1=a_1$, let $x_k=a_k-a_{k-1}$ for $k=2,\\ldots,6$, and let $x_7=49-a_6$. Then you want the number of integer solutions to $$x_1+x_2+x_3+x_4+x_5+x_6+x_7=49$$ subject to the conditions $1\\le x_k$ for", "1 instead of 1 and 0. What are the first ten terms of the Lucas sequence? 2. Zeckendorf’s Theorem states that every positive integer can be represented uniquely as a sum of nonconsecutive Fibonacci numbers. What is the Zeckendorf representation of the number 50 and the number 100? 3. Consider the following pattern generating rule: If the last number is odd, multiply it by 3 and add 1. If the last number is even, divide the number by 2. Repeat. Try a few different starting numbers and see if you can state what you think always happens. 1. $2,1,3,4,7,11,18,29,47,76$ 2. $50=34+13+3;\\ 100=89+8+3$ 3. You can choose any starting positive integer you like. Here are the sequences that start with 7 and 15. $7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1\\ldots$ $15,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1,4,2,1\\ldots$ You could make the conjecture that any starting number will eventually lead to the repeating sequence 4, 2, 1. This problem is called the Collatz Conjecture and is an unproven statement in mathematics. People have used computers to try all the numbers up to $5\\times2^{60}$ and many mathematicians believe it to be true, but since all natural numbers are infinite in number, this test does not constitute a proof. #### Practice Write a recursive definition for each of the following sequences. 1. $3,7,11,15,19,\\ldots$ 2. $3,9,27,81,\\ldots$ 3. $3,6,9,12,15,\\ldots$ 4. $3,6,12,24,48,\\ldots$ 5. $1,4,16,64,\\ldots$ 6. Find the", "$2$'s into groups, such that each group has at least one $2$. By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$, is not needed for the remaining calculations.) Then, to get $D(96)$, in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five. The resulting number of possible sequences is $3 \\cdot 1 + 5 \\cdot 4 + 7 \\cdot 6 + 9 \\cdot 4 + 11 \\cdot 1 = \\boxed{\\textbf{(A) }112}$. ~emerald_block", "too many too easy questions in too little time. • +1 for Remark Programmers know this as the fencepost or one-off problem. It is one of the most common coding mistakes – Fred Kline Oct 1 '15 at 20:23 Whenever you want to count the number of things, align it to the sequence $1, 2, \\ldots$. In the sequence $1,\\ldots,n$, there are exactly $n$ items. This should be abundantly clear. In the sequence $0,\\ldots,n$, there are exactly $n+1$ items. This should be abundantly clear once you understand the first statement. When you ask \"How many numbers are there between 30 and 40?\" and you answer \"10\" instinctively, you've done subtraction. Subtraction starts you at 0, not 1 -- if we subtract 30 from every number between 30 and 40, the sequence becomes $0, \\ldots, 10$. Instead of subtracting the first number, I always subtract the number that gets me to 1 -- here, 29 (i.e., 30-29=1). And 40-29 is 11. Or, if you prefer, just add 1 to 10, in parallel to the $0,\\ldots,10$ argument. You can use the idea of Arithmetic Progression. $$t_n=a+(n-1)d$$ in any arithmetic progression where $$t_n$$ is the $$n-th$$ term which readily implies that $$n\\in\\mathbb Z$$. For your first example. Between $$250$$ and $$350$$ the first multiple of $$3$$ is $$252$$ and the last", "# A New Method to divide Numbers by 8 (Eight)! Consider the following procedure for dividing the three-digit number $375$ by $8$: • Write down the number formed by the first two digits, namely $37$. • Multiply this by $2$ to get $74$. • Add to this, the units digit of $375$ (the original number), obtaining $74+5=79$. • Then divide it by $8$ to get $9$ with a remainder of $7$. • Add this result ($9$, remainder $7$) with $37$ (the first two digits of the original number) to get your answer: $46$, remainder $7$. Thus $375$ divided by $8$ equals $46$ with a remainder of $7$. Does this method always work for three-digit numbers? Why, or why not? ×", "digits, transforming number $\\textbf{22}3$ into $\\textbf{33}3$, and then make a -1 operation on the last two digits, transforming $3\\textbf{33}$ into $3\\textbf{22}$. It's also possible to do these operations in reverse order, which makes another correct answer. In the last example, one can show that it's impossible to transform $35$ into $44$.", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$." ]

[ "-44 & 15 & -47 \\ldots \\\\ \\hline -41 & 200 & -92 & -72 & 17 & -48 & 38 & -35 & 0 & 0 & 0 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1 \\\\ \\hline \\ldots 10 & -144 & 136 & -80 & -126 & 189 & -14 & 27 & 60 & 3 & 0 & 0 \\end{array}$$ Adding up the numbers below the line, we end up with $-19$, which modulo $97$ equals $78$, as expected. An advantage of this method is that, once you have those $24$ numbers, the computations are really simple and require only really small numbers. You can even do this method by hand, as I actually did for the example above.", "4, 16, 37, 58........ You can find a lot of 'happy numbers' such as: 310, 70, 86, 130, 7, 5555, 1212, 13, 1000 and other numbers which are variations of these, e.g. swapping the numbers the other way around (13, 31) and adding zeros onto the end of numbers (31, 310). Also you get numbers that add up to other happy numbers e.g. 5555 - four fives squared = 100. The following sequences are the patterns of fixed points and cycles in base eight that I have found. These are fixed points so they are happy numbers and any numbers which start sequences ending like this are happy numbers. 64, 64, 62, 64, 64, $\\ldots$ 24, 24, 24, 24, 24, $\\ldots$ 1, 1, 1, 1, ,1, $\\ldots$ These are 2-cycles: 32, 15, 32, 15, 32, 15, $\\ldots$ 20, 4, 20, 4, 20, 4, $\\ldots$ This is a 3-cycle: 31, 12, 5, 31, 12, 5, 31, 12, 5, $\\ldots$ Pen Areecharoenlert, also from The Mount School, York found some more happy numbers and cycles in base eight. Pen says \"There does not appear to be any pattern in which ones are happy, but it looks as though there are 3 fixed points: 1, 24 and 64.\" This is Pen's list. 1, 1, 1, $\\ldots$ so 1 is happy. 2,", "10404, 10816, \\ldots$ for $x=49,\\ldots,53$ and beyond. But that doesn't make for a very satisfying puzzle, so I suspect that's not the answer you're looking for. Next number could be 530 (Which is square of 23 plus 1) As, the first 5 numbers were obtained by adding 4 to the squares of even numbers- 2,4,6,8 and 10. Then, 1 is getting added to the squares of odd numbers - 11, 13, 15, 17, 19, 21. So the next number could be square of 23 that is 529 and 1 added to it. • Not bad, but there is a more elegant way, which doesn't discriminate between the first five numbers and the remaining numbers. Jun 12, 2018 at 0:12 It looks like the answer is 530 Because the difference increases by 8 each time although that breaks down for 104, 122, 170 - so unless that's a mistake my answer is wrong", "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "# Another Think Out Of The Boxes(The Id-\"1\"-- ) What is the next number in the sequence $1, 11, 21, 1211, \\ldots?$ ×", "\\mod 97: k = 0, 1, \\ldots, 23\\}$ such as these given below: $$1, 10, 3, 30, 9, -7, 27, -21, -16, 34, -48, 5, -47, 15, -44, 45, -35, 38, -8, 17, -24, -46, 25, -41.$$ For instance, in your example, you can put the $24$-digit number $182\\ldots100$ above and these $24$ numbers (in reversed order) below to get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 & -44 & 15 & -47 \\ldots \\\\ \\\\ \\ldots 2 & 3 & 4 & 5 & 6 & 7 & 2 & 3 & 2 & 1 & 0 & 0 \\\\ \\ldots 5 & -48 & 34 & -16 & -21 & 27 & -7 & 9 & 30 & 3 & 10 & 1\\end{array}$$ Calculating the product of those pairs of numbers above and below, you get $$\\begin{array}{cccccccccccc} 1 & 8 & 2 & 3 & 1 & 6 & 1 & 1 & 0 & 0 & 0 & 1 \\ldots \\\\ -41 & 25 & -46 & -24 & 17 & -8 & 38 & -35 & 45 &", "# Definition:Lucky Number ## Definition $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, \\ldots$ Remove every $2$nd number from $2$ onwards (that is, all the even integers): $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, \\ldots$ The $2$nd term is $3$. Starting from the $3$rd element in this list (which is $5$), remove every $3$rd element from what is left: $1, 3, 7, 9, 13, 15, 19, 21, 25, \\ldots$ The $3$rd number is now $7$. Starting from the $7$th element in this list (which is $19$), remove every $7$th element from what is left: $1, 3, 7, 9, 13, 15, 21, 25, \\ldots$ The numbers remaining are the lucky numbers. ### Sequence of Lucky Numbers The sequence of lucky numbers begins: $1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, 51, 63, 67, 69, 73, 75, 79, 87, 93, 99, \\ldots$ ## Also see • Results about lucky numbers can be found here.", "3 are there from 100 to 500? Again, let's start by listing some of them out: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots$ At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots 492,\\, 495,\\, 498$ Now let's divide each number in the list by 3: $34,\\, 35,\\, 36,\\, 37,\\, \\ldots 164,\\, 165,\\, 166$ See what happened? The numbers are now consecutive. Now the question becomes, \"How many numbers are there from 34 to 166?\" If you're not sure, ask yourself how many numbers there are from 5 to 10. There are 6. Now how did you get that? Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1: $10 - 5 + 1 = 6$ So to get the number of numbers from 34 to 166, we do the same thing: $166 - 34 + 1 = 133$ Finally, we have the answer, 133. Although this problem may seem long and complicated, the method we used to solve it applies to many other questions like this (namely, multiple questions). Once you understand it, solving similar questions is a breeze. Hopefully, you now have a good grasp", "3 are there from 100 to 500? Again, let’s start by listing some of them out: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots$ At this point, it’s not hard to see that this will take forever. Here’s the trick—instead of listing out all the numbers in the middle, let’s list out a few at the end: $102,\\, 105,\\, 108,\\, 111,\\, \\ldots 492,\\, 495,\\, 498$ Now let’s divide each number in the list by 3: $34,\\, 35,\\, 36,\\, 37,\\, \\ldots 164,\\, 165,\\, 166$ See what happened? The numbers are now consecutive. Now the question becomes, “How many numbers are there from 34 to 166?” If you’re not sure, ask yourself how many numbers there are from 5 to 10. There are 6. Now how did you get that? Perhaps you didn’t realize it, but you subtracted 5 from 10 and then added 1: $10 – 5 + 1 = 6$ So to get the number of numbers from 34 to 166, we do the same thing: $166 – 34 + 1 = 133$ Finally, we have the answer, 133. Although this problem may seem long and complicated, the method we used to solve it applies to many other questions like this (namely, multiple questions). Once you understand it, solving similar questions is a breeze. Hopefully, you now have a good grasp", ". . . . . . . . . . . . . . . . . . $138$ Since the largest number is 145, the sequence looks like this: . . $\\begin{array}{cccccccccccc}131 & 133 & 135 & 137 & 139 & 141 & 143 & 145 \\end{array}$ . - . - . . . . . . . . . . $\\uparrow$ Therefore, the smallest integer is 131.", "us to find all the values taking $\\pm 1$ in the three following recurrent linear sequences satisfying $a_n = 5a_{n-1} - 7a_{n-2}$, • $1,2,3,1,-16,-87,-323,\\ldots$ • $0,1,5,18,55,149,360,\\ldots$ • $1,3,8,19,39,62,37,\\ldots$ Looking at the recurrence relation mod $18$, we have $a_n \\equiv 5a_{n-1}+11a_{n-2} \\equiv 5(5a_{n-2}+11a_{n-3}) + 11a_{n-2} \\equiv a_{n-3}$. Looking at the first three terms of each sequence modulo $18$, we can already discard $2/3$ of all the terms and also remove the possibility of a $-1$ appearing, which reduces the problem to finding the $1$ values in the linear recurrent sequences satisfying $a_n = 20a_{n-1}-343_{n-2}$ • $1,1,-323,-6803,\\ldots$ • $1,55,757,-3725,\\ldots$ • $1,19,37,-5777,\\ldots$ Now each sequence is a linear combination of the coefficient sequences of $(2-\\omega)^{3n} = (1-18\\omega)^n = (10 - 9 \\sqrt{-3})^n$, that is, the two sequences • $1,10,-143,-6290,\\ldots$ • $0,-9,-180,-513, \\ldots$ (with integer coefficents in all three cases) With the binomial theorem, we can expand $(1+9(1-\\sqrt{-3}))^n = 1 + 9(1-\\sqrt{-3})n + 81(1-\\sqrt{-3})^2n(n-1)/2 + \\ldots$. If we place ourselves in $\\Bbb Z_3[\\sqrt{-3}]$ and we notice that $v_3(9^k/k!) >= 3k/2 \\to \\infty$, we can reorder the summation and get $1 + [9(1-\\sqrt{-3}) + O(81)]n + O(81)n^2 + \\ldots$ i.e. each coefficient sequence can be extended as a function $\\Bbb Z_3 \\mapsto \\Bbb Z_3$ which is a power series in $n$ with coefficients in $\\Bbb Z_3$ Now, if a power series is of", "# Sequence of Powers of 16 The sequence of powers of $16$ begins: $1, 16, 256, 4096, 65 \\, 536, 1 \\, 048 \\, 576, 16 \\, 777 \\, 216, 268 \\, 435 \\, 456, \\ldots$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "opposite direction. So we need to create similar imbalances in the Subsets $A_1$ to $A_{117}$, so that we could add the above $117$ numbers to those imbalances to keep the total sum unchanged. Now start swapping the $1st$ number of each pair that we added to subsets $A_1$ to $A_{117}$ to create the above imbalances: Swap $1$ from $A_1$ with $59$ from $A_{59}$ - this creates an imbalance of $+58$ and $-58$ Swap $2$ from $A_2$ with $58$ from $A_{58}$ - this creates an imbalance of $+56$ and $-56$ $\\ldots$ Swap $29$ from $A_{29}$ with $31$ from $A_{31}$ - creates an imbalance of $+2$ and $-2$ Leave $30$ in $A_{30}$ as it is - $0$ imbalance. Similarly, Swap $60$ from $A_{60}$ with $117$ from $A_{117}$ - this creates an imbalance of $+57$ and $-57$ Swap $61$ from $A_{61}$ with $116$ from $A_{116}$ - this creates an imbalance of $+55$ and $-55$ $\\ldots$ Swap $88$ from $A_{88}$ with $89$ from $A_{89}$ - this creates an imbalance of $+1$ and $-1$ Now start adding the $117$ numbers from $(1)$ to the subsets $A_1$ to $A_{117}$ to counter the imbalances $-58$ to $+58$ so that the sum remains unchanged. Notice that each subset now has $17$ elements with total sum = $8*1990 + 995 = 16915$. $- Kris17$", "# What is the next item in this pattern: 100, 81, 64, 49,....? Oct 5, 2016 The sequence is ${i}^{2}$ for $i = 10 , \\ldots .$ ${10}^{2} , {9}^{2} , {8}^{2} , {7}^{2} , \\ldots .$ Hence the next element is: ${6}^{2} = 36$", "find whole number values of $a$ and $b$ satisfying $aF(n-2) + bF(n-1) = 196$. and so we need to find whole number solutions to these equations for $n = 4, 5, 6, \\ldots12$. We shall consider one remaining case and leave the rest to the reader. For $n = 6$ we seek values of $a$ and $b$ such that $3a + 5b = 196$ There are no solutions for $a = 1$ because then b would not be a whole number. We have already seen that $a = 2$ and $b = 38$ gives $196$ as the sixth term. For larger values of $a$ we have to take smaller values of $b$. For $a = 3$ or $4$ or $5$ or $6$ there are again no solutions because b has to be a whole number. For $a = 7$ we have: $21 + 5b$ $=$ $196$ $5b$ $=$ $175$ $b$ $=$ $35$ giving the sequence $7, 35, 42, 77, 119, 196, \\ldots$ To find the remaining solutions for $n = 6$ we increase $a$ by steps of $5$ and decrease $b$ by steps of $3$. There are five solutions for $n = 6$, which are: • $2, 38, \\ldots$ • $7, 35, \\ldots$ • $12, 32, \\ldots$ • $17, 29,\\ldots$ • $22, 26,\\ldots$ You can use the same", "13 & 169,171,\\ldots,193,195\\\\ 118 & 15 & 226,228,\\ldots,252,254\\\\ 151 & 16 & 256,257,\\ldots,287,288\\\\ 169 & 17 & 289,291,\\ldots,321,323\\\\ 188 & 19 & 362,364,\\ldots,396,398\\\\ 229 & 20 & 400,401,\\ldots,439,440\\\\ 251 & 21 & 441,443,\\ldots,481,483\\\\ 274 & 23 & 530,532,\\ldots,572,574\\\\ 323 & 24 & 576,577,\\ldots,623,624\\\\ 349 & 25 & 625,627,\\ldots,673,675\\\\ 376 & 27 & 730,732,\\ldots,780,782\\\\ 433 & 28 & 784,785,\\ldots,839,840\\\\ 463 & 29 & 841,843,\\ldots,897,899\\\\ 483 & 31 & 962,964,\\ldots,998,1000\\\\ \\end{array}$$ Notes • $\\color{blue}{[1]}$ Above analysis implies none of the $n$ with $2^2 \\le n < 3^2$ satisfies the condition. This is different from what you claim in the question. I've verified by hand and by wolfram alpha that your claim is invalid. $$\\lfloor\\sqrt x\\rfloor=\\begin{cases}1,~x\\in[1,4)\\\\ 2,~x\\in[4,9)\\\\ 3,~x\\in[9,16)\\\\ \\vdots\\\\ 31,~x\\in[961,1000]\\end{cases}$$ Use this to rewrite the integral as, $$I=\\left(\\sum_{i=1}^{p-1}\\int\\limits_{i^2}^{(i+1)^2}ix\\,\\mathrm dx\\right)+\\int\\limits_{p^2}^npx\\,\\mathrm dx$$ where $p\\in\\Bbb{Z}$ such that $p^2\\leq n\\leq (p+1)^2$ $$I=\\sum_{i=1}^{p-1}\\left(\\frac{i}{2}(4i^3+6i^2+4i+1)\\right)+\\frac{p}{2}(n^2-p^4)$$ • I'm not sure if this will result in a nice simplification or not, but you can try to examine it and take it further. – Prasun Biswas Apr 16 '15 at 23:04 Let $k$ be the greatest integer $k$ such that $k^2\\leq n$ henece: $$\\int_1^n x \\lfloor \\sqrt x \\rfloor dx=\\sum_{i=1}^{k-1}i\\int_{i^2}^{(i+1)^2}xdx+k\\int_{k^2}^nxdx=\\sum_{i=1}^{k-1}i\\frac{(i+1)^4-i^4}{2}+\\frac{k(n^2-k^4)}{2}$$ • That's what I was thinking of too! But I don't think it'll be $(2i+1)$. – Prasun Biswas Apr 16 '15 at 23:00 • Because, on evaluation of that part, it's", "whose digital sum is 14. This only works up to a certain number, of course, because you'll hit something like $1,234,567,890$ to $1,234,567,990$, failing. – Christopher Marley Jan 2 at 1:01 One of these 100 consecutive integers $$a$$ will end in 00. Say that the remainder of $$a$$ divided by 14 is $$v\\in\\{0,1,\\ldots,13\\}$$. Case 1. There exist among these 100 consecutive integers at least 49 larger than $$a$$. In such case the remainders of the sum of the digits of $$a,a+1,\\ldots,a+49$$ will range between $$a\\!\\!\\!\\!\\!\\!\\mod\\!\\! 14,\\,\\, a+1\\!\\!\\!\\!\\!\\!\\mod\\!\\! 14,\\ldots, a+49\\!\\!\\!\\!\\!\\!\\mod\\!\\! 14,$$ and hence will cover the whole of $$\\{0,1,\\ldots,13\\}$$. Case 2. There exist among these 100 consecutive integers at least 50 smaller than $$a$$. Then we shall 50 numbers with two last digits ranging for 50 to 99 and all the other digits identical, and if the remainder the sum of the digits of the one ending in 50 is $$w$$ then we shall have as remainders $$w\\!\\!\\!\\!\\!\\!\\mod\\! 14, w+1\\!\\!\\!\\!\\!\\!\\mod 14,\\ldots, w+49\\!\\!\\!\\!\\!\\!\\mod 14$$ which again will cover the whole of $$\\{0,1,\\ldots,13\\}$$. • Note the OP asked for the sum of the digits being divisible by $14$, not the number itself. Your proof is basically correct, but there are $2$ things you should change. First, have $v$ and $w$ be the remainders of the number when divided by $14$. Second,", "# What is the max of $n$ such that $\\sum_{i=1}^n\\frac{1}{a_i}=1$ where $2\\le a_1\\lt a_2\\lt\\cdots\\lt a_n\\le 99$? What is the max of $n$ such that $$\\sum_{i=1}^n\\frac{1}{a_i}=1$$ where $a_{i}\\ (i=1,2,\\cdots,n)$ are integers which satisfy $2\\le a_1\\lt a_2\\lt\\cdots\\lt a_n\\le 99$ ? Also, I need how to prove that the $n$ you get is the maximum. My approach: I'm going to represent $\\sum_{i=1}^n\\frac{1}{a_i}$ as $(a_1,a_2,\\cdots,a_n).$ $(2,3,6)\\rightarrow(4,5,7,9, 12, 15,18,30,42,45,90)$ $\\rightarrow(8,9,10, 12, 14, 15, 16, 18,22,27, 30, 35, 40, 42, 45, 48, 54, 56, 60, 72, 90, 99)$ (here I used $(3)=(4,12), (6)=(7,42)$ etc.) This is the $n=22$ case. Update: I've just got the following $n=42$ case. I don't know if this is the max. $(8,9,10, 12, 14, 15, 16, 18,22,27, 30, 35, 40, 42, 45, 48, 54, 56, 60, 72, 90, 99)\\rightarrow(15,17,20,21,22,26,27,30,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,75,76,77,78,80,84,85,88,90,91,95,96,99)$ Here I used $(10)=(17,34,85), (14)=(28,44,77)$ etc. Update 2 : I've just got another $n=42$ case. $(17,18,20,21,22,24,26,27,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,72,75,76,77,78,80,84,85,88,91,95,96,99)$ Here I used $(15,30,90)=(18,24,72)$. Update 3 : I crossposted to MO. https://mathoverflow.net/questions/142300/what-is-the-max-of-n-such-that-sum-i-1n-frac1a-i-1-where-2-le-a-1-l • I think $n_{\\max}=22$, and you can see math.stackexchange.com/questions/128371/… – math110 Sep 9 '13 at 7:36 • @math110: Thank you, but I need a proof. – mathlove Sep 9 '13 at 7:39 • I generated all transformation rules with up to 4 numbers $(a)=(b,c,d,e)$ and among these there are 35 which never pop up on the right hand side: 23,25,29,31,37,41,42,43,46,47,49,51,53,58,59,61,62,64,67,68,69,71,73,74,79,81,82,83,86,87,89,92,93,94,97,98. There is only", "reasoning for the other equivalent classes. Also notice that $[i] \\cap \\left\\{ 1, \\ldots, 300 \\right\\}$ has the same cardinality for $i=1,\\ldots, 6$. Hence the maximum number is $1+258/2=130$ Remark: Let's count how many numbers are in $[6]_7 \\cap \\left\\{1, \\ldots, 300 \\right\\}$, the first number is $6$ and the last number is $300$. They can be written as $7i+6$ where $i=0, \\ldots 42$, hence there are 43 numbers which is a mistake that you make. Another mistake is you forgot to exclude additional multiple of $7$'s. Notice that there are 42 elements in $[0]_7 \\cap \\left\\{1, \\ldots, 300 \\right\\}.$ $$300-3 \\times 43-42+1=130$$" ]

[ "ROUGH DRAFT authorea.com/99863 # The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats How many shirt, pants, and hat combinations are there? Let’s list off what we know: There are 7 shirts There are 5 pairs of pants There are 12 hats We know that shirt and pants", "## College Algebra (10th Edition) As the shirts and ties are totally independent, meaning the choice of the shirt doesn't affect the tie, the number of outfits will be the variation of the two. A man has 5 shirts and with each shirt, the man can wear 3 ties. $5\\times 3= 15$", "# The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats", "the number of possible combinations 3 x 3 = 9 So, Abigail can make 9 different kinds of outfits. Problem2: Abigail has a green and purple shirt to match with either a white or red pair of pants We need to find the number of different outfits that Abigail can make I drew a table to show the number of possible combinations 2 x 2 = 4 So, Abigail can make 4 different kinds of outfits. Scroll to Top", "# A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and pair of slacks, how many different outfits can the man make? 150 Explanation Total number of jackets=3Total number of shirts=10 Total number of pair of slacks=5 Ways of selecting one jacket$$=3 \\mathrm{C}_1=3$$ Ways of selecting one shirt $$= 10 \\mathrm{C}_1=10$$ Ways of selecting one pair of slack $$= 5 \\mathrm{C}_1=5$$ Number of different outfits that can be made$$=3\\times10\\times5=150$$ Number of different outfits that can be made=150", "pants, all different? Same for the shirts and shoes. And what exactly are we trying to count? All possible outfits for the 4 people? What I mean is that the 4 people have already decided their outfits, but the department store still needs to give each of the 4 people what they wanted. The question is asking, how many ways can the store give these people their outfits while still adhering to the rule that they must give a person pants first, then shirt, then shoes Sorry. The problem makes no sense to me for the reasons I already specified. The store as exactly 4 of each item, all distinct The order makes no difference. I don't understand why they're emphasizing that. Yea it does 10:51 PM OK, so if you and I both want brown pants, the store can't make us both happy. Why does the order matter? No no. The store has the exact quantity of what we each want. Its just a matter of how many different orders can the store give out the materials to the people So if theres just two people, the store could either do: [Pants1 , Shirt1, Shoes1, Pants2, Shirt2, Shoes2] or [Pants1, Pants2, Shirt1, Shirt2, Shoes2, Shoes1] or [Pants2, Shirt2, Pants1, Shoes2, Shirt1, Shoes1] , etc. So we don't", "of 78 and HCF of 13. Option (d) is correct Ans . 51 and 34 1. Explanation : Solve using options. Option (d) 51 and 34 satisfies the required conditions. Ans . all 1. Explanation : 2,4,6,8,0 at the end makes number divisible by 2 Ans . 3 1. Explanation : the last digit being 1 can never be obtained if x is 2,5,6. Ans . Exactly 11 1. Explanation : Ties < Trousers < Shirts. Since each of the three is minimum 11, the total would be a minimum of 33 (for all 3). The remaining 5 need to be distributed amongst ties, trousers and shirts so that they can maintain the inequality Ties < Trousers < Shirts This can be achieved with 11 ties, and the remaining 27 pieces of clothing distributed between trousers and shirts such that the shirts are greater than the trousers. This can be done in at least 2 ways: 12 trousers and 15 shirts; 13 trousers and 14 shirts. If you try to go for 12 ties, the remaining 26 pieces of clothing need to be distributed amongst shirts and trousers such that the shirts are greater than the trousers and both are greater than 12. With only 26 pieces of clothing to be distributed between shirts and trousers this is", "## lgbasallote Group Title A person has 5 new shirts and 3 new pants. If an outfit consists of a shirt and a pant, how many distinct outfits can the person have? 2 years ago 2 years ago 15 2. lgbasallote why is the solution to this 5 * 3? why doesn't it use combination? one pair goes with 5 different shirts to make 5 outfits while another pants pair makes another 5 outfits while the last pants pair makes 5 more outfits which im pretty sure leads to 5+5+5=15 if there is another way you are suppose to do ithe problem i don't know it 4. LolWolf It is order-dependent. Since each shirt-pants combination is unique, then we have that it is simply the amount of ways in which one can pick any numbering of shirts with any numbering of pants, and have a different combination in each case. I.e. Pants 1, Shirt 2 is a different combination than Pants 2, Shirt 1. 5. lgbasallote yes....but why is it 5 * 3 and not combination? 6. lgbasallote the question is the number of ways you can pick...so combination should be used right? 7. LolWolf We use combinations on order-independent events. E.g. cards. Saying we have a hand of cards 1, 2, 3 is the same as saying", "# how many different outfits can be put together if you may choose one of 7 jeans, one of 4 shirts and one of 3 belts? texelaare 2021-02-08 Answered how many different outfits can be put together if you may choose one of 7 jeans, one of 4 shirts and one of 3 belts? You can still ask an expert for help Expert Community at Your Service • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the price of one coffee • Available 24/7 • Math expert for every subject • Pay only if we can solve it ## Expert Answer StrycharzT Answered 2021-02-09 Author has 102 answers The Fundamental Counting Principle states that if there are pp outcomes for one event and qq outcomes for another event, then there are p×qp×q ways of both events occurring. Therefore, if there are 7 ways to choose a pair of jeans, 4 ways to choose a shirt, and 3 ways to choose a belt, there are $7×4×3=28×3=84$ different outfits possible. We have step-by-step solutions for your answer! Expert Community at Your Service • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the", "choosing a shirt, a tie and four accessories seperately, each and every choice will be different from each other. That's why we can directly apply product rule as Useful (nevermind his current name :D) stated.", "# Making Outfits in 2015 John has some number of distinct shirts, some number of distinct pairs of pants, and some number of distinct pairs of shoes. With these, he can make $$2015$$ different outfits consisting of a shirt, a pair of pants, and a pair of shoes. • If he gets one more pair of shoes, he can make $$403$$ more different outfits. • If he gets two more shirts, he can make $$130$$ more different outfits. • If he gets three more pairs of pants, he can make $$465$$ more outfits. What is the combined number of shirts, pants, and shoes that John currently has? ×", "To start, it's worth noting that you CANNOT have a 'fraction' of a piece of clothing. We're told that the ratio of shirts:dresses:jackets is 9:4:5, so the number of shirts MUST be a multiple of 9, the number of dresses MUST be an equivalent multiple of 4 and the number of jackets MUST be an equivalent multiple of 5. I'm going to put together a quick list of the first few potential possibilities... There COULD be... 9 shirts/4 dresses/5 jackets 18 shirts/8 dresses/10 jackets 27 shirts/12 dresses/15 jackets 36 shirts/16 dresses/20 jackets Etc. We're also told that there are MORE than 7 dresses. We're asked for the total number of articles of clothing. (1) The total number of shirts and jackets in the closet is less than 30. With this Fact, we can look at our notes and find whatever options fit this information (AND include MORE than 7 dresses). There's only one... 18 shirts/8 dresses/10 jackets Fact 1 is SUFFICIENT. (2) The total number of shirts and dresses in the closet is 26. With this Fact, we can again look at the available options. There's only one... 18 shirts/8 dresses/10 jackets Fact 2 is SUFFICIENT. GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at [email protected] ### Best Conversation Starters 1 lheiannie07 84 topics 2 LUANDATO", "# Create all possible combinations taking one item from each group I'm not a mathematician nor is this a mathematics question per se, instead it's a real life problem I need the solution for. I have three groups of different items let's say Group 1: T-Shirt, Solid Shirt, Patterned Shirt Group 2: Pants, Jeans Group 3: Blazers, Jackets, Coats, Sweaters I need to make a list of all possible unique combinations taking one item from each group. Is there a formula of doing it? For instance: Tshirt + Pants + Sweaters; Solid Shirt + Jeans + Blazers • What are your thoughts? – Wojciech Karwacki Jan 5 '16 at 14:45 • Well I'm writing a blog on men's fashion and variety in fashion that can be created like Solid Shirt + Jeans + Blazer is a unique outfit. How many more like these can be created? – Vipul K Jan 5 '16 at 14:48 You can chose from the first group in $3$ way, from the second one in $2$ ways, from the third one in $4$ ways. Every choice is indipendent. Thus you can make $3 \\times 2 \\times 4 = 24$ combinations. As a general formula, if you have $n$ groups, indicating with $|Group_k|$ the number of elements of the k-th group, you can make this", "that it would be okay to wear more than two ties to your clown college interview. 1. You must select some of your ties to wear. Everything is okay, from no ties up to all ties. How many choices do you have? 2. If you want to wear at least one regular tie and one bow tie, but are willing to wear up to all your ties, how many choices do you have for which ties to wear? 3. How many choices of which ties to wear do you have if you wear exactly 2 of the 3 regular ties and 3 of the 5 bow ties? 4. Once you have selected 2 regular and 3 bow ties, in how many orders could you put the ties on, assuming you must have one of the three bow ties on top? Solution. 1. $$2^8 = 256$$ choices. You have two choices for each tie: wear it or don’t. 2. You have 7 choices for regular ties (the 8 choices less the “no regular tie” option) and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total you have $$7 \\cdot 31 = 217$$ choices. 3. $${3\\choose 2}{5\\choose 3} = 30$$ choices. 4. Select one of the 3 bow ties to go on top. There", "8 shirts - 2 specific avoided - 6 shirts 6C1 * 6C1 = 36 (select one from each for a combination) What is wrong ? Rgds, TGC ! hi TGC, Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants? it means he is going to wear all pants and shirts except few combinations he is going to avoid. so there are 2 groups now 6 shirts ==>they all have choices of 9 pants therefore total ways = $$6C1*9C1 = 54$$ NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = $$2C1*6C1 =12$$ hence total choice = $$54 +12 =66$$ what you are doing is that you are just removing those shirts and pants. hope it helps I think you didn't get my question. My question is that what is wrong in my method? Rgds, TGC ! You are interpreting question in wrong way. let say there are 9 pants numbered :1 2 3 4 5 6 7 8 9 AND 8 shirts numbered : 10 11 12 13 14 15 16 17 now question is saying you cannot wear pants numbered : 1, 2, 3 with shirt 10 and 11", "1. ## Counting Problem You have more than one pair of shoes, more pairs of pants than pairs of shoes, and more shirts than pairs of pants. You notice that you can combine those items to obtain 105 distinct outfits, each consisting of one pair of shoes, one pair of pants, and one shirt. You deduce that you have _______ pairs of shoes, _______ pairs of pants, and _______ shirts. I know this is probably easier than it seems, can I get some help. Thank you 2. Hello, thebristolsound! You have more than one pair of shoes, more pairs of pants than pairs of shoes, and more shirts than pairs of pants. You notice that you can combine those items to obtain 105 distinct outfits, each consisting of one pair of shoes, one pair of pants, and one shirt. You deduce that you have ___ pairs of shoes, ___ pairs of pants, and ___ shirts. We are told: . $1 \\:<\\:\\text{shoes} \\:<\\:\\text{pants} \\:<\\:\\text{shirts}$ We find that: . $105 \\:=\\:3\\cdot5\\cdot7$ Therefore: . $\\begin{Bmatrix}\\text{3 shoes} \\\\ \\text{5 pants} \\\\ \\text{7 shirts} \\end{Bmatrix}$", "Ratio Question Wed Jul 13, 2016 10:14 am The only articles of clothing in a certain closet are shirts, dresses, and jackets. The ratio of the number of shirts to the number of dresses to the number of jackets in the closet is 9:4:5, respectively. If there are more than 7 dresses in the closet, what is the total number of articles of clothing in the closet? (1) The total number of shirts and jackets in the closet is less than 30. (2) The total number of shirts and dresses in the closet is 26. ### GMAT/MBA Expert Scott@TargetTestPrep GMAT Instructor Joined 25 Apr 2015 Posted: 891 messages Followed by: 5 members Upvotes: 43 Fri Dec 08, 2017 7:39 am mv2019 wrote: The only articles of clothing in a certain closet are shirts, dresses, and jackets. The ratio of the number of shirts to the number of dresses to the number of jackets in the closet is 9:4:5, respectively. If there are more than 7 dresses in the closet, what is the total number of articles of clothing in the closet? (1) The total number of shirts and jackets in the closet is less than 30. (2) The total number of shirts and dresses in the closet is 26. We are given that the ratio of shirts to", "has four ties, twelve shirts, and three belts. If each day he we [#permalink] ### Show Tags 20 Jul 2018, 03:14 Bunuel wrote: John has four ties, twelve shirts, and three belts. If each day he wears exactly one tie, one shirt, and one belt, what is the maximum number of days he can go without repeating a particular combination? Number of ways to select a Tie = 4C1 = 4 Number of ways to select a Shirt = 12C1 = 12 Number of ways to select a Belt = 3C1 = 3 Therefore, Possible Number of ways combined = 4*12*3 = 12*12 = 144 Hence, D. _________________ The few, the fearless ! Thanks Intern Joined: 17 May 2018 Posts: 31 Re: John has four ties, twelve shirts, and three belts. If each day he we [#permalink] ### Show Tags 20 Jul 2018, 03:30 Why use combinations? Simply multiply the number of items. For each of the four ties, you can choose among 12 shirts = 4 x 12 possible combinations. For each of these combinations, you can choose among three belts so le't multiply the previous number by 3 = 4 x 12 x 3. _________________ ¿Tienes que presentar el GMAT y no sabes por dónde empezar? ¡Visita GMAT para Principiantes y recibe el curso completo gratis!", "make consisting of 1 shirt, 1 pair of pants, and 1 pair of shoes is A. 40 B. 60 C. 240 D. 120 Ed has 5 shirts, 4 pairs of pants, and 3 pairs of shoes. The total number of outfits Ed can make consisting of 1 shirt, 1 pair of pants, and 1 pair of shoes is A. 40 B. 60 C. 240 D. 120... ### In a survey a bunch of students were asked to name their favorite elective class. There were 15 students who chose \"other\". How many students participated? In a survey a bunch of students were asked to name their favorite elective class. There were 15 students who chose \"other\". How many students participated?... ### The state of_______ has upper and lower peninsulas.Michigan Minnesota ohio The state of_______ has upper and lower peninsulas.Michigan Minnesota ohio... ### Pls help me as soon as possible. pls help me as soon as possible.... ### Think about how Mattie feels when she first sees the notebooks in the excerpt from a northern light. Write a response analyzing how Mattie’s reaction to the notebooks advances the plot of the story. Use details from the story to support your ideas Think about how Mattie feels when she first sees the notebooks in the excerpt from a northern light. Write", "4 : 5, then there are infinitely many scenarios: - There are 9 shirts, 4 dresses and 5 jackets - There are 18 shirts, 8 dresses and 10 jackets - There are 27 shirts, 12 dresses and 15 jackets - There are 36 shirts, 16 dresses and 20 jackets - There are 45 shirts, 20 dresses and 25 jackets etc... Also given: There are more than 7 dresses in the closet This means the first scenario (9 shirts, 4 dresses and 5 jackets) is not possible. This leaves us with the following cases: case i: 18 shirts, 8 dresses and 10 jackets case ii 27 shirts, 12 dresses and 15 jackets case iii 36 shirts, 16 dresses and 20 jackets case iv 45 shirts, 20 dresses and 25 jackets . . . etc Statement 1: The total number of shirts and jackets in the closet is less than 30 When we check the possible cases, we see that only case i meets this condition. This means there MUST be 18 shirts, 8 dresses and 10 jackets So, the answer to the target question is the total number of articles of clothing = 18 + 8 + 10 = 36 Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The total number of" ]

[ "8 shirts - 2 specific avoided - 6 shirts 6C1 * 6C1 = 36 (select one from each for a combination) What is wrong ? Rgds, TGC ! hi TGC, Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants? it means he is going to wear all pants and shirts except few combinations he is going to avoid. so there are 2 groups now 6 shirts ==>they all have choices of 9 pants therefore total ways = $$6C1*9C1 = 54$$ NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = $$2C1*6C1 =12$$ hence total choice = $$54 +12 =66$$ what you are doing is that you are just removing those shirts and pants. hope it helps I think you didn't get my question. My question is that what is wrong in my method? Rgds, TGC ! You are interpreting question in wrong way. let say there are 9 pants numbered :1 2 3 4 5 6 7 8 9 AND 8 shirts numbered : 10 11 12 13 14 15 16 17 now question is saying you cannot wear pants numbered : 1, 2, 3 with shirt 10 and 11", "## College Algebra (10th Edition) As the shirts and ties are totally independent, meaning the choice of the shirt doesn't affect the tie, the number of outfits will be the variation of the two. A man has 5 shirts and with each shirt, the man can wear 3 ties. $5\\times 3= 15$", "Many Different Outfits The point is this: There are 2 possibilities for pants. For each of those, there are two possibilities for shirt, giving 4 = 2 * 2 possibilities for pants and shirt. For each of those, there are two possibilities for tie, giving 8 = 2 * 2 * 2 possibilities for pants, shirt, and tie. For each of those, there are two possibilities for jacket, giving 16 = 2 * 2 * 2 * 2 possibilities for pants, shoes, tie, and jacket. And if there were also two belts, there would be 32 = 2 * 2 * 2 * 2 * 2 possibilities. The sequence is $2^n \\ne n^2$ except by accident in the special case where n = 4 because $2^4 = 16 = 4^2.$ 9. Re: How Many Different Outfits And this special case is why you see it on tests, you know they love to throw you off when WE dont know the basics, as stated, thank you for the explaination, I have not seen a good basic explaination of this scenario, I still find myself drawing pictures of all permutations when the numbers are small, but now I think I have it so thanks. If you have a deck of cards, 52, what are the chances of getting a 3", "ROUGH DRAFT authorea.com/99863 # The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats How many shirt, pants, and hat combinations are there? Let’s list off what we know: There are 7 shirts There are 5 pairs of pants There are 12 hats We know that shirt and pants", "2015, 5:48 p.m. Would it be 1- (99 choose 1) (199 choose 1) all over (200)(100)? My reasoning would be that the chances of choosing at least one blue item would be one minus the ways of choosing anything but blue. weisbart: Feb. 8, 2015, 8:44 p.m. Working with the complement is a good idea. I suppose I just want to point out that 6 comes up in the following way: 1 blue shirt paired with 2 non blue pants, 3 non blue shirts paired with 1 blue pair of pants, finally, 1 blue shirt and 1 blue pair of pants. This gives $1\\times 2 + 3\\times 1 + 1\\times 1 = 6$ appropriate outfits. I just want to point out that it is very easy to over count the doubled blue outfit.", "can choose from 14 plain blue dresses or 10 blue embroidered dresses. So, the total numbers of option are $14+10 = 24$ The Counting Principle and Rule of Sum We have discussed the problems related to the fundamental counting principle and the rule of sum. What if a problem has to be solved using both the rules? Let us discuss a few examples. Example 5: Harris went to a shop to buy some items of clothing. He can choose a shirt from 3 different colors or a T-shirt from 2 different colors. Also, he can either get a pair of pants from 3 available colors or a pair of jeans from three available colors. How many possible outfits can Harris have? Solution: Harris can take one shirt from the given 3 colors or T-shirts from the available 2 colors. So number of available shirts $= 3+2 =5$. Similarly, he can choose a pair of pants from either 3 options or a pair of jeans from three options. The number of available pairs of jeans/pants $3+3=6$. So, we have $5$ different shirts and $6$ different pants, and the total number of possible outfits will be $5 \\times 6= 30$. Example 6: Steve wants to go to a job interview in another city. He can choose from $4$ local bus", "+0 What are all the possible Combinations? 0 137 1 +123 In his closet, John has one red shirt, one blue shirt, one pair of shorts, one pair of jeans, and a pair of slacks. What are the possible shirt-and-pant combinations Nov 27, 2018 #1 +2340 +1 For shirts, John has two options. He can either choose to wear the red shirt or the blue one. For pants, John has a total of three options. One pair of shorts, jeans, and slacks equals three options. The total number of combinations can be found by multiplying the number of independent options together. In this case, picking a shirt has no impact on which pair of pants John wears, so we can just multiply them together. $$2*3=6 \\text{ shirt-and-pants combinations}$$ . Nov 27, 2018", "get one $\\spadesuit$ Therefore, there are: . $715 \\times 13 \\:=\\:9295$ hands with 4 $\\diamondsuit \\text{s and 1 }\\spadesuit.$ 4. John has five shirts, three ties, and seven suits. How many possible outfits have 1 shirt, 1 tie, and 1 suit? Another \"fundamental\" problem. He has: . $5 \\times 3 \\times 7 \\:=\\:105$ possible outfits. 5. A locker combination system uses three digits from 0 to 9. How many different three-digit combinations with no digit repeated are possible? Yet another \"fundamental\" problem. The 1st digit has 10 choices. The 2nd digit has 9 choices. The 3rd digit has 8 choices. Therefore, there are: . $10 \\times 9 \\times 8 \\:=\\:720$ possible combinations.", "# The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats", "allowed, the answer is $10^4\\text{.}$ Without repetition, you would have $P(10,4)\\text{.}$ 25. ${10 \\choose 4}$ since that is equal to ${9 \\choose 4} + {9 \\choose 3}\\text{.}$ 26. Neither. It will be a complicated (possibly PIE) counting problem. ##### 1.7.3 Solution 1. $2^8 = 256$ choices. You have two choices for each tie: wear it or don't. 2. You have 7 choices for regular ties (the 8 choices less the “no regular tie” option) and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total you have $7 \\cdot 31 = 217$ choices. 3. ${3\\choose 2}{5\\choose 3} = 30$ choices. 4. Select one of the 3 bow ties to go on top. There are then 4 choices for the next tie, 3 for the tie after that, and so on. Thus $3\\cdot 4! = 72$ choices. ##### 1.7.4 Solution You own 8 purple bow ties, 3 red bow ties, 3 blue bow ties and 5 green bow ties. How many ways can you select one of each color bow tie to take with you on a trip? $8 \\cdot 3 \\cdot 3 \\cdot 5$ ways. How many choices do you have for a single bow tie to wear tomorrow? $8 + 3 + 3 + 5$ choices. ##### 1.7.5 Solution 1. $4^5$ numbers.", "# Multiple Category Combination Problem I need to figure out the total number of combinations for three categories of items which must be combined, and I'm trying to determine the appropriate way to do so. Let's say the categories are as follows: Shirts: 3 Total, must choose 1 Ties: 18 Total, must choose 1 Additional Accessories: 20 Total, must choose 4 I must choose 1 shirt, 1 tie, and 4 accessories, which constitutes an outfit. I understand how to develop a \"normal\" combinatorics problem (have 20, choose 4), but I'm not sure I understand the appropriate way to develop the total number of \"outfits\" from this example...What is the appropriate way to determine how many outfits can be created? Thank you in advance, • You have 3 choices for the shirt, 18 choices for the tie, and $\\binom{20}{4}$ possible combinations of accessories. So by the product rule, what is your total? – Mauve Jan 24 '18 at 21:31 • Well, I think, by rule of product it would be 261,630...That's multiplying through the combinations for each. Is that correct? I'm trying to determine if that's #1, appropriate, and #2, correct...Is this the correct application of Rule of Product? – jules325 Jan 24 '18 at 21:33 The answer is $$\\binom{3}{1}\\binom{18}{1}\\binom{20}{4} = 261630$$ because when we make an outfit by", "# A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and pair of slacks, how many different outfits can the man make? 150 Explanation Total number of jackets=3Total number of shirts=10 Total number of pair of slacks=5 Ways of selecting one jacket$$=3 \\mathrm{C}_1=3$$ Ways of selecting one shirt $$= 10 \\mathrm{C}_1=10$$ Ways of selecting one pair of slack $$= 5 \\mathrm{C}_1=5$$ Number of different outfits that can be made$$=3\\times10\\times5=150$$ Number of different outfits that can be made=150", "that it would be okay to wear more than two ties to your clown college interview. 1. You must select some of your ties to wear. Everything is okay, from no ties up to all ties. How many choices do you have? 2. If you want to wear at least one regular tie and one bow tie, but are willing to wear up to all your ties, how many choices do you have for which ties to wear? 3. How many choices of which ties to wear do you have if you wear exactly 2 of the 3 regular ties and 3 of the 5 bow ties? 4. Once you have selected 2 regular and 3 bow ties, in how many orders could you put the ties on, assuming you must have one of the three bow ties on top? Solution. 1. $$2^8 = 256$$ choices. You have two choices for each tie: wear it or don’t. 2. You have 7 choices for regular ties (the 8 choices less the “no regular tie” option) and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total you have $$7 \\cdot 31 = 217$$ choices. 3. $${3\\choose 2}{5\\choose 3} = 30$$ choices. 4. Select one of the 3 bow ties to go on top. There", "of just 17 clothing items. Better yet is 3 hats, 3 pairs of shoes, 3 scarves, 3 pairs of pants and 5 shirts for a total 405 days of fashion. 405 was the best I could do in my head, but I wrote a Python script to try it out, and it came up with 2 pairs of shoes, 3 shirts, 3 pairs of pants, 3 hats, 3 scarves and 3 belts for a total of 486 days on only 17 items clothing. [footnote 3 for that script] If you want to know how to make n outfits out of least number of clothing items, factor n completley: $\\dpi{120} n = A^aB^bC^cD^d...$ and your total number of clothing items is m: $\\dpi{120} m = A a + Bb + Cc + Dd...$ What's the strategy for for finding the optimal number? I'm not sure. Is 17 the least number of clothing items needed to go a whole year without repeat outfits? I think so, but I can't prove it. Yes: We already know that 17 is a candidate for the least number of clothing items. That means any n will have to be in the form: $\\dpi{120} \\\\ n = 2^a \\cdot 3^b \\cdot 5^c \\cdot 7^d \\cdot 11^e \\cdot 13^f \\cdot 17^g \\\\ a<9,b<6,c<4,d<3,e<2,f<2,g<2$ If our optimal", "the other? • The five different choices of shirt and eight different choices of pairs of jeans can form $$5 \\times 8$$ different arrangements at the ends of the rail. • There are $$2!$$ different ways to arrange a shirt at one end and a pair of jeans on the other: S - - - - - - - - - - - J and J - - - - - - - - - - - S • If a shirt is at one end and a pair of jeans at the other, there remains $$11!$$ different arrangements of the remaining clothing items. Therefore, there are: $$2 \\times 8 \\times 5 \\times 11! = \\text{3 193 344 000} \\text{ different ways to arrange the clothing}$$ The probability of a clothing arrangement with a shirt at one end and a pair of jeans at the other $$= \\frac{\\text{3 193 344 000}}{\\text{6 227 020 800}} = \\text{0,513}$$ A photographer places eight chairs in a row in his studio in order to take a photograph of the debating team. The team consists of three boys and five girls. In how many ways can the debating team be seated? $8! = \\text{40 320}$ What is the probability that a particular boy and a particular girl sit next to each other? Regard", "+0 +1 117 2 +69 Jimmy is going on a quick vacation, and is packing some clothes. He has twelve shirts, and wants to take three of them with him. How many ways can Jimmy choose his shirts? rohitaop May 18, 2018 #1 +87604 +2 He can choose any 3 of the 12 shirts = C (12,3) = 220 different choices CPhill May 18, 2018 #2 +2766 +1 This is combinations because he can choose any 3 shirts from the 12 total. We use the formula, $$\\frac{n!}{r!*(n-r)!}$$ Plugging the values in, we have: $$\\frac{12!}{3!*(9!)}=\\frac{12*11*10}{3*2*1}=44*5=\\boxed{220}$$ ways. tertre May 19, 2018", "number of outfits without listing all possibilities. Rashid has 3 choices for a shirt, and for each of the 3 choices of a shirt he has 2 choices for pants, and for each shirt/pants combination he has 2 choices for socks. So he has a total of $$3 \\times 2 \\times 2 = 12$$ different outfits he can put together. 2. There are two combinations that include all blue items: light blue shirt/blue jeans/blue socks dark blue shirt/blue jeans/blue socks 3. Since there are 12 different outfits and 2 of them contain all blue items, $$\\tfrac{2}{12}$$, or equivalently $$\\tfrac{1}{6}$$, of the total outfits contain all blue items. 4. All 4 of the combinations that have a burgundy shirt will include something burgundy. As well, half of the combinations that have a light blue shirt will have burgundy pants, so 2 of these outfits include something burgundy. Similarly, there are 2 outfits that will have a dark blue shirt that include something burgundy. This is a total of $$4+2+2=8$$ outfits that will include a burgundy item. This means that $$\\tfrac{8}{12}$$, or equivalently $$\\tfrac{2}{3}$$, of the total outfits include at least one burgundy item. ### Teacher’s Notes This problem asks for the number of different outfits Rashid could wear. Mathematically, this is asking to find a number of combinations. Assuming", "## Archive for September, 2010 ### How to Count…Better This post is an introduction to combinatorics, which is how to count without actually counting. For example, you have 12 pairs of shoes, 16 shirts, and 8 pairs of pants in your closet. How many different outfits can you make out of these, assuming that you have to wear a pair of shoes, a shirt, and a pair of pants (and assuming everything matches :)). Listing out all of the combinations would be tedious! Better is to use the following method. There are 12 choices for shoes and 16 shirts. So there are $12 \\cdot 16$ shirt-pant combinations! If this is unclear, think about drawing a grid with 12 slots on one side and 16 on the other. Each box represents a different combination, and there are $12 \\cdot 16$ boxes. This method generalizes to more than 2 things; there are $12 \\cdot 16 \\cdot 8 = 1536$ shoe-shirt-pant combinations. Don’t try listing that out 😛 Now suppose that there are 20 people in a class, and you need to pick a class president, vice president, and secretary. How many ways are there to do this? This is an application of our method. It doesn’t matter, but say you pick the president first. There are 20 choices for the", "with 3 specific pants? it means he is going to wear all pants and shirts except few combinations he is going to avoid. so there are 2 groups now 6 shirts ==>they all have choices of 9 pants therefore total ways = $$6C1*9C1 = 54$$ NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = $$2C1*6C1 =12$$ hence total choice = $$54 +12 =66$$ what you are doing is that you are just removing those shirts and pants. hope it helps I think you didn't get my question. My question is that what is wrong in my method? Rgds, TGC ! You are interpreting question in wrong way. let say there are 9 pants numbered :1 2 3 4 5 6 7 8 9 AND 8 shirts numbered : 10 11 12 13 14 15 16 17 now question is saying you cannot wear pants numbered : 1, 2, 3 with shirt 10 and 11 but this doesnt mean that you cannot wear pants numbered : 1, 2, 3 with shirt 12 13 14 15 16 17 ===>you are missing this case. what you are doing is that you are just removing pants 1,2,3 and shirts 10,11. thats why you are having 6 shirts and 6 pants. hope", "of 78 and HCF of 13. Option (d) is correct Ans . 51 and 34 1. Explanation : Solve using options. Option (d) 51 and 34 satisfies the required conditions. Ans . all 1. Explanation : 2,4,6,8,0 at the end makes number divisible by 2 Ans . 3 1. Explanation : the last digit being 1 can never be obtained if x is 2,5,6. Ans . Exactly 11 1. Explanation : Ties < Trousers < Shirts. Since each of the three is minimum 11, the total would be a minimum of 33 (for all 3). The remaining 5 need to be distributed amongst ties, trousers and shirts so that they can maintain the inequality Ties < Trousers < Shirts This can be achieved with 11 ties, and the remaining 27 pieces of clothing distributed between trousers and shirts such that the shirts are greater than the trousers. This can be done in at least 2 ways: 12 trousers and 15 shirts; 13 trousers and 14 shirts. If you try to go for 12 ties, the remaining 26 pieces of clothing need to be distributed amongst shirts and trousers such that the shirts are greater than the trousers and both are greater than 12. With only 26 pieces of clothing to be distributed between shirts and trousers this is" ]

[ "## College Algebra (10th Edition) As the shirts and ties are totally independent, meaning the choice of the shirt doesn't affect the tie, the number of outfits will be the variation of the two. A man has 5 shirts and with each shirt, the man can wear 3 ties. $5\\times 3= 15$", "ROUGH DRAFT authorea.com/99863 # The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats How many shirt, pants, and hat combinations are there? Let’s list off what we know: There are 7 shirts There are 5 pairs of pants There are 12 hats We know that shirt and pants", "# A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and pair of slacks, how many different outfits can the man make? 150 Explanation Total number of jackets=3Total number of shirts=10 Total number of pair of slacks=5 Ways of selecting one jacket$$=3 \\mathrm{C}_1=3$$ Ways of selecting one shirt $$= 10 \\mathrm{C}_1=10$$ Ways of selecting one pair of slack $$= 5 \\mathrm{C}_1=5$$ Number of different outfits that can be made$$=3\\times10\\times5=150$$ Number of different outfits that can be made=150", "# The Rule of Product Math 381 Blog # 1 How many shirt and pants combinations can I wear if I have 7 shirts and 5 pairs of pants? This and other questions can be answered by the basic concepts of combinatorics. We will be using something called The Rule of Product. According to Arthur T. Benjamin, “The rule of product states that if an action can be performed by making A choices followed by B choices, then it can be performed A*B ways.” Let’s look at the shirt and pants problem first. Again, we have 7 shirts (A shirts): and 5 pairs of pants (B pairs of pants) Forget fashion for a minute. How many possible outfits could you have? Well, think about this. If we pick the pink shirt, then we have 5 choices of pants to wear with that shirt. Since there are 7 shirts and 5 pairs of pants, we must use the rule of products (A*B) to give us $$7\\cdot 5=35$$ different outfit combinations. Now let’s think of another problem. Say I have 12 hats", "Many Different Outfits The point is this: There are 2 possibilities for pants. For each of those, there are two possibilities for shirt, giving 4 = 2 * 2 possibilities for pants and shirt. For each of those, there are two possibilities for tie, giving 8 = 2 * 2 * 2 possibilities for pants, shirt, and tie. For each of those, there are two possibilities for jacket, giving 16 = 2 * 2 * 2 * 2 possibilities for pants, shoes, tie, and jacket. And if there were also two belts, there would be 32 = 2 * 2 * 2 * 2 * 2 possibilities. The sequence is $2^n \\ne n^2$ except by accident in the special case where n = 4 because $2^4 = 16 = 4^2.$ 9. Re: How Many Different Outfits And this special case is why you see it on tests, you know they love to throw you off when WE dont know the basics, as stated, thank you for the explaination, I have not seen a good basic explaination of this scenario, I still find myself drawing pictures of all permutations when the numbers are small, but now I think I have it so thanks. If you have a deck of cards, 52, what are the chances of getting a 3", "can choose from 14 plain blue dresses or 10 blue embroidered dresses. So, the total numbers of option are $14+10 = 24$ The Counting Principle and Rule of Sum We have discussed the problems related to the fundamental counting principle and the rule of sum. What if a problem has to be solved using both the rules? Let us discuss a few examples. Example 5: Harris went to a shop to buy some items of clothing. He can choose a shirt from 3 different colors or a T-shirt from 2 different colors. Also, he can either get a pair of pants from 3 available colors or a pair of jeans from three available colors. How many possible outfits can Harris have? Solution: Harris can take one shirt from the given 3 colors or T-shirts from the available 2 colors. So number of available shirts $= 3+2 =5$. Similarly, he can choose a pair of pants from either 3 options or a pair of jeans from three options. The number of available pairs of jeans/pants $3+3=6$. So, we have $5$ different shirts and $6$ different pants, and the total number of possible outfits will be $5 \\times 6= 30$. Example 6: Steve wants to go to a job interview in another city. He can choose from $4$ local bus", "that it would be okay to wear more than two ties to your clown college interview. 1. You must select some of your ties to wear. Everything is okay, from no ties up to all ties. How many choices do you have? 2. If you want to wear at least one regular tie and one bow tie, but are willing to wear up to all your ties, how many choices do you have for which ties to wear? 3. How many choices of which ties to wear do you have if you wear exactly 2 of the 3 regular ties and 3 of the 5 bow ties? 4. Once you have selected 2 regular and 3 bow ties, in how many orders could you put the ties on, assuming you must have one of the three bow ties on top? Solution. 1. $$2^8 = 256$$ choices. You have two choices for each tie: wear it or don’t. 2. You have 7 choices for regular ties (the 8 choices less the “no regular tie” option) and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total you have $$7 \\cdot 31 = 217$$ choices. 3. $${3\\choose 2}{5\\choose 3} = 30$$ choices. 4. Select one of the 3 bow ties to go on top. There", "8 shirts - 2 specific avoided - 6 shirts 6C1 * 6C1 = 36 (select one from each for a combination) What is wrong ? Rgds, TGC ! hi TGC, Question is: Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants? it means he is going to wear all pants and shirts except few combinations he is going to avoid. so there are 2 groups now 6 shirts ==>they all have choices of 9 pants therefore total ways = $$6C1*9C1 = 54$$ NOW the left 2 shirts is going to avoid 3 pants therefore we have 6 pants hence total ways = $$2C1*6C1 =12$$ hence total choice = $$54 +12 =66$$ what you are doing is that you are just removing those shirts and pants. hope it helps I think you didn't get my question. My question is that what is wrong in my method? Rgds, TGC ! You are interpreting question in wrong way. let say there are 9 pants numbered :1 2 3 4 5 6 7 8 9 AND 8 shirts numbered : 10 11 12 13 14 15 16 17 now question is saying you cannot wear pants numbered : 1, 2, 3 with shirt 10 and 11", "of 78 and HCF of 13. Option (d) is correct Ans . 51 and 34 1. Explanation : Solve using options. Option (d) 51 and 34 satisfies the required conditions. Ans . all 1. Explanation : 2,4,6,8,0 at the end makes number divisible by 2 Ans . 3 1. Explanation : the last digit being 1 can never be obtained if x is 2,5,6. Ans . Exactly 11 1. Explanation : Ties < Trousers < Shirts. Since each of the three is minimum 11, the total would be a minimum of 33 (for all 3). The remaining 5 need to be distributed amongst ties, trousers and shirts so that they can maintain the inequality Ties < Trousers < Shirts This can be achieved with 11 ties, and the remaining 27 pieces of clothing distributed between trousers and shirts such that the shirts are greater than the trousers. This can be done in at least 2 ways: 12 trousers and 15 shirts; 13 trousers and 14 shirts. If you try to go for 12 ties, the remaining 26 pieces of clothing need to be distributed amongst shirts and trousers such that the shirts are greater than the trousers and both are greater than 12. With only 26 pieces of clothing to be distributed between shirts and trousers this is", "# Multiple Category Combination Problem I need to figure out the total number of combinations for three categories of items which must be combined, and I'm trying to determine the appropriate way to do so. Let's say the categories are as follows: Shirts: 3 Total, must choose 1 Ties: 18 Total, must choose 1 Additional Accessories: 20 Total, must choose 4 I must choose 1 shirt, 1 tie, and 4 accessories, which constitutes an outfit. I understand how to develop a \"normal\" combinatorics problem (have 20, choose 4), but I'm not sure I understand the appropriate way to develop the total number of \"outfits\" from this example...What is the appropriate way to determine how many outfits can be created? Thank you in advance, • You have 3 choices for the shirt, 18 choices for the tie, and $\\binom{20}{4}$ possible combinations of accessories. So by the product rule, what is your total? – Mauve Jan 24 '18 at 21:31 • Well, I think, by rule of product it would be 261,630...That's multiplying through the combinations for each. Is that correct? I'm trying to determine if that's #1, appropriate, and #2, correct...Is this the correct application of Rule of Product? – jules325 Jan 24 '18 at 21:33 The answer is $$\\binom{3}{1}\\binom{18}{1}\\binom{20}{4} = 261630$$ because when we make an outfit by", "of just 17 clothing items. Better yet is 3 hats, 3 pairs of shoes, 3 scarves, 3 pairs of pants and 5 shirts for a total 405 days of fashion. 405 was the best I could do in my head, but I wrote a Python script to try it out, and it came up with 2 pairs of shoes, 3 shirts, 3 pairs of pants, 3 hats, 3 scarves and 3 belts for a total of 486 days on only 17 items clothing. [footnote 3 for that script] If you want to know how to make n outfits out of least number of clothing items, factor n completley: $\\dpi{120} n = A^aB^bC^cD^d...$ and your total number of clothing items is m: $\\dpi{120} m = A a + Bb + Cc + Dd...$ What's the strategy for for finding the optimal number? I'm not sure. Is 17 the least number of clothing items needed to go a whole year without repeat outfits? I think so, but I can't prove it. Yes: We already know that 17 is a candidate for the least number of clothing items. That means any n will have to be in the form: $\\dpi{120} \\\\ n = 2^a \\cdot 3^b \\cdot 5^c \\cdot 7^d \\cdot 11^e \\cdot 13^f \\cdot 17^g \\\\ a<9,b<6,c<4,d<3,e<2,f<2,g<2$ If our optimal", "# how many different outfits can be put together if you may choose one of 7 jeans, one of 4 shirts and one of 3 belts? texelaare 2021-02-08 Answered how many different outfits can be put together if you may choose one of 7 jeans, one of 4 shirts and one of 3 belts? You can still ask an expert for help Expert Community at Your Service • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the price of one coffee • Available 24/7 • Math expert for every subject • Pay only if we can solve it ## Expert Answer StrycharzT Answered 2021-02-09 Author has 102 answers The Fundamental Counting Principle states that if there are pp outcomes for one event and qq outcomes for another event, then there are p×qp×q ways of both events occurring. Therefore, if there are 7 ways to choose a pair of jeans, 4 ways to choose a shirt, and 3 ways to choose a belt, there are $7×4×3=28×3=84$ different outfits possible. We have step-by-step solutions for your answer! Expert Community at Your Service • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the", "get one $\\spadesuit$ Therefore, there are: . $715 \\times 13 \\:=\\:9295$ hands with 4 $\\diamondsuit \\text{s and 1 }\\spadesuit.$ 4. John has five shirts, three ties, and seven suits. How many possible outfits have 1 shirt, 1 tie, and 1 suit? Another \"fundamental\" problem. He has: . $5 \\times 3 \\times 7 \\:=\\:105$ possible outfits. 5. A locker combination system uses three digits from 0 to 9. How many different three-digit combinations with no digit repeated are possible? Yet another \"fundamental\" problem. The 1st digit has 10 choices. The 2nd digit has 9 choices. The 3rd digit has 8 choices. Therefore, there are: . $10 \\times 9 \\times 8 \\:=\\:720$ possible combinations.", "allowed, the answer is $10^4\\text{.}$ Without repetition, you would have $P(10,4)\\text{.}$ 25. ${10 \\choose 4}$ since that is equal to ${9 \\choose 4} + {9 \\choose 3}\\text{.}$ 26. Neither. It will be a complicated (possibly PIE) counting problem. ##### 1.7.3 Solution 1. $2^8 = 256$ choices. You have two choices for each tie: wear it or don't. 2. You have 7 choices for regular ties (the 8 choices less the “no regular tie” option) and 31 choices for bow ties (32 total minus the “no bow tie” option). Thus total you have $7 \\cdot 31 = 217$ choices. 3. ${3\\choose 2}{5\\choose 3} = 30$ choices. 4. Select one of the 3 bow ties to go on top. There are then 4 choices for the next tie, 3 for the tie after that, and so on. Thus $3\\cdot 4! = 72$ choices. ##### 1.7.4 Solution You own 8 purple bow ties, 3 red bow ties, 3 blue bow ties and 5 green bow ties. How many ways can you select one of each color bow tie to take with you on a trip? $8 \\cdot 3 \\cdot 3 \\cdot 5$ ways. How many choices do you have for a single bow tie to wear tomorrow? $8 + 3 + 3 + 5$ choices. ##### 1.7.5 Solution 1. $4^5$ numbers.", "2015, 5:48 p.m. Would it be 1- (99 choose 1) (199 choose 1) all over (200)(100)? My reasoning would be that the chances of choosing at least one blue item would be one minus the ways of choosing anything but blue. weisbart: Feb. 8, 2015, 8:44 p.m. Working with the complement is a good idea. I suppose I just want to point out that 6 comes up in the following way: 1 blue shirt paired with 2 non blue pants, 3 non blue shirts paired with 1 blue pair of pants, finally, 1 blue shirt and 1 blue pair of pants. This gives $1\\times 2 + 3\\times 1 + 1\\times 1 = 6$ appropriate outfits. I just want to point out that it is very easy to over count the doubled blue outfit.", "To start, it's worth noting that you CANNOT have a 'fraction' of a piece of clothing. We're told that the ratio of shirts:dresses:jackets is 9:4:5, so the number of shirts MUST be a multiple of 9, the number of dresses MUST be an equivalent multiple of 4 and the number of jackets MUST be an equivalent multiple of 5. I'm going to put together a quick list of the first few potential possibilities... There COULD be... 9 shirts/4 dresses/5 jackets 18 shirts/8 dresses/10 jackets 27 shirts/12 dresses/15 jackets 36 shirts/16 dresses/20 jackets Etc. We're also told that there are MORE than 7 dresses. We're asked for the total number of articles of clothing. (1) The total number of shirts and jackets in the closet is less than 30. With this Fact, we can look at our notes and find whatever options fit this information (AND include MORE than 7 dresses). There's only one... 18 shirts/8 dresses/10 jackets Fact 1 is SUFFICIENT. (2) The total number of shirts and dresses in the closet is 26. With this Fact, we can again look at the available options. There's only one... 18 shirts/8 dresses/10 jackets Fact 2 is SUFFICIENT. GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at [email protected] ### Best Conversation Starters 1 lheiannie07 84 topics 2 LUANDATO", "the number of possible combinations 3 x 3 = 9 So, Abigail can make 9 different kinds of outfits. Problem2: Abigail has a green and purple shirt to match with either a white or red pair of pants We need to find the number of different outfits that Abigail can make I drew a table to show the number of possible combinations 2 x 2 = 4 So, Abigail can make 4 different kinds of outfits. Scroll to Top", "pants, all different? Same for the shirts and shoes. And what exactly are we trying to count? All possible outfits for the 4 people? What I mean is that the 4 people have already decided their outfits, but the department store still needs to give each of the 4 people what they wanted. The question is asking, how many ways can the store give these people their outfits while still adhering to the rule that they must give a person pants first, then shirt, then shoes Sorry. The problem makes no sense to me for the reasons I already specified. The store as exactly 4 of each item, all distinct The order makes no difference. I don't understand why they're emphasizing that. Yea it does 10:51 PM OK, so if you and I both want brown pants, the store can't make us both happy. Why does the order matter? No no. The store has the exact quantity of what we each want. Its just a matter of how many different orders can the store give out the materials to the people So if theres just two people, the store could either do: [Pants1 , Shirt1, Shoes1, Pants2, Shirt2, Shoes2] or [Pants1, Pants2, Shirt1, Shirt2, Shoes2, Shoes1] or [Pants2, Shirt2, Pants1, Shoes2, Shirt1, Shoes1] , etc. So we don't", "1. ## Counting Problem You have more than one pair of shoes, more pairs of pants than pairs of shoes, and more shirts than pairs of pants. You notice that you can combine those items to obtain 105 distinct outfits, each consisting of one pair of shoes, one pair of pants, and one shirt. You deduce that you have _______ pairs of shoes, _______ pairs of pants, and _______ shirts. I know this is probably easier than it seems, can I get some help. Thank you 2. Hello, thebristolsound! You have more than one pair of shoes, more pairs of pants than pairs of shoes, and more shirts than pairs of pants. You notice that you can combine those items to obtain 105 distinct outfits, each consisting of one pair of shoes, one pair of pants, and one shirt. You deduce that you have ___ pairs of shoes, ___ pairs of pants, and ___ shirts. We are told: . $1 \\:<\\:\\text{shoes} \\:<\\:\\text{pants} \\:<\\:\\text{shirts}$ We find that: . $105 \\:=\\:3\\cdot5\\cdot7$ Therefore: . $\\begin{Bmatrix}\\text{3 shoes} \\\\ \\text{5 pants} \\\\ \\text{7 shirts} \\end{Bmatrix}$", "# Create all possible combinations taking one item from each group I'm not a mathematician nor is this a mathematics question per se, instead it's a real life problem I need the solution for. I have three groups of different items let's say Group 1: T-Shirt, Solid Shirt, Patterned Shirt Group 2: Pants, Jeans Group 3: Blazers, Jackets, Coats, Sweaters I need to make a list of all possible unique combinations taking one item from each group. Is there a formula of doing it? For instance: Tshirt + Pants + Sweaters; Solid Shirt + Jeans + Blazers • What are your thoughts? – Wojciech Karwacki Jan 5 '16 at 14:45 • Well I'm writing a blog on men's fashion and variety in fashion that can be created like Solid Shirt + Jeans + Blazer is a unique outfit. How many more like these can be created? – Vipul K Jan 5 '16 at 14:48 You can chose from the first group in $3$ way, from the second one in $2$ ways, from the third one in $4$ ways. Every choice is indipendent. Thus you can make $3 \\times 2 \\times 4 = 24$ combinations. As a general formula, if you have $n$ groups, indicating with $|Group_k|$ the number of elements of the k-th group, you can make this" ]

[ "+0 # Fractions 0 82 1 Express as a fraction in lowest terms: $0.\\overline{12} + 0.\\overline{001}$ May 25, 2022", "+0 # Express $4.\\overline{054}$ as a common fraction in lowest terms. 0 126 1 Express $4.\\overline{054}$ as a common fraction in lowest terms. Aug 31, 2020 #1 0 If we expand it out, we get 4.054054054..... Setting this equal to x, we have: 1000x = 4054.054054054... x = 4.054054054... Subtracting the equations, we get: 999x = 4050 x = 4050/999 x = 450/111 x = 150/37 Jan 14, 2021", "+0 Decimals 0 49 1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ May 15, 2022 #1 +13731 +1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ Hello Guest! $$x=0.\\overline{81}\\\\ 100x=81.\\overline{81}\\\\ 99x=81\\\\ x=\\dfrac{9}{11}$$ $$\\color{blue}0.\\overline{81}=\\dfrac{9}{11}$$ ! May 15, 2022", "a fraction! #### Examples ##### Question 1 Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. ##### Question 2 Starting with $x=0.\\overline{216}$x=0.216, express $x$x as a fraction in simplest form. ##### Question 3 Starting with $x=0.6\\overline{87}$x=0.687, express $x$x as a fraction in simplest form.", "# Express it in fraction Express $$0.461538461538...$$ as a fraction. (In other words, 461538 repeats forever which is called 0 point 461538 recurring.) ×", "# Convert recurring decimals to fractions ## Interactive practice questions Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. Easy Approx a minute Starting with $x=2.\\overline{3}$x=2.3, express $x$x as a fraction in simplest form. Starting with $x=1.\\overline{7}$x=1.7, express $x$x as a fraction in simplest form. Starting with $x=3.\\overline{6}$x=3.6, express $x$x as a fraction in simplest form.", "+0 # help asap 0 111 3 Express $$0.3\\overline{2}$$ as a fraction in simplest form. May 31, 2022 #1 0 $$0.3 \\overline{2} = \\dfrac{32}{99}$$ . May 31, 2022 #3 0 That's wrong!!! Guest Jun 1, 2022 #2 +2448 0 Let $$x = 0.3 \\overline 2$$. This means $$10x = 3. \\overline 2$$ and $$100x = 32. \\overline 2$$ Subtracting $$10x$$ from $$100x$$ gives us: $$90x=29$$, because the repeating part cancels each other out. Can you find the value of x from here? May 31, 2022", "downloads about What Is Rewrite 4 As A Fraction In Simplest Form. Find answers researching ebooks, papers or essays. ... Express as a fraction in simplest form: a) 0.806 b) 0.0256 c) 11.375 6. Rewrite as a decimal: a) 9 16 b) 17 40 c) 52 464 25i d) 7 12 e) 3 7 f) 24 11 ...", "+0 # Pls help 0 92 1 Compute:$$0.\\overline{7}-0.\\overline{4}+0.\\overline{2}$$ . Express your answer as a common fraction. Sep 22, 2020", "the fraction $k/p$ is not in lowest terms.", "HomeEnglishClass 7MathsChapterDecimals Express each of the following ... # Express each of the following as fraction is simplest form : <br> 0.345 Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, Text Solution 69/200", "Math Help - Fraction 1. Fraction Express as a simple fraction: [(1/2(x+h))-(2/3x)] / h 2. Is this what you're trying to show? $[1/2(x+h) - 2/3x]/h$ 3. Yeah, now I need to figure out where to start though.", "+0 # decimals 0 75 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jun 19, 2021 #1 0 0.4888... = 22/45 0.3777... = 17/45 22/45 + 17/45 = 39/45 Jun 20, 2021", "Express $\\dfrac{14}{-5}$ as a rational number with numerator $56$. • A $\\dfrac{56}{-20}$ • B $\\dfrac{56}{-5}$ • C $\\dfrac{56}{-10}$ • D $\\dfrac{56}{-15}$", "+0 # Decimals 0 35 5 Express as a common fraction the reciprocal of $0.\\overline{72}$. Jun 28, 2022 #1 +1 Express as a common fraction the reciprocal of $0.\\overline{72}$. I can give you the answer. The decimal equals 8 / 11 so its reciprocal would be 11 / 8. I can't tell you how to calculate it though. I guessed and got lucky. I guessed 8/9 ... that was too big, so I guessed 8/11 and that was it. . Jun 28, 2022 #2 +2275 +1 Let $$x = 0. \\overline{72}$$ This means $$100x = 72.\\overline{72}$$ Subtracting the 2 gives $$99x = 72$$, meaning $$x = {72 \\over 99} ={ 8 \\over 11}$$. The reciprocal of this, as Guest found, is $$\\color{brown}\\boxed{11 \\over 8}$$ Jun 28, 2022 #3 +69 +2 Just a fun fact, if it repeats every one digit, then the fraction version is [digit] / 9. If it repeats every two digits, then it is [digit] / 99. You can see the pattern here. ⚡⚡⚡ ShockFish Jun 28, 2022 edited by ShockFish Jun 28, 2022 #4 +2275 +1 Hi Shockfish, nice to see you here BuilderBoi Jun 28, 2022 #5 +69 +2 Heyyy just wondering if you were here too ShockFish Jun 28, 2022", "+0 # Subtract ​ 0. repeating decimal 02from 0.02 and express the result as a fraction. 0 23 1 +40 Subtract $$0.\\overline{02}$$ from 0.02 and express the result as a fraction. Jan 11, 2021", "a fraction, $0.\\overline{72}$, is equal to $\\dfrac{8}{11}$. 10. Which of the following shows the conversion of $0.48$ to fraction? 11. Which of the following shows the conversion of $5.25$ to fraction?", "Question Express $150 as a percentage of$500 (please give the answer in a sum way so that it is easy to understand) ## 30% Step-by-step explanation: Express $150 as a percentage of$500 (please give the answer in a sum way so that it is easy to understand) • solution with an equation 150 : 500 = x : 100 x = 150 * 100 : 500 x = 30 ————— check 150 : 500 = 30 : 100 0.3 = 0.3", "$v$ and $\\rho$ and then express $\\rho$ in terms of $z$.)", "+0 # Subtract ​ 0. repeating decimal 02from 0.02 and express the result as a fraction. 0 100 1 +236 Subtract $$0.\\overline{02}$$ from 0.02 and express the result as a fraction. Jan 11, 2021" ]

[ "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "+0 # pls help me 0 85 3 What is $\\frac{0.\\overline{72}}{0.\\overline{27}}$? Express your answer as a common fraction in lowest terms. Jul 27, 2022 #1 -1 The expression reduces to 2/9. Jul 27, 2022 #2 +2448 0 Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1... BuilderBoi Jul 28, 2022 #3 +2448 0 I'll do the numerator, and you can do the denominator. For the numerator, let $$x = 0.{ \\overline{72}}$$. This means $$100x = {72. \\overline{72}}$$ (move the decimal point 2 places to the right) Subtracting the 2 yields us $$99x = 72$$ (the repeating parts cancel out) Dividing by 99, we find that $$x = {72 \\over 99}$$ Jul 28, 2022", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "# How do you find the repeating decimal 0.82 with 82 repeated as a fraction? ##### 1 Answer Dec 8, 2016 $0. \\overline{82} = \\frac{82}{99}$ #### Explanation: In case you have not encountered it, let me introduce you to some notation: A repeating decimal can be represented using a bar placed over the repeating pattern of decimals. So instead of writing $0.828282 \\ldots$, you can write $0. \\overline{82}$ If we multiply $0. \\overline{82}$ by $\\left(100 - 1\\right)$ we get an integer: $\\left(100 - 1\\right) 0. \\overline{82} = 100 \\cdot 0. \\overline{82} - 1 \\cdot 0. \\overline{82}$ $\\textcolor{w h i t e}{\\left(100 - 1\\right) 0. \\overline{82}} = 82. \\overline{82} - 0. \\overline{82}$ $\\textcolor{w h i t e}{\\left(100 - 1\\right) 0. \\overline{82}} = 82$ Notice that the $100$ shifts the number two places to the left - the length of the repeating pattern. Then the $- 1$ cancels out the repeating tail. Next divide both ends by $\\left(100 - 1\\right)$ and simplify: $0. \\overline{82} = \\frac{82}{100 - 1} = \\frac{82}{99}$ $82$ and $99$ have no common factor, so this is in simplest terms. $\\textcolor{w h i t e}{}$ Alternative method An alternative method recognises that: $0. \\overline{82} = 0.82 + 0.0082 + 0.000082 + \\ldots$ is a geometric series, with initial term $a = 0.82$ and common ratio $r =", "# What is 0.8959595... as a fraction ? Jan 9, 2017 $0.8 \\overline{95} = \\frac{887}{990}$ #### Explanation: Given: $0.8 \\overline{95}$ First multiply by $10 \\left(100 - 1\\right) = 1000 - 10$ to get an integer. The first multiplier $10$ is to shift the repeating pattern to just after the decimal point. Then $\\left(100 - 1\\right)$ shifts the number $2$ places to the left and subtracts the original value to cancel out the repeating tail: $\\left(1000 - 10\\right) \\cdot 0.8 \\overline{95} = 895. \\overline{95} - 8. \\overline{95} = 887$ Then divide both ends by $\\left(1000 - 10\\right)$ to find: $0.8 \\overline{95} = \\frac{887}{1000 - 10} = \\frac{887}{990}$ $887$ and $990$ have no common factor larger than $1$, so this fraction is in simplest form.", "# Why is ⅓ = 0.3333....? • ## Why is it true that $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ ? I wish I had a very simple and clever way to show this, but all the simple and clever ways usually use some idea that is already a step more advanced. For example, you could say that $$0.\\overline{999}\\ldots = 1$$ and then divide both sides by $$3$$ (but knowing the fact that $$0.\\overline{999}\\ldots = 1$$ is even harder to prove!). Or you could start from the fact that $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ and multiply both sides by $$3,$$ but wait, isn't that just the reverse of what we did to get to this place? We were originally trying to show $$\\frac{1}{9} = 0.\\overline{111}\\ldots$$ in the first place, assuming that we knew the decimal representation of $$\\frac{1}{3}!$$ (Round and round we go!) I think the best way to show why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is just to show the long division. Then you'll find that $$1$$ divided by $$3$$ equals $$0.\\overline{333}\\ldots.$$ . This isn't hard, but there are some basic concepts to lay down first. : place value and the idea of division as a reverse-multiplication operation. ## Place value I have some affectionate names that I personally like to use when referring to place value digits. You can make up your own names,", "# Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \\overline{6}$<br/> <br/> (ii) $0.4 \\overline{7}$ <br/> <br/>(iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$", "Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. (i) $0 . \\overline{6}$ (ii) $0.4 \\overline{7}$ (iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$ Editor", "\\;=\\; \\frac{0.28}{0.40} \\;=\\;0.70$ $(b)\\;\\;P(\\overline{A}\\,|\\,\\overline{B}) \\;=\\;\\frac{P(\\overline{A} \\wedge \\overline{B})}{P(\\overline{B})} \\;=\\; \\frac{0.32}{0.40} \\;=\\;0.80$ 4. Originally Posted by Soroban Hello, jordanrs! Place the information in a chart . . . . . $\\begin{array}{c||c|c||c|} & P(B) & P\\left(\\overline{B}\\right) & \\text{Total} \\\\ \\hline \\hline P(A) & 0.32 & & 0.40 \\\\ \\hline \\\\[-4mm] P\\left(\\overline{A}\\right) & & & \\\\ \\hline \\hline \\text{Total} & 0.60 & & 1.00 \\\\ \\hline \\end{array}$ Fill in the empty cells: . . $\\begin{array}{c||c|c||c|} & P(B) & P\\left(\\overline{B}\\right) & \\text{Total} \\\\ \\hline \\hline P(A) & 0.32 & 0.08 & 0.40 \\\\ \\hline \\\\[-4mm] P\\left(\\overline{A}\\right) & 0.28 & 0.32 & 0.60 \\\\ \\hline \\hline \\text{Total} & 0.60 & 0.40 & 1.00 \\\\ \\hline \\end{array}$ $(a)\\;\\;P(B\\,|\\,\\overline{A}) \\;=\\;\\frac{P(B \\wedge \\overline{A})}{P(\\overline{A})} \\;=\\; \\frac{0.28}{0.40} \\;=\\;0.70$ $(b)\\;\\;P(\\overline{A}\\,|\\,\\overline{B}) \\;=\\;\\frac{P(\\overline{A} \\wedge \\overline{B})}{P(\\overline{B})} \\;=\\; \\frac{0.32}{0.40} \\;=\\;0.80$", "# Are numbers with repeating patterns in their decimal expansion (e.g. $0.123123123\\ldots$) rational? There's a question that I've been thinking about for quite some time now. We all know that numbers with infinite decimal expansion such as $0.\\overline{3}$ or $0.\\overline{1}$ are not necessarily irrational. $$0.\\overline{3}=0.333333\\dots=\\frac{1}{3}$$ $$0.\\overline{1}=0.111111\\ldots=\\frac{1}{9}$$ Therefore we can say that for instance $0.\\overline{4}$ is rational because we have: $$0.\\overline{4}=4\\cdot 0.\\overline{1}=\\frac{4}{9}$$ My question is: can we generalize these results to any number that has a repeating pattern in its decimal expansion? Can we claim for instance that $0.\\overline{123}=0.123123123\\ldots$ is rational? This question boils down to asking whether or not expressions like $0.\\overline{01}$ and $0.\\overline{001}$ etcetera are rational. If so, in the case of my example we could say: $$0.\\overline{123}=123\\cdot 0.\\overline{001}$$ A naive thought of mine was that $0.\\overline{01}$ is simply $\\frac{1}{10}\\cdot 0.\\overline{1}$ but this is of course not true. Then I thought that maybe we can represent $0.\\overline{01}$ as $\\sum\\limits_{i=1}^{\\infty}\\frac{1}{10^{2i}}$ which converges by the ratio test. However this does not tell us if it is rational or not. This is not for homework or anything, just something I've been thinking about and I was wondering if any of you have some knowledge about this issue. Thanks. - Yes, all repeating decimal numbers are rational. I'm sure someone here will give a good explanation but here's one link that'll", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}", "\\bar{3} = 0.33333 \\ldots,$$ and $$0.296 \\overline{412}$$ means “repeat the 412 forever”: $$0.296 \\overline{412} = 0.296412412412412 \\ldots$$ Now we’re in a position to give a perhaps more satisfying answer to the question $$1024 \\div 3$$. In the example above, we found the answer to be $$1024 \\div 3 = 341\\; \\text{R} 1 \\ldotp$$ But now we know we can keep dividing that last stubborn dot by 3. Remember, that represents a single dot in the ones place, so if we keep dividing by three it really represents $$\\frac{1}{3}$$. So we have: $$1024 \\div 3 = 341\\; \\text{R} 1 = 341 \\frac{1}{3} = 341.3333333 \\ldots = 341. \\bar{3} \\ldotp$$ Example: 6/7 As another (more complicated) example, here is the work that converts the fraction $$\\frac{6}{7}$$ to an infinitely long repeating decimal. Make sure to understand the steps one line to the next. With this 6 in the final right-most box, we have returned to the very beginning of the problem. (Do you see why? Remember, we started with a six in the ones box!) This means that we will simply repeat the work we have done and obtain the same sequence of answers: $$857142$$. And then again, and then again, and then again. We have: $$\\begin{split} \\frac{6}{7} &= 0.857142857142857142857142 \\ldots \\\\ &= 0. \\overline{857142} \\ldotp \\end{split}$$ 1. Compute $$\\frac{4}{7}$$", "feasible, perhaps dividing by $99$ could work? After all, the question does tell us to consider the fraction $\\frac{1}{99^2}$ and not $\\frac{1}{9801}$, so $\\frac{1}{99}$ should have some significance. As we can see from the working above, $\\frac{1}{99} = 0.\\overline{01}$. That seems nice enough! Armed with this fact, how can we obtain the decimal expression for $\\frac{1}{99^2}$? Whenever we encounter infinity, it’s a good idea to try small cases to get an idea of what the “infinite” case would look like. In this case, small cases would refer to truncating the decimal expansion of $0.\\overline{01}$ to see if we can find a pattern. Below we have 3 examples of this, where we have multiplied $0.\\overline{01}$ by 0.01, 0.0101 and 0.010101: Looks like we have a pattern! If we were to extrapolate our pattern, the decimal part of $0.\\overline{01} \\times 0.\\overline{01}$ would be 00, followed by 01, followed by 02, followed by 03, … There’s just one slight snag: since each step is allotted 2 digits in the decimal expansion, what would it look like when we hit “followed by 100”? Well, if we add it up, we get the following: The red 9 & 1 add up to the red 0 at the bottom and carry a 1 over to the blue 9. This in return results in the", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =", "1$ Hence, $62$ and $81$ are co-primes. iii) $69, 92$ HCF of $69, 92 = 23$ Hence, $69$ and $92$ are not co-primes. $\\\\$ Question 5: Find the greatest number that exactly divides $105, 1001$ and $2436$. We need to find the HCF of $105, 1001$ and $2436$. $\\begin{array}{r r r r r r r r } 1001) & \\overline{2436} & (2 & & & & & \\\\ & 2002 & & & & & & \\\\ & \\overline{434)} & \\overline{1001} & (2 & & & & \\\\ & & 868 & & & & & \\\\ & & \\overline{133)} & \\overline{434} & (3 & & & \\\\ & & & 399 & & & & \\\\ & & & \\overline{35)} & \\overline{133} & (3 & & \\\\ & & & & 105 & & & \\\\ & & & & \\overline{28)} & \\overline{35} & (1 & \\\\ & & & & & 28 & & \\\\ & & & & & \\overline{7)} & \\overline{28} & (4 \\\\ & & & & & & 28 & \\\\ & & & & & & \\overline{0} & \\end{array}$ $\\begin{array}{r r r } 7) & \\overline{105} & (15 \\\\ & 105 & \\\\ & \\overline{0} & \\end{array}$ Hence the HCF of the three numbers is $7$. That means, $7$ is the", "+0 # help asap 0 111 3 Express $$0.3\\overline{2}$$ as a fraction in simplest form. May 31, 2022 #1 0 $$0.3 \\overline{2} = \\dfrac{32}{99}$$ . May 31, 2022 #3 0 That's wrong!!! Guest Jun 1, 2022 #2 +2448 0 Let $$x = 0.3 \\overline 2$$. This means $$10x = 3. \\overline 2$$ and $$100x = 32. \\overline 2$$ Subtracting $$10x$$ from $$100x$$ gives us: $$90x=29$$, because the repeating part cancels each other out. Can you find the value of x from here? May 31, 2022", "(i) from (ii), we get: 9x = 12 = x = $\\frac{4}{3}$ $1.\\overline{)3}=\\frac{4}{3}$ (iii) $0.\\overline{)34}$ Let x$0.\\overline{)34}$ x = 0.3434... ...(i) 100x = 34.3434... ...(ii) On subtracting (i) from (ii), we get: 99x = 34 = x = $\\frac{34}{99}$ $0.\\overline{)34}=\\frac{34}{99}$ (iv) $3.\\overline{)14}$ Let x = $3.\\overline{)14}$ x = 3.1414... ...(i) 100x = 314.1414... ...(ii) On subtracting (i) from (ii), we get: 99x = 311 = x = $\\frac{311}{99}$ $3.\\overline{)14}$ = $\\frac{311}{99}$ (v) $0.\\overline{)324}$ Let x = $0.\\overline{)324}$ x = 0.324324... ...(i) 1000x = 324.324324... ...(ii) On subtracting (i) from (ii), we get: 999x = 324 = x = $\\frac{324}{999}$=$\\frac{12}{37}$ $0.\\overline{)324}$= $\\frac{12}{37}$ (vi) $0.1\\overline{)7}$ Let x = $0.1\\overline{)7}$ x = 0.1777... 10x = 1.777... ...(i) 100x = 17.777.... ...(ii) On subtracting (i) from (ii). we get: 90x = 16 = x = $\\frac{8}{45}$ $0.1\\overline{)7}$ = $\\frac{8}{45}$ (vii) $0.5\\overline{)4}$ Let x = $0.5\\overline{)4}$ x = 0.5444... 10x = 5.4444... ...(i) 100x = 54.4444... ...(ii) On subtracting (i) from (ii), we get: 90x = 49 = x = $\\frac{49}{90}$ $0.5\\overline{)4}$= $\\frac{49}{90}$ (viii) $0.1\\overline{)63}$ Let x$0.1\\overline{)63}$ x = 0.16363... 10x = 1.6363... ...(i) 1000x = 163.6363... ...(ii) On subtracting (i) from (ii), we get: 990x = 162 = x = $\\frac{162}{990}=\\frac{9}{55}$ $0.1\\overline{)63}$ = $\\frac{9}{55}$ #### Question 4: Write, whether the given statement is true or false. Give reasons. (i) Every natural number", "yields $$\\begin{matrix} 2&9&\\not 6^5&{}^1 2&.&9&\\not6^5&{}^1 2&9&\\not6^5&{}^12&... \\\\ & & 2&9&.&6&2&9&6&2&9&... \\\\\\hline \\\\ 2 & 9 & 3 & 3 & . & 3 & 3 & 3 & 3 & 3 & 3 & \\cdots \\end{matrix}$$ (I hope that is how they still notate subtraction these days) and so you have $$1000 x - 10 x = 2933 + \\frac{1}{3}$$ and $$1000 x - 10 x = 990 x$$ and so we've derived $$990 x = 2933 + \\frac{1}{3}$$ Of course, it would have been easier to compare $1000x$ to $x$.... - Hint: there's a very easy way to transform periodic into decimals without any calculation. Here's an example: $$0.123\\overline{4567}=\\frac{1234567-123}{9999000}.$$ In your case it's just $$2\\frac{962}{999}.$$ - The number $2.962962$ certainly is rational, because $$2.962962 = \\frac {2962962}{1000000}$$ BUT... if you mean $2.962962\\ldots$, supposed to mean $2.962\\overline{962} = 2.\\overline{962}$, then $$1000\\cdot 2.\\overline{962} = 2962.\\overline{962}$$ so $$1000\\cdot 2.\\overline{962} - 2.\\overline{962} = 2960$$ thus $$999\\cdot 2.\\overline{962} = 2960$$ and $$2.\\overline{962} = \\frac {2960}{999}$$ -", "first by $10^{s}$, where $s$ is the length of the period of the pure recurring decimal $x$. Hence, $$10^{s}x = q_{1}q_{2}\\ldots q_{s-1}q_{s}.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + 0.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + x.$$ Therefore, $$x = \\frac{q_{1}q_{2}\\ldots q_{s}}{10^{s} - 1} = \\frac{q_{1}q_{2}\\ldots q_{s}}{999....9},$$ where the denominator in the last expression is consisted only of $9$'s and it contains the same number of $9$'s as that of the length $s$ of the period of the pure recurring decimal. If $d = g.c.d((q_{1}q_{2}\\ldots q_{s}), (999....9))$, then $$x = \\frac{\\frac{q_{1}q_{2}\\ldots q_{s}}{d}}{\\frac{999....9}{d}},$$ gives the equivalent irreducible proper fraction form of the recurring decimal $x$. Let us illustrate this through an example. Consider $x = 0.\\overline{285714}$, then $x = \\frac{285714}{10^{6} - 1} = \\frac{285714}{999999} = \\frac{2}{7}.$ Case II (Mixed Recurring Decimal) : Let, $0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}}$ be the mixed recurring decimal and we need to find out its equivalent irreducible proper fraction form. $x = 0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}},$ then we will multiply $x$ first by $10^{t},$ $t$ is the length of the starting non-repeating part of the decimal representation. Hence, $$10^{t}x = (r_{1}r_{2}\\ldots r_{t}) + 0.\\overline{q_{1}q_{2}\\ldots q_{s}}.$$ The above expression will then be multiplied by $10^{s},$ where $s$ is the length of the period of the mixed recurring decimal $x$. We, therefore, obtain $$10^{s + t}x = 10^{s}(r_{1}r_{2}\\ldots r_{t}) + q_{1}q_{2}\\ldots q_{s}" ]

[ "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "+0 # pls help me 0 85 3 What is $\\frac{0.\\overline{72}}{0.\\overline{27}}$? Express your answer as a common fraction in lowest terms. Jul 27, 2022 #1 -1 The expression reduces to 2/9. Jul 27, 2022 #2 +2448 0 Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1... BuilderBoi Jul 28, 2022 #3 +2448 0 I'll do the numerator, and you can do the denominator. For the numerator, let $$x = 0.{ \\overline{72}}$$. This means $$100x = {72. \\overline{72}}$$ (move the decimal point 2 places to the right) Subtracting the 2 yields us $$99x = 72$$ (the repeating parts cancel out) Dividing by 99, we find that $$x = {72 \\over 99}$$ Jul 28, 2022", "# Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \\overline{6}$<br/> <br/> (ii) $0.4 \\overline{7}$ <br/> <br/>(iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "+0 # help asap 0 111 3 Express $$0.3\\overline{2}$$ as a fraction in simplest form. May 31, 2022 #1 0 $$0.3 \\overline{2} = \\dfrac{32}{99}$$ . May 31, 2022 #3 0 That's wrong!!! Guest Jun 1, 2022 #2 +2448 0 Let $$x = 0.3 \\overline 2$$. This means $$10x = 3. \\overline 2$$ and $$100x = 32. \\overline 2$$ Subtracting $$10x$$ from $$100x$$ gives us: $$90x=29$$, because the repeating part cancels each other out. Can you find the value of x from here? May 31, 2022", "# Why is ⅓ = 0.3333....? • ## Why is it true that $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ ? I wish I had a very simple and clever way to show this, but all the simple and clever ways usually use some idea that is already a step more advanced. For example, you could say that $$0.\\overline{999}\\ldots = 1$$ and then divide both sides by $$3$$ (but knowing the fact that $$0.\\overline{999}\\ldots = 1$$ is even harder to prove!). Or you could start from the fact that $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ and multiply both sides by $$3,$$ but wait, isn't that just the reverse of what we did to get to this place? We were originally trying to show $$\\frac{1}{9} = 0.\\overline{111}\\ldots$$ in the first place, assuming that we knew the decimal representation of $$\\frac{1}{3}!$$ (Round and round we go!) I think the best way to show why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is just to show the long division. Then you'll find that $$1$$ divided by $$3$$ equals $$0.\\overline{333}\\ldots.$$ . This isn't hard, but there are some basic concepts to lay down first. : place value and the idea of division as a reverse-multiplication operation. ## Place value I have some affectionate names that I personally like to use when referring to place value digits. You can make up your own names,", "# How do you find the repeating decimal 0.82 with 82 repeated as a fraction? ##### 1 Answer Dec 8, 2016 $0. \\overline{82} = \\frac{82}{99}$ #### Explanation: In case you have not encountered it, let me introduce you to some notation: A repeating decimal can be represented using a bar placed over the repeating pattern of decimals. So instead of writing $0.828282 \\ldots$, you can write $0. \\overline{82}$ If we multiply $0. \\overline{82}$ by $\\left(100 - 1\\right)$ we get an integer: $\\left(100 - 1\\right) 0. \\overline{82} = 100 \\cdot 0. \\overline{82} - 1 \\cdot 0. \\overline{82}$ $\\textcolor{w h i t e}{\\left(100 - 1\\right) 0. \\overline{82}} = 82. \\overline{82} - 0. \\overline{82}$ $\\textcolor{w h i t e}{\\left(100 - 1\\right) 0. \\overline{82}} = 82$ Notice that the $100$ shifts the number two places to the left - the length of the repeating pattern. Then the $- 1$ cancels out the repeating tail. Next divide both ends by $\\left(100 - 1\\right)$ and simplify: $0. \\overline{82} = \\frac{82}{100 - 1} = \\frac{82}{99}$ $82$ and $99$ have no common factor, so this is in simplest terms. $\\textcolor{w h i t e}{}$ Alternative method An alternative method recognises that: $0. \\overline{82} = 0.82 + 0.0082 + 0.000082 + \\ldots$ is a geometric series, with initial term $a = 0.82$ and common ratio $r =", "# What is 0.8959595... as a fraction ? Jan 9, 2017 $0.8 \\overline{95} = \\frac{887}{990}$ #### Explanation: Given: $0.8 \\overline{95}$ First multiply by $10 \\left(100 - 1\\right) = 1000 - 10$ to get an integer. The first multiplier $10$ is to shift the repeating pattern to just after the decimal point. Then $\\left(100 - 1\\right)$ shifts the number $2$ places to the left and subtracts the original value to cancel out the repeating tail: $\\left(1000 - 10\\right) \\cdot 0.8 \\overline{95} = 895. \\overline{95} - 8. \\overline{95} = 887$ Then divide both ends by $\\left(1000 - 10\\right)$ to find: $0.8 \\overline{95} = \\frac{887}{1000 - 10} = \\frac{887}{990}$ $887$ and $990$ have no common factor larger than $1$, so this fraction is in simplest form.", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "+0 # Can I get help with 1 question? 0 107 1 +37 Express $$\\frac{0.\\overline{12}}{0.\\overline{321}}$$as a common fraction. Thanks for the help Jun 19, 2021 #1 +2250 +1 12 with line = 12/99 321 with line = 321/999 (12/99)/(321/999) Can you take it from here? =^._.^= Jun 19, 2021", "# Are numbers with repeating patterns in their decimal expansion (e.g. $0.123123123\\ldots$) rational? There's a question that I've been thinking about for quite some time now. We all know that numbers with infinite decimal expansion such as $0.\\overline{3}$ or $0.\\overline{1}$ are not necessarily irrational. $$0.\\overline{3}=0.333333\\dots=\\frac{1}{3}$$ $$0.\\overline{1}=0.111111\\ldots=\\frac{1}{9}$$ Therefore we can say that for instance $0.\\overline{4}$ is rational because we have: $$0.\\overline{4}=4\\cdot 0.\\overline{1}=\\frac{4}{9}$$ My question is: can we generalize these results to any number that has a repeating pattern in its decimal expansion? Can we claim for instance that $0.\\overline{123}=0.123123123\\ldots$ is rational? This question boils down to asking whether or not expressions like $0.\\overline{01}$ and $0.\\overline{001}$ etcetera are rational. If so, in the case of my example we could say: $$0.\\overline{123}=123\\cdot 0.\\overline{001}$$ A naive thought of mine was that $0.\\overline{01}$ is simply $\\frac{1}{10}\\cdot 0.\\overline{1}$ but this is of course not true. Then I thought that maybe we can represent $0.\\overline{01}$ as $\\sum\\limits_{i=1}^{\\infty}\\frac{1}{10^{2i}}$ which converges by the ratio test. However this does not tell us if it is rational or not. This is not for homework or anything, just something I've been thinking about and I was wondering if any of you have some knowledge about this issue. Thanks. - Yes, all repeating decimal numbers are rational. I'm sure someone here will give a good explanation but here's one link that'll", "Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. (i) $0 . \\overline{6}$ (ii) $0.4 \\overline{7}$ (iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$ Editor", "# Express $3.72444\\ldots$ as a fraction using the formula for geometric progressions I did the following: $$3.72\\overline{4} = 3.724+\\left(\\left(\\frac{4}{10^4}\\right)+\\left(\\frac{4}{10^5}\\right)+\\left(\\frac{4}{10^6}\\right)+\\cdots\\right)$$ where $a=3.724$ and $r=\\dfrac{1}{10}$ Using $S=\\dfrac{a}{1-r}$ I get $\\dfrac{37.24}{9}\\neq3.72\\overline{4}$ The answer given by the text is $\\dfrac{838}{225}$ I've been staring at this for an hour and can't figure out where I'm going wrong. • Your first term in the series is $3.724$, but your next term is not $3.724\\times\\frac{1}{10}$. Not all of your fraction will actually be part of the series. – Steven Stadnicki Jul 6 '14 at 18:24 Start your geometric series after $3.72$: $$3.72\\overline{4} = 3.72+\\left(\\left(\\frac{4}{10^3}\\right)+\\left(\\frac{4}{10^4}\\right)+\\left(\\frac{4}{10^5}\\right)+\\cdots\\right)$$ Then you have $a=\\dfrac{4}{10^3}$ and $r=\\dfrac 1{10}$. The ultimate sum is then $$3.72 + \\frac{a}{1-r}.$$ First of all, your $a$ is incorrect. I think reviewing the formula for a geometric series would be helpful. The way to do these kinds of problems is to split it up into two parts: one involving the part without the repeated pattern (in this case $3.72=\\frac{372}{100}$) and the other part involving the repeated part (in this case $.00444\\ldots$) which can be rewritten as a geometric series. Hint: You consider only the geometric progression, it's $((\\frac{4}{10^4})+(\\frac{4}{10^5})+(\\frac{4}{10^6})+\\cdots)$. For this sequence $q=\\frac{1}{10}$ and $a=\\frac{4}{10^4}$, not $a=3.724$. Only part of this is geometric: $$3.72\\bar{4}= 3.72 +0.00\\bar 4=\\frac{372}{100}+\\frac{4}{1000}\\left(1+\\frac{1}{10}+\\frac{1}{10^2}+\\frac{1}{10^3}+\\cdots\\right)$$ $$=\\frac{372}{100}+\\frac{4}{1000}\\cdot\\frac{1}{1-\\frac{1}{10}} = \\frac{372}{100}+\\frac{4}{1000}\\cdot\\frac{10}{9} =\\frac{372}{100}+\\frac{4}{900}$$ $$=\\frac{4\\cdot93}{4\\cdot25}+\\frac{4\\cdot1}{4\\cdot225}=\\frac{9\\cdot93}{9\\cdot25} +\\frac{1}{225}=\\frac{837}{225}+\\frac{1}{225}=\\boxed{\\dfrac{838}{225}}$$", "& \\overline{1856} & (2 & \\\\ & & 1792 & & \\\\ & & \\overline{64)} & \\overline{896} & (14 \\\\ & & & 896 & \\\\ & & & \\overline{0} & \\end{array}$ $\\begin{array}{r r r } 64) & \\overline{1344} & (21 \\\\ & 1344 & \\\\ & & \\overline{0} \\end{array}$ HCF is $64$. Hence $64$ is the largest number which can divide $1354, 1866$ and $2762$ leaving the same remainder $10$ in each case. $\\\\$ Question 8: Find the greatest number that will divide $1305, 4665$ and $6905$ so as to leave the same remainder in each case. Required number $=$ HCF of $(4665 - 1305), (6905 - 4665)$ and $(6905 - 1305)$ $=$ HCF of $3360, 2240, 5600$. $\\begin{array}{r r r r r } 3360) & \\overline{5600} & (1 & & \\\\ & 3360 & & & \\\\ & \\overline{2240)} & \\overline{3360} & (1 & \\\\ & & 2240 & & \\\\ & & \\overline{1120)} & \\overline{2240} & (2 \\\\ & & & 2240 & \\\\ & & & \\overline{0} & \\end{array}$ $\\begin{array}{r r r } 1120) & \\overline{3360} & (3 \\\\ & 3360 & \\\\ & \\overline{0} & \\end{array}$ HCF is $1120$. The required number is $112$ that will divide $1305, 4665$ and $6905$ so as to leave the same remainder in each case. $\\\\$ Question", "of simple fractions as decimals: $\\frac 12 = 0.5$ $\\frac 13 = 0.333333\\ldots = 0.\\overline{3}$ $\\frac 23 = 0.666666\\ldots = 0.\\overline{6}$ $\\frac 14 = 0.25$ $\\frac 34 = 0.75$ $\\frac 15 = 0.2$ $\\frac 16 = 0.1666666\\ldots = 0.1\\overline{6}$ $\\frac 17 = 0.14285714285714\\ldots = 0.\\overline{142857}$ $\\frac 18=0.125$ $\\frac 38 = 0.375$ $\\frac 78 = 0.875$ $\\frac 19 = 0.111111\\ldots = 0.\\overline{1}$ $\\frac 29 = 0.222222\\ldots = 0.\\overline{2}$ $\\frac 1{10} = 0.1$ $\\frac 1{11} = 0.09090909\\ldots = 0.\\overline{09}$ $\\frac 1{12} = 0.08333333\\ldots = 0.08\\overline{3}$ A few things can be observed: • Some decimals are short and to the point. • Others have an infinite number of digits (students like to say: \"they are infinite\"), in a repeating pattern. • There can be a single digit that repeats, or a string of two or more digits repeating. • The repeating pattern (period) does not always start right after the decimal point. The long period of $\\frac 17$ sticks out like a sore thumb. It has a few crazy cousins further along: $\\frac 1{13} = 0.\\overline{076923}$ $\\frac 1{17} = 0.\\overline{0588235294117647}$ $\\frac 1{19} = 0.\\overline{052631578947368421}$ There are also the weird $\\frac 1{81} = 0.\\overline{012345679}$ (sic!) and the mysterious $$\\frac 1{9801} = 0.0001020304050607080910111213\\ldots$$ It is also worthwhile to notice this pattern (for later): $$\\frac 19 = 0.\\overline{1}$$ $$\\frac 1{99} = 0.\\overline{01}$$ $$\\frac 1{999}", "r r r } 148) & \\overline{370} & (1 & \\\\ & 296 & & \\\\ & \\overline{74)} & \\overline{148} & (2 \\\\ & & 148 & \\\\ & & \\overline{0} & \\end{array}$ Therefore HCF $= 74$ vi) $1701, 2106, 2754$ $\\begin{array}{r r r r r } 2106) & \\overline{2754} & (1 & & \\\\ & 2106 & & & \\\\ & \\overline{648)} & \\overline{2106} & (3 & \\\\ & & 1944 & & \\\\ & & \\overline{162)} & \\overline{648} & (4 \\\\ & & & 648 & \\\\ & & & \\overline{0} & \\end{array}$ HCF of $2754$ and $2106 = 162$ $\\begin{array}{r r r r } 162) & \\overline{1701} & (10 & \\\\ & 1620 & & \\\\ & \\overline{81)} & \\overline{162} & (2 \\\\ & & 162 & \\\\ & & \\overline{0} & \\end{array}$ Therefore HCF $= 81$ $\\\\$ Question 3: Reduce each of the following fractions to its lowest terms: i) $\\frac{682}{868}$ ii) $\\frac{777}{1147}$ iii) $\\frac{1095}{1168}$ i) $\\frac{682}{868}$ $\\begin{array}{r r r r r r} 682) & \\overline{868} & (1 & & & \\\\ & 682 & & & & \\\\ & \\overline{186)} & \\overline{682} & (3 & & \\\\ & & 558 & & & \\\\ & & \\overline{124)} & \\overline{186} & (1 & \\\\ & & & 124 & & \\\\ & & &", "to 500. Since, HCF must be a factor of LCM, but 500 is not a factor of 1200. #### Question 18: Express $0.\\overline{4}$ as a rational number in simplest form. Let x be $0.\\overline{4}$. $x=0.\\overline{4}$ .....(1) Multiplying both sides by 10, we get $10x=4.\\overline{4}$ .....(2) Subtracting (1) from (2), we get $10x-x=4.\\overline{4}-0.\\overline{4}\\phantom{\\rule{0ex}{0ex}}⇒9x=4\\phantom{\\rule{0ex}{0ex}}⇒x=\\frac{4}{9}$ Thus, simplest form of $0.\\overline{4}$ as a rational number is $\\frac{4}{9}$. #### Question 19: Express $0.\\overline{23}$ as a rational number in simplest form. ​Let be $0.\\overline{23}$. $x=0.\\overline{23}$ .....(1) Multiplying both sides by 100, we get $100x=23.\\overline{23}$ .....(2) Subtracting (1) from (2), we get $100x-x=23.\\overline{23}-0.\\overline{23}\\phantom{\\rule{0ex}{0ex}}⇒99x=23\\phantom{\\rule{0ex}{0ex}}⇒x=\\frac{23}{99}$ Thus, simplest form of $0.\\overline{23}$ as a rational number is $\\frac{23}{99}$. #### Question 20: Explain why 0.15015001500015 ... is an irrational number. Irrational numbers are non-terminating non-recurring decimals. Thus, 0.15015001500015 ... is an irrational number. #### Question 21: Show that $\\frac{\\sqrt{2}}{3}$ is irrational. Let $\\frac{\\sqrt{2}}{3}$ is a rational number. ∴ $\\frac{\\sqrt{2}}{3}=\\frac{p}{q}$, where p and q are some integers and HCF(p, q) = 1 ....(1) $⇒\\sqrt{2}q=3p\\phantom{\\rule{0ex}{0ex}}⇒{\\left(\\sqrt{2}q\\right)}^{2}={\\left(3p\\right)}^{2}\\phantom{\\rule{0ex}{0ex}}⇒2{q}^{2}=9{p}^{2}$ p2 is divisible by 2 ⇒ p is divisible by 2 ....(2) Let p = 2m, where m is some integer. ∴ $\\sqrt{2}q=3p$ $⇒\\sqrt{2}q=3\\left(2m\\right)\\phantom{\\rule{0ex}{0ex}}⇒{\\left(\\sqrt{2}q\\right)}^{2}={\\left(3\\left(2m\\right)\\right)}^{2}\\phantom{\\rule{0ex}{0ex}}⇒2{q}^{2}=4\\left(9{p}^{2}\\right)\\phantom{\\rule{0ex}{0ex}}⇒{q}^{2}=2\\left(9{p}^{2}\\right)$ ⇒ q2 is divisible by 2 ⇒ q is divisible by 2 ....(3) From (2) and (3), 2 is a common factor of both p and q, which contradicts (1). Hence, our assumption", "Thus, the number of time 9 to be repeated in the denominator becomes three. $$\\begin{array}{l}0.\\overline{125} = \\frac{125}{999}\\end{array}$$ ### Case 2: Fraction of the type $$\\begin{array}{l}0.ab..\\overline{cd}\\end{array}$$ The formula to convert this type of repeating decimal to the fraction is given by: $$\\begin{array}{l}0.ab..\\overline{cd} =\\frac{(ab….cd…..) – ab……}{Number \\; of \\; time \\; 9’s \\; the \\; repeating \\; term \\; followed \\; by \\; the \\; number \\; of \\; times \\; 0’s \\; for \\; the \\; non-repeated \\; terms }\\end{array}$$ Example 3: Convert $$\\begin{array}{l}0.12\\overline{34}\\end{array}$$ to the fractional form. Solution: In the given decimal number, 12 is a non-repeated decimal value, and 34 is in the repeating form. Thus denominator becomes 9900. $$\\begin{array}{l}0.12\\overline{34} = \\frac{1234 – 12 }{9900} = \\frac{1222}{9900}\\end{array}$$ Example 4: Convert $$\\begin{array}{l}0.00\\overline{69}\\end{array}$$ in the fraction form. Solution: In the given decimal number, the number 00 is a non-repeated decimal value, and 69 is in the repeating form. Thus, the denominator becomes 9900. $$\\begin{array}{l}0.00\\overline{69} = \\frac{0069}{9900} = \\frac{69}{9900}\\end{array}$$ It is all about the conversion of repeating decimal to the fraction form. To learn more interesting topics in Maths, download BYJU’S – The Learning App and learn with ease. Test your Knowledge on Repeating Decimal to Fraction", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =", "claim: -10¾ + 1/2(120) = 49¼ 19¼ + 1/3(90) = 49¼ 29¼ + 1/4(80) = 49¼ 34¼ + 1/5(75) = 49¼ 37¼ + 1/6(72) = 49¼. Justification for the rote procedure: Let $$x_i =$$ the money that man $$i$$ carries $${\\overline{x_i}}=$$ the money that the other 4 men carry $$T=x_i +{\\overline{x_i}}=$$ the total money carried $$c=$$ the cost of the piece of cloth $$x = T – c =$$ the excess money. Taking man 5 as an example, we start with the given condition: $x_5 +\\frac{1}{6}{\\overline{x_5}}=c$ $x_5 +{\\overline{x_5}}=c+\\frac{5}{6}{\\overline{x_5}}$ $T=c+\\frac{5}{6}{\\overline{x_5}}$ $x=\\frac{5}{6}{\\overline{x_5}}$ ${\\frac{x}{\\frac{5}{6}}}={\\overline{x_5}}.$ Similarly, ${\\frac{x}{\\frac{4}{5}}}={\\overline{x_4}},\\,\\,{\\rm etc.}$ Summing all 5 such equations, we get ${\\frac{x}{\\frac{1}{2}}}+\\cdots+{\\frac{x}{\\frac{5}{6}}}={\\overline{x_1}}+\\cdots+{\\overline{x_5}}.$ Since the money that any one man carries is included in 4 of the 5 terms on the right-hand side, the total on the right is $$4T:$$ ${\\frac{x}{\\frac{1}{2}}}+\\cdots+{\\frac{x}{\\frac{5}{6}}}=4T$ $\\frac{1}{4}\\left[{\\frac{x}{\\frac{1}{2}}}+\\cdots+{\\frac{x}{\\frac{5}{6}}}\\right]=T$ The parameter $$x$$ is an arbitrary multiplicative constant, whose choice determines each $$x_i$$ and thus their total, $$T.$$ To determine the money that any man carries, say man 5, $x_5=T-{\\overline{x_5}}$ $x_5={\\frac{1}{4}}{\\left[{\\frac{x}{\\frac{1}{2}}}+\\cdots+{\\frac{x}{\\frac{5}{6}}}\\right]}-{\\frac{x}{\\frac{5}{6}}},$ and to determine the cost, $c=T-x$ $c=\\frac{1}{4}\\left[{\\frac{x}{\\frac{1}{2}}}+\\cdots+{\\frac{x}{\\frac{5}{6}}}\\right]-x.$ Solution B: matrix algebra Let $$x_i=$$ the money that man $$i$$ carries and $$c =$$ the cost of the piece of cloth. The given conditions can be thought of as a system of 5 linear equations and 5 unknowns, with a common parameter on the right-hand side: ${\\begin{array}{r} x_1+\\frac{1}{2}\\left(x_2+x_3+x_4+x_5\\right)=c\\\\" ]

[ "Math Help - need help on \"integral part\" [x] 1. need help on \"integral part\" [x] Using definition 3.3 on NZM For real x, the symbol [x] denotes the greatest integer less than or equal to x. Why is [-15/2]=-8. 2. $-\\frac{15}{2} = -7.5$ What's the largest integer that's smaller than $-7.5$ ? 3. Originally Posted by o_O $-\\frac{15}{2} = -7.5$ What's the largest integer that's smaller than $-7.5$ ? -8, so [-3/2]=-2, since $-\\frac{3}{2} = -1.5$ and the the largest integer that is smaller -1.5 is -2? 4. Exactly.", "largest possible value? It follows by Cauchy Schwarz: $\\sum_i a_i \\cdot b_i \\le \\sqrt{\\sum_i a_i^2} \\sqrt{\\sum_i b_i^2}$. Let $b_i = 1$ and $a_i$ be the Fourier coefficients. – Manu May 17 '16 at 10:35", "$H$, in your case greater than $-4$. $T$ should just be large enough (your $H$ has an obvious decay for $\\Im(s) \\to \\infty$, you could try to work out estimates on the error for the truncation of the integral).", "• zmudz Find the value of the largest of $$A$$, $$B$$, and $$C$$ in the partial fractions expansion of $$\\frac{2(x^2+x-1)}{x(x^2-1)} = \\frac{A}{x} + \\frac{B}{x-1} + \\frac{C}{x+1}.$$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "If the maximum value of $$k$$ is $$\\frac{-A}{B}$$ for which $$kx^{2}-2kx+3x-6$$ is positive for $$2015$$ integral values of $$x$$.Find $$A+B$$", "Find the number of integral solutions of $\\large 35x + 63y + 45z = 1$ for $$\\left| x \\right| <9$$, $$\\left| y \\right| <5$$, and $$\\left| z \\right| <7$$. ×", "# A very big number Find the maximum integral value of $n$ such that $2^n$ divides $\\Large 3^{2^{2017}}-1.$ ×", "## anonymous 3 years ago find the integral 1. anonymous |dw:1358617791231:dw| 2. anonymous i attempted to solve it but i can't get the answer.. which is 24 3. anonymous $\\frac{1}{3}\\ln(x^3+3x)$ 4. anonymous just guessed 5. anonymous yeah i got up to there but i can't get the answer :/ 6. rishabh.mission 1st take x2=t nd find is dx/dy 7. rishabh.mission its easy but i dont have pen/pencil let me try give few min. 8. anonymous $\\huge{\\ln(x^2+3x)}\\huge{|_{1}^{3}}={\\ln(3^2+3(3))-\\ln(1^2+3(1))}$ 9. anonymous sorry i forgot$\\frac{1}{3}$ on boh sides", "the k4, but I tried to do this, and I got a huge number not even close to 4. Can anyone help me out, please? Find the largest positive integer solution of $\\frac{(b-a)^3}{12 n^2} \\, \\text{max}_{[a, b]} f''(x) \\leq \\frac{1}{100}$ where $a = 2, \\, b = 5, \\, f(x) = x^{-1/4}$ and $\\text{max}_{[a, b]} f''(x)$ is the maximum value of f(x) over the interval [a, b] of integration.", "+0 # algebra 0 35 1 Find the largest possible value of x in the simplified form x=\\frac{a+b\\sqrt{c}}{d} if 5x/6 +1= 3/x - 1 + 5 + 2x/3, where a,b,c,\\$ and d are integers. What is acd/b? Oct 23, 2022", "to get a similar integral that shows $$355/113$$ is bigger than $$\\pi$$: $$0 \\lt \\int_0^1\\frac{x^8\\left(1-x\\right)^8\\left(25+816x^2\\right)}{3164\\left(1+x^2\\right)}\\,dx=\\frac{355}{113}-\\pi.$$ This was done by Stephen Lucas here: For more, go here:", "An algebra problem by Florin Celoiu Algebra Level 3 $\\Large n^{5^{ (\\log_{5}2) + (\\log_{5}x )+ \\log_{5}\\left(1-\\frac{9}{2x}\\right) }}=n^{7^{(\\log_{7}56) - \\log_{7}x}}$ For $$n \\geqslant 2$$, what is the positive integral value of $$x$$ which satisfies the above equation? ×", "evaluating values $x$ that are infinitely close together. So $f(x) = (1 + 2x)^2$ Substitute: What are the bounds? The bounds are the smallest input and the largest input into the variable. Since the integrand is $f(x) = (1 + 2x)^2$, where Let's look at the smallest and largest value of $x$. The smallest value of $x$ is $\\frac{1}{n}, \\text{ and }\\lim\\limits_{n\\rightarrow \\infty }\\frac{1}{n}=0$ The largest value of $x$ is So the bounds are form $0$ to $1$. Substitute:", "## integral and its largest value Along what curve of the family y=x^n does the integral int{(25xy-8y^2)dx} attain its largest value? and the boundaries for the integral is from (0,0) to (1,1) thank you.. PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks Recognitions: Homework Help Science Advisor Any thoughts on this problem? Maybe just evaluating the integral along the curve x^n? Recognitions: Gold Member Staff Emeritus Along what curve of the family y=x^n does the integral int{(25xy-8y^2)dx} attain its largest value? \\ doesn't mean a whole lot to me. Do you mean \"where does x^n intersect the integral of 26xy- 8y^2 dx with the largest y value? ## integral and its largest value Quote by nepenthe hello..could you please help me to solve this problem? Along what curve of the family y=x^n does the integral int{(25xy-8y^2)dx} attain its largest value? and the boundaries for the integral is from (0,0) to (1,1) thank you.. Is this what you're looking for ?? $$\\begin{gathered} y = x^n \\Rightarrow 25xy - 8y^2 = 25x^{n + 1} - 8x^{2n} \\hfill \\\\ \\frac{d} {{dn}}\\left[ {\\int\\limits_0^1 {\\left( {25x^{n + 1} - 8x^{2n} } \\right)dx} } \\right] = \\frac{{16}}{{\\left( {2n + 1} \\right)^2 }} -", "Limit Integral Difficult Problem - 4547 Evaluate $\\displaystyle\\lim_{n\\to\\infty}S_n$ where $$S_n = 1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\cdots + (-1)^{n-1}\\frac{1}{n}$$ The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "• # question_answer If ${{\\left( \\frac{1+i}{1-i} \\right)}^{m}}=1,$then the least integral value of $m$ is [IIT 1982; MNR 1984; UPSEAT 2001; MP PET 2002] A) 2 B) 4 C) 8 D) None of these $\\frac{1+i}{1-i}=\\frac{1+i}{1-i}\\times \\frac{1+i}{1+i}=\\frac{{{(1+i)}^{2}}}{2}=\\frac{2i}{2}=i$ $\\therefore$${{\\left( \\frac{1+i}{1-i} \\right)}^{m}}={{i}^{m}}=1$ (as given) So the least value of $m=4$$\\{\\because {{i}^{4}}=1\\}$", "Least integral value! Algebra Level 2 If $$x$$ is positive, what is the least value of $x+\\frac{25}{x}?$ ×", "1975 USAMO Problems/Problem 1 Problem (a) Prove that $[5x]+[5y]\\ge [3x+y]+[3y+x]$, where $x,y\\ge 0$ and $[u]$ denotes the greatest integer $\\le u$ (e.g., $[\\sqrt{2}]=1$). (b) Using (a) or otherwise, prove that $\\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$ is integral for all positive integral $m$ and $n$. Solution 1975 USAMO (Problems • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions", "# Small looking integrals have really big solutions Calculus Level 3 $\\large \\int _{ 0 }^{ 1 }{ \\frac { { ( 1 -16{ x }^{ \\frac { 1 }{ 4 } } ) }^{ 2 } }{ { ( { 2 }^{ 8 }\\ x ) }^{ \\frac { 7 }{ 8 } } } \\, dx }$ If the integral above can be written as $$\\frac{a}{b}$$ for positive coprime integers $$a, b$$. Find $$a + b$$. ×", "# Algebraic Identities Come In Handy! $\\large \\dfrac { x+y }{ { x }^{ 3 }+{ y }^{ 3 } } =\\frac { 1 }{ 10 }$ How many integral solutions does the above equation have? ×" ]

[ "# Fractions: Overview If $$13 < 7x + 1 < 25$$, how many integer values of $$x$$ are there? Correct. [[Snippet]] Subtract 1 from the inequality to isolate $$7x$$. >$$13 - 1 < 7x + 1 - 1 < 25 - 1$$ >$$12 < 7x < 24$$ Divide the inequality by 7 to isolate $$x$$. >$$\\frac{12}{7} < \\frac{7x}{7} < \\frac{24}{7}$$ >$$\\frac{12}{7} < x < \\frac{24}{7}$$ Based on this, $$x$$ is more than 1 ($$\\frac{7}{7} = 1$$, so $$\\frac{12}{7}$$ is greater) and less than 4 ($$\\frac{28}{7}=4$$, so $$\\frac{24}{7}$$ is smaller). In other words, the integer values of $$x$$ are 2 and 3. Hence, there are two possible integer values for $$x$$. Incorrect. [[snippet]] Carefully check your calculations. Incorrect. [[snippet]] Carefully check your work. Incorrect. [[snippet]] Solve the given inequality for $$x$$. Incorrect. [[snippet]] You need to solve the inequality $$13 < 7x + 1 < 25$$ for $$x$$. 1 2 3 4 5", "$$x = \\frac{(2)(3)(4)(5)(7)(11)(13)}{(3)(13)y}$$ Simplify: $$x = \\frac{(2)(4)(5)(7)(11)}{y}$$ In order for x to be an integer, y must be a divisor of $$(2)(4)(5)(7)(11)$$ 28 is a divisor of (2)(4)(5)(7)(11) Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Re: QOTD # 16 If x and y are integers and which of the followin [#permalink] 12 Jun 2019, 07:44 Display posts from previous: Sort by", "Re: The sum of two integers is 27. The larger integer is 25% greater than [#permalink] ### Show Tags 26 Jun 2018, 05:38 Solution Given: • The sum of two integers is 27 • The larger integer is 25% greater than the smaller integer To find: • The positive difference between the two integers Approach and Working: If we assume the smaller integer to be n, then • The larger integer = $$n + \\frac{n}{4} = \\frac{5n}{4}$$ As their sum is 27, we can write • $$n + \\frac{5n}{4} = 27$$ Or, $$\\frac{9n}{4} = 27$$ Or, n = $$\\frac{27*4}{9} = 12$$ The smaller integer = 12 Therefore, the larger integer = $$12 + \\frac{12}{4} = 12 + 3 = 15$$ • positive difference =$$15 – 12 = 3$$ Hence, the correct answer is option A. _________________ BSchool Forum Moderator Joined: 05 Jul 2017 Posts: 513 Location: India GMAT 1: 700 Q49 V36 GPA: 4 Re: The sum of two integers is 27. The larger integer is 25% greater than [#permalink] ### Show Tags 26 Jun 2018, 05:55 Let the smaller integer be x therefore, the larger integer will be 1.25x Given that, x+1.25x = 27 2.25x = 27 x = 12 larger integer = 15 difference between larger and smaller integer = 15-12 =3 _________________ Board of Directors", "Re: If A is an integer, what is the value of A? [#permalink] ### Show Tags 30 Apr 2020, 03:00 1 Bunuel wrote: If A is an integer, what is the value of A? (1) $$1,000,000 < 23^A < 10,000,000$$ (2) $$\\frac{7^{A+2}−7^A}{48}=7^A$$ Project PS Butler Are You Up For the Challenge: 700 Level Questions (Statement1): $$10^{6} < 23^{A} < 10^{7}$$ $$23^{2}= 529$$ $$23^{4}= 529*529 > \\frac{10^{3}}{2}* \\frac{10^{3}}{2}= \\frac{10^{6}}{4} < 10^{6}$$ $$23^{5} ≈ \\frac{10^{6}}{4}* 23= 5.75* 10^{6} > 10^{6}$$ $$23^{6} ≈ 5.75* 10^{6}* 23= 5.75*2.3* 10^{7} > 10^{7}$$ Well, we got only one value of A satisfying the inequality A=5 Sufficient (Statement2): $$\\frac{7^{A+2}−7^A}{48}=7^A$$ $$\\frac{7^{A} (49 -1)}{48}= 7^{A}$$ --> $$7^{A} = 7^{A}$$ A could be any integer Insufficient Re: If A is an integer, what is the value of A? [#permalink] 30 Apr 2020, 03:00", "+0 0 66 1 What is the value of $\\frac1x + \\frac1y$ if $x = \\frac23$ and $y = \\frac34$? Express your answer as a mixed number. Jan 1, 2021 #1 0 I found the solution: $\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{2/3} + \\frac{1}{3/4} = \\frac32 + \\frac43 = \\frac{9}{6} + \\frac{8}{6} = \\frac{17}{6}.$ Writing $\\frac{17}{6}$ as a mixed number gives $\\boxed{2\\frac{5}{6}}$. Jan 1, 2021", "+0 # Find (x,y).. 0 65 1 Find the number of pairs of integers (x,y) with 0 < x, y < 10 that satisfy $$\\frac{1}{1-\\frac{10}{x}} > 1 - \\frac{5}{y}.$$ Feb 3, 2019 #1 +100439 +1 We can simplify this as x / [ x - 10 ] > [ y - 5] / y Since 0 < x, y < 10.....the left side will always be negative And the right side will be negative when y < 5 So when x = 1 The possible integer values for y are 1, 2, 3 and 4 When x = 2 The possible integer values are 1, 2, 3 When x = 3 The possible integer values of y are 1,2,3 When x = 4 The possible integer values for y are 1, 2 When x = 5 The possible integer values for y are 1,2 When x = 6 The possible values for y are 1 When x = 7 The possible values for y are 1 When x = 8...there is no value of y that makes the inequality true So....there are 16 pairs of (x, y) that make the inequality true Feb 3, 2019 edited by CPhill Feb 3, 2019 edited by CPhill Feb 3, 2019 edited by CPhill Feb 3, 2019", "Skipping a little ahead because of the latex, the inequality we want is: $$\\sum_{i>0}[\\frac{n+a}{p}]$$ greater than or equal to:$$\\sum_{i>0}[\\frac{n}{p}]+\\sum_{i>0}[\\frac{a}{p}]$$ for all p. This comes about because let n=kp+r for r less than p and positive or zero, and let a=jp+y, y same conditions as above, then greatest integer in n+a =k+j + greatest integer in (r+y)/p which is greater than or equal to k+J. Last edited: Oct 3, 2004 Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Have something to add?", "value of integer x? (1) 2x^2+9<9x --> factor qudratics: $$(x-\\frac{3}{2})(x-3)<0$$ --> roots are $$\\frac{3}{2}$$ and 3 --> \"<\" sign indicates that the solution lies between the roots: $$1.5<x<3$$ --> since there only integer in this range is 2 then $$x=2$$. Sufficient. $$(x-\\frac{3}{2})(x-3)<0$$ I agree one solution of this inequality is $$(x-\\frac{3}{2})$$ > 0 , (x-3)<0 => $$(\\frac{3}{2})$$ < x < 3 However, Don't u think this can also resort to $$(x-\\frac{3}{2})$$ < 0 , (x-3)>0 => x < $$(\\frac{3}{2})$$ , x > 3 and in that case x can have infinite values. and if that is the case Stmt-1 alone would not be sufficient. The second case of $$(x-\\frac{3}{2})<0$$ and $$(x-3)>0$$ is not possible. This condition leads to $$x<\\frac{3}{2}$$ and $$x>3$$, but $$x$$ can not be simultaneously less than $$\\frac{3}{2}$$ (to make $$x-\\frac{3}{2}$$ negative) and more than 3 (to make $$x-3$$ positive). For more on how to solve such kind of inequalities check: x2-4x-94661.html#p731476 (Check this first) inequalities-trick-91482.html everything-is-less-than-zero-108884.html?hilit=extreme#p868863 Hope it helps. _________________ Manager Joined: 07 Sep 2010 Posts: 246 Re: The Discreet Charm of the DS [#permalink] ### Show Tags 27 Jun 2012, 08:03 Hi Bunuel, Can we solve this question using graphical approach. Since (x-a)^2+(y-b)^2 = r^2 ----(A) Equation 1 represents equation of circle with radius 1. Now, if we can maximize the value of both", "# One integer is 15 more than 3/4 of another integer. The sum of the integers is greater than 49. How do you find the least values for these two integers? Jun 11, 2018 The 2 integers are 20 and 30. #### Explanation: Let x be an integer Then $\\frac{3}{4} x + 15$ is the second integer Since the sum of the integers is greater than 49, $x + \\frac{3}{4} x + 15 > 49$ $x + \\frac{3}{4} x > 49 - 15$ $\\frac{7}{4} x > 34$ $x > 34 \\times \\frac{4}{7}$ $x > 19 \\frac{3}{7}$ Therefore, the smallest integer is $20$ and the second integer is $20 \\times \\frac{3}{4} + 15 = 15 + 15 = 30$.", "there are too many cases to consider. A much more convenient conception of the integers is the Narcissist Yellow sale manchester great sale new arrival cheap price MuWnSZ construction. The essential observation is that every integer can be expressed (not uniquely) as the difference of two natural numbers, so we may as well define an integer as the difference of two natural numbers. Addition is then defined to be compatible with subtraction: Addition of New Look Wide Fit Satin Mid Heeled Boot sale online store outlet best store to get clearance looking for cheap popular kHpyB6E can be computed using the least common denominator , but a conceptually simpler definition involves only integer addition and multiplication: As an example, the sum $\\frac{3}{4}+\\frac{1}{8}=\\frac{3×8+4×1}{4×8}=\\frac{24+4}{32}=\\frac{28}{32}=\\frac{7}{8}$ .", "an integer? [#permalink] 18 Apr 2018, 01:42 1 KUDOS let $$x$$be the first integer than 2nd integer is$$x+1$$ There average will be $$\\frac{2x + 1}{2} = x + \\frac{1}{2}$$ which is not a integer For option B 1st integer = $$x$$ 2nd integer = $$x+1$$ 3rd integer = $$x+2$$ Hence there average $$= \\frac{3x +3}{3}$$ or, $$\\frac{3(x+1)}{3} = x + 1$$ since x is an integer $$x +1$$ is also a integer option B _________________ This is my response to the question and may be incorrect. Feel free to rectify any mistakes Re: Which one of the following could be an integer? [#permalink] 18 Apr 2018, 01:42 Display posts from previous: Sort by", "more examples by taking any number with this property and multiplying it by a power of ten. For example, 36 = 3 * 6 + 3 * 6 and thus 360 = 3 * 60 + 3 * 60. This shows that there is an infinity of numbers with this property. More families of these numbers are given by 12500...08 = 1 * 2500...08 + 12500...0 * 8 1600...0625 = 1600...0 * 625 + 1 * 600...0625 65454...5455 = 6 * 5454...5455 + 65454...545 * 5 These facts are not too difficult to prove. We will prove the third equality here which is the most interesting one, as it has no zero in its decimal expansion. Numbers with zero are a bit more common, although there are not too many of either. First, let’s express $5454\\ldots54$ in form that is easier to manipulate. This way we easily find similar representations for $65454\\ldots$, $5454\\ldots5455$ and $65454\\ldots545$. Let $k$ be the amount of $5{}4$‘s in this number. Now if $x = 0.545454\\ldots$, then $10^{2k}x = 5454\\ldots54.5454\\ldots$, implying $(10^{2k}-1)x = 5454\\ldots54$. What is $x$ as a fraction? Since $100x = 54.5454\\ldots$, we get $99x = 54$ and so $x = \\frac{54}{99} = \\frac{6}{11}$. Thus $5454\\ldots54 = (10^{2k}-1)\\frac{6}{11}$. Then by multiplying by $100$ (which adds two zeros at the end) and adding", "# When you take my value and multiply it by -8, the result is an integer greater than -220. If you take the result and divide it by the sum of -10 and 2, the result is my value. I am a rational number. What is my number? Jan 11, 2018 Your value is any rational number greater than $27.5$, or $\\frac{55}{2}$. #### Explanation: We can model these two requirements with an inequality and an equation. Let $x$ be our value. $- 8 x > - 220$ $\\frac{- 8 x}{- 10 + 2} = x$ We will first attempt to find the value of $x$ in the second equation. $\\frac{- 8 x}{- 10 + 2} = x$ $\\frac{- 8 x}{-} 8 = x$ $x = x$ This means that regardless of the initial value of $x$, the second equation will always be true. Now to work out the inequality: $- 8 x > - 220$ $x < 27.5$ So, the value of $x$ is any rational number greater than $27.5$, or $\\frac{55}{2}$.", "(1) $$x^{(x+3)}=(2x)^{(x-1)}$$ --> $$x^{(x+3)}=2^{(x-1)}*x^{(x-1)}$$ --> $$x^{(x+3-x+1)}=2^{(x-1)}$$ --> $$x^4=2^{(x-1)}$$ --> $$x=1$$. $$x^4=2^{(x-1)}$$ --> $$x=1$$---------How do we arrive at this. How to solve these????? just by testing values vaguely or is there any particular mathematics we can do with this????? I believe Bunuel has already explained this. Just another way to look at it is this: We end up with: $$x^4=2^{(x-1)} \\to x = 2^{\\frac{(x-1)}{4}}$$ . We know that if x is an integer, the power of 2,i.e. $$\\frac{x-1}{4}$$ has to be a non-negative integer. I.$$\\frac{x-1}{4}>0$$ As x equals some positive power of 2, x is even.Also, the only way x can be an integer is if the exponent of 2 is raised to an integer$$\\to$$ x is a multiple of 4. If (x-1) is a multiple of 4, then x has to be odd and this contradicts out previous statement, that x is even. II. $$\\frac{x-1}{4}=0$$$$\\to x=1$$ and this satisfies the equality. _________________ Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 3 Kudos [?]: 165 [0], given: 134 Re: How many factors does the integer X have? [#permalink] ### Show Tags 10 Jul 2013, 12:37 How many factors does the integer x have? (1) x^(x+3) = (2x)^(x-1) x^(x+3) = (2x)^(x-1) x^(x+3) = x^(x-1) 2^(x-1) (x^(x-1)) / x^(x+3) = x^(x-1) 2^(x-1) / (x^(x-1)) x^(x+3)-(x-1) = 2^(x-1) x^(4) = 2^(x-1)", "A 8 B 9 C 10 A ### PCAT Quantitative Practice Questions -7 1. For how many positive integers, $n$, is true that $n^{2} \\leq 3n$ 1. $2$ 2. $3$ 3. $4$ 4. $5$ 2. If $a^{4}=16$, then $3^{a}$ 1. $3$ 2. $9$ 3. $16$ 4. $27$ 3. $\\sqrt{20}\\sqrt{5}=$ 1. $2\\sqrt{5}$ 2. $10$ 3. $4\\sqrt{5}$ 4. $5\\sqrt{10}$ 4. The sum of three positive consecutive even integers is x. What is the value of the middle of the three integers? 1. $\\frac{x}{3}+2$ 2. $3x$ 3. $\\frac{x-2}{3}$ 4. $\\frac{x}{3}$ 5. What is the average of $5^{10}$, $5^{20}$, $5^{30}$, $5^{40}$ and $5^{50}$? 1. $5^{9}+5^{19}+5^{29}+5^{39}+5^{49}$ 2. $5^{30}$ 3. $5^{149}$ 4. $150$ 6. Which of the following is equal to $(5^{6} \\times 5^{9})^{10}$? 1. $25^{150}$ 2. $25^{540}$ 3. $5^{540}$ 4. $5^{150}$ 7. What is the value of $3^{\\frac{1}{3}} \\times 3^{\\frac{2}{3}} \\times 3^{\\frac{3}{3}}$? 1. $3$ 2. $9$ 3. $27$ 4. $30$ 8. How many integers satisfy the inequality $|x| < 2 \\pi$. 1. $3$ 2. $4$ 3. $7$ 4. More than $7$ 9. What is the average of $5^{a} \\times 5^{b}=5^{300}$ 1. $50$ 2. $100$ 3. $150$ 4. $200$ 10. If $5^{a}5^{b}=\\frac{5^{c}}{5^{d}}$, what is d in terms of $a$, $b$ and $c$? 1. $\\frac{c}{a+b}$ 2. $c+ab$ 3. $c-a-b$ 4. $c+a-b$ 1 B 2 B 3 B 4 D 5 A 6 D 7 B", "solution 2. $$7B=3A$$ $$(x-44)C=4A$$ $$(x-y)C=(y-50)B$$ Our goal is to try to isolate $y$ into an inequality. The first equation gives $A=\\frac{7}{3}B$, which we plug into the second equation to get $$(x-44)C=\\frac{28}{3}B$$ To eliminate $x$, subtract equation 3 from equation 2: $$(x-44)C-(x-y)C=\\frac{28}{3}B-(y-50)B$$ $$(y-44)C=(\\frac{178}{3}-y)B$$ In order for the coefficients to be positive, $\\[44 Thus, the greatest integer value is $y=59$, choice $(E)$.", "in the inequality $$25 - 4n < 0$$. Adding $$4n$$and dividing by 4, we find $$n>6.25$$. The probability that I chose one of the numbers 7, 8, 9, or 10 is $$\\boxed{\\frac{2}{5}}$$ . Aug 11, 2018 #5 +809 +2 Thank you, both CPhill and azsun for wonderful solutions! Aug 11, 2018 #6 +98173 +1 OK, mathtoo !!!!! CPhill Aug 11, 2018", "+0 # HEEEEELLLLLPPPPP 0 105 1 Given that $x$, $\\frac{1}{x}$, $y$, $\\frac{1}{y}$, $z$ and $\\frac{1}{z}$ are all integers, how many distinct values of $x+ y+ z$ are possible? Feb 17, 2021 #1 +32894 +2 I think x y and z can each be 1 or -1 1 + 1 +1 = 3 -1 + 1 +1 =1 -1 -1 + 1 = -1 -1 -1 -1 = -3 4 values Feb 17, 2021", "to more detail. The question asks what is the range of possible values for “$$x$$” when $$\\frac{13!}{2^x}$$ is an integer. In other words, you’re looking to find the range of values of \"x\" that make $$2^x$$ divide $$13!$$ evenly. This is possible only when the number of factors of $$2$$ in the numerator is greater than or equal to the factors of $$2$$ in the denominator. Finding the lower end of the range is easy. We simply observe that if \"x\" is zero, then we have $$2^0 = 1$$, and any integer divided by 1 is an integer. $$13!/1$$ is an integer. Therefore, zero is the lower end of the range. Now we have to find the upper value in the range. Given what we've said above (that the number of factors of $$2$$ in the numerator must be greater than or equal to the factors of $$2$$ in the denominator), we see that what we’re really trying to find is how many factors of “$$2$$” does $$13!$$ have? To begin, remember that $$13!$$ is just the sequence: $$(13*12*11*10* … *4*3*2*1.)$$ So first, how many multiples of “$$2$$” are on that list? Well, if we take $$13$$ divided by $$2$$, we get the count of numbers that are a multiple of $$2$$. $$13/2 = 6$$. So there are", "integer 1. ${\\;\\;}$ $\\frac {61}{50} \\qquad \\frac {\\sqrt {6}} {2} \\qquad \\frac {16}{13} \\qquad \\sqrt {1.6}$ 1. $\\frac {2}{3}$ 2. $\\frac {9}{5}$ 3. $\\frac {9}{4}$ 4. $\\frac {34}{9}$ Feb 22, 2012 Sep 28, 2014" ]

[ "Math Help - need help on \"integral part\" [x] 1. need help on \"integral part\" [x] Using definition 3.3 on NZM For real x, the symbol [x] denotes the greatest integer less than or equal to x. Why is [-15/2]=-8. 2. $-\\frac{15}{2} = -7.5$ What's the largest integer that's smaller than $-7.5$ ? 3. Originally Posted by o_O $-\\frac{15}{2} = -7.5$ What's the largest integer that's smaller than $-7.5$ ? -8, so [-3/2]=-2, since $-\\frac{3}{2} = -1.5$ and the the largest integer that is smaller -1.5 is -2? 4. Exactly.", "# Difference between revisions of \"2006 UNCO Math Contest II Problems/Problem 4\" ## Problem Determine all positive integers $n$ such that $n^2+3$ divides evenly (without remainder) into $n^4-3n^2+10$ ? By polynomial division, $\\frac{n^4 - 3n^2 + 10}{n^2 + 3} = n^2 - 6 + \\frac{28}{n^2 + 3}$. By default, $n^2 - 6 \\in \\mathbb{N}$ when $n \\in \\mathbb{N}$, so we must find all positive integral values of $n$ for which $\\frac{28}{n^2 + 3}$ is also an integer. Listing out the factors of $28$, we see that $n^2 + 3 \\in \\{1, 2, 4, 7, 14, 28\\}$. Looking for positive integral values that satisfy these conditions yields $\\boxed{n \\in \\{1, 2, 5\\}}$, as desired. $\\blacksquare$", "4(2x - 1) + 5(2x - 1) = 75.$ 22.) What is the largest integral value of x If “=” in Item 21 is replaced by “<”? 23.) Solve for x in the equation |3 – 2x| = 3. 24.) Solve for x in the inequality |3 – 2x| > 2. 25.) In 2005, Cora was three times as old as her son Jayson. Four years back then, she was four times as old as Jayson. In what year was Jayson born? 1.) 0 2.) $\\frac{1}{6}$ 3.) 6 4.) -5 and -4 5.) $68 cm^2$ 6.) 0 7.) {11} 8.) {2, 4, 6, 8} 9.) 2 10.) 144 11.) 6 12.) $3x^2 - 5x - 3$ 13.) $2x^2 - 3x^2 - 2x + 4 3$ 14.) -x – 4 15.) x + 4 16.) 4027 17.) $\\sqrt{85} cm$ 18.) $8x^2 + 24xy + 18y^2$ 19.) $\\frac{60000x}{y}$ 20.) (x + 6)cm 21.) 3 22.) 2 23.) 0 and 3 24.) $x < \\frac{1}{2}$ or $x > \\frac{5}{2}$ 25.) 1993 View Part 2 here: 2014 Grade 8 Math Challenge Elimination Questions with answers – Part 2 This entry was posted in Grade 7-8 and tagged , , , . Bookmark the permalink.", "value of 'a' for which $$x^2 + (2a+1)x + a^2+1$$ is always greater than 0 for all values of x. Question 6 How many integers solve the inequality $$\\frac{|Y^2+6Y+2|}{|Y^2+4Y+5|} > 4$$ Question 7 For how many integral values of x does the following inequality hold true : $$\\frac{(x^2+7x-44)}{(x^2+9x-136)} < 0$$ Question 8 If K = $$\\frac{y^4+\\frac{1}{y^4}+1}{y^2+\\frac{1}{y^2}+1}$$, which of the following is not a valid value for K for any value real value of y? Question 9 Find the sum of all the integral solutions of the equation:$$\\frac{6x^2-19x+8}{x^2+x+7}<0$$ Backspace 789 456 123 0.- Clear All Question 10 How many integers satisfy the inequality $$|X^2-20| < 6$$ Question 11 The number of positive integral solutions for the equation $$\\log_{2}\\frac{3x-7}{2x+3} \\leq 0$$ Question 12 Find the number of integral solutions of the equation $$\\frac{x^2-13|x|+30}{x^2-18x+81}<0$$ Question 13 How many negative integral values does ‘x’ take: ||x - |x - 2| + 3| - 4| < 3 Backspace 789 456 123 0.- Clear All Question 14 If y is a real number, what is the difference in the maximum and minimum values obtained by $$\\frac{y+5}{y^2+5y+25}$$ ? Question 15 Find the range of values of x which satisfy the condition: $$\\frac{x}{(x^2 + x - 56)} > 0$$? Question 16 If a,b and c are positive real numbers, what is the minimum value of a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b)", "$k=\\boxed{\\sqrt{26}}$ is the highest possible value. ### Fourth Solution By AM-QM, $\\frac{2x + 3y}{13} = \\frac{4 \\cdot \\frac{x}{2} + 9 \\cdot \\frac{y}{3}}{13} \\le \\sqrt{\\frac{4 \\cdot \\left(\\frac{x}{2}\\right)^2 + 9 \\left(\\frac{y}{3}\\right)^2}{13}} = \\sqrt{\\frac{2}{13}}$, so $2x + 3y \\le \\sqrt{26}$, equality when $\\frac x2 = \\frac y3$.", "$5 \\times5$ $3$ numbers that add up to ten: $3.333 \\times3.333 \\times3.33$ 4... 5... and the largest of these products is: $2 \\frac{1}{2} \\times 2 \\frac{1}{2} \\times 2 \\frac{1}{2} \\times 2 \\frac{1}{2}= 39.0625$. It soon becomes apparent that when $10$ is split into $n$ parts (where $n$ is a whole number), the largest product is given by $(\\frac{10}{n})^n$ . So the problem is in fact to find the largest value of $(\\frac{10}{n})^ n$ . The largest value of $(\\frac{10}{n})^ n$ is when $n=4$ that is $39.0625$. This is the largest product. Thomas Hu from A Y Jackson school used his knowledge of the euler number ($e=2.71828...$) to find the maximal solution for all values of $x$ Given the product $(\\frac{x}{n})^n$ (where $x$ is the number in question and $n$ in the number of parts it is being divided into). It is already clear that repeated multiplication of the same number $(2.5^4)$ is greater than that of two different numbers $(4*6)$ due to maximization and difference of squares $(x-a)(x+a)= x^2-a^2$. Although the optimal $n$ for $10$ was stated to be $4$, that is only true if one assumes that $n$ must be an integer. Otherwise, $n = 3.7, 3.68, 3.679, 3.6788$ each provides increasingly larger products. But these cannot work as the number must still add to the value", "# Fractions: Overview If $$13 < 7x + 1 < 25$$, how many integer values of $$x$$ are there? Correct. [[Snippet]] Subtract 1 from the inequality to isolate $$7x$$. >$$13 - 1 < 7x + 1 - 1 < 25 - 1$$ >$$12 < 7x < 24$$ Divide the inequality by 7 to isolate $$x$$. >$$\\frac{12}{7} < \\frac{7x}{7} < \\frac{24}{7}$$ >$$\\frac{12}{7} < x < \\frac{24}{7}$$ Based on this, $$x$$ is more than 1 ($$\\frac{7}{7} = 1$$, so $$\\frac{12}{7}$$ is greater) and less than 4 ($$\\frac{28}{7}=4$$, so $$\\frac{24}{7}$$ is smaller). In other words, the integer values of $$x$$ are 2 and 3. Hence, there are two possible integer values for $$x$$. Incorrect. [[snippet]] Carefully check your calculations. Incorrect. [[snippet]] Carefully check your work. Incorrect. [[snippet]] Solve the given inequality for $$x$$. Incorrect. [[snippet]] You need to solve the inequality $$13 < 7x + 1 < 25$$ for $$x$$. 1 2 3 4 5", "? a) 93 b) 73 c) 85 d) 81 Question 6: Find the largest integral value of x which satisfies the inequality (x – 4)(|x| – 5) < 0. a) 5 b) -6 c) -5 d) -8 Question 7: How many negative integral values does ‘x’ take: ||x – |x – 2| + 3| – 4| < 3 a) None b) None c) None d) None Question 8: Suppose A and B represent natural numbers less than 10 and the number ‘9874A67B’ is divisible by both 9 and 11. What is the product of A and B? a) 36 b) 63 c) 40 d) None of these Question 9: For how many positive integral values of x is the inequality |4-3x|<-1 true? a) 0 b) 1 c) 2 d) 3 Question 10: What is the solution set of the inequality ||x|-3| > 4? a) $$(-\\infty, -7) \\bigcup (7, \\infty)$$ b) $$(-\\infty, -7)$$ c) $$(7, \\infty)$$ d) None of these Inequalities and Modulus Answers & Solutions: 1) Answer (A) The given cubic equation is a simple cubic with integral roots. On factorizing we get, $$(x-1)(x-2)(x-2) > 0$$ Thus, the solution set is $$1 < x < 2$$ or $$x > 2$$ Thus only the integers 3, 4, 5, 6, 7, 8 and 9 satisfy the given inequality. 2) Answer", "solution 2. $$7B=3A$$ $$(x-44)C=4A$$ $$(x-y)C=(y-50)B$$ Our goal is to try to isolate $y$ into an inequality. The first equation gives $A=\\frac{7}{3}B$, which we plug into the second equation to get $$(x-44)C=\\frac{28}{3}B$$ To eliminate $x$, subtract equation 3 from equation 2: $$(x-44)C-(x-y)C=\\frac{28}{3}B-(y-50)B$$ $$(y-44)C=(\\frac{178}{3}-y)B$$ In order for the coefficients to be positive, $\\[44 Thus, the greatest integer value is $y=59$, choice $(E)$.", "Ramanujan's $\\pi(x)^{2}<{\\frac {ex}{\\log x}}\\pi {\\bigg (}{\\frac {x}{e}}{\\bigg )}$ \"In his well-known notebooks, Ramanujan[23] proves that the inequality $$\\pi(x)^{2}<{\\frac {ex}{\\log x}}\\pi {\\bigg (}{\\frac {x}{e}}{\\bigg )}$$ holds for all sufficiently large values of $x$.\" It is the \"holds for all sufficiently large values of $x$\" part that I need some help. I do not have the book to verify the proof (Is is correct?), and I wonder if someone has computed the minimum value of x? This is discussed by Axler in arXiv:1703.02407; see Theorems 1.3, 1.4 and surrounding discussion. In particular, if the Riemann hypothesis is true, then $$\\pi(x)^{2}<{\\frac {ex}{\\log x}}\\pi {\\bigg (}{\\frac {x}{e}}{\\bigg )} \\qquad\\mbox{ for all }x\\ge 38 358 837 683.$$ Regardless of the Riemann hypothesis, Axler proves that $$\\pi(x)^{2}<{\\frac {ex}{\\log x}}\\pi {\\bigg (}{\\frac {x}{e}}{\\bigg )} \\qquad\\mbox{ for all }x\\ge e^{9032}.$$ The largest known integer $x$ for which the inequality fails is $$x=38 358 837 682;$$ we have $$\\pi(x)^2=2704950040057325824, \\qquad {\\frac {ex}{\\log x}}\\pi {\\bigg (}{\\frac {x}{e}}{\\bigg )}= 2704950040042588896.1001\\ldots$$", "# Math Help - [SOLVED] Sequence involving largest integer function. 1. ## [SOLVED] Sequence involving largest integer function. Let $(a_n)$be the sequence defined by $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? 2. Hello wizard654zzz Originally Posted by wizard654zzz Let $(a_n)$be the sequence defined $a_n = [\\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\\sum_{n=1}^{30}a_n$ ?? Divide and get quotient and remainder: $\\frac{n^2 + 8n +10}{n+9}=(n-1) + \\frac{19}{n+9}$ Now for $1\\le n\\le 10, \\;2>\\frac{19}{n+9}\\ge 1$. Hence $a_n = n$. But for $n > 10, \\;\\frac{19}{n+9}<1$. Hence $a_n = n-1$. $1+2+3+...+10+10+11+12+...+29 = 445$", "A 8 B 9 C 10 A ### PCAT Quantitative Practice Questions -7 1. For how many positive integers, $n$, is true that $n^{2} \\leq 3n$ 1. $2$ 2. $3$ 3. $4$ 4. $5$ 2. If $a^{4}=16$, then $3^{a}$ 1. $3$ 2. $9$ 3. $16$ 4. $27$ 3. $\\sqrt{20}\\sqrt{5}=$ 1. $2\\sqrt{5}$ 2. $10$ 3. $4\\sqrt{5}$ 4. $5\\sqrt{10}$ 4. The sum of three positive consecutive even integers is x. What is the value of the middle of the three integers? 1. $\\frac{x}{3}+2$ 2. $3x$ 3. $\\frac{x-2}{3}$ 4. $\\frac{x}{3}$ 5. What is the average of $5^{10}$, $5^{20}$, $5^{30}$, $5^{40}$ and $5^{50}$? 1. $5^{9}+5^{19}+5^{29}+5^{39}+5^{49}$ 2. $5^{30}$ 3. $5^{149}$ 4. $150$ 6. Which of the following is equal to $(5^{6} \\times 5^{9})^{10}$? 1. $25^{150}$ 2. $25^{540}$ 3. $5^{540}$ 4. $5^{150}$ 7. What is the value of $3^{\\frac{1}{3}} \\times 3^{\\frac{2}{3}} \\times 3^{\\frac{3}{3}}$? 1. $3$ 2. $9$ 3. $27$ 4. $30$ 8. How many integers satisfy the inequality $|x| < 2 \\pi$. 1. $3$ 2. $4$ 3. $7$ 4. More than $7$ 9. What is the average of $5^{a} \\times 5^{b}=5^{300}$ 1. $50$ 2. $100$ 3. $150$ 4. $200$ 10. If $5^{a}5^{b}=\\frac{5^{c}}{5^{d}}$, what is d in terms of $a$, $b$ and $c$? 1. $\\frac{c}{a+b}$ 2. $c+ab$ 3. $c-a-b$ 4. $c+a-b$ 1 B 2 B 3 B 4 D 5 A 6 D 7 B", "power of $n$. So for $n$ large enough, $$\\frac{\\sqrt{2n-1}\\ln(4n+2)}{n(n+1)} < \\frac{\\sqrt{2n}*n^{1/10}}{n^2}<\\frac{2}{n^{14/10}}.$$ Use the Comparison Test (then the Integral Test). You may have to justify the first inequality, depending on who's asking. -", "# Differences This shows you the differences between two versions of the page. javascript_math [2013/02/10 22:14]will created javascript_math [2015/02/02 08:28] (current) 2013/02/10 22:15 will 2013/02/10 22:14 will created Next revision Previous revision 2013/02/10 22:15 will 2013/02/10 22:14 will created Line 4: Line 4: The largest exact integral value is $2^{53} = 9007199254740992$,​ integer values larger than this are represented inexactly. The smallest exact integral value is similarly $-2^{53}$. The largest exact integral value is $2^{53} = 9007199254740992$,​ integer values larger than this are represented inexactly. The smallest exact integral value is similarly $-2^{53}$. + + [algorithm javascript-math]", "have $$\\frac{15}{18}$$ ÷ $$\\frac{3}{18}$$ = $$\\frac{15}{18}$$ X $$\\frac{18}{3}$$ = $$\\frac{15 X 18}{18 X 3}$$ = 5, Matches with bit B. 5 pieces. Spiral Review Question 10. What is the absolute value of -8? _________________ 8, Explanation: The absolute value is the non-negative value of a real number without regard for its sign so the absolute value of -8 is 8. Question 11. Write an inequality to compare the integers -5 and -6. _________________ -5 > -6, Explanation: As – 5 is greater than -6 so an inequality to compare the integers -5 and -6 is -5 > -6. Question 12. Find the product: $$\\frac{2}{3}$$ × $$\\frac{3}{8}$$. $$\\frac{1}{4}$$, $$\\frac{2}{3}$$ X $$\\frac{3}{8}$$ so it is $$\\frac{2 X 3}{3 X 8}$$ = $$\\frac{1}{4}$$.", "# Absolute value equation infinite solutions $$|3-x|+4x=5|2+x|-13$$ One of the solutions is $[3,\\infty)$ I'm not familiar with interval solutions for absolute equations. How to solve for this interval? • I don't think that interval can be a solution: you wrote an equation, not an inequality ... – DonAntonio Jun 14 '17 at 22:35 • @DonAntonio One simple example is $|x-1| + |x+1| = 2$. All $x\\in[-1,1]$ are solutions. – peterwhy Jun 14 '17 at 22:41 • @DonAntonio No...every $x$ in that interval is a solution. Indeed, in that range $|3-x|=x-3$ and $|2+x|=2+x$ so the equation reads $x-3+4x=10+5x-13$ which is true. – lulu Jun 14 '17 at 22:42 • @peterwhy Excellent point, thanks. – DonAntonio Jun 14 '17 at 22:42 • @lulu Excellent comment, thank you. – DonAntonio Jun 14 '17 at 22:42 The first thing to do is to notice that if $x$ is large and positive, then $|3-x|$ is the same as $x-3$, so the equation becomes $$(x-3)+4x = 5(x+2) -13 \\\\ 5x-3 = 5x -3$$ and so any time $x$ is that large, the equation is automatically true. How large is \"that large\"? Well, as long as $x\\geq 3$ the essential property that $|3-x| = x-3$ holds. So the equation is autmatically satisfied for $x\\geq 3$, that is, on the intervale $[3,+\\infty)$. (By the way, the", "value of integer x? (1) 2x^2+9<9x --> factor qudratics: $$(x-\\frac{3}{2})(x-3)<0$$ --> roots are $$\\frac{3}{2}$$ and 3 --> \"<\" sign indicates that the solution lies between the roots: $$1.5<x<3$$ --> since there only integer in this range is 2 then $$x=2$$. Sufficient. $$(x-\\frac{3}{2})(x-3)<0$$ I agree one solution of this inequality is $$(x-\\frac{3}{2})$$ > 0 , (x-3)<0 => $$(\\frac{3}{2})$$ < x < 3 However, Don't u think this can also resort to $$(x-\\frac{3}{2})$$ < 0 , (x-3)>0 => x < $$(\\frac{3}{2})$$ , x > 3 and in that case x can have infinite values. and if that is the case Stmt-1 alone would not be sufficient. The second case of $$(x-\\frac{3}{2})<0$$ and $$(x-3)>0$$ is not possible. This condition leads to $$x<\\frac{3}{2}$$ and $$x>3$$, but $$x$$ can not be simultaneously less than $$\\frac{3}{2}$$ (to make $$x-\\frac{3}{2}$$ negative) and more than 3 (to make $$x-3$$ positive). For more on how to solve such kind of inequalities check: x2-4x-94661.html#p731476 (Check this first) inequalities-trick-91482.html everything-is-less-than-zero-108884.html?hilit=extreme#p868863 Hope it helps. _________________ Manager Joined: 07 Sep 2010 Posts: 246 Re: The Discreet Charm of the DS [#permalink] ### Show Tags 27 Jun 2012, 08:03 Hi Bunuel, Can we solve this question using graphical approach. Since (x-a)^2+(y-b)^2 = r^2 ----(A) Equation 1 represents equation of circle with radius 1. Now, if we can maximize the value of both", "to find the expected value of the largest length, we need to integrate over $(\\frac{1}{3},1)$ as the largest piece needs to be greater than $\\frac{1}{3}$. Hence $$\\begin{split} E[Z] = \\int_{\\frac{1}{3}}^{1} z f_Z(z) dz = \\int_{\\frac{1}{3}}^{1} z (9z-3) dz = \\frac{14}{9} \\end{split}$$ The result is obviously wrong as it needs to be something between $0$ and $1$, however after going over the solution multiple times, and checking the calculations with Wolfram, I cannot seem to figure out what went wrong. • Offhand, it looks like your setup allows $1-X-Y$ to be negative. – Barry Cipra Jun 14 '18 at 15:54 • What are $X,Y$ supposed to represent? If they're the lengths of two of the parts, those are not uniformly distributed. I would set $X,Y$ to the coordinates of the two points, and then find the expectation of $\\max(\\min(X,Y), |X-Y|, 1-\\max(X,Y))$. – Daniel Schepler Jun 14 '18 at 15:54 • Note that $F_Z(1)=\\frac 12\\times 4=2>1$. For that matter $F_Z(0)=\\frac 12$ which is already absurd. As a suggestion, walk through the calculation of $F_Z(0)$ to see where you go awry. – lulu Jun 14 '18 at 15:55 • @DanielSchepler $X$ is the length of the segment from $0$ to the first cut, and $Y$ is the length from the first cut to the second cut. – Andrei Crisan Jun 14", "Integral UConn Intermediate 2014 Problem - 4538 Evaluate $$I=\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{3}}\\frac{1}{\\tan\\theta +\\cot\\theta}d\\theta$$ The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "# How do you determine the largest integer value of x in the solution for 5x-9<=7? Jan 4, 2017 Re-write to get $x$´s and numbers on opposite sides #### Explanation: Add 9 to both sides: $5 x - \\cancel{9} + \\cancel{9} \\le 7 + 9 \\to 5 x \\le 16 \\to$ Divide by 5: $x \\le \\frac{16}{5} \\mathmr{and} x \\le 3 \\frac{1}{5}$ The largest integer value would be $x = 3$" ]

[ "and round to the nearest tenth as needed 37. mystique2626 well the answer ended up being $2.1\\times10^{13}$ 38. jim_thompson5910 hmm there's probably some formula used to get that...not sure which one", "your final answer to the nearest dollar.$", "(3.9) (7) = 13.65 = 13.7 rounded up to nearest tenth.", "to the nearest tenth, so the answer is 36.6. …round to the nearest tenth, so the answer is 18.2.", "and round to the nearest tenth as needed 37. anonymous well the answer ended up being $2.1\\times10^{13}$ 38. jim_thompson5910 hmm there's probably some formula used to get that...not sure which one", "# How do you round 8,428 to the nearest hundred? Oct 26, 2016 $8 , 428$ rounded to the nearest hundred is $8 , 000$ #### Explanation: We have the number $8 , 428$ and we want to round it to the nearest hundred. In order to round it to the nearest hundred, though, let's go over some rules... 1) If the number is more than $500$, then you round up. If the number is less than $500$, then you round down. In this case, the $428$ in $8 , 428$ is less than $500$, so $8 , 428$ rounded to the nearest hundred is $8 , 000$ Hope this helps!", "# How do you round 4,278,003 to the nearest 10,000? Round $4 , 278 , 003 \\to 4 , 280 , 000$ Since you are rounding to the nearest 10,000, you have to focus on the digit in the ten-thousands and thousands place in $4 , 278 , 003.$ Then change all the digits to the right of the rounded 8 to 0, giving you $4 , 280 , 000.$", "round up). (c) 43.653 to the nearest unit. ##### Answer (c) 44 (6 in the tenths place indicates we round up). (d) 752.245 to the nearest tenth. ##### Answer (d) 752.2 (the 4 in the hundredths place indicates we leave the digit in the tenths place unchanged). (e) 10.995 to 1 decimal place. ##### Answer (e) 11.0 (the 9 in the second decimal place indicates that we round up 9 to 10). (f) 15.5435 to 3 decimal places. ##### Answer (f) 15.544 (the 5 in the fourth decimal place indicates that we up to 3 to 4). (g) There is an advert in your local newspaper for a used car. The advertised price is$10,649. Estimate the price to the nearest $100 and$1000. $10,600 (to the nearest$100). $11,000 (to the nearest$1000). ## 2.5 Quiz Time ### Practice Quiz Now that you have taken the time to work through these sections, do this short quiz. It will help you to monitor your progress, particularly if you took the quiz at the start of the unit as well. The quiz does not check all the topics in the unit, but it should give you some idea of the areas you may need to spend more time on. Remember, it doesn’t matter if you get some, or even all of the questions", "# What is $4068.80 rounded to the closest cent? ##### 2 Answers Jan 28, 2018 #### Answer: It is already! #### Explanation: Assuming you are asking for $4068.80: $4068.80 to the nearest cent is the same thing as rounding $4068.80$to the nearest hundredth (because 1c=$0.01) $4068.80$ to the nearest hundredth is $4068.80$ :.$4068.80 to the nearest cent is $4068.80. Jan 28, 2018 See below #### Explanation: For calculating the nearest cent ,you need to see the digit in the ten's place. If the digit is less than 5, then decrease the number by hundred. Ex- Let the number be 524.5 So, in this case answer is 500. If the digit is greater than or equal to 5, increase the number by hundred. Ex- Let the number be 5789.8 So, in this case the answer is 5800. In your case, the digit is 6 means greater than five.", "number $21.75$ rounded to the nearest tenth is $21.8$. The number $72.33$ rounded to the nearest tenth is $72.3$. This is how to round to the nearest tenth. And this is precisely what our round to the nearest tenth calculator will do for you. ## Using our round to the nearest tenth calculator So to start with the tool, the first thing that you need to do is: • Input the decimal number which you want to round to its nearest tenth; • And that's it. That's the only thing that you need to do. The rest is up to the tool: it will display the rounded number. For instance, if you enter $88.56$ it will use the number at the hundredth position, which is $6$, to round the number at the tenth position, which is $5$ so the answer will be $\\bold {88.6}$. The rounding-off rules that we studied in elementary are applied here exactly. That is, if a number is greater than $4$ you add $1$ to the next digit while rounding up. ## All around the rounding! As mentioned earlier, \"the round to the nearest tenth calculator\" is not the only tool we have in our vault. In fact, we created an entire ecosystem around the rounding of numbers at various place values. Take a", "and 413. multiply the two numbers. 53 x 413 = 21889. so the product is 21889. Question 9. 906 × 57 Answer: The product is 51,642. Explanation: In the above-given question, given that, the two numbers are 906 and 57. multiply the two numbers. 906 x 57 = 51,642. so the product is 51,642. Question 10. 1,037 × 80 Answer: The product is 82,960. Explanation: In the above-given question, given that, the two numbers are 1037 and 80. multiply the two numbers. 1037 x 80 = 82,960. so the product is 82,960. Round Decimals Round each number to the nearest tenth. Question 11. 842.121 Answer: The number to the nearest tenth = 842.10. Explanation: In the above-given question, given that, the number is 842.121. to round a number to the nearest tenth look at the number of ones. if this is 5 or more round up. if it is 4 or less round down. 842.10. so the number to the nearest tenth = 842.10. Question 12. 10,386.145 Answer: The number to the nearest tenth = 10386.10. Explanation: In the above-given question, given that, the number is 10,386.145. to round a number to the nearest tenth look at the number of ones. if this is 5 or more round up. if it is 4 or less round down. 10386.10.", "cost is rounded to the tenth of a cent which means 3 places after the decimal. But the cost is given to us as $7.65. Since we need 3 places, the cost must be$7.650 (which will be written as $7.65) When will 7.6xy7 get rounded to 7.650? Now this is the tricky part of the question – from 7.6xy7, you need to drop the 7 and round up y. When you do that, you get 7.650. This means 7.6xy7 must have been 7.6497. Only in this case, when we drop the 7, we round up the 9 to make 10, carry the 1 over to 4 and make it 5. This is the only way to get 7.650 on rounding 7.6xy7 to the tenth of a cent. Hence x must be 4 and y must be 9. This statement alone is sufficient to answer the question. Answer (B) Hi Bunuel Need a clarification as I am not very thorough with rounding off concept. What if y=0 & x =5, 7.6507. Wouldn't this no. also be rounded to 7.65 ? 7.6507 rounded to the nearest tenth of a cent is 7.651. _________________ Manager Joined: 11 Aug 2019 Posts: 83 Re: The exact cost price to make each unit of a widget is$7.6xy7, where x [#permalink] ### Show Tags 27", "4. 1.63 11. What is 94,092.76 rounded to the nearest ones place? 1. 95,092.00 2. 94,093.00 3. 94,092.80 4. 94,000.00 12. Which number is 16.584 rounded to the nearest whole number? 1. 16 2. 16.5 3. 16.6 4. 17 13. Which shows 54.436 correctly rounded to the nearest hundredth? 1. 54.43 2. 54.00 3. 54.44 4. 54.437 14. Round$3.976 to the nearest cent. 1. $3.90 2.$3.98 3. $4.00 4.$4.90 15. Round 83.436 to the nearest hundredth. 1. 83.4 2. 83.44 3. 83.43 4. 83.433 16. Round 1,345,765.997 to the nearest whole number. 1. 2,000,000.0 2. 1,345,765.9 3. 1,345,766 4. 1,345,800 17. What is 34.678 rounded to the nearest hundredth? 1. 34.70 2. 34.67 3. 34.68 4. 34.60 18. A gas station sold 658.5849 gallons of gas in a day. How many gallons of gas did the gas station sell, rounded to the nearest tenth? 1. 660 2. 658.6 3. 700 4. 658.58 19. Which TWO statements about rounding decimals are correct? 1. The number $3.053$ rounded to the nearest hundredth is $3.05$. 2. The number $3.807$ rounded to the nearest hundredth is $3.80$. 3. The number $3.102$ rounded to the nearest hundredth is $3.11$. 4. The number $3.769$ rounded to the nearest hundredth is $3.76$. 5. The number $3.296$ rounded to the nearest hundredth is $3.30$. 20. Ms.", "to the nearest cent.)$", "# How do you round 2,220 to the nearest hundred? $2200.$ It's $2200$, this is the closest 100 to the number. The number lies between $2200 < 2200 < 2300$, thus it is $2200$.", "number $76$ to the nearest ten. Is $76$ closer to $70$ or $80$ on the number line? Now consider the number $72$. Find $72$ in Figure 1.8. How do we round $75$ to the nearest ten? Find $75$ in Figure 1.9. So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, $80$. So $75$ rounded to the nearest ten is $80$. Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than $5$, round down. If it is greater than or equal to $5$, round up. So, for example, to round $76$ to the nearest ten, we look at the digit in the ones place. The digit in the ones place is a $6$. Because $6$ is greater than or equal to $5$, we increase the digit in the tens place by one. So the $7$ in the tens place becomes an $8$. Now, replace any digits to the right of the $8$ with zeros. So, $76$ rounds to $80$. Let’s look again at rounding $72$ to the nearest ten. Again, we look to the ones", "impact on the answer. • Round each number greater than one to the nearest whole number, ten, hundred or thousand. • Round any number less than one to the nearest fractional amount. • Round all numbers before performing the calculation. For example, estimate the size of the answer to 39 × 4.85. 39 rounds up to 40. Multiples of 10 are generally easier to multiply. 4.85 rounds up to 5. Five is chosen as it is also an easier number to multiply. Since 4 × 5 = 20, 40 × 5 = 200. We need to add another zero. Here is an example of estimating the total cost of a shopping list. When estimating money, round each amount to the nearest whole number. •$0.90 rounds up to $1. •$1.25 rounds down to $1. •$2.87 rounds up to $3. •$6.10 rounds down to $6. •$3.22 rounds down to $3. Adding up the total we have$1 + $1 +$3 + $6 +$3 = $14. #### How to Estimate an Addition To estimate an addition, round all numbers to their first digit before adding them. To do this, look at the second digit of each number. If this digit is 5 or more, round up. If it is 4 or less, round down. For example, estimate the addition of 48 +", "$187.54,$218.89, $174.74,$104.76, $189.75, and$228.64. By estimating each of the amounts of the checks to the nearest ten dollars, come up with a reasonable estimate for Melissa's total expenditure for groceries. Possible Answers: Correct answer: Explanation: Round each of the amounts to the nearest ten dollars as follows: $187.54 rounds to$190. $218.89 rounds to$220. $174.74 rounds to$170. $104.76 rounds to$100. $189.75 rounds to$190. $228.64 rounds to$230. Add the rounded figures: ### Example Question #60 : Number Concepts And Operations Estimate the product by rounding each factor to the nearest unit, then multiplying. Possible Answers: Correct answer: Explanation: 8.39 rounded to the nearest unit is 8 because 0.39 is less than 0.5. 7.34 rounded to the nearest unit is 7 because 0.34 is less than 0.5. 3.52 rounded to the nearest unit is 4 because 0.52 is greater than 0.5. Multiply the three whole numbers to get the desired estimate:", "So 3,978 rounded to the nearest hundred is 4,000. Round to the nearest hundred: $17,852.$ 17,900 Round to the nearest hundred: $4,951.$ 5,000 Round each number to the nearest thousand: 1. $147,032$ 2. $29,504$ Solution ⓐ Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 0. Since 0 is less than 5, we do not change the digit in the thousands place. We then replace all digits to the right of the thousands pace with zeros. So 147,032 rounded to the nearest thousand is 147,000. ⓑ Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 5. Since 5 is greater than or equal to 5, round up by adding 1 to the 9. Then replace all digits to the right of the thousands place with zeros. So 29,504 rounded to the nearest thousand is 30,000. Notice that in part , when we add $1$ thousand to the $9$ thousands, the total is $10$ thousands. We regroup this as $1$ ten thousand and $0$ thousands. We add the $1$ ten thousand to the $3$ ten thousands and put a $0$ in the thousands place. Round to the nearest thousand:", "round off 198.287 to the nearest 10 cents, we have to round it off to 1 place of decimal. =$ 198.30 (iii) Round off 782.58 to the nearest dollar. To round off 782.58 to the nearest dollar, we have to round it off to the nearest whole number. Therefore, $782.58 =$ 783 rounded to the nearest rupee dollar. (iv) Round off 475.095 to the nearest cents. To round off 475.095 to the nearest cents, we have to round it off to 2 places of decimal. 475.095 rounded off to the nearest cents is 475.10. (v) Round off 18.066 cm to the nearest mm. Since, 1 mm = 0.1 cm, so to round it off to the nearest mm, we have to round it off to one decimal places. 18.066 cm = 18.1 cm rounded off to the nearest mm or 181 mm. (vi) Round off 53.4278 m to the nearest cm. Since, 1 cm = 0.01 m, we round off 53.4278 m to the nearest cm. We have to round it off to two places of decimal. 53.427 m = 53.43 m rounded off to the nearest cm or 5343 cm. (vii) 0.00737 kg to nearest g. Since, 1 g = 0.001 kg so to round off 0.00737 kg to the nearest g, we have to round it" ]

[ "tenths d. hundredths 7. Compare: 0.57 $\\begin{array}{|c|}\\hline \\\\ \\hline\\end{array}$ 0.54 a. + b. > c. = d. < 8. In which place is the digit 9 in the number 7.09? a. ones b. tenths c. hundredths d. thousands 9. In which place is the digit 1 in the number 8.18? a. ones b. hundredths c. thousands d. tenths", "to the nearest tenth. Use a non-calculator approach. Solution: To round to the tenths place, we must know if the digit in the hundredths place is $\\,5\\,$ or greater, or less than $\\,5\\,.$ As above, first determine that $\\,\\sqrt{130}\\,$ is between $\\,11\\,$ and $\\,12\\,$. Then, use long multiplication: $\\,11.5^2 = 132.25\\,$, so $\\,11.5\\,$ is a bit too big $\\,11.4^2 = 129.96\\,$, so $\\,11.4\\,$ is a bit too small Thus, $\\,\\sqrt{130}\\,$ lies between $\\,11.4\\,$ and $\\,11.5\\,$. Again using long multiplication: $\\,11.45^2 = 131.1025\\,$, so $\\,11.45\\,$ is a bit too big. Thus, the digit in the hundredths place must be less than $\\,5\\,$, and so the square root is closer to $\\,11.4\\,$. Thus, $\\,\\sqrt{130}\\approx 11.4\\,$. ## Practice For this exercise, you must list the ‘lies between’ integers from least (farthest left) to greatest (farthest right). lies between and", "and $$8$$ is greater than $$5$$. • Thus, the digit of tenths place increases by $$1$$ and the digit at the hundredths place become zero. • So, the nearest tenths to $$49.38$$ after round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$", "0.98 ________ The hundredths Explanation: As 2 < 5, We round 0.982 to 0.98. The place value of the digit 8 is hundredths. The hundredths Question 5. 3.695 to 4 ________ The ones Explanation: As 6 > 5, We round 3.695 to 4. The place value of the digit 3 is ones. The ones Question 6. 7.486 to 7.5 ________ The tenths Explanation: As 8 > 5, We round 7.486 to 7.5. The place value of the digit 4 is tenths. The tenths Write the place value of the underlined digit. Round each number to the place of the underlined digit. Question 7. 0.592 Place value: ________ Round: ________ Place value: 5 tenths = 0.5 Round: 0.6 Explanation: 0.592 (0 x 1) + (5 x $$\\frac{1}{10}$$) + (9 x $$\\frac{1}{100}$$) + (2 x $$\\frac{1}{1000}$$) Place Value: 5 x $$\\frac{1}{10}$$ = 5 tenths = 0.5 0.592 9 > 5 0.6 Question 8. 6.518 Place value: ________ Round: ________ Place value: 6 ones = 6 Round: 7 Explanation: 6.518 (6 x 1) + (5 x $$\\frac{1}{10}$$) + (1 x $$\\frac{1}{100}$$) + (8 x $$\\frac{1}{1000}$$) Place Value: 6 x 1 = 6 ones = 6 6.518 5 = 5 7 Question 9. 0.809 Place value: ________ Round: ________ Place value: 0 hundredths = 0 Round: 0.8 Explanation: 0.809 (0 x 1)", "round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$ D $$56.0$$ Option A is Correct # Rounding Rule for Decimal Number (Thousandths Place) • To round off a decimal to the nearest hundredth place, analyze the digit at the thousandths place. •", "column to the left, making the hundredths place $10 − 4 = 6$Changes: top row, fourth column is 0, last column is 10. Added to answer row has, 6 in last column. Now the tenths place is $0 − 0 = 0$Added to Answer row: 0 in fourth column. Subtract the numbers in the ones place. Added to Answer row: 2 in second column, decimal in third column. Subtract the numbers in the tens place. Added to answer row: 6 in first column.", "you have to multiply your number by 10³. Now you have $\\boxed{\\begin{array}{lcr}1000x&=&2345.345345345... \\\\ x&=&2.345345345...\\end{array}}$ Subtract and you have $999x = 2343~\\iff~x = \\frac{2343}{999}$", "# What is 1345/99 rounded to the nearest integer? Jul 20, 2017 See a solution process below: #### Explanation: $\\frac{1345}{99} = 13.58 \\ldots$ To round to the nearest integer we need to look at the digit in the tenths position, just to the right of the decimal place. If the digit in the tenths position is greater than or equal to $5$ we need to add $1$ to the digit in the one's position (the number just to the left of the decimal point.) If the digit in the tenths position is less than $5$ we just need to remove the decimal portion of the number. For this problem, the digit in the tenths position is a $5$ which is greater than or equal to $5$ so we need to add $1$ to the digit in the ones position: $\\frac{1345}{99} = 1 \\textcolor{red}{3} .58 \\ldots = 1 \\textcolor{red}{4}$ rounded to the nearest integer.", "the tenths place, we must know if the digit in the hundredths place is $\\,5\\,$ or greater, or less than $\\,5\\,$. As above, first determine that $\\,\\sqrt{130}\\,$ is between $\\,11\\,$ and $\\,12\\,$. Then, use long multiplication: $\\,11.5^2 = 132.25\\,$, so $\\,11.5\\,$ is a bit too big $\\,11.4^2 = 129.96\\,$, so $\\,11.4\\,$ is a bit too small Thus, $\\,\\sqrt{130}\\,$ lies between $\\,11.4\\,$ and $\\,11.5\\,$. Again using long multiplication, $\\,11.45^2 = 131.1025\\,$, so $\\,11.45\\,$ is a bit too big. Thus, the digit in the hundredths place must be less than $\\,5\\,$, and so the square root is closer to $\\,11.4\\,$. Thus, $\\,\\sqrt{130}\\approx 11.4\\,$. Master the ideas from this section When you're done practicing, move on to: Renaming Square Roots The number lies between and CONCEPT QUESTIONS EXERCISE: PROBLEM TYPES: 1 2 3 4 5 AVAILABLE MASTERED IN PROGRESS (MAX is 5; there are 5 different problem types.)", "$$10.6055$$. #### Add $$16.375+5.25+11.5$$ A $$16511.375255$$ B $$32.405$$ C $$33.125$$ D $$33.405$$ × Write $$5.25$$ just below $$16.375$$ and $$11.5$$ just below $$5.25$$, so that the decimal points are lined up exactly one above the other. Add zeros, so that all the numbers have the same place values. The given decimal numbers can be written as: $$\\begin{array} {c} \\hline &\\text{Tens}&\\text{Ones}&\\text{Decimal point}&\\text{Tenths}&\\text{Hundredths}&\\text{Thousandths}\\\\ \\hline &1&6&\\cdot&3&7&5\\\\ &0&5&\\cdot&2&5&0\\\\ +&1&1&\\cdot&5&0&0\\\\ \\hline &&&\\cdot&&&\\\\ \\hline \\end{array}$$ First, we add the digits of thousandths column. $$5+0+0=5$$ $$5$$ is to be written in the thousandths column. $$\\begin{array} {c} &1&6&\\cdot&3&7&5\\\\ &0&5&\\cdot&2&5&0\\\\ +&1&1&\\cdot&5&0&0\\\\ \\hline &&&\\cdot&&&5\\\\ \\hline \\end{array}$$ Now, add the digits of hundredths column. $$7+5+0=12$$ $$2$$ is to be written in the hundredths column and $$10$$ is carried to the next column (tenths). $$\\begin{array} {c} &1&6&\\cdot&^\\color{red}13&7&5\\\\ &0&5&\\cdot&\\;\\,2&5&0\\\\ +&1&1&\\cdot&\\;\\,5&0&0\\\\ \\hline &&&\\cdot&&2&5\\\\ \\hline \\end{array}$$ Add the digits of tenths column. $$1+3+2+5=11$$ $$1$$ is to be written in the tenths column and $$10$$ is carried to the next column (ones). $$\\begin{array} {c} &1&^\\color{red}16&\\cdot&3&7&5\\\\ &0&\\;\\,5&\\cdot&2&5&0\\\\ +&1&\\;\\,1&\\cdot&5&0&0\\\\ \\hline &&&\\cdot&1&2&5\\\\ \\hline \\end{array}$$ Now, add the digits of ones column. $$1+6+5+1=13$$ $$3$$ is to be written in the ones column and $$10$$ is carried to the next column (tens). $$\\begin{array} {c} &^\\color{red}11&6&\\cdot&3&7&5\\\\ &\\;\\,0&5&\\cdot&2&5&0\\\\ +&\\;\\,1&1&\\cdot&5&0&0\\\\ \\hline &&3&\\cdot&1&2&5\\\\ \\hline \\end{array}$$ Add the digits of tens column. $$1+1+0+1=3$$ $$3$$ is to be written in the tens column. $$\\begin{array} {c}", "$$12.4568$$ $$1$$:tens, $$2$$: ones, $$4$$: tenths, $$5$$: hundredths, $$6$$: thousandths, $$8$$: tens thousandths • Step 2: To round a decimal, find the place value you’ll round to. • Step 3: Find the digit to the right of the place value you’re rounding to. If it is $$5$$ or bigger, add $$1$$ to the place value you’re rounding to and remove all digits on its right side. If the digit to the right of the place value is less than $$5$$, keep the place value and remove all digits on the right. ### Rounding Decimals – Example 1: Round $$1.9287$$ to the thousandth place value. Solution: First look at the next place value to the right, (tens thousandths). It’s $$7$$ and it is greater than $$5$$. Thus add $$1$$ to the digit in the thousandth place. The thousandth place is $$8$$. $$→ 8 \\ + \\ 1=9$$, then, the answer is $$1.929$$ ### Rounding Decimals – Example 2: Round $$9.4126$$ to the nearest hundredth. Solution: First, look at the next place value to the right of the hundredth place. It’s $$2$$ and it is less than $$5$$, thus removing all the digits to the right. Then, the answer is $$9.41$$. ### Rounding Decimals – Example 3: Round $$2.1837$$ to the tenth place value. Solution: First look at the next", "lesser than 5), 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even), 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even). Scientific notation allows orders of magnitude to be more easily compared. Quantities are characterized with regard to accuracy (closeness to a true or accepted value) and precision … (a) $\\begin{array}{l}\\\\ \\begin{array}{l}\\hfill \\\\ \\frac{\\begin{array}{c}\\phantom{\\rule{1.4em}{0ex}}1.0023 g\\\\ \\text{+ 4.383 g}\\end{array}}{\\phantom{\\rule{1.5em}{0ex}}5.3853 g}\\hfill \\end{array}\\end{array}$, Answer is 5.385 g (round to the thousandths place; three decimal places), (b) $\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\\\ \\frac{\\begin{array}{l}\\text{}\\phantom{\\rule{0.8em}{0ex}}486 g\\hfill \\\\ -421.23 g\\hfill \\end{array}}{\\phantom{\\rule{1.3em}{0ex}}64.77 g}\\end{array}$, Answer is 65 g (round to the ones place; no decimal places). Here and in the lecture the capital U is used to denote a generic uncertainty estimate. Start counting sig figs at the first non-zero number and continue to the end of the number. I have no doubt about it. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 2). The number 0.000122300 still has only six significant figures (the zeros before the 1 are not significant). To express a number in scientific notation, you move the decimal place to the right if the number is less than zero or to the left if the number is greater than zero.For example,", "The ones place digit $$4$$ is less than $$5$$ and also $$24$$ is closer to $$20$$. So, we will round off $$24$$ to $$20$$. Hence, option (A) is correct. ### Round off the number $$24$$. A $$20$$ . B $$30$$ C $$25$$ D $$40$$ Option A is Correct # Rounding Rules for Decimal Number (Hundredths Place) • To round off the decimal to the nearest tenth's place, analyze the digit after decimal point i.e. at the hundredths place. • If the hundredths place value is $$5$$ or greater than $$5$$ then digit of tenths place increases by $$1$$ and the digit at hundredths place and thereafter become zero. • If the hundredths place value is less than $$5$$ then digit of tenths place remains same and the digit at hundredths place and thereafter becomes zero. Rounding off decimal (hundredths place): • Let's take an example to understand it. Example (1): $$49.38$$ • Let's use place value chart. Tens Ones Tenths Hundredths 4 9 . 3 8 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$8$$ and $$8$$ is greater than $$5$$. • Thus, the digit of tenths place increases by $$1$$ and the digit at the hundredths place become zero. • So, the nearest tenths to $$49.38$$ after", "If the digit at the thousandths place is $$5$$ or greater than $$5$$ , then the digit at the hundredth place increases by $$1$$ and the digits at the thousandth place and thereafter become zero. • If the digit at the thousandths place is less than $$5$$ , then the digit at hundredth place remains same and the digits at the thousandth place and thereafter become zero. Rounding off Decimal Number (Thousandths place): Rounding up to hundredths place or thousandths place is similar to tenth place. Let's look at an example to understand it. Example: $$23.759$$ Let's use the place value chart. Tens Ones Tenths Hundredths Thousandths 2 3 . 7 5 9 • Analyze the digit at the thousandths place in the given decimal number. • Here the digit at the thousandths place is greater than $$5$$ • Thus, the digit at the hundredths place increases by $$1$$ and the digit at the thousandth place becomes zero. • So, the nearest hundredths to $$23.759$$ after round off is $$23.76$$. • So, the nearest hundredths to $$23.759$$ after round off is $$23.76$$ . #### What is the round off value of $$28.39173$$ up to hundredth place? A $$28.38$$ B $$28.391$$ C $$28.3$$ D $$28.39$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths Thousandths", ". . . . . . . . . . . . . . . . . . $138$ Since the largest number is 145, the sequence looks like this: . . $\\begin{array}{cccccccccccc}131 & 133 & 135 & 137 & 139 & 141 & 143 & 145 \\end{array}$ . - . - . . . . . . . . . . $\\uparrow$ Therefore, the smallest integer is 131.", "Ten thousandths Millionths 2 8 . 3 9 1 7 3 Analyze the place value chart. Here, the digit at the thousandths place is $$1$$ which is less than $$5$$. Thus, the digit at the hundredths place i.e. $$9$$ remains same and the digits at the thousandths place and thereafter are removed. So, the nearest hundredth of $$28.39173$$ after round off is $$28.39$$. Hence, option (D) is correct. ### What is the round off value of $$28.39173$$ up to hundredth place? A $$28.38$$ . B $$28.391$$ C $$28.3$$ D $$28.39$$ Option D is Correct # Rounding Rule for hundreds place Rounding • Rounding a number means to change its value slightly. It is not as exact as the original value but it is approximately close to the original value. A number can be rounded to a certain place values like tens, hundreds, thousands etc. Rounding Rules: The rules which are used to change the number to the nearest power of ten are known as Rounding Rules. Rounding Rules for hundreds place: • To round off a whole number to the nearest hundreds place, analyze the digit at the tens place. • If the digit at the tens place is $$5$$ or greater than $$5$$, round the number up to the nearest hundreds place. • If the digit at", "the answer is 10. We have a five-digit number: .$$\\displaystyle \\square\\;\\square\\;\\square\\;\\square\\;\\square$$ The even digits must be in the order 2,4. The odd digits must be in the order 1,3,5. Choose two spaces for the two even digits. . . $$\\displaystyle \\text{There are: }\\:_5C_2 \\:=\\:{5\\choose2} \\:=\\:\\frac{5!}{2!\\,3!} \\:=\\:10\\,\\text{ choices.}$$ Once you choose the two spaces, the two even digits are inserted in 2-4 order. Then the three odd digits are inserted in the remaining spaces in 1-3-5 order. So there are 10 numbers with odds and evens in ascending order. In case you're skeptical, here they are: . . $$\\displaystyle \\begin{array}{ccccc}2&4&1&3&5 \\\\ 2&1&4&3&5 \\\\ 2&1&3&4&5 \\\\ 2&1&3&5&4 \\\\ 1&2&4&3&5 \\\\ 1&2&3&4&5 \\\\ 1&2&3&5&4 \\\\ 1&3&2&4&5 \\\\ 1&3&2&5&4 \\\\ 1&3&5&2&4 \\end{array}$$ Last edited:", "$$157$$ $$160$$ Exercise $$\\PageIndex{16}$$ Round to the nearest ten: $$884$$ $$880$$ Example $$\\PageIndex{9}$$: round a whole number Round each number to the nearest hundred: 1. $$23,658$$ 2. $$3,978$$ Solution Locate the hundreds place. The digit of the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. So 23,658 rounded to the nearest hundred is 23,700. Locate the hundreds place. Underline the digit to the right of the hundreds place. The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by added 1 to the 9. Then replace all digits to the right of the hundreds place with zeros. So 3,978 rounded to the nearest hundred is 4,000. Exercise $$\\PageIndex{17}$$ Round to the nearest hundred: $$17,852$$. $$17,900$$ Exercise $$\\PageIndex{18}$$ Round to the nearest hundred: $$4,951$$. $$5,000$$ Example $$\\PageIndex{10}$$: Round a whole number Round each number to the nearest thousand: 1. $$147,032$$ 2. $$29,504$$ Solution Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the", "\\\\ &&&& 1 & 0 & 0 & 6 & 0 \\\\ &&&&&&&& 0 \\\\ &&&& -- & -- & -- & -- & -- \\\\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\\\ &&&&& 9 & 8 & 9 & 9 & 4 \\\\ &&&&& -- & -- & -- & -- & -- \\\\ \\end{array}$ . - - . . . . . . . . . . . . . . . $\\begin{array}{cccccccccccc} 1 & 6 & 0 & 6 & 0 \\\\ 1 & 4 & 1 & 4 & 2 \\\\ -- & -- & -- & -- & -- \\\\ & 1 & 9 & 1 & 8 & 0 & \\hdots \\end{array}$ Pretty tedious, dividing by a 5-digit number. . . (Worse, if we had more decimal places.) Rationalizing, we would have: . $\\frac{\\sqrt{2}}{2}$ And I would much prefer to do: . $1.414213562... \\div 2$ . . Wouldn't you? And Quacky's view on combining fractions is excellent! • February 6th 2010, 04:25 PM Quacky Wow, very impressive answer, did you deduce that from logic, or did you learn it somewhere? (Surprised) • February 6th 2010, 06:48 PM Vicky1997 Quote: Originally Posted by Soroban Hello, Vicky! In the Old Days, say 100 B.C. (Before Calculators), we had", "of accuracy. Now we want to add ten numbers: [1000, 1, 1, 1, 1, 1, 1, 1, 1, 1] Of course the exact answer is 1009, but we can't get that in our 3 digit format. Rounding to 3 digits, the most accurate answer we get get is 1010. If we add smallest to largest, on each loop we get: Loop Index s 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 1009 -> 1010 So we get the most accurate answer possible for our format. Now lets assume that we add from largest to smallest. Loop Index s 1 1000 2 1001 -> 1000 3 1001 -> 1000 4 1001 -> 1000 5 1001 -> 1000 6 1001 -> 1000 7 1001 -> 1000 8 1001 -> 1000 9 1001 -> 1000 10 1001 -> 1000 Since the floating point numbers are rounded after each operation, all of the additions are rounded away, increasing our error from 1 to 9 from the exact. Now imagine if your set of numbers to add had a 1000, and then a hundred 1's, or a million. Note that to be truly accurate, you would want to sum the smallest two numbers, then resort the result into your set of numbers." ]

[ "and $$8$$ is greater than $$5$$. • Thus, the digit of tenths place increases by $$1$$ and the digit at the hundredths place become zero. • So, the nearest tenths to $$49.38$$ after round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$", "to the nearest tenth. Use a non-calculator approach. Solution: To round to the tenths place, we must know if the digit in the hundredths place is $\\,5\\,$ or greater, or less than $\\,5\\,.$ As above, first determine that $\\,\\sqrt{130}\\,$ is between $\\,11\\,$ and $\\,12\\,$. Then, use long multiplication: $\\,11.5^2 = 132.25\\,$, so $\\,11.5\\,$ is a bit too big $\\,11.4^2 = 129.96\\,$, so $\\,11.4\\,$ is a bit too small Thus, $\\,\\sqrt{130}\\,$ lies between $\\,11.4\\,$ and $\\,11.5\\,$. Again using long multiplication: $\\,11.45^2 = 131.1025\\,$, so $\\,11.45\\,$ is a bit too big. Thus, the digit in the hundredths place must be less than $\\,5\\,$, and so the square root is closer to $\\,11.4\\,$. Thus, $\\,\\sqrt{130}\\approx 11.4\\,$. ## Practice For this exercise, you must list the ‘lies between’ integers from least (farthest left) to greatest (farthest right). lies between and", "round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$ D $$56.0$$ Option A is Correct # Rounding Rule for Decimal Number (Thousandths Place) • To round off a decimal to the nearest hundredth place, analyze the digit at the thousandths place. •", "number of places (two places) to the right. Note that we must add two trailing zeros in the dividend. Thus, the problem becomes: $314 \\overline{ )2000}\\nonumber$ We need to round to the nearest hundredth. This requires that we carry the division one additional place to the right of the hundredths place (i.e., to the thousandths place). For the final step, we must round 6.369 to the nearest hundredth. In the schematic that follows, we’ve boxed the hundredths digit (the “rounding digit”) and the “test digit” that follows the “rounding digit.” Because the “test digit” is greater than or equal to 5, we add 1 to the “rounding digit,” then truncate. Therefore, to the nearest hundredth of a foot, the diameter of the circle is approximately $d ≈ 6.37 \\text{ ft.}\\nonumber$ Exercise Dylan has a circular dog pen with circumference 100 feet. Find the radius of the pen, correct to the nearest tenth of a foot. Use π ≈ 3.14. 15.9 feet ## Exercises In Exercises 1-16, solve the equation. 1. $$5.57x − 2.45x = 5.46$$ 2. $$−0.3x − 6.5x = 3.4$$ 3. $$−5.8x + 0.32 + 0.2x = −6.96$$ 4. $$−2.2x − 0.8 − 7.8x = −3.3$$ 5. $$−4.9x + 88.2 = 24.5$$ 6. $$−0.2x − 32.71 = 57.61$$ 7. $$0.35x − 63.58 = 55.14$$ 8. $$−0.2x −", "43/82, and round your answer to the nearest tenth. 79. Compute the quotient 51/59, and round your answer to the nearest tenth. 80. Compute the quotient 17/69, and round your answer to the nearest tenth. 81. Compute the quotient 5/74, and round your answer to the nearest hundredth. 82. Compute the quotient 3/41, and round your answer to the nearest hundredth. 83. Compute the quotient 5/94, and round your answer to the nearest hundredth. 84. Compute the quotient 3/75, and round your answer to the nearest hundredth. 85. Compute the quotient 7/72, and round your answer to the nearest hundredth. 86. Compute the quotient 4/57, and round your answer to the nearest hundredth. 87. Compute the quotient 16/86, and round your answer to the nearest tenth. 88. Compute the quotient 21/38, and round your answer to the nearest tenth. In Exercises 89-100, simplify the given expression. 89. $$\\frac{7.5 \\cdot 7.1 − 19.5}{0.54}$$ 90. $$\\frac{1.5(−8.8) − (−18.6)}{1.8}$$ 91. $$\\frac{17.76 − (−11.7)}{0.52}$$ 92. $$\\frac{−14.8 − 2.1}{2.62}$$ 93. $$\\frac{−18.22 − 6.7}{14.75 − 7.75}$$ 94. $$\\frac{1.4 − 13.25}{−6.84 − (−2.1)}$$ 95. $$\\frac{−12.9 − (−10.98)}{0.5^2}$$ 96. $$\\frac{5.1 − (−16.5)}{(−1.5)^2}$$ 97. $$\\frac{−9.5 \\cdot 1.6 − 3.7}{−3.6}$$ 98. $$\\frac{6.5(−1.6) − 3.35}{−2.75}$$ 99. $$\\frac{−14.98 − 9.6}{17.99 − 19.99}$$ 100. $$\\frac{−5.6 − 7.5}{−5.05 − 1.5}$$ 101. Given a = −2.21, c = 3.3, and d = 0.5,", "# Chapter 1 - Section 1.11 - Rounding Numbers - Exercises - Page 67: 21 $$Hundred=600$$ $$Ten=650$$ $$Unit=650$$ $$Tenth=649.9$$ $$Hundredth=649.90$$ $$Thousandth=649.900$$ #### Work Step by Step Hundreds place = nearest 100 (in the whole numbers) So, 649.8995 rounded to the hundreds place is $600$ Tens place = nearest 10 (in the whole numbers) So, 649.8995 rounded to the tens place is $650$. Unit = nearest 1 (in the whole numbers) So, 649.8995 rounded to the unit is $650$. Tenths place = to the nearest 10 (in the decimals) So, 649.8995 rounded to the tenth place is $649.9$ Hundredths place = to the nearest 100 (in the decimals) So, 649.8995 rounded to the hundredths place is $649.90$ Thousandths place = to the nearest 1000 (in the decimals) So, 649.8995 rounded to the thousandths place is $649.900$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "0.98 ________ The hundredths Explanation: As 2 < 5, We round 0.982 to 0.98. The place value of the digit 8 is hundredths. The hundredths Question 5. 3.695 to 4 ________ The ones Explanation: As 6 > 5, We round 3.695 to 4. The place value of the digit 3 is ones. The ones Question 6. 7.486 to 7.5 ________ The tenths Explanation: As 8 > 5, We round 7.486 to 7.5. The place value of the digit 4 is tenths. The tenths Write the place value of the underlined digit. Round each number to the place of the underlined digit. Question 7. 0.592 Place value: ________ Round: ________ Place value: 5 tenths = 0.5 Round: 0.6 Explanation: 0.592 (0 x 1) + (5 x $$\\frac{1}{10}$$) + (9 x $$\\frac{1}{100}$$) + (2 x $$\\frac{1}{1000}$$) Place Value: 5 x $$\\frac{1}{10}$$ = 5 tenths = 0.5 0.592 9 > 5 0.6 Question 8. 6.518 Place value: ________ Round: ________ Place value: 6 ones = 6 Round: 7 Explanation: 6.518 (6 x 1) + (5 x $$\\frac{1}{10}$$) + (1 x $$\\frac{1}{100}$$) + (8 x $$\\frac{1}{1000}$$) Place Value: 6 x 1 = 6 ones = 6 6.518 5 = 5 7 Question 9. 0.809 Place value: ________ Round: ________ Place value: 0 hundredths = 0 Round: 0.8 Explanation: 0.809 (0 x 1)", "$$157$$ $$160$$ Exercise $$\\PageIndex{16}$$ Round to the nearest ten: $$884$$ $$880$$ Example $$\\PageIndex{9}$$: round a whole number Round each number to the nearest hundred: 1. $$23,658$$ 2. $$3,978$$ Solution Locate the hundreds place. The digit of the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. So 23,658 rounded to the nearest hundred is 23,700. Locate the hundreds place. Underline the digit to the right of the hundreds place. The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by added 1 to the 9. Then replace all digits to the right of the hundreds place with zeros. So 3,978 rounded to the nearest hundred is 4,000. Exercise $$\\PageIndex{17}$$ Round to the nearest hundred: $$17,852$$. $$17,900$$ Exercise $$\\PageIndex{18}$$ Round to the nearest hundred: $$4,951$$. $$5,000$$ Example $$\\PageIndex{10}$$: Round a whole number Round each number to the nearest thousand: 1. $$147,032$$ 2. $$29,504$$ Solution Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the", "5$ rounded to the nearest unit is $136$. We have that $136 \\cdotp 5$ is half way between $136$ and $137$, and the convention is to round to the nearest even digit. $2 \\cdotp 484$ to Nearest Hundredth $2 \\cdotp 484$ rounded to the nearest hundredth is $2 \\cdotp 48$. This is because $2 \\cdotp 484$ is closer to $2 \\cdotp 48$ than $2 \\cdotp 49$. $0 \\cdotp 0435$ to Nearest Thousandth $0 \\cdotp 0435$ rounded to the nearest thousandth is $0 \\cdotp 044$. We have that $0 \\cdotp 0435$ is half way between $0 \\cdotp 043$ and $0 \\cdotp 044$, and the convention is to round to the nearest even digit. $4 \\cdotp 500 \\, 01$ to Nearest Unit $4 \\cdotp 500 \\, 01$ rounded to the nearest unit is $5$. This is because $4 \\cdotp 500 \\, 01$ is closer to $5$ than $4$.", "Figure 1.8. Figure 1.8 We can see that $72 72$ is closer to $70 , 70 ,$ so $72 72$ rounded to the nearest ten is $70 . 70 .$ How do we round $7575$ to the nearest ten. Find $7575$ in Figure 1.9. Figure 1.9 The number $75 75$ is exactly midway between $70 70$ and $80 . 80 .$ So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, $80.80.$ So, $7575$ rounded to the nearest ten is $80.80.$ Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than $5,5,$ round down. If it is greater than or equal to $5,5,$ round up. So, for example, to round $7676$ to the nearest ten, we look at the digit in the ones place. The digit in the ones place is a $6.6.$ Because $66$ is greater than or equal to $5,5,$ we increase the digit in the tens place by one. So the $77$ in the tens place becomes an $8.8.$ Now, replace any digits to the right of the $88$ with zeros. So,", "rounding decimals are correct? 1. The number $3.053$ rounded to the nearest hundredth is $3.05$. 2. The number $3.807$ rounded to the nearest hundredth is $3.80$. 3. The number $3.102$ rounded to the nearest hundredth is $3.11$. 4. The number $3.769$ rounded to the nearest hundredth is $3.76$. 5. The number $3.296$ rounded to the nearest hundredth is $3.30$. Grade 5 :: Place Value by LBeth Which statement correctly compares the two values? 1. The value of the 4 in 7.49 is 10 times the value of the 4 in 4.85. 2. The value of the 4 in 7.49 is $1/10$ the value of the 4 in 4.85. 3. The value of the 4 in 7.49 is 100 times the value of the 4 in 4.85. 4. The value of the 4 in 7.49 is $1/100$ the value of the 4 in 4.85. Grade 5 :: Place Value by desmondsmithsr1 266,661.56 Which six digit is one-tenth of the six digit in the hundreds place? 1. six hundredths 2. sixty 3. six thousands 4. sixty thousands Grade 5 :: Place Value by valdon19 Grade 5 :: Place Value by valdon19 Write the number 678,425 in expanded form. 1. 600,000 + 70,000 + 8,000 + 400 + 20 + 5 2. 60,000 + 70,000 + 8,000 + 400 + 20", "# What is 1345/99 rounded to the nearest integer? Jul 20, 2017 See a solution process below: #### Explanation: $\\frac{1345}{99} = 13.58 \\ldots$ To round to the nearest integer we need to look at the digit in the tenths position, just to the right of the decimal place. If the digit in the tenths position is greater than or equal to $5$ we need to add $1$ to the digit in the one's position (the number just to the left of the decimal point.) If the digit in the tenths position is less than $5$ we just need to remove the decimal portion of the number. For this problem, the digit in the tenths position is a $5$ which is greater than or equal to $5$ so we need to add $1$ to the digit in the ones position: $\\frac{1345}{99} = 1 \\textcolor{red}{3} .58 \\ldots = 1 \\textcolor{red}{4}$ rounded to the nearest integer.", "### Home > MC1 > Chapter 9 > Lesson 9.3.2 > Problem9-137 9-137. When writing numbers that are approximate, it is possible to use different degrees of accuracy. If you are instructed to be accurate to the nearest tenth, your answer should be the closest approximation using tenths. For example, you would approximate $5.248962$ as $5.2$, because the digit in the hundredths place is less than 5. You would approximate $10.396$ as $10.4$, because the digit in the hundredths place is greater than or equal to $5$. If you are instructed to be accurate to the nearest hundredth, your answer should be the closest approximation using hundredths. For example, you would approximate $5.248962$ as $5.25$ and $10.396$ as $10.40$, based on the digit in the thousandths place. Express each of the following numbers as approximations to the given degrees of accuracy. 1. Write $12.6243258$ accurately to the nearest tenth. You only need to look at the hundredths place, since that will determine how to round the tenths place. Since the hundredths place is a $2$, you round down. The correct answer is $12.6$. 2. Write $0.119834$ accurately to the nearest hundredth. What number is in the thousandths place? What is the nearest hundredth? $0.12$ 3. Write $π$ accurately to the nearest ten-thousandth. The ten-thousandth place is the fourth", "that $2.72$ is closer to $3$ than to $2$. So, $2.72$ rounded to the nearest whole number is $3$. (b) We see that $2.72$ is closer to $2.70$ than $2.80$. So we say that $2.72$ rounded to the nearest tenth is $2.7$. Can we round decimals without number lines? Yes! We use a method based on the one we used to round whole numbers. ### Round a decimal. 1. Locate the given place value and mark it with an arrow. 2. Underline the digit to the right of the given place value. 3. Is this digit greater than or equal to $5?$ • Yes – add $1$ to the digit in the given place value. • No – do not change the digit in the given place value 4. Rewrite the number, removing all digits to the right of the given place value. ### Tip For Success If the digit to the right of your given place value is 5 or above, give your given place value a shove. If the digit to the right of your given place value is 4 or less, let your given place value rest. ### example Round $18.379$ to the nearest hundredth. Solution $18.379$ Locate the hundredths place and mark it with an arrow. Underline the digit to the right of the", "$$12.4568$$ $$1$$:tens, $$2$$: ones, $$4$$: tenths, $$5$$: hundredths, $$6$$: thousandths, $$8$$: tens thousandths • Step 2: To round a decimal, find the place value you’ll round to. • Step 3: Find the digit to the right of the place value you’re rounding to. If it is $$5$$ or bigger, add $$1$$ to the place value you’re rounding to and remove all digits on its right side. If the digit to the right of the place value is less than $$5$$, keep the place value and remove all digits on the right. ### Rounding Decimals – Example 1: Round $$1.9287$$ to the thousandth place value. Solution: First look at the next place value to the right, (tens thousandths). It’s $$7$$ and it is greater than $$5$$. Thus add $$1$$ to the digit in the thousandth place. The thousandth place is $$8$$. $$→ 8 \\ + \\ 1=9$$, then, the answer is $$1.929$$ ### Rounding Decimals – Example 2: Round $$9.4126$$ to the nearest hundredth. Solution: First, look at the next place value to the right of the hundredth place. It’s $$2$$ and it is less than $$5$$, thus removing all the digits to the right. Then, the answer is $$9.41$$. ### Rounding Decimals – Example 3: Round $$2.1837$$ to the tenth place value. Solution: First look at the next", ", 27.11.2020 09:30, russboys3 # • With money, you often round to the nearest dollar. • If the number of cents is 50¢ or more, round up. $0.00$0.50 $1.00 • If the number of cents less than 50¢, round down.$7.35 rounded to the nearest dollar is $7.00.$24.86 rounded to the nearest dollar is $25.00. 6. Round to the nearest dollar. a)$5.48 b) $2.97 c)$0.85 f) $14.98 d)$9.99 e) \\$1.29 + ### Other questions on the subject: Mathematics Mathematics, 21.06.2019 13:40, NasirKA7372 Use the distance formula to determine the distance that point q is from r on the hypotenusebased on a 2-3 ratio. i need asap ! With number 11 ( show how you got the answer)", "rounded up to $$2,000$$. #### Round off the number: $$2,875$$ A $$3,000$$ B $$2,000$$ C $$1,000$$ D $$4,000$$ × Given: $$2,875$$ The nearest thousands to $$2,875$$ are $$2,000$$ and $$3,000$$. Hundreds place digit $$8$$, is greater than $$5$$ and also $$3,000$$ is closer to $$2,875$$. So, we will round the number up to $$3,000$$. Hence, option (A) is correct. ### Round off the number: $$2,875$$ A $$3,000$$ . B $$2,000$$ C $$1,000$$ D $$4,000$$ Option A is Correct # Rounding Rules for Ten-thousands Place • To round off a whole number to the nearest ten-thousands place, analyze the digit at the thousands place. • If the digit at the thousands place is $$5$$ or greater than $$5$$, round the number up to the nearest ten-thousands place. • If the digit at the thousands place is less than $$5$$, round the number down to the nearest ten-thousands place. Let's take an example. Round off the whole number $$23,750$$ to the nearest ten-thousands place. • We know that the nearest ten thousands to $$23,750$$ are $$20,000$$ and $$30,000$$. • Thousands place digit $$3$$ is less than $$5$$ and also $$23,750$$ is closer to $$20,000$$. • So, we will round the number $$23,750$$ down to $$20,000$$. Note: The values from $$20,001$$ to $$24,999$$ are rounded down to $$20,000$$ and the value", "# How do you divide 0.359 by 0.044 and round the quotient to the nearest hundredth? Jun 2, 2018 $8.16$ #### Explanation: $\\frac{0.359}{0.044} = 8.15 \\overline{90}$ Rounded to nearest 100th: $8.16$ Jun 2, 2018 $8.16$ #### Explanation: It is uncomfortable to divide by a decimal. $\\frac{0.359}{0.044} \\times \\frac{1000}{1000} = \\frac{359}{44}$ Now divide $359 \\div 44$ to get $8.15909 \\ldots .$ The 'nearest hundredth' = means $2$ decimal places The third decimal place is more than $5$, so we round up. $8.159090 . . \\approx 8.16$", "column to the left, making the hundredths place $10 − 4 = 6$Changes: top row, fourth column is 0, last column is 10. Added to answer row has, 6 in last column. Now the tenths place is $0 − 0 = 0$Added to Answer row: 0 in fourth column. Subtract the numbers in the ones place. Added to Answer row: 2 in second column, decimal in third column. Subtract the numbers in the tens place. Added to answer row: 6 in first column.", "Noelanijd 2022-07-26 1. Round the following numberls to the nearest ten. 527, 104, 955, 423 2. Round the following numerals to the nearest hundred. 689, 527, 1.365, 421 3. Round the following numerals to the nearest thousand. 1.365, 1.748, 5.231, 4.522. 4. Round the following numerals to the nearest tenth. 527.63, 289.34, 671.57, 15.85. encoplemt5 Expert Step 1 1. Round the following numerals to the nearest 10. $527=530\\phantom{\\rule{0ex}{0ex}}104=100\\phantom{\\rule{0ex}{0ex}}955=960\\phantom{\\rule{0ex}{0ex}}423=420$ Step 2 2. Round the numerals to the nearest hundred. $689=700\\phantom{\\rule{0ex}{0ex}}527=500\\phantom{\\rule{0ex}{0ex}}1365=1400\\phantom{\\rule{0ex}{0ex}}421=400$ Step 3 3. Round the numerals to nearest thousand. $1365=1400\\phantom{\\rule{0ex}{0ex}}1748=1700\\phantom{\\rule{0ex}{0ex}}5231=5200\\phantom{\\rule{0ex}{0ex}}4522=4500$ Step 4 4 . Round the numerals to nearest tenth. $527.63=527.6\\phantom{\\rule{0ex}{0ex}}289.34=289.3\\phantom{\\rule{0ex}{0ex}}671.57=671.6\\phantom{\\rule{0ex}{0ex}}15.85=15.9$ Do you have a similar question?" ]

[ "of possible combinations\"", "# Combinations How many arrangements are there of the letters of BANANA such that no two N's appear in adjacent positions? ×", "form one, two, three, ... distinct arrangements Semiprimes - problem 187 >> more about me can be found on my homepage. some names mentioned on this site may be trademarks of their respective owners. thanks to the KaTeX team for their great typesetting library !", "\"How many arrangements of this word are there so that \"cat\" AND \"dog\" do not appear?\", i.e. neither cat nor dog appear, i.e. $A\\cap B$. • I think this is what the OP intended -- the quantity he is looking for is the number of arrangements in which neither \"cat\" nor \"dog\" appear. – Y. Forman Jan 26 '18 at 19:57 • @Y.Forman Quoted from the OP \"How many arrangements of this word are there so that \"cat\" OR \"dog\" do not appear?\" This is a different phrase than \"How many arrangements... so that NEITHER \"cat\" NOR \"dog\" appear\" The one is $A\\cup B$ using his notation. The other is $A\\cap B$. – JMoravitz Jan 26 '18 at 19:59 • I read \"OR\" as \"nor.\" What OP wrote was also a different phrase than \"...so that either \"cat\" does not appear, or \"dog\" does not appear.\" I think the OP should clarify. – Y. Forman Jan 26 '18 at 20:01 • I meant it as either cat does not appear or dog does not appear. – rachelhoward Jan 26 '18 at 20:54 • I will choose this as the answer to my post because it points out that my set C does contain dog or cat but not both. I had overlooked this. – rachelhoward Jan 26 '18", "display 10 different brands of dog food in a row. In... A store manager wishes to display 10 different brands of dog food in a row. In how many different ways can this be done? How many such arrangements are possible if he can use only 7 out of the 10 different dog food types?... ##### (15 points) Let W be the region determined by 12 + y? + 22 < 4 and 2 0. Evaluate IILso\" +v+*)ddyda (15 points) Let W be the region determined by 12 + y? + 22 < 4 and 2 0. Evaluate IILso\" +v+*)ddyda... ##### In a double bond how many pi and sigma bonds are there? In a double bond how many pi and sigma bonds are there?...", "pairs of words.", "there? b) There are 3 times as many circles as there are diamonds. There is a total of 8 shapes altogether. How many circles are there? c) There are 25 pigs on the farm. There are 5 times as many pigs as horses. How many horses are there? d) There are 6 times as many red cars as blue cars in the mall parking lot. There are 40 more red cars as blue cars in the mall parking lot. How many blue cars are there? • 5. Solving equations: word problems Write an equation for the word problem and solve. a) There are 5 brown-haired dogs and 7 blonde-haired dogs at the pet adoption center. 1. How many dogs are there in total? 2. Some of the dogs are adopted later that day. If there are only 8 dogs left, how many dogs were adopted? b) There are 12 red jelly beans. There are 3 times as many green jelly beans as red jelly beans. 1. How many green jelly beans are there? 2. If there are 2 times as many green jelly beans as yellow jelly beans, how many yellow jelly beans are there? 3. How many jelly beans are there in total? c) Alice is 4 years younger than Benji. Benji is twice as old as", "Solution: The possible arrangements of the letters in the word FCP can be organized as shown below: FCP, FPC, CPF, CFP, PFC and PCF So, the number of words that can be formed with the letters of the word FCP is 6.", "distinct arrangements (n1,n2,n3,n4,n5) is ?", "# Combinations How many arrangements are there of the letters of BANANA such that no two N's appear in adjacent positions? × Problem Loading... Note Loading... Set Loading...", "form the word “DOG” using ‘D’ from the first cube, ‘O’ from the third and ‘G’ from the second. Note that if we had used the second cube to get ‘D’ instead, we would be missing a ‘G’.", "• Combinations of the above.", "Combinations Probability Level 2 How many arrangements are there of the letters of BANANA such that no two N's appear in adjacent positions? ×", "## Algebra 1: Common Core (15th Edition) There are 7 dogs and 21 treats; how many treats can each dog have? $7\\times x=21$ Now, we can divide both sides by 7 to find $x$: $\\frac{7x}{7}=\\frac{21}{7}$ $x=3$", "(((mouse,cat),elephant))dog; (((elephant,mouse),cat))dog; (((elephant,cat),mouse))dog;", "is after all, just a reversal of the multiplication process. If we say \"3 dogs times 5 = 15 dogs\" then by division we should have the eqivalent expressions that \"15 dogs divided by 3 = 5 dogs.\" and just as naturally \"15 dogs divided by 5dogs = 3\" . Students who have learned (I've been in England too long, I just had to edit \"learnt\") that \"eighths\" and \"fifths\" are just units like \"dogs\" and \"kittens\" should then undertand that 5 eighths divided by three eighths is just as clearly 5/3. ## Monday, 29 December 2008 ### Viete on Pythagorean Triples Reading The Analytic Art by Francois Viete, or at least the T R Witmer translation, and came across an interesting way of combining the legs of any two Pythagorean triples to create two others. Viete calls the two methods synaeresis and diaeresis, which seem to be language terms Viete appropriated. Synaeresis is cramming two vowel sounds together to make one... like the way people in New Orleans say \"Nor\"leans. I think the official term is diphthong, but check with an English major for confirmation. The actual Greek roots mean “a joining or bringing together\" or something similar Diaeresis is stretching one vowel out into two....and you can find your own example... To illustrate Viete's approach, we", "first test case, there are 3 \"cat\" and 1 \"dog\" in the string, so the answer is 4. For the second test case, there is only 1 \"cat\" and no \"dog\" in the string, so the answer is 1. None", "- word How many ways can the letters F, A, I, R be arranged?", "ones, d white ones. The question is, how many different pairs (a,b,c,d) are there such that a+b+c+d=9. 20. Mar 14, 2010", "3 such variations, one place out of three has a different symbol." ]

[ "$$6\\times 5=30$$ ways to place the $$E$$ and the $$G$$. Then, for each of the $$30$$ possible placements of the $$E$$ and $$G$$, there are $$4$$ ways to place the $$O$$. There are then $$30\\times 4=120$$ ways to place the $$E$$, the $$G$$, and the $$O$$. Then, for each of the $$120$$ possible placements of the $$E$$, $$G$$, and $$O$$, there are $$3$$ ways to place the $$R$$. There are then $$120\\times 3=360$$ ways to place the $$E$$, the $$G$$, the $$O$$, and the $$R$$. For each of the $$360$$ ways to place the $$E$$, $$G$$, $$O$$, and $$R$$, the $$2~D$$s must go in the remaining two empty spaces in $$1$$ way. Therefore, there are $$360\\times 1=360$$ ways to place the $$E$$, the $$G$$, the $$O$$, the $$R$$, and the $$2~D$$s. Thus, there are $$360$$ possible “words” that Nalan can make. 2. Determine how many “words” contain $$RED$$. There are $$4$$ ways to place the word $$RED$$ in the six spaces. The word $$RED$$ could start in the first, second, third, or fourth position. For each placement of the word $$RED$$, there are $$6$$ ways to place the letters of the word $$DOG$$ in the remaining three spaces: $$DOG$$, $$DGO$$, $$GDO$$, $$GOD$$, $$ODG$$ and $$OGD$$. So there are $$4 \\times 6=24$$ “words” containing $$RED$$. 3. Determine how many", "that “doggy” is a bad word we want to filter. We’re going to make the following assumptions: • People will use anywhere between 0 and 3 “trash” characters between each letter, and won’t always use the same number between letters. So if “.” is a trash character, we could see these: doggy d.og.gy d.o..g…g.y …252 others… dog…gy • (This is a conservative count) There are only 6 different characters that people will use as “trash”, and they are: [” “, “.”, “,”, “*”, “^”, “|”] Which in text is: space, period, comma, asterisk, carrot, pipe. • To keep the word recognizable, people will always keep the first and last letters in their spot, but will shuffle the middle letters. For doggy, these are: doggy dgogy dggoy • And of course, for each of these they will put a variable number of various trash characters between some or all letters. ### Separator Combinations Let’s grab an estimate for the number of possibilities. To start, we need to know how many ways they can place spacers. If they will use between 0 and 3 spacers between each character, then we have $4^4$ choices- each space can have 4 different “trash counts” and there are 4 spaces, and order does matter to us (so we’re looking at permutations, not combinations). so", "form the word “DOG” using ‘D’ from the first cube, ‘O’ from the third and ‘G’ from the second. Note that if we had used the second cube to get ‘D’ instead, we would be missing a ‘G’.", "for the first unfree letter from the the $3$ available to it. Two cases: • choose one of the two $T$ start positions, in which case the other unfree letter has only two choices, or • choose the other unfree letter start position, in which case that letter is free. Remaining $n$ free letters go as $n!$. Over all we have $\\binom 42(2\\cdot2\\cdot 2! + 1\\cdot 3!) = 6\\cdot (8+6) = 84$. Steampunk Solutions Inc", "“words” contain $$DOG$$. There are $$4$$ ways to place the word $$DOG$$ in the six spaces. The word $$DOG$$ could start in the first, second, third, or fourth position. For each placement of the word $$DOG$$, there are $$6$$ ways to place the letters of the word $$RED$$ in the remaining three spaces: $$DER$$, $$DRE$$, $$EDR$$, $$ERD$$, $$RDE$$ and $$RED$$. So there are $$4 \\times 6=24$$ “words” containing $$DOG$$. 4. Determine how many “words” contain both $$RED$$ and $$DOG$$. There are $$4$$ “words” that contain both $$RED$$ and $$DOG$$. They are as follows. $REDDOG\\quad\\quad DOGRED \\quad\\quad REDOGD\\quad\\quad DREDOG$ These $$4$$ “words” have been double-counted, as they would have been counted in case 2 and in case 3. Thus in total, there are $$24+24-4=44$$ “words” that contain $$RED$$ or $$DOG$$ (or both). Since there are $$360$$ possible “words” Nalan can make, we can subtract $$44$$ from this to determine the number of these “words” that do not contain $$RED$$ or $$DOG$$ (or both). Therefore, $$360-44=316$$ “words” can be formed in which the letters $$R$$, $$E$$ and $$D$$ are not adjacent to each other and in that order and the letters $$D$$, $$O$$ and $$G$$ are not adjacent to each other and in that order.", "into 2 groups of 4, then you can find a word for the first 4 and a word for the second 4 in our code book. How many ways are there to do that? Well there are 8! ways to order the cards and if we plop a divider in the middle we recognize there are 2! ways of ordering those two groups and that order is irrelevant to us (it doesn't matter which cards are in the first group or the second), but there are also 4! ways of ordering the cards within each group, and we don't care about that ordering either. Dividing out these irrelevant orderings, we get: ${ 8! \\over { 2! 4! 4! } } = { 40320 \\over { 2 \\cdot 24 \\cdot 24 } } = 35$ For any given board, there are 35 possible ways of assigning two words (in any order) and keep in mind these two words must be different (by design). Since there are only 25 words excluded from use as a codeword in any given game, that means we can always find a way to use our codebook in the two turn play.", "consider what the first and last symbol are. There are $$2$$ choices for the first, two for the last, and then the middle symbols can be chosen in $$d_{n-1}$$ ways. Therefore, $$d_{n}=2\\cdot 2\\cdot d_{n-1},\\tag{n\\ge 1}$$ with the base case $$d_0=2$$. Can you get a similar recursion for $$e_n$$?", "many different ways can the letters of the word MATH be rearranged to form a four-letter code word? Solution This problem is a bit different. Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the same category (the category is: the letters of the word MATH) and each time we choose an item we do not replace it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T). Thus there are $$4 \\cdot 3 \\cdot 2 \\cdot 1=24$$ ways to spell a code worth with the letters MATH. In this example, we needed to calculate $$n \\cdot(n-1) \\cdot(n-2) \\cdots 3 \\cdot 2 \\cdot 1$$. This calculation shows up often in mathematics, and is called the factorial, and is notated $$n$$! ##### Factorial $$n !=n \\cdot(n-1) \\cdot(n-2) \\cdots 3 \\cdot 2 \\cdot 1$$ ##### Example 26 How many ways can five different door prizes be distributed among five people? Solution There are 5 choices of prize for the first person, 4 choices", "$2$ ways to serve their meals. Thus, there are $(3\\cdot2)^3=\\boxed{216}$ ways to serve their meals. Note: This solution gets the correct answer through coincidence and should not be used. ~ dolphin7", "counts as 1 step. After the pressing, you could begin to spell the next character in the key (next stage), otherwise, you've finished all the spelling. Example: Input: ring = \"godding\", key = \"gd\" Output: 4 Explanation: For the first key character 'g', since it is already in place, we just need 1 step to spell this character. For the second key character 'd', we need to rotate the ring \"godding\" anticlockwise by two steps to make it become \"ddinggo\". Also, we need 1 more step for spelling. So the final output is 4. Note: 1. Length of both ring and key will be in range 1 to 100. 2. There are only lowercase letters in both strings and might be some duplcate characters in both strings. 3. It's guaranteed that string key could always be spelled by rotating the string ring. Seen this question in a real interview before? When did you encounter this question? Which stage? Which company?", "ending with “g,” such as “dig,” “dug,” and “dog.” 就像星号一样，您可以在单词的任何地方使用问号。 例如，搜索“ d？g”将找到所有以“ d”开头并以“ g”结尾的三个字母词，例如“ dig”，“ dug”和“ dog”。 You can also use multiple question marks together to stand in for multiple letters. For example, searching for “d??g” will find all four letter words starting with “d” and ending with “g”, such as “doug” and “dang.” 您还可以一起使用多个问号来代表多个字母。 例如，搜索“ d ?? g”将找到所有以“ d”开头并以“ g”结尾的四个字母词，例如“ doug”和“ dang”。 You can even use them in different places in your search phrase. For example, searching for “d?n?” would find four letter words where the first letter is “d” and the third letter is “n,” such as “dang” and “ding.” 您甚至可以在搜索短语的不同位置使用它们。 例如，搜索“ d？n？” 会找到四个字母词，其中第一个字母为“ d”，第三个字母为“ n”，例如“ dang”和“ ding”。 ### 使用At符号(@)和花括号({和})查找上一个字符的出现 (Use the At Sign (@) and Curly Brackets ({ and}) to Find Occurrences of the Previous Character) You can use the at sign (@) to specify one or more occurrences of the previous character. For example, searching for “ro@t” would find all words that begin with “ro” and end with “t” and that have any number of the letter “o” following that first occurrence. So, the search would find words like “rot,” “root,” and even “roooooot.” 您可以使用at符号(@)指定一个或多个出现的前一个字符。 例如，搜索“ ro @ t”将找到所有以“ ro”开头，以“ t”结尾并且在首次出现之后具有任意数量的字母“ o”的单词。 因此，搜索将找到“ rot”，“ root”甚至“ roooooot”之类的词。 For even more control over finding previous characters, you can use curly brackets to specify the exact number", "triple : , , , , , , , . In the second example there are 14 ways to choose the triple : , , , , , , , , , , , , , .", "problem. In problems like this, we should use stars and bars. There are $\\binom{5}{2}=10$ ways to assign three 2s to three distinct letters, and there are $\\binom{3}{2}=3$ ways to assign one 251 to three distinct letters. Multiplying, we get $3\\cdot10=\\boxed{30}$.", "for all $i = 1,\\ldots,n$, and we want to know which postman the dog can catch first. We present two solutions to these problems. Both solutions use deterministic data structures.", "ways of posting your eight letters into 12 letter boxes. If the order matters then there are 128 = 429,981,696 ways to do it: there are twelve letters boxes to choose from for your first letter, twelve for your second, …, twelve for your eighth and final letter. ~~ Dr Dec (Talk) ~~ 11:38, 8 October 2009 (UTC) You are choosing eight things from 12, with repeats: this would not make any sense, since then all 8 letters would be present in every sequence, i.e., the probability the OP is asking for would be trivially 1. Moreover, the OP explicitly describes the random choice being made as \"one sequence of 12\". We are choosing 12 things from 8, with repeats. — Emil J. 12:17, 8 October 2009 (UTC) • Why doesn't \"choosing eight things from 12, with repeats\" make any sense? Let me rephrase what I thought was the OP's question: You go shopping and you need to buy eight tins of dog food. There are twelve types available. What is the probability that you pick all the same type of dog food? You are making eight selections from 12 varieties, and you can repeat, i.e. buy the same type more than once. Maybe I'm misunderstanding what the OP was asking, but I'm quite sure that my answer", "of length $7$ from $\\{a,b,c,d,e\\}$ where we can't have more that $3$ of a single letter. There are $7^5$ strings without any restrictions. Now we need to subtract out the number of strings with $4, 5, 6,$ or $7$ of the same letter. A word with $4$ of the same letter will be a permutation of a set of one of the following forms: $$\\{w,w,w,w,x,y,z\\} \\text{ or } \\{w,w,w,w,x,x,y\\} \\text{ or } \\{w,w,w,w,x,x,x\\}$$ where $w,x,y,z$ represent some choice of four letters from $\\{a,b,c,d,e\\}$. Now for the first form, there are $5$ choices for $w$, $\\binom{4}{3}$ choices for $x,y,z$, and $\\binom{7}{4}$ permutations of $\\{w,w,w,w,x,y,z\\}$. (See here for details on the last part of the previous sentence). Thus the number of strings with exactly $4$ of the same letter is $$5\\cdot\\binom{4}{3}\\cdot\\binom{7}{4,1,1,1} = \\frac{5 \\cdot 4! \\cdot 7!}{3! \\cdot 4!} = 5 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 = 4200$$ Similarly, for $\\{w,w,w,w,x,x,y\\}$ and $\\{w,w,w,w,x,x,x\\}$ we have $$5\\cdot\\binom{4}{2}\\cdot\\binom{7}{4,2,1} = \\frac{5 \\cdot 4! \\cdot 7!}{2! \\cdot 4! \\cdot 2!} = 5 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 3 \\cdot 2 = 6300$$ and $$5\\cdot\\binom{4}{1}\\cdot\\binom{7}{4,3} = \\frac{5 \\cdot 4 \\cdot 7!}{4! \\cdot 3!} = 5 \\cdot 4 \\cdot 7 \\cdot 5 = 700$$ Putting this all together so far, $$5 \\cdot \\sum_{i=1}^3 \\binom{4}{i}\\binom{7}{4,4-i}$$ [... Running out of time for now.", "it were selected by s2 but not by s3. Note that the three_letter option has been deprecated in favour of the output_mode option, which provides the options \"oneletter\", \"threeletter\", \"basename\", or \"fullname\". Examples of each are given below. Note that the distinctions are particularly important for noncanonicals: some noncanonicals, such as ornithine, don't have a separate three-letter code for the D-equivalent. Output type Example oneletter RSTLNEXXYYS threeletter ARG,SER,THR,LEU,ASN,GLU,ORN,ORN,DTY,TYR,DSE basename ARG,SER,THR,LEU,ASN,GLU,ORN,DORN,DTYR,TYR,DSER fullname ARG:NtermProteinFull,SER,THR,LEU,ASN,GLU,ORN,DORN,DTYR,TYR,DSER:CtermProteinFull", "$1$. There are $\\dbinom{2005}{k}$ ways this can be done. For the remaining $2005-k$ people, there are one of two groups they can be in, namely group $2$ and group $3$. This means that there are $2^{2005-k}$ ways this can be done. There are $k$ ways to determine who is cool, and $k$ ways to determine who is cool. This is $k^2$. As $k$ ranges from $1$ to $2005$, we will get all such scenarios. This means that the number of ways that this can be done is also $3^{2005}S$. Another way to count this is two split it up into two cases. Case 1: $1$ person is both cool and smart. There are $2005$ ways to choose this person. The remaining $2004$ people have a choice of one of $3$ groups, making $2005 \\cdot 3^{2004}$ ways. Case 2: $1$ person is cool, and another person in smart. There are $2005$ ways to choose who is cool, and $2004$ ways to choose who is smart. The remaining $2003$ people have a choice of one of $3$ groups, making $2005 \\cdot 2004 \\cdot 3^{2003}$. Thus, we have: $$3^{2005}S=2005 \\cdot 3^{2003}(3+2004) \\implies S=2005 \\cdot \\dfrac{2007}{9}=2005 \\cdot 223 \\equiv 115 \\ (\\text{mod} \\ 1000)$$ Thus, the answer is $\\boxed{115}$", "in any order). Using the same logic above, the number of possible different distinguishable ways to order the string of letters $$RRRRRUUUUUUU$$ in which the first 7 letters must contain 4 R's and 3 U's, is equal to $${7 \\choose 3} \\cdot {5 \\choose 1} = 175$$ ways. That is the number of ways that Marvin will meet the frog and get eaten. Since we don't want that to happen, just subtract the total number of ways he is going to get to (5, 7) by the number of ways he is going to get eaten, which is equal to $$792-175=\\boxed{617}$$ ways. textot Jan 31, 2021 #2 +64 +1 THANK YOU SO MUCH! I really appreciate it! :) Jan 31, 2021", "d,o,g. How many possibilities exist to draw exactly one letter out those 3 letters? For sure, there are 3 possibilities to draw exactly 1 letter: “d”, “o”, or “g”. Thus we can write all the combinations as $${3 \\choose 1}$$. What about two letters? How many combinations exist to draw exactly two letters out of 3 letters? The combinations are “do”, “dg”, and “og” (Please note that for example “og” and “go” are counted only as one combinations, as order does not matter right now). Accordingly, the answer is 3 which can be written as $${3 \\choose 2}$$. One final question: How many combinations exist to draw exactly three letters out of 3 letters? Let us ask R! First, we implement a naive approach, by applying the formula from above. n <- 3 # the word \"dog\" has 3 letters k <- 3 # we draw exactly three letters factorial(n)/(factorial(k)*(factorial(n-k))) ## [1] 1 However, we may as well apply the built-in function called choose() to calculate the number of combinations. We do that for $$k = 1,2,3$$. n <- 3 choose(n, 1) ## [1] 3 choose(n, 2) ## [1] 3 choose(n, 3) ## [1] 1 Now, that we are familiar with the concept let us consider a more complex example: The Department of Earth Sciences of FU Berlin" ]

[ "form the word “DOG” using ‘D’ from the first cube, ‘O’ from the third and ‘G’ from the second. Note that if we had used the second cube to get ‘D’ instead, we would be missing a ‘G’.", "there? b) There are 3 times as many circles as there are diamonds. There is a total of 8 shapes altogether. How many circles are there? c) There are 25 pigs on the farm. There are 5 times as many pigs as horses. How many horses are there? d) There are 6 times as many red cars as blue cars in the mall parking lot. There are 40 more red cars as blue cars in the mall parking lot. How many blue cars are there? • 5. Solving equations: word problems Write an equation for the word problem and solve. a) There are 5 brown-haired dogs and 7 blonde-haired dogs at the pet adoption center. 1. How many dogs are there in total? 2. Some of the dogs are adopted later that day. If there are only 8 dogs left, how many dogs were adopted? b) There are 12 red jelly beans. There are 3 times as many green jelly beans as red jelly beans. 1. How many green jelly beans are there? 2. If there are 2 times as many green jelly beans as yellow jelly beans, how many yellow jelly beans are there? 3. How many jelly beans are there in total? c) Alice is 4 years younger than Benji. Benji is twice as old as", "by calling cat all one letter and dog all one letter, we arrange CvsD in $4!$ ways and so the number of ways that we can arrange catvsdog without the word cat and without the word dog is $8!-4!$. In actuality, this counts the number of arrangements without both cat and dog appearing, i.e. the number of arrangements with at least one of them missing, not the number of arrangements with both of them missing. You should actually have this ($8!-4!$) as the amount you were originally looking for and the amount you intended to calculate ($|C|$ where both are missing) is harder to find. Using your notation that $A$ as the set of arrangements without cat and $B$ the set without dog: $\\begin{array}{c|c|c} A^c&\\text{has a cat}&6!\\\\ A&\\text{has no cat}&8!-6!\\\\ B&\\text{has no dog}&8!-6!\\\\ A^c\\cap B^c&\\text{has a cat and has a dog}&4!\\\\ (A\\cup B)^c&\\text{complement of has no cat or has no dog}&4!\\\\ A\\cup B&\\text{has no cat or has no dog}&8!-4!\\\\ A\\cap B&\\text{has no cat and has no dog}&(8!-6!)+(8!-6!)-(8!-4!)\\\\ \\end{array}$ The phrasing you used \"How many arrangements of this word are there so that \"cat\" OR \"dog\" do not appear?\" reads to me that you are looking for $A\\cup B$, i.e. cat does not appear or dog does not appear. Your answer was for the related but different question of", "\"How many arrangements of this word are there so that \"cat\" AND \"dog\" do not appear?\", i.e. neither cat nor dog appear, i.e. $A\\cap B$. • I think this is what the OP intended -- the quantity he is looking for is the number of arrangements in which neither \"cat\" nor \"dog\" appear. – Y. Forman Jan 26 '18 at 19:57 • @Y.Forman Quoted from the OP \"How many arrangements of this word are there so that \"cat\" OR \"dog\" do not appear?\" This is a different phrase than \"How many arrangements... so that NEITHER \"cat\" NOR \"dog\" appear\" The one is $A\\cup B$ using his notation. The other is $A\\cap B$. – JMoravitz Jan 26 '18 at 19:59 • I read \"OR\" as \"nor.\" What OP wrote was also a different phrase than \"...so that either \"cat\" does not appear, or \"dog\" does not appear.\" I think the OP should clarify. – Y. Forman Jan 26 '18 at 20:01 • I meant it as either cat does not appear or dog does not appear. – rachelhoward Jan 26 '18 at 20:54 • I will choose this as the answer to my post because it points out that my set C does contain dog or cat but not both. I had overlooked this. – rachelhoward Jan 26 '18", "“words” contain $$DOG$$. There are $$4$$ ways to place the word $$DOG$$ in the six spaces. The word $$DOG$$ could start in the first, second, third, or fourth position. For each placement of the word $$DOG$$, there are $$6$$ ways to place the letters of the word $$RED$$ in the remaining three spaces: $$DER$$, $$DRE$$, $$EDR$$, $$ERD$$, $$RDE$$ and $$RED$$. So there are $$4 \\times 6=24$$ “words” containing $$DOG$$. 4. Determine how many “words” contain both $$RED$$ and $$DOG$$. There are $$4$$ “words” that contain both $$RED$$ and $$DOG$$. They are as follows. $REDDOG\\quad\\quad DOGRED \\quad\\quad REDOGD\\quad\\quad DREDOG$ These $$4$$ “words” have been double-counted, as they would have been counted in case 2 and in case 3. Thus in total, there are $$24+24-4=44$$ “words” that contain $$RED$$ or $$DOG$$ (or both). Since there are $$360$$ possible “words” Nalan can make, we can subtract $$44$$ from this to determine the number of these “words” that do not contain $$RED$$ or $$DOG$$ (or both). Therefore, $$360-44=316$$ “words” can be formed in which the letters $$R$$, $$E$$ and $$D$$ are not adjacent to each other and in that order and the letters $$D$$, $$O$$ and $$G$$ are not adjacent to each other and in that order.", "Solving by Combining Like Terms Solving by Using the Distributive Property Expressions Equation Grab Bag Word Problems 5h + 2 + 2h = 23 h = 3 2(x + 5) = 14 x = 2 ### 100 Somewhere over the Rainbow Wave goodbye Half Baked Thumbs Up ### 200 Troy has more than two dogs at home. All of them are corgis, except for two. All of them are pugs, except for two. All of them are labs, except for two. What kinds of dogs and how many of each kind does Troy have? 3 Dogs: one corgi, one pug, one lab ### 200 Simplify. 19x - 24x -5x x + 3x = -8 x = -2 The underdog ### 300 3a + 12 - 6a = -9 a = 7 ### 300 Bump in the Night ### 300 Simplify. 3 + 5(a + 4) 5a + 23 ### 300 What do an island and the letter “t” have in common? They are both in the middle of water. ### 300 Tyler has 4 boxes of apples. Each box has the same number of apples. After Tyler eats 3 apples, there are 109 apples left in the boxes. How many apples were in each box? a - # of apples in each box 4a - 3 = 109", "# Accenture previous placement papers - 2 1. One dog tells the other that there are two dogs in front of me.The other one also shouts that he too had two behind him. How many are they? Explanation: Dog Dog Dog So there are 3 dogs. They are in circle. 2. In the following questions, the following letters indicate mathematical operations as indicated below: A: Addition; V: Equal to; S: Subtraction; W:Greater than; M: Multiplication; X: Less than; D: Division Find Out of the four alternatives given in these questions, only one is correct according to the above letter symbols. Identify the correct one. See the options given below A) 6 S 7 A 2 M 3 W 0 D 7 B) 6 A 7 S 2 M 3 W 0 A 7 C) 6 S 7 M 2 S 3 W 0 M 7 D) 6 M 7 S 2 A 3 X 0 D Explanation:By BODMAS rule 6 – 7 + 3 × 2 > 0/7 3. What will be the output of the following code statements? Integer a = 10, b = 35, c = 5 print (a × b / c) – c Explanation: Apply BODMAS rule 10 × 35 = $\\dfrac{{350 }}{{5}}$ =70 –5 =65 4. Usually the room tariff in this hotel", "How many ways are there to put two slashes between letters ?One-to-one correspondence. A dog trainer wants to buy $8$ dogs all of which are either cocker spaniels,Irish setters,or Russian wolfhounds.In how many ways can she make the choice ? The answer is given as ${ 10 \\choose 2 } =45$ but I am not really understanding the undergoing logic. Let me explain. When I think about the problem I see how it's a matter of choosing how many sequences of the type ddd/dd/ddd (where the number of d's before the first / is the number of Irish,the number of d's in the middle represents the number of setters,and the number after the second / the number of wolfhounds ) we can have. I understand the one-to-one correspondence the author is talking about,as for every sequence we have exactly one choice of dogs.But when he explains his answer ${ 10 \\choose 2}$ he says that this is because we have $10$ positions (because we have $8$ d's and $2$ /'s ),and we need to count the number of ways to choose 2 positions for the two slashes out of 10. But I don't understand why he counts $10$ positions ,because we can put a / only between the d's and there are exactly $7$ places for that,and after", "ways of posting your eight letters into 12 letter boxes. If the order matters then there are 128 = 429,981,696 ways to do it: there are twelve letters boxes to choose from for your first letter, twelve for your second, …, twelve for your eighth and final letter. ~~ Dr Dec (Talk) ~~ 11:38, 8 October 2009 (UTC) You are choosing eight things from 12, with repeats: this would not make any sense, since then all 8 letters would be present in every sequence, i.e., the probability the OP is asking for would be trivially 1. Moreover, the OP explicitly describes the random choice being made as \"one sequence of 12\". We are choosing 12 things from 8, with repeats. — Emil J. 12:17, 8 October 2009 (UTC) • Why doesn't \"choosing eight things from 12, with repeats\" make any sense? Let me rephrase what I thought was the OP's question: You go shopping and you need to buy eight tins of dog food. There are twelve types available. What is the probability that you pick all the same type of dog food? You are making eight selections from 12 varieties, and you can repeat, i.e. buy the same type more than once. Maybe I'm misunderstanding what the OP was asking, but I'm quite sure that my answer", "walk with 2 dogs? List the ways. 3. On Wednesday Cesar has 4 dogs-Looie, Huey, Dewey, and Stewie–but now has 3 leashes. How many different ways can Cesar take a walk with 3 dogs? List the ways. Take a few minutes to discuss your findings with a partner. Share your method of finding all of the possible combinations. IV. Evaluate Combinations Using Combination Notation We can use a formula to help us to calculate combinations. This is very similar to the work that you did in the last section with factorials and permutations. Example Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation you write this as: ${_5}C_3 \\Longleftarrow 5 \\ \\text{items taken 3 at a time}$ In general, combinations are written as: ${_n}C_r \\Longleftarrow n \\ \\text{items taken} \\ r \\ \\text{at a time}$ To compute ${_n}C_r$ use the formula: ${_n}C_r = \\frac{n!}{r!(n - r)!}$ This may seem a bit confusing, but it isn’t. Notice that the factorial symbol is used with the number of object $(n)$ and the number taken at any one time $(r)$. This helps us to understand which value goes where in the", "at 20:58 The word cat could be in $6$ possible places \\begin{eqnarray*} cat \\underbrace{*****}_{5!} \\end{eqnarray*} so there are $6!$ permutations with the word cat & similarly for dog. There are $6$ ways to have cat followed by dog \\begin{eqnarray*} catdog \\underbrace{**}_{2!} \\end{eqnarray*} and $2!$ ways to arrange the last two characters (times this by $2$ for dog followed by cat) $24$. Now inclusion exclusion gives $8!- 2 \\times 6! + 24$. So your logic is sound. Your work seem perfectly good to me.", "step uses : 4) $Animal(F(x)) ∨ Loves(G(x),x)$. In order to unify it with 3) we have simply to perform the substitution $\\{ x/Jack \\}$ and then apply resolution to get : 5) $Loves(G(Jack),Jack)$ and we can simplify it because $p \\lor p \\equiv p$.", "first test case, there are 3 \"cat\" and 1 \"dog\" in the string, so the answer is 4. For the second test case, there is only 1 \"cat\" and no \"dog\" in the string, so the answer is 1. None", "only bears are chosen. 2. Find the probability that 2 bears and 3 dogs are chosen. 3. Find the probability that at least 2 dogs are chosen. ### Solution 1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $C\\left(6,5\\right)$ ways to choose 5 bears. There are 14 toys, so there are $C\\left(14,5\\right)$ ways to choose any 5 toys. $\\frac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{6}{2\\text{,}002}=\\frac{3}{1\\text{,}001}$ 2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $C\\left(6,2\\right)$ ways to choose 2 bears. There are 5 dogs, so there are $C\\left(5,3\\right)$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $C\\left(6,2\\right)\\cdot C\\left(5,3\\right)$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability. $\\frac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{15\\cdot 10}{2\\text{,}002}=\\frac{75}{1\\text{,}001}$ 3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or", "and 6 bears. 1. Find the probability that only bears are chosen. 2. Find the probability that 2 bears and 3 dogs are chosen. 3. Find the probability that at least 2 dogs are chosen. 1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\\text{\\hspace{0.17em}}C\\left(6,5\\right)\\text{\\hspace{0.17em}}$ ways to choose 5 bears. There are 14 toys, so there are $\\text{\\hspace{0.17em}}C\\left(14,5\\right)\\text{\\hspace{0.17em}}$ ways to choose any 5 toys. $\\text{\\hspace{0.17em}}\\frac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{6}{2\\text{,}002}=\\frac{3}{1\\text{,}001}\\text{\\hspace{0.17em}}$ 2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\\text{\\hspace{0.17em}}C\\left(6,2\\right)\\text{\\hspace{0.17em}}$ ways to choose 2 bears. There are 5 dogs, so there are $\\text{\\hspace{0.17em}}C\\left(5,3\\right)\\text{\\hspace{0.17em}}$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $\\text{\\hspace{0.17em}}C\\left(6,2\\right)\\cdot C\\left(5,3\\right)\\text{\\hspace{0.17em}}$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability. $\\text{\\hspace{0.17em}}\\frac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{15\\cdot 10}{2\\text{,}002}=\\frac{75}{1\\text{,}001}\\text{\\hspace{0.17em}}$ 3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either", "and 6 bears. 1. Find the probability that only bears are chosen. 2. Find the probability that 2 bears and 3 dogs are chosen. 3. Find the probability that at least 2 dogs are chosen. 1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\\text{\\hspace{0.17em}}C\\left(6,5\\right)\\text{\\hspace{0.17em}}$ ways to choose 5 bears. There are 14 toys, so there are $\\text{\\hspace{0.17em}}C\\left(14,5\\right)\\text{\\hspace{0.17em}}$ ways to choose any 5 toys. $\\text{\\hspace{0.17em}}\\frac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{6}{2\\text{,}002}=\\frac{3}{1\\text{,}001}\\text{\\hspace{0.17em}}$ 2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $\\text{\\hspace{0.17em}}C\\left(6,2\\right)\\text{\\hspace{0.17em}}$ ways to choose 2 bears. There are 5 dogs, so there are $\\text{\\hspace{0.17em}}C\\left(5,3\\right)\\text{\\hspace{0.17em}}$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $\\text{\\hspace{0.17em}}C\\left(6,2\\right)\\cdot C\\left(5,3\\right)\\text{\\hspace{0.17em}}$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability. $\\text{\\hspace{0.17em}}\\frac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{15\\cdot 10}{2\\text{,}002}=\\frac{75}{1\\text{,}001}\\text{\\hspace{0.17em}}$ 3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either", "probability that at least 2 dogs are chosen. 1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are$\\,C\\left(6,5\\right)\\,$ways to choose 5 bears. There are 14 toys, so there are$\\,C\\left(14,5\\right)\\,$ways to choose any 5 toys. $\\,\\frac{C\\left(6\\text{,}5\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{6}{2\\text{,}002}=\\frac{3}{1\\text{,}001}\\,$ 2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are$\\,C\\left(6,2\\right)\\,$ways to choose 2 bears. There are 5 dogs, so there are$\\,C\\left(5,3\\right)\\,$ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are$\\,C\\left(6,2\\right)\\cdot C\\left(5,3\\right)\\,$ways to choose 2 bears and 3 dogs. We can use this result to find the probability. $\\,\\frac{C\\left(6\\text{,}2\\right)C\\left(5\\text{,}3\\right)}{C\\left(14\\text{,}5\\right)}=\\frac{15\\cdot 10}{2\\text{,}002}=\\frac{75}{1\\text{,}001}\\,$ 3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are$\\,C\\left(9,5\\right)\\,$ways to choose toys from the 9 toys", "# Simple combinations question Justin needs to pick 17 toy animals to give to children at a party. The animals come in 5 kinds: dogs, dinosaurs, cows, lizards, and fish. How many different ways can he choose his set of toys Ok the answer for this is 21 choose 4. However, I dont understand how they arrived at this answer. If there are 17 slots, and for each slot, you can choose between 5 animals, then you should have $5^{17}$. Where has my reasoning gone wrong? - Presumably you cannot distinguish between 2 dogs, between 2 dinosaurs, etc. Your method counts dog-dog-dog-dog-dinosaur as distinct from dog-dinosaur-dog-dog-dog.. – Jonas Meyer Dec 14 '10 at 1:35 There is something missing from this question; where is the 21 coming from? – Hans Parshall Dec 14 '10 at 1:42 It says: 17+5-1 choose 5-1 – maq Dec 14 '10 at 1:43 Zaricuse made it clear below, sorry. – Hans Parshall Dec 14 '10 at 1:46 Check out Sivaram's explanation here: math.stackexchange.com/questions/11468/… – VelvetThunder Dec 14 '10 at 7:58 As Jonas Meyer indicated, the way you counted will include separately different orderings of the same numbers of each animal. Instead, you can have 21 slots. Four slots are to be filled with \"dividers\". You put dogs to the slots (if any) to the", "# Permutations of finishing a race Four cats and six dogs are in a race. How many ways exist if a dog must come first, second, and third. If a dog must come in the first three positions of a ten position race then the number of ways the first three positions can be ordered is $\\frac{6!}{(6-3)!}$ or 120 ways. My question is are the next seven positions expressed as $7!$ or $3!4!$? • Are the dogs distinguishable? Dec 13 '17 at 19:47 • I believe that was the author's intention, yes. Dec 13 '17 at 20:22 So once we've ordered the leaders in $_6P_3 = 120$ ways, we can order the rest in $7!$ ways, for a total of $_6P_3 \\cdot 7! = 604800$ ways.", "is after all, just a reversal of the multiplication process. If we say \"3 dogs times 5 = 15 dogs\" then by division we should have the eqivalent expressions that \"15 dogs divided by 3 = 5 dogs.\" and just as naturally \"15 dogs divided by 5dogs = 3\" . Students who have learned (I've been in England too long, I just had to edit \"learnt\") that \"eighths\" and \"fifths\" are just units like \"dogs\" and \"kittens\" should then undertand that 5 eighths divided by three eighths is just as clearly 5/3. ## Monday, 29 December 2008 ### Viete on Pythagorean Triples Reading The Analytic Art by Francois Viete, or at least the T R Witmer translation, and came across an interesting way of combining the legs of any two Pythagorean triples to create two others. Viete calls the two methods synaeresis and diaeresis, which seem to be language terms Viete appropriated. Synaeresis is cramming two vowel sounds together to make one... like the way people in New Orleans say \"Nor\"leans. I think the official term is diphthong, but check with an English major for confirmation. The actual Greek roots mean “a joining or bringing together\" or something similar Diaeresis is stretching one vowel out into two....and you can find your own example... To illustrate Viete's approach, we" ]

[ "# All in a Single File How many positive perfect squares less than $$10^{10}$$ have their digits appearing in non-decreasing order when represented in decimal notation? Eg : $$4, 16, 169, 1225$$ are valid numbers. But, $$64, 100, 196$$ are not valid. ×", "How many positive four-digit integers have their digits in ascending order from thousands to units? A. 6 B. 7 C. 84 D. 126 E. 210 _________________ Re: How many positive four-digit integers have their digits in ascending [#permalink] 24 Dec 2018, 05:01 Display posts from previous: Sort by", "# Too odd Out of the first 100 positive integers, how many odd primes are there? Details and assumptions You may use the fact that there are 25 prime numbers from 1 to 100. ×", "# Digit Sum 18 How many 3 digit positive integers $$N= \\overline{abc}$$ are there, such that the digit sum of $$N$$ (i.e. $$a+b+c$$) is equal to 18? Details and assumptions The number $$012=12$$ is a 2-digit number. ×", "# 2016 AMC 10 B #24 (video solution) How many four-digit integers $abcd$, with $a \\neq 0$, have the property that the three two-digit integers $ab < bc < cd$ form an increasing arithmetic sequence? One such number is $4692$, where $a=4, b=6, c=9,$ and $d=2$.", "# \"Double\" Divisors If a three-digit number $$\\overline{abc}$$ has 28 positive divisors, how many positive divisors does the six-digit number $$\\overline{abcabc}$$ have? ×", "all 3 digit numbers that leave a remainder of '2' when divided by 3? A. 897 B. 1,64,850 C. 1,64,749 D. 1,49,700 Answer: Option B… ## How many 2-digit positive integers are.. How many 2-digit positive integers are divisible by 4 or 9? A. 32 B. 22 C. 30 D. 34 Answer: Option C Show Answer Solution(By Apex Team) Number of 2-digit…", "# Four Digit Number - Can You Find It? I am thinking of a four digit positive integer with distinct digits. Of course, there's a total of $$4!-1=23$$ ways to rearrange the digits to form a new 4 digit positive integer. If the sum of these other 23 numbers is 157193, what is the number that I was thinking of? ×", "# Only Two is Allowed How many 2 digit positive integers are there? Details and assumptions The number $$12 = 012$$ is a 2 digit number, not a 3 digit number. ×", "# Digit product Probability Level 3 How many positive integers less than $1,000,000,000$ have the product of their digits equal to 49? ×", "# Digit product Discrete Mathematics Level 3 How many positive integers less than $$1,000,000,000$$ have the product of their digits equal to 49? ×", "1. ## Easy combinatorics problem How many four-digit positive integers are there that contain the digit 1? 2. ## Re: Easy combinatorics problem Originally Posted by alexmahone How many four-digit positive integers are there that contain the digit 1? To be clear $0109=109$ is not a four digit number, right?? There are $9\\cdot 10^3$ four-digit numbers. Of those there are $8\\cdot 9^3$ that do not contain any 1's. So how many contain at least one 1? 3. ## Re: Easy combinatorics problem hi alexmahone should the digits be the same or must be different? if they are the same then there are: 1000+900+900+900=2800 if they are not the same then there are: 720+648+648+648=2664", "# Digit product How many positive integers less than $$1,000,000,000$$ have the product of their digits equal to 49? ×", "# Three digit numbers How many three digit positive integers have $$0$$ as (at least) one of its digits? Details and assumptions The number $$12 = 012$$ is a two digit number, not a three digit number. ×", "# Only Two is Allowed Discrete Mathematics Level 1 How many 2 digit positive integers are there? Details and assumptions The number $$12 = 012$$ is a 2 digit number, not a 3 digit number. ×", "• KingGeorge [SOLVED] While I'm not terribly busy, have this as a challenge problem. What are the next two numbers in this sequence of four digit positive integers? $8741, 7632, 6552, 9963, 6642, \\_\\_\\_\\_, \\_\\_\\_\\_$ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "GMAT CLUB'S CHALLENGE QUESTIONS: How many positive four-digit integers have their digits in ascending order from thousands to units? A. 6 B. 7 C. 84 D. 126 E. 210 Hi.. an easier, short and logical way.. choose 4 out of 10... there can be ONLY 1 way the 4 can be arranged in ascending order, so selection is what we are looking for and NOT arrangements.. $$10C4 = \\frac{10!}{6!4!} = \\frac{10*9*8*7}{4*3*2} = 210$$ BUT it contains numbers with 0 in thousands digit making these 3-digit numbers so $$9C3= \\frac{9!}{6!3!} = 84$$ $$ans = 210-84=126$$ D Great explanation chetan2u we can reduce 1 step here because $$0$$ cannot be part of any number with ascending digits so we simply need to select $$4$$ digits out of $$9$$ digits Hence $$9_C_4=\\frac{9!}{4!5!}=126$$ Math Expert Joined: 02 Sep 2009 Posts: 52173 Re: How many positive four-digit integers have their digits in ascending [#permalink] ### Show Tags 24 Dec 2018, 04:01 Bunuel wrote: GMAT CLUB'S CHALLENGE QUESTIONS: How many positive four-digit integers have their digits in ascending order from thousands to units? A. 6 B. 7 C. 84 D. 126 E. 210 _________________ Re: How many positive four-digit integers have their digits in ascending &nbs [#permalink] 24 Dec 2018, 04:01 Display posts from previous: Sort by # How many positive four-digit integers have", "# Primes N^4 + 4 Number Theory Level 2 How many positive integers $$N$$ are there such that $$N^4 + 4$$ is prime? ×", "their 5 26 Nov 2015, 06:33 1 How many different positive integers can be formed 5 24 Feb 2015, 11:42 How many different four-digit integers can be composed of digits 0, 2, 2 30 Sep 2014, 01:26 25 How many four-digit positive integers can be formed by using 20 22 May 2012, 09:35 1 What is the greatest number of four-digit integers whose 5 15 Sep 2008, 04:15 Display posts from previous: Sort by", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×" ]

[ "tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$? The first term of a sequence is $2005$. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the $2005^\\text{th}$ term of the sequence? Each term in the sequence that begins 13, 9, 18, $\\cdots$ is the sum of three times the tens digit and two times the units digit of the previous term. What is the greatest value of any term in this sequence? Define a sequence of positive integers $s_1, s_2, . . . , s_{10}$ to be $terrible$ if the following conditions are satisfied for any pair of positive integers $i$ and $j$ satisfying $1 \\le i < j \\le 10$: - $s_i > s_j$ - $j - i + 1$ divides the quantity $s_i + s_{i+1} + \\cdots + s_j$ Determine the minimum possible value of $s_1 + s_2 + \\cdots + s_{10}$ over all terrible sequences.", "# 500th term The number sequence $$2,3,5,6,7,10,11, \\ldots$$ consists of positive integers that are neither squares nor cubes. Find the 500th term of this sequence. × Problem Loading... Note Loading... Set Loading...", "Except for the first two terms, each term of the sequence $$1000, x, 1000 - x,\\ldots$$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encountered. What positive integer $$x$$ produces a sequence of maximum length? (第十六届AIME1998 第8题)", "value to \\$10000? Month Interestearned Total amountof moneyin the account $0$ $\\dollar0.00$ $\\dollar5000.00$ $1$ $\\dollar33.33$ $\\dollar5033.33$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ # Subsection8.1.1Sequences As our discussion in the introduction and Preview Activity 8.1.1 illustrate, many discrete phenomena can be represented as lists of numbers (like the amount of money in an account over a period of months). We call these any such list a sequence. In other words, a sequence is nothing more than list of terms in some order. To be able to refer to a sequence in a general sense, we often list the entries of the sequence with subscripts, \\begin{equation*} s_1, s_2, \\ldots, s_n \\ldots \\text{,} \\end{equation*} where the subscript denotes the position of the entry in the sequence. More formally, ##### Definition8.1.3 A sequence is a list of terms $s_1, s_2, s_3, \\ldots$ in a specified order. As an alternative to Definition 8.1.3, we can also consider a sequence to be a function $f$ whose domain is the set of positive integers. In this context, the sequence $s_1\\text{,}$ $s_2\\text{,}$ $s_3\\text{,}$ $\\ldots$ would correspond to the function $f$ satisfying $f(n) = s_n$ for each positive integer $n\\text{.}$ This alternative view will be be useful in many situations. We will often write the sequence \\begin{equation*} s_1, s_2, s_3, \\ldots \\end{equation*}", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # 2005 AIME I Problems/Problem 2 ## Problem For each positive integer $k,$ let $S_k$ denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is $k.$ For example, $S_3$ is the sequence $1,4,7,10,\\ldots.$ For how many values of $k$ does $S_k$ contain the term 2005? ## Solution Suppose that the nth term of the sequence $S_k$ is 2005. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\\cdot 3\\cdot 167.$ The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002$, $(3,668)$, $(4,501)$, $(6,334)$, $(12,167)$, $(167,12)$,$(334,6)$, $(501,4)$, $(668,3)$, $(1002,2)$, and $(2004,1)$. Thus the requested number of values is 12.", "+0 # Help! 0 326 1 1. The sequence 2, 3, 5, 6, 7, 10, 11, $\\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $400^{\\mathrm{th}}$ term of the sequence? 2. The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string? Jun 3, 2018", "## goformit100 one year ago Consider the sequence 2;3;5;6;7;8;10;::: of all positive integers that are not perfect squares. Determine the 2011th term of this sequence. 1. goformit100 @oldrin.bataku 2. oldrin.bataku Find the number of perfect squares under 2011 -- this is the difference between 2011 and our number. 3. oldrin.bataku Well, we know $$\\sqrt{2011}\\approx45$$ with rounding (since $$44^2=1936$$, $$45^2=2025$$). This means we have skipped $$45$$ square numbers at this point, so our 2011th term is $$2011+45=2056$$", "# 500th term The number sequence $$2,3,5,6,7,10,11, \\ldots$$ consists of positive integers that are neither squares nor cubes. Find the 500th term of this sequence. ×", "= 9.25238 - (-0.05)^t \\times 9.25238 \\qquad t = 0, 1, \\cdots$ 4. $W_{100} \\approx 9.25238$ Exercise 3 a 1. $W_0, \\cdots ,W_4$ = 0, 1,0 , 1, 0. 2. The sequence continues to alternate between 0 and 1. c 1. $W_0, \\cdots ,W_4$ = 0, 1, 2, 3, 4. 2. The sequence is the sequence of non-negative integers. e 1. $W_0, \\cdots ,W_4$ = 0, 1, 3, 7, 15. 2. The sequence is $W_t = 2^t - 1$.", "each block of $$1000$$ integers contains one which ends in three $$0's$$. Starting from any integer $$k$$, we go to the next multiple of $$1000$$ (a gap of at most $$999$$). We then add whatever three (or fewer) digit integer we need to \"correct\" the sum of the digits $$\\pmod {27}$$, which takes, at most, another $$999$$. Thus, every block of $$2\\times 999$$ consecutive integers contains at least one for which the sum of the digits is a multiple of $$27$$. A similar argument works for any desired divisor. I expect the bound could be tightened considerably, but at least this shows that a bound exists. • Simple insight, great help, thank you:) – Martund Jul 27 at 13:57 Let $$Q(x)$$ denote the digit sum of $$x$$. $$999$$ is the smallest $$n$$ such that among any $$n$$ consecutive positive integers, at least one has digit sum a multiple of $$27$$. First note that none of the $$998$$ consecutive integers $$1,2,\\ldots ,998$$ has digit sum a multiple of $$27$$. Any sequence of $$999$$ consecutive integers is either of the form $$\\tag11000N+1,\\ldots, 1000N+999$$ or $$\\tag21000N+k+2,\\ldots, 1000N+999,1000(N+1),\\ldots, 1000(N+1)+k$$ with $$0\\le k\\le 997$$. In $$(1)$$, the digit sums are $$Q(N)+Q(i)$$ with $$i$$ running from $$1$$ to $$999$$ and hence $$Q(i)$$ covering all values from $$1$$ to $$27$$. We conclude that $$(1)$$ contains", "fraction. Let $\\{a_k\\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general, $a_k=\\begin{cases}(0.\\underbrace{20101\\cdots 0101}_{k+2\\text{ digits}})^{a_{k-1}}\\qquad\\text{if }k\\text{ is odd,}\\\\(0.\\underbrace{20101\\cdots 01011}_{k+2\\text{ digits}})^{a_{k-1}}\\qquad\\text{if }k\\text{ is even.}\\end{cases}$ Rearranging the numbers in the sequence $\\{a_k\\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\\{b_k\\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\\le k \\le 2011$, such that $a_k=b_k?$ Let $(a_1,a_2, \\dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \\le i \\le 10$ either $a_i+1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there? In the eight term sequence $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, the value of $C$ is $5$ and the sum of any three consecutive terms is $30$. What is $A+H$? The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x - y|$? Let $T_1$ be a triangle with sides $2011$, $2012$, and $2013$. For $n \\geq 1$, if $T_n = \\Delta ABC$ and $D, E$, and $F$ are the points of tangency of the incircle of $\\Delta ABC$ to the sides $AB$, $BC$, and $AC$, respectively, then $T_{n+1}$ is a triangle", "$2$'s into groups, such that each group has at least one $2$. By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$, is not needed for the remaining calculations.) Then, to get $D(96)$, in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five. The resulting number of possible sequences is $3 \\cdot 1 + 5 \\cdot 4 + 7 \\cdot 6 + 9 \\cdot 4 + 11 \\cdot 1 = \\boxed{\\textbf{(A) }112}$. ~emerald_block", "# What is the sum of the first 100 consecutive positive integers? May 8, 2016 $5050$ #### Explanation: The sum is: number of terms $\\times$ average term. The number of terms in our example is $100$ The average term is the same as the average of the first and last term (since this is an arithmetic sequence), namely: $\\frac{1 + 100}{2} = \\frac{101}{2}$ So: $1 + 2 + \\ldots + 99 + 100 = 100 \\times \\frac{1 + 100}{2} = 50 \\times 101 = 5050$ Another way of looking at it is: $1 + 2 + \\ldots + 99 + 100$ $= \\left.\\begin{matrix}\\textcolor{w h i t e}{00} 1 + \\textcolor{w h i t e}{00} 2 + \\ldots + \\textcolor{w h i t e}{0} 49 + \\textcolor{w h i t e}{0} 50 + \\\\ 100 + \\textcolor{w h i t e}{0} 99 + \\ldots + \\textcolor{w h i t e}{0} 52 + \\textcolor{w h i t e}{0} 51\\end{matrix}\\right.$ $= \\left.{\\underbrace{101 + 101 + \\ldots + 101 + 101}}_{\\text{50 times}}\\right.$ $= 101 \\times 50 = 5050$", "question from BMO1 from ~10 years ago that is quite good to use to 'get back into' the maths mode. 325* For each positive integer , define the sequence () by: for each Determine all the values of for which 2000 is a term of the sequence. 8. (Original post by Smaug123) Most of us can't :P By the way, your LaTeX is interesting… I assume you meant: Yep, that's what I meant... I'm new to this Latex stuff Posted from TSR Mobile 9. (Original post by Pterodactyl) A neat question from BMO1 from ~10 years ago that is quite good to use to 'get back into' the maths mode. 325* For each positive integer , define the sequence () by: for each Determine all the values of for which 2000 is a term of the sequence. First, note that , because otherwise . Next, consider the sequence mod 2. If , then all terms of the sequence are odd, so 2000 isn't a term of the sequence. So k must be odd. The remaining 1000 cases can easily be checked, and are left as an exercise to the reader. ETA: Suppose that a_1 = 2000. Then k = 2001. Suppose that a_2 = 2000. Then 2k+(k-1) = 2000, and k=667. Suppose that a_3 = 2000. Then 3k-(2k+(k-1))", "do not have any number that has 10 in it...NOT possible E). 1000.... 10*10^2....1000:2000:3000:4000 So D Yes, Got it now ! Thanks for the explanantion ! VP Joined: 07 Dec 2014 Posts: 1067 If there are four positive integers a, b, c and d in the ratio 1:2:3:4 [#permalink] ### Show Tags 22 Jul 2018, 11:12 chetan2u wrote: If there are four positive integers a, b, c and d in the ratio 1:2:3:4 and their sum is a perfect square, what cannot be the value of any of the four number? (A) 10 (B) 30 (C) 90 (D) 100 (E) 1000 sum of sequence=10a, so a=1/10 of sum, or perfect square 10 is first term of sequence summing to 100, or 10^2 NO 30 is third term of sequence summing to 100, or 10^2 N0 90 is first term of sequence summing to 900, or 30^2 N0 1000 is first term of sequence summing to 10,000, or 100^2 N0 by default, 100 cannot be the value of any of the four numbers D Director Joined: 09 Mar 2016 Posts: 762 If there are four positive integers a, b, c and d in the ratio 1:2:3:4 [#permalink] ### Show Tags 27 Jul 2018, 06:42 pushpitkc wrote: chetan2u wrote: If there are four positive integers a, b, c and d", "# Some problems from Estonian Mathematical Olympiad: 2000 Problem 1: How many positive integers less than 20002001 and not containing digits other than 0 and 2 are there? Solution 1: The set of integers under consideration consists of all integers with up to 7 digits containing only digits 0 and 2, all 8 digit integers of the form 20000***, and the integer 20002000. There are $2^{k-1}$ integers with exactly k digits 0 and 2, and $2^{3}$ integers of the form 20000***. So, the required number of integers is $(2^{0}+2^{1}+2^{2}+ \\ldots + 2^{6})+8+1=(2^{7}-1)+9=136$ Problem 2: Find the last two digits of the number: $1! + 2! + 3! +4! + \\ldots +2000!$ Solution 2: This is a classic type of question: the product $1.2 \\ldots 10$ has 2, 5, and 10 as factors, therefore being divisible by 100. Hence, the last two digits of n! are zeros for any $n \\geq 10$ and it suffices to find the last two digits of $1!+2! + 3! + \\ldots + 9!$. The last two digits of the summands are 01, 02, 06, 24, 20, 20, 10, 20 and 80, yielding 13 as the answer. Problem 3: Real numbers x and y satisfy the system of equations: $x + y + \\frac{x}{y}=10$ $\\frac{x(x+y)}{y}=20$ Find the sum of all possible values of the", "+0 # help 0 83 1 How many integers belong to the arithmetic sequence 13, 20, 27, 34, ..., 2008? Jun 1, 2020 #1 +40 +2 To find the number of terms in an arithmetic sequence we can use formula $n = \\frac{a_n - a_1}{d}$. In this case, n is the number of terms in the sequence so $$a_n$$ is the last term of the sequence. $$d$$ is the common difference between each term. Using this formula we plug in the terms we already know $$d = 7, a_n =2008, a_1 = 13$$. $n = \\frac{2008-13}{7} +1= \\frac{1995}{7} +1 = 285 +1 = \\boxed{286}$ Jun 1, 2020", "Decimal Sums of Successive Integers What comes next in the following sequence? $1 \\times 9 + 2 = 11$ $12 \\times 9 + 3 = 111$ $123 \\times 9 + 4 = 1111$ $1234 \\times 9 + 5 = 11111$ $12345 \\times 9 + 6 = 111111$ $123456 \\times 9 + 7 = 1111111$ $1234567 \\times 9 + 8 = 11111111$ $12345678 \\times 9 + 9 = 111111111$ $123456789 \\times 9 + 10 = 1111111111$ A road to answer to this question lies in the distinction between arithmetic and algebra - specific and general [Beiler, 57-58]. The trick is to express all nine of the above identites as specific cases of a more general one. We can do that! All nine appear to be instances of $(10 - 1) \\times (1\\cdot 10^{n-1}+2\\cdot 10^{n-2} + \\ldots + r\\cdot 10^{n-r}+\\ldots + n) + (n+1)$. Multiply out and collect the terms with the same powers of 10 (these will obviously be in the form $(r+1)-r=1$: $10^{n}+10^{n-1} + \\ldots + 1 = \\frac{10^{n+1}-1}{10-1}=111\\ldots 1$, with $n+1$ successive $1$'s. So, we may now add entries to the above table: $1234567900 \\times 9 + 11 = 11111111111$ $12345679011 \\times 9 + 12 = 111111111111$ $123456790122 \\times 9 + 13 = 1111111111111$ $1234567901233 \\times 9 + 14 = 11111111111111$ $12345679012344 \\times 9 + 15 =", "# Computing the nth term of a sequence when n is really large How to find the, say, 28383rd term of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2,.... ? EDIT: The sequence is the sequence of digits of positive integers in order. thanks, - I assume the sequence consists of the digits of the sequence of all positive integers in order? In any case, you should clarify this. – Ted Sep 4 '11 at 20:21 Yes, you got it right. I'm sorry if I missed any detail.:) – yati sagade Sep 4 '11 at 20:34 @yati sagade, Please take a look at this thread: math.stackexchange.com/questions/626216/… – Artem Jan 3 '14 at 20:15 Hint: think about how many terms are produced by the 1 digit numbers, then how many terms are produced by the 2 digit numbers, etc. That will allow you to get that you are in the $m$ digit numbers and the end of the $m-1$ digit numbers is $p$. So now you want the $28383-p$ term of the $m$ digit numbers, and each one contributes $m$ terms. - great. here's what I can corroborate(tell me If I go wrong) - 9 terms from 1-digiters, 180 from 2-digiters, 1800 from 3 digiters, 18000 from 4 digiters. That", "given: 11626 ### Show Tags 18 Sep 2012, 01:16 1 KUDOS Expert's post dandarth1 wrote: Bunuel wrote: RuslanMRF wrote: Series A(n) is such that i∗A(i)=j∗A(j) for any pair of positive integers (i,j). If A(1) is a positive integer, which of the following is possible? I. 2A(100)=A(99)+A(98) II. A(1) is the only integer in the series III. The series does not contain negative numbers I only II only I and III only II and III only I, II, and III New edition of this question with a solution: The sequence $$a_1$$, $$a_2$$, $$a_3$$, ..., $$a_n$$, ... is such that $$i*a_i=j*a_j$$ for any pair of positive integers $$(i, j)$$. If $$a_1$$ is a positive integer, which of the following could be true? I. $$2*a_{100}=a_{99}+a_{98}$$ II. $$a_1$$ is the only integer in the sequence III. The sequence does not contain negative numbers A. I only B. II only C. I and III only D. II and III only E. I, II and III Given that the sequence of numbers $$a_1$$, $$a_2$$, $$a_3$$, ... have the following properties: $$i*a_i=j*a_j$$ and $$a_1=positive \\ integer$$, so $$1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \\ integer$$. We should determine whether the options given below can occur (notice that the question is which of the following COULD be true, not MUS be true). I. $$2a_{100}=a_{99}+a_{98}$$ --> since $$100a_{100}=99a_{99}=98a_{98}$$, then $$2a_{100}=\\frac{100}{99}a_{100}+\\frac{100}{98}a_{100}$$ --> reduce" ]

[ "+0 # Help! 0 326 1 1. The sequence 2, 3, 5, 6, 7, 10, 11, $\\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $400^{\\mathrm{th}}$ term of the sequence? 2. The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string? Jun 3, 2018", "\\binom{5}{1} \\cdot 5 \\cdot 5^4 = 4 \\binom{5}{1} 5^5$$ Case 2: The leading digit is odd. Two of the remaining five positions must be filled with even digits. There are $\\binom{5}{2}$ ways to select the positions of the even digits and five ways to fill each of those positions with an even digit. There are five ways to fill each of the four other positions with an odd digit. $$\\binom{5}{2} \\cdot 5^2 \\cdot 5^4 = \\binom{5}{2} 5^6$$ Total: Since the above cases are mutually exclusive and exhaustive, the number of six-digit positive integers that contain exactly two even digits is $$4 \\binom{5}{1} 5^5 + \\binom{5}{2} 5^6$$ In how many five-digit positive integers are there digits that appear more than once? Hint: Subtract the number of five-digit positive integers in which all the digits are distinct from the total number of five-digit positive integers. The total number of five-digit positive integers is $$9 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10 = 90000$$ since the leading digit can be chosen in nine ways (as $0$ is prohibited) and each of the remaining digits can be selected in $10$ ways. Alternatively, the largest five-digit number is $99999$ and the largest number with fewer than five digits is $9999$, so the number of five-digit positive integers is $$99999 - 9999 =", "# How many positive, three-digit integers contain at least one $3$ How many positive, three-digit integers contain at least one $3$ as a digit but do not contain $5$ as a digit? I have an answer for that which is $215$ ,is that right ? If its wrong then ,how to solve it? - I think your number is a little high. Take the number of 3-digit integers that contain no 5 and subtract the number of 3-digit numbers that contain no 5 and no 3. - Let us consider the number of three-digit integers that do not contain $3$ and $5$ as digits; let this set be $S$. For any such number, there would be $7$ possible choices for the hundreds digit (excluding $0,3$, and $5$), and $8$ possible choices for each of the tens and ones digits. Thus, there are $7 \\cdot 8 \\cdot 8 = 448$ three-digit integers without a $3$ or $5$. Now, we count the number of three-digit integers that just do not contain a $5$ as a digit; let this set be $T$. There would be $8$ possible choices for the hundreds digit, and $9$ for each of the others, giving $8 \\cdot 9 \\cdot 9 = 648$. By the complementary principle, the set of three-digit integers with at least one $3$", "each block of $$1000$$ integers contains one which ends in three $$0's$$. Starting from any integer $$k$$, we go to the next multiple of $$1000$$ (a gap of at most $$999$$). We then add whatever three (or fewer) digit integer we need to \"correct\" the sum of the digits $$\\pmod {27}$$, which takes, at most, another $$999$$. Thus, every block of $$2\\times 999$$ consecutive integers contains at least one for which the sum of the digits is a multiple of $$27$$. A similar argument works for any desired divisor. I expect the bound could be tightened considerably, but at least this shows that a bound exists. • Simple insight, great help, thank you:) – Martund Jul 27 at 13:57 Let $$Q(x)$$ denote the digit sum of $$x$$. $$999$$ is the smallest $$n$$ such that among any $$n$$ consecutive positive integers, at least one has digit sum a multiple of $$27$$. First note that none of the $$998$$ consecutive integers $$1,2,\\ldots ,998$$ has digit sum a multiple of $$27$$. Any sequence of $$999$$ consecutive integers is either of the form $$\\tag11000N+1,\\ldots, 1000N+999$$ or $$\\tag21000N+k+2,\\ldots, 1000N+999,1000(N+1),\\ldots, 1000(N+1)+k$$ with $$0\\le k\\le 997$$. In $$(1)$$, the digit sums are $$Q(N)+Q(i)$$ with $$i$$ running from $$1$$ to $$999$$ and hence $$Q(i)$$ covering all values from $$1$$ to $$27$$. We conclude that $$(1)$$ contains", "positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \\cup A_2 \\cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is $$|A_1 \\cup A_2 \\cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A \\cap A_3| - |A_2 \\cap A_3| + |A_1 \\cap A_2 \\cap A_3|$$ $|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \\cdot 10 \\cdot 10 = 400$. $|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \\cdot 5 \\cdot 10 = 450$. $|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose", "it really quite vexing that the first digits don’t go $1,2,3,4$. Furthermore, $400 \\div 13 \\approx 31$, so it takes a long time before you see a 4 at all, and that seemed really unfair. ### The OEIS now contains 300,000 integer sequences The Online Encyclopedia of Integer Sequences just keeps on growing: at the end of last month it added its 300,000th entry. Especially round entry numbers are set aside for particularly nice sequences to mark the passing of major milestones in the encyclopedia’s size; this time, we have four nice sequences starting at A300000. These were sequences that were originally submitted with indexes in the high 200,000s but were bumped up to get the attention associated with passing this milestone.", "such as $$1000$$, $$1,000,000$$ or any number we like to think of, then there are more than $$n$$ positive integers. Thus, if the number we think of is $$1,000,000$$, there are obviously at least $$1,000,001$$ positive integers. Similarly the class of rational numbers, or of real numbers, is infinite. It is convenient to express this by saying that there are an infinite number of positive integers, or rational numbers, or real numbers. But the reader must be careful always to remember that by saying this we mean simply that the class in question has not a finite number of members such as $$1000$$ or $$1,000,000$$.", "GMAT CLUB'S CHALLENGE QUESTIONS: How many positive four-digit integers have their digits in ascending order from thousands to units? A. 6 B. 7 C. 84 D. 126 E. 210 Hi.. an easier, short and logical way.. choose 4 out of 10... there can be ONLY 1 way the 4 can be arranged in ascending order, so selection is what we are looking for and NOT arrangements.. $$10C4 = \\frac{10!}{6!4!} = \\frac{10*9*8*7}{4*3*2} = 210$$ BUT it contains numbers with 0 in thousands digit making these 3-digit numbers so $$9C3= \\frac{9!}{6!3!} = 84$$ $$ans = 210-84=126$$ D Great explanation chetan2u we can reduce 1 step here because $$0$$ cannot be part of any number with ascending digits so we simply need to select $$4$$ digits out of $$9$$ digits Hence $$9_C_4=\\frac{9!}{4!5!}=126$$ Math Expert Joined: 02 Sep 2009 Posts: 52173 Re: How many positive four-digit integers have their digits in ascending [#permalink] ### Show Tags 24 Dec 2018, 04:01 Bunuel wrote: GMAT CLUB'S CHALLENGE QUESTIONS: How many positive four-digit integers have their digits in ascending order from thousands to units? A. 6 B. 7 C. 84 D. 126 E. 210 _________________ Re: How many positive four-digit integers have their digits in ascending &nbs [#permalink] 24 Dec 2018, 04:01 Display posts from previous: Sort by # How many positive four-digit integers have", "+0 counting question 0 1 39 1 How many 4-digit positive integers exist that satisfy the following conditions: (A) Each of the first two digits must be 1, 4, or 5, and (B) the last two digits cannot be the same digit Dec 30, 2020 #1 +80 +2 How many 4-digit positive integers exist that satisfy the following conditions: (A) Each of the first two digits must be 1, 4, or 5, and (B) the last two digits cannot be the same digit, and (C) each of the last two digits must be 5, 7, or 8? (A) Each of the first two digits must be 1, 4, or 5 ? $$\\begin{array}{|lllc|c|} \\hline & x&x &xx \\\\ \\hline &1 \\cdots &\\begin{Bmatrix} 1\\\\4\\\\5 \\end{Bmatrix}\\cdots &\\begin{Bmatrix} 00\\\\ \\vdots \\\\ 99 \\end{Bmatrix} & 3\\times 100 \\\\\\\\ &4 \\cdots &\\begin{Bmatrix} 1\\\\4\\\\5 \\end{Bmatrix}\\cdots &\\begin{Bmatrix} 00\\\\ \\vdots \\\\ 99 \\end{Bmatrix} & +3\\times 100 \\\\\\\\ &5 \\cdots &\\begin{Bmatrix} 1\\\\4\\\\5 \\end{Bmatrix}\\cdots &\\begin{Bmatrix} 00\\\\ \\vdots \\\\ 99 \\end{Bmatrix} & +3\\times 100 \\\\\\\\ \\hline & & & & = 3\\times 3\\times 100 = {\\color{red}900} \\\\ \\hline \\end{array}$$ There are 900 4-digit positive integers that satisfy the condition. (B) The last two digits cannot be the same digit ? $$\\begin{array}{|lcc|c|} \\hline & xx &xx \\\\ \\hline &\\begin{Bmatrix} 10\\\\ \\vdots \\\\ 99 \\end{Bmatrix}\\cdots & 00 & 90 \\\\\\\\ &\\begin{Bmatrix} 10\\\\", "I am working on some AMC-8 questions. I am not sure how to approach this one. 279 views 0 19 months ago by Anonymous How many integers between 1000 and 9999 have four distinct digits? 2 19 months ago by The question can be rephrased to \"How many four-digit positive integers have four distinct digits,\" since numbers between 1000 and 9999 are four-digit integers. There are 9 choices for the first number, since it cannot be 0, There are only 9 choices left for the second number since it must differ from the first, 8 choices for the third number, since it must differ from the first two, and 7 choices for the fourth number, since it must differ from all three. This means there are $9\\times9\\times8\\times7=4536$9×9×8×7=4536 integers between 1000 and 9999 with four distinct digits.", "are $11\\cdot 2^{50}$ numbers with exactly $52$ bits of precision-that have a $1$ in the leading and trailing places. The factor $11$ comes because the highest order bit can be in $11$ places. Similarly there are $12 \\cdot 2^{49}$ numbers with $51$ bits of precision, $13 \\cdot 2^{48}$ with exactly $50$ bits, ... $61$ with $2$ bits of precision and $62$ with a single bit of precision. Alpha tells me this is $54043195528445950$ including the negatives. Adding your count of floating point numbers to $2^{64}$ signed integers and deducting the double count gives $36830437752635916294$ floats and signed integers. To this we have to add the positive signed integers that cannot be otherwise represented. They have to have a $1$ in the most significant bit-otherwise they fit in a signed integer. This would give $2^{63}$ of them, but we have to deduct the ones that have twelve trailing zeros because they can be represented by floats. That gives $2^{63}-2^{51}$ of them. Adding these in to the previous count gives $46051557989677006854$ This compares with $5.534E19$ if we could use all $3\\cdot2^{64}$ bit patterns times formats. We are only $9E18$ short of that. Of course, some languages have other ways of representing numbers that have many more possibilities. • I believe you meant 8-byte signed integers. – karakusc Sep 23 '14", "# Computing the nth term of a sequence when n is really large How to find the, say, 28383rd term of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2,.... ? EDIT: The sequence is the sequence of digits of positive integers in order. thanks, - I assume the sequence consists of the digits of the sequence of all positive integers in order? In any case, you should clarify this. – Ted Sep 4 '11 at 20:21 Yes, you got it right. I'm sorry if I missed any detail.:) – yati sagade Sep 4 '11 at 20:34 @yati sagade, Please take a look at this thread: math.stackexchange.com/questions/626216/… – Artem Jan 3 '14 at 20:15 Hint: think about how many terms are produced by the 1 digit numbers, then how many terms are produced by the 2 digit numbers, etc. That will allow you to get that you are in the $m$ digit numbers and the end of the $m-1$ digit numbers is $p$. So now you want the $28383-p$ term of the $m$ digit numbers, and each one contributes $m$ terms. - great. here's what I can corroborate(tell me If I go wrong) - 9 terms from 1-digiters, 180 from 2-digiters, 1800 from 3 digiters, 18000 from 4 digiters. That", "## goformit100 one year ago Consider the sequence 2;3;5;6;7;8;10;::: of all positive integers that are not perfect squares. Determine the 2011th term of this sequence. 1. goformit100 @oldrin.bataku 2. oldrin.bataku Find the number of perfect squares under 2011 -- this is the difference between 2011 and our number. 3. oldrin.bataku Well, we know $$\\sqrt{2011}\\approx45$$ with rounding (since $$44^2=1936$$, $$45^2=2025$$). This means we have skipped $$45$$ square numbers at this point, so our 2011th term is $$2011+45=2056$$", "# For positive integers between 999 and 100 inclusive, how many contain the digit 5? The question comes in two parts: 1. For positive integers between 999 and 100 inclusive, how many contain the digit 5 at least once? 2. For positive integers between 999 and 100 inclusive, how many contain the digit 5 exactly once? Question 1 Numbers between 999 and 100 inclusive: 999 - 99 = 900 Numbers between 999 and 100 inclusive that do not contain digit 5: 8 * 9 * 9 = 648 (The 8 is because the first digit can't be 5 or 0) Numbers that contain digit 5 at least once: 900 - 648 = 252 Question 2 Total numbers that contain digit 5 exactly once: (1*9*9) + (8*1*9) + (8*9*1) = 225 I think I got it wrong but not sure which part. • Well, a good way to check is to count the number with exactly two $5's$. $55X$ gives us $9$, $5X5$ gives us $9$, $X55$ gives us $8$. Thus there are $9+9+8=26$ such numbers. Of course $555$ is the unique number in the range with three $5's$. We check that $252=225+26+1$ which seems like pretty good evidence that you are right. – lulu Nov 1 '18 at 16:34 • @DavidG.Stork The OP is counting numbers with no", "1. ## Easy combinatorics problem How many four-digit positive integers are there that contain the digit 1? 2. ## Re: Easy combinatorics problem Originally Posted by alexmahone How many four-digit positive integers are there that contain the digit 1? To be clear $0109=109$ is not a four digit number, right?? There are $9\\cdot 10^3$ four-digit numbers. Of those there are $8\\cdot 9^3$ that do not contain any 1's. So how many contain at least one 1? 3. ## Re: Easy combinatorics problem hi alexmahone should the digits be the same or must be different? if they are the same then there are: 1000+900+900+900=2800 if they are not the same then there are: 720+648+648+648=2664", "A positive integer is said to be \"tri-factorable\" if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We'll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let's not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, \"Ok it happened to work for this one question. What do you do when the list keeps on going and going?\" When that happens, making a list will still help, but another step or tactic will often be necessary. Here's one example: 3. How many multiples of", "+0 # How many positive integers less than 500 are the product of exactly 0 1166 1 How many positive integers less than 500 are the product of exactly two distinct (meaning different) primes, each of which is greater than 10? Guest Mar 5, 2015 #1 +18835 +5 How many positive integers less than 500 are the product of exactly two distinct (meaning different) primes, each of which is greater than 10 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "# How many different words are needed to name all numbers in a given base? I read What is Mathematics? by Herbert Robbins, and there is this exercise: Consider the question of representing integers with the base $a$. In order to name the integers in this system we need words for the digits $0, 1, ..., a-1$ and for the various powers of $a$: $a, a^2, a^3, ...$. How many different number words are needed to name all numbers from zero to one thousand, for $a = 2, 3, 4, 5, ..., 15$? Which base requires the fewest? For example, if $a = 10$, we need ten words for the digits, plus words for 10, 100, and 1000, making a total of 13. For $a = 20$, we need twenty words for the digits, plus words for 20 and 400, making a total of 22. If $a = 100$, we need 100 plus 1. Well, I come up with the formula for the total of number words from $0$ to $x$ with base $a$: $$a + \\lfloor{\\log_a x}\\rfloor$$ With this formula, I found that base of 4 requires the fewest number words from zero to thousand: 8. I couldn't find any information on the web about this exercise. Can you please refer me to a more precise formula,", "A positive integer is said to be “tri-factorable” if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We’ll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let’s not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, “Ok it happened to work for this one question. What do you do when the list keeps on going and going?” When that happens, making a list will still help, but another step or tactic will often be necessary. Here’s one example: 3. How many multiples of", "# I exist sometimes Given a list of integers $$1\\cdots n$$, return the count of the number that have the number $$k$$ in them. For $$n\\leq10^{9}$$ how many numbers have the number $$7$$ in them. Details and Assumptions For $$n\\leq10$$, the number of occurrence of $$7$$ is one. For $$n\\leq100$$ the number of occurrence of $$7$$ is $$19$$...$$7,17,27,37,47,57,67,70,71\\cdots ,79,87,97$$ ×" ]

[ "# Mysterious Polygon Geometry Level 1 Given that the measure of each external angle is $$24^{\\circ}$$, find the number of sides of the mystery polygon. ×", "# Can you determine the sides? Geometry Level 1 Find the number of sides of the polygon if the sum of it's interior angles is equal to $$1260^\\circ$$. ×", "### Home > INT2 > Chapter 8 > Lesson 8.3.1 > Problem8-58 8-58. The exterior angle of a regular polygon is $20°$. 1. What is the measure of an interior angle of this polygon? Show how you know. The interior and exterior angles are supplementary. 2. How many sides does this polygon have? Show all work. The sum of the measures of the exterior angles is $360°$. $18$", "### Find the measure of each interior angle of a regular polygon of $10$ sides. $144 ^ \\circ$ 1. We know that the sum of all exterior angles of a polygon is $360 ^ \\circ.$ 2. Each exterior angle of a regular polygon of $10$ sides $= { \\Bigg ( \\dfrac{ 360 } { 10 } \\Bigg ) } ^ \\circ = 36 ^ \\circ.$ 3. Thus, each interior angle of a regular polygon of $10$ sides $= 180 ^ \\circ - 36 ^ \\circ = 144 ^ \\circ$.", "### Home > INT2 > Chapter 8 > Lesson 8.2.2 > Problem8-39 8-39. A regular polygon has an interior angle measuring $135°$. How many sides does it have? Determine the number of sides using two different strategies. Show all work. Which strategy was most efficient? Homework Help ✎ Use the formula in the Math Notes box. $135°=\\frac{180°(n-2)}{n}$ What is the measure of an exterior angle? The exterior angles of a regular polygon sum to $360°$.", "How many right angles are in a regular pentagon? If you are interested, it is easy to prove that regular $N$-sided polygon has each interior angle equal to $\\phi = {180}^{o} \\cdot \\frac{N - 2}{N}$ (degrees.) So in case of a regular $5$-sided polygon (pentagon) the interior angles measure at $\\phi = {180}^{o} \\cdot \\frac{5 - 2}{5} = {108}^{o}$", "# Can you determine the sides? Geometry Level 1 Find the number of sides of the polygon if the sum of it's interior angles is equal to $$1260^\\circ$$. × Problem Loading... Note Loading... Set Loading...", "a polygon? ... 15 30 50 per page", "# Q5. What is a regular polygon? State the name of a regular polygon of (iii) 6 sides All angles of the hexagon are $120\\degree$ each.", "# Triangles and circles but not geometry part -2 Discrete Mathematics Level 5 There is a $$193$$ sided regular polygon. Now triangles are formed by joining the vertices of the polygon. Find the number of obtuse angled triangles formed. ×", "shrinks smaller and smaller, the angles converge at the center, forming $$360^{\\circ}.$$ Do these proofs work for any polygon with any number of sides? Yes, they do! If the polygon had more sides, there would be more colored angles, on average smaller than the ones shown above. They would still add to $$360^{\\circ}.$$ Which proof of the Sum of Exterior Angles of a Polygon is your favorite? • @debbie Thanks", "Free Version Difficult # Regular Convex Polygon GEOM-NHMLXC What regular convex polygon has an interior angle measure of $135 ^{\\circ}$? A octagon B decagon C hexagon D heptagon", "# An exterior angle of a regular polygon is 15 degrees. How many sides does the polygon have? $24$ The exterior angles of a polygon always add up to $360$ degrees. Given this a regular polygon, all the angles are equal and all the sides are equal. The number of sides is therefore $\\frac{360}{15} = 24$", "Free Version Moderate Interior Angle Sum SATSTM-FZ4JCP How many sides does a polygon have where the sum of the interior angles is $1440$ degrees? A $8$ B $10$ C $12$ D $14$ E $16$", "# For a regular 20-sided polygon, what is the degree of rotation? The 20-sided polygon, AKA \"icosagon\", has $18$ degree rotational symmetry about the center.", "### Home > GC > Chapter 9 > Lesson 9.2.3 > Problem9-83 9-83. West High School has a math building in the shape of a regular polygon. When Mrs. Woods measured an interior angle of the polygon (which was inside her classroom), she got $135°$. 1. How many sides does the math building have? Show how you got your answer. Draw a diagram. What does it show? Refer to Math Note Box 8.1.5 for the formula for calculating the interior angle of a polygon. The building has $8$ sides. 2. If Mrs. Wood’s ceiling is $10$ feet high and the length of one side of the building is $25$ feet, find the volume of West High School’s math building.", "Two polygons fit together so that the exterior (orange) angle at each end of their shared side is $81^\\circ$. If both shapes now have to be regular polygons, but do not need to be the same, and each polygon can have any number of sides, could the orange angle still be $81^\\circ$, and if that is possible how many sides would each polygon have? Find solutions for when the orange angle is $27^\\circ$ and when it is $54^\\circ$.", "3 Tutor System Starting just at 265/hour # 3.How many sides does a regular polygon have if the measure of an exterior angle is $$24^{\\circ}$$? We know that sum of all exterior angles of a regular polygon = $$360^{\\circ}$$ So we have,Number of sides=$$\\frac{360^{\\circ}}{measure\\; of \\;an\\;angle}$$", "+1 vote For a regular polygon having $10$ sides, the interior angle between the sides of the polygon, in degrees, is: 1. $396$ 2. $324$ 3. $216$ 4. $144$ recategorized +1 vote The interior angle between the sides of the polygon $= \\frac{(n-2) \\times 180^{\\circ}}{n},$ where $n =$ number of sides of the polygon. Here, $n = 10,$ therefore the interior angle between the sides of the polygon $= \\frac{(10-2) \\times 180^{\\circ}}{10}$ $\\qquad = 8 \\times 18^{\\circ} = 144^{\\circ}.$ So, the correct answer is $(D).$ by (3.7k points) selected by +1 vote", "# A regular polygon has 21 sides. Find the size of each interior angle? Apr 22, 2018 $\\textcolor{b l u e}{{162.86}^{\\circ}}$ 2 d.p. #### Explanation: The sum of the interior angles of a regular polygon are given by: ${180}^{\\circ} n - {360}^{\\circ}$ Where $n$ is the number of sides. $\\therefore$ $180 \\left(21\\right) - 360 = {3420}^{\\circ}$ Each angle is therefore: $\\frac{3420}{21} = {162.86}^{\\circ}$ 2 d.p." ]

[ "47°, 119°, 90°, x°, and x° The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2) = 180° (3) = 540° So, 47° + 119° + 90° + x° + x° = 540° 256° + 2x° = 540° 2x° = 540° – 256° 2x° = 284° x° = $$\\frac{284}{2}$$ x° = 142° Hence, ∠X = ∠Y = 142° Question 21. m∠X = m∠Y = 100.5° Explanation: Question 22. ∠X = ∠Y = 140° Explanation: The given figure is: We can observe that, the polygon has 6 sides. We know that, if the angles are not mentioned in a polygon, the consider that angles as equal angles. The given angle measures of a polygon with 6 sides are: 110°, 149°, 91°, 100°, x°, and x° The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2) = 180° (4) = 720° So, 110° + 149° + 91° + 100 + x° + x° = 720° 440° + 2x° = 720° 2x° = 720° – 440° 2x° = 280° x° = $$\\frac{280}{2}$$ x° = 140° Hence, ∠X = ∠Y = 140° In Exercises 23-26, find the value of x. Question 23. X = 111° Explanation: The sum of the", "\\frac{180a}{\\pi}$ $\\therefore \\frac{(a - d)\\pi}{180a} = \\frac{1}{120} \\Rightarrow \\frac{(\\frac{\\pi}{3} - d)\\pi}{3\\frac{\\pi}{3}} = \\frac{1}{2}$ $d = \\frac{\\pi}{3} - \\frac{1}{2}$ Thus, angles are $\\frac{1}{2}, \\frac{\\pi}{3}, \\frac{2\\pi}{3}- \\frac{1}{2}$ radians. 48. Let us solve these one by one: 1. We know that if polygon has $n$ sides then sum of angles is $(n - 2)\\pi$ radians or $(n - 2)180^\\circ$ For pentagon sum of angles $= 3\\pi$ or $540^\\circ$ Measure of one interior angle $= \\frac{3\\pi}{5}$ or $108^\\circ$ 2. Measure of one interior angle $= \\frac{5\\pi}{7}$ or $\\frac{900^\\circ}{7}$ 3. Measure of one interior angle $= \\frac{3\\pi}{4}$ or $135^\\circ$ 4. Measure of one interior angle $= \\frac{5\\pi}{6}$ or $150^\\circ$ 5. Measure of one interior angle $= \\frac{15\\pi}{17}$ or $\\frac{2700^\\circ}{17}$ 49. Let there be $n$ sides in one polygon and $2n$ in another. Angles will be $\\frac{(n - 2)\\pi}{n}$ and $\\frac{(2n - 2)\\pi}{2n} - \\frac{(n - 1)\\pi}{n}$ Given ratio of angles $\\frac{3}{2} = \\frac{n -1}{n - 2} \\Rightarrow 3n - 6 = 2n -2 \\Rightarrow n = 4.$ So one polygon is a square while the other is an octagon. 50. Let number of sides in one polygon be $n$ then number of sides in another $\\frac{5n}{4}.$ The angles will be $\\frac{(n - 2)}{n}180^\\circ, \\frac{(\\frac{5n}{4} - 2)}{\\frac{5n}{4}}180^\\circ.$ Given that difference in angles is $9^\\circ$ Solving for difference we find $n = 8$ and thus", "a decagon (10 sides) -- when you find the sum of the interior angles you use (n - 2) * 180 so now you do the reverse. #6: [ (5-2) + (7-2) + (9-2)] x 180 = (3 + 5 + 7) x 180 = 5 x 3 x 180 = 2700 degrees. #7:$$\\frac{720}{180}$$= 4 and 4 + 2 = 6 so this is a regular hexagon and one of its interior is $$\"\\frac{720}{6}$$ = 120 degrees. Or you can also do $$\\frac{720 + 360}{180}$$ = 6 because all the interior angles and their exterior angles are supplementary and the sum of any exterior angle of a convex polygon is 360 degrees. Using this method, you get how many sides instantly. #8: This one is tricky because there are actually two different polygon pairs that the total number of degrees add up to 1980 degrees. One is a regular hexagon and a nonagon. (720 + 1260) Another is a heptagon and an octagon (900 + 1080). Looking at the sum of their diagonals, the latter is the right pair. (14 + 20 = 34) so the answer is 8 - 7 = 1 #9: 360 - 108 - 120 = 132 degrees. #10: To find the central angle B, you do $$\\frac{360}{5}$$ = 72 degrees. BA and BC", "An = nr2 n-+00 Exterior angle of a regular polygon having n sides = $$\\dfrac{360^\\circ}{n}$$ Interior angle of a regular polygon having n sides = $$180^\\circ$$ - Exterior angle; Apothem falls on the midpoint of a side dividing it into two equal parts. For example, consider the polygon shown below: This polygon can be divided into a combination of triangles and trapezium. To get the area of the whole polygon, just add up the areas of all the little triangles (\"n\" of them): Area of Polygon = n × side × apothem / 2. Let’s work out a few example problems about area of a regular polygon. The area of this polygon is n times the area of triangle, since n triangles make up this polygon. 2 π r = n × a. where r = radius of circle, a = side of polygon with n sides. π is a mathematical constant. Given the radius (circumradius) If you know the radius (distance from the center to a vertex, see figure above): where r is the radius (circumradius) n is the number of sides sin is the sine function calculated in degrees (see Trigonometry Overview) . As we know, Area (A) = ½ x p x a, here p = 44 cm and a = 10 cm = ½", "180° (5 – 2) = 180° (3) = 540° So, 47° + 119° + 90° + x° + x° = 540° 256° + 2x° = 540° 2x° = 540° – 256° 2x° = 284° x° = $$\\frac{284}{2}$$ x° = 142° Hence, from the above, We can conclude that ∠X = ∠Y = 142° Question 21. Question 22. The given figure is: From the given figure, We can observe that The polygon has 6 sides We know that, If the angles are not mentioned in a polygon, the consider that angles as equal angles Now, The given angle measures of a polygon with 6 sides are: 110°, 149°, 91°, 100°, x°, and x° So, The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2) = 180° (4) = 720° So, 110° + 149° + 91° + 100 + x° + x° = 720° 440° + 2x° = 720° 2x° = 720° – 440° 2x° = 280° x° = $$\\frac{280}{2}$$ x° = 140° Hence, from the above, We can conclude that ∠X = ∠Y = 140° In Exercises 23-26, find the value of x. Question 23. Question 24. The given figure is: From the given figure, We can observe that The polygon has 7 sides The given angle measures", "box with the volume of the cube. Volume of rectangular box = L × B × H = 20cm × 6cm × 4cm = 480cm3 Volume of cube = s3 = 2cm3 = 8cm3 => Number of cubes that will fit into the box = $\\frac{480cm^3}{8cm^3}$ = $\\frac{\\cancel{480}^{60}\\cancel{cm^3}}{\\cancel8_1\\cancel{cm^3}}$ = 60 29. The interior angle of a regular polygon is 120°. How many sides has this polygon? A. 3 B. 4 C. 5 D. 6 D. 6 Solution Interior angle of regular polygon = $\\frac{180°(n-2)}{n}$ =>120° = $\\frac{180°(n-2)}{n}$ =>120°× $n$ = 180°$(n-2)$ =>120°$n$ = 180°$n$ – 360° =>360° = 180°$n$ – 120°$n$ =>360° = 60°$n$ =>$\\frac{360°}{60°}$ = $\\frac{60°n}{60°}$ =>$\\frac{\\cancel{360°}^6}{\\cancel{60°}_1}$ = $\\frac{\\cancel{60°}n}{\\cancel{60°}}$ =>n = 6 30. In the diagram above, length of PS = length of SQ and angle SQR = 112°. Find the value of $x$. A. 68° B. 56° C. 46° D. 44° D. 44° Solution ∠PQS + ∠SQR = 180° => ∠PQS + 112° = 180° => ∠PQS = 180° – 112° => ∠PQS = 68° △PSQ is an equilateral triangle because |PS| = |SQ|. This means that ∠PQS = ∠SPQ = 68°. For △PSQ, ∠PQS + ∠QSP + ∠SPQ = 180° => 68° + $x$ + 68° = 180° => $x$ + 136° = 180° => $x$ = 180° – 136° => $x$ = 44° 31.", "and get $\\alpha=108^\\circ$. 3) and 4) The sum of all interior angles in an n-sided regular polygon is $n\\alpha = (n-2)\\cdot 180^\\circ$. For a 7-sided regular polygon we set n=7 and get $7\\alpha = (7-2)\\cdot 180^\\circ = 5 \\cdot 180^\\circ = 900^\\circ$. 5) If we make n bigger and bigger, the value for $\\alpha$ approaches 180°. We can express this with the limit: $\\lim_{n\\rightarrow \\infty} \\alpha$ $= \\lim_{n\\rightarrow \\infty} \\frac{(n-2)}{n} \\cdot 180^\\circ$ $= \\lim_{n\\rightarrow \\infty} \\left(1-\\frac{2}{n}\\right) \\cdot 180^\\circ$ $= 180^\\circ$ ## Wallstreet Journal articles Posted by Ed on November 9, 2012 Recently, I’ve been reading the Wallstreet Journal. I particularly enjoy reading the section on life, health and work at the end. Here are two articles that I like: Some Don’t Live to Earn, but Earn to Live By RHEA WESSEL Self-Help for Skeptics By ELIZABETH BERNSTEIN", "Is it possible to have a regular polygon, each of whose interior angles is a. $${45^ \\circ }$$ b. $${60^ \\circ }$$ c. $${75^ \\circ }$$ Ans: a. Given interior angle $$= {45^ \\circ }$$ We know that the interior angle of a polygon of $$n$$ number of sides is given by $$\\frac{{\\left({n – 2} \\right) \\times {{180}^ \\circ }}}{n}.$$ $$\\Rightarrow {45^ \\circ } = \\frac{{\\left({n – 2} \\right){{180}^ \\circ }}}{n}$$ $$\\Rightarrow n \\times {45^ \\circ } = n \\times {180^ \\circ } – {360^ \\circ }$$ $$\\Rightarrow 360 = 180\\,n – 45\\,n$$ $$\\Rightarrow 360 = 135\\,n$$ $$\\Rightarrow n = \\frac{{360}}{{135}}$$ $$\\Rightarrow n = \\frac{8}{3} = 2\\frac{2}{3}$$ which is not a whole number. So, it is impossible to have a regular polygon, each of whose interior angles is $${45^ \\circ }.$$ b. Given interior angle $$= {60^ \\circ }$$ We know that the interior angle of a polygon of $$n$$ number of sides is given by $$\\frac{{\\left({n – 2} \\right) \\times {{180}^ \\circ }}}{n}.$$ $$\\Rightarrow {60^ \\circ } = \\frac{{\\left({n – 2} \\right){{180}^ \\circ }}}{n}$$ $$\\Rightarrow n \\times {60^ \\circ } = n \\times {180^ \\circ } – {360^ \\circ }$$ $$\\Rightarrow 360 = 180\\,n – 60\\,n$$ $$\\Rightarrow 360 = 120\\,n$$ $$\\Rightarrow n = \\frac{{360}}{{120}}$$ $$\\Rightarrow n = 3$$ So, it is possible to have a regular polygon,", "– 2) Let the number of sides of a new polygon be x So, 180° (x – 2) + 540° = 180° (n – 2) 180x – 360° + 540° = 180n – 360° 180x – 180n = -540° 180 (x – n) = -540° x – n = $$\\frac{540}{180}$$ x – n = -3 n – x = 3 n = x + 3 Hence, from the above, We can conclude that We have to add 3 more sides to the original convex polygon Maintaining Mathematical Proficiency Find the value of x. Question 53. Question 54. The given figure is: From the given figure, We can observe that 113° and x° are the corresponding angles We know that, According to the “Corresponding Angles Theorem”, the corresponding angles are congruent i.e., equal So, x° = 113° Hence, from the above, We can conclude that the value of x is: 113° Question 55. Question 56. The given figure is: From the given figure, We can observe that (3x + 10)° and (6x – 19)° are the corresponding angles We know that, According to the “Corresponding angles Theorem”, the corresponding angles are congruent i.e., equal So, (3x + 10)° = (6x – 19)° 6x – 3x = 19° + 10° 3x° = 29° x° = $$\\frac{29}{3}$$ x° = 9.6° Hence,", "$\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)}$ i) $30$ $180-30=150^{\\circ}$ $n=$ $\\frac{360}{180-150}$ $=12$ ii) $36$ $180-36=144^{\\circ}$ $n=$ $\\frac{360}{180-144}$ $=10$ iii) $40$ $180-40=140^{\\circ}$ $n=$ $\\frac{360}{180-140}$ $=9$ iv) $18$ $180-18=162^{\\circ}$ $n=$ $\\frac{360}{180-162}$ $=20$ $\\\\$ Question 11: Is it possible to have a regular polygon whose interior angles measure $130^{\\circ}$ No. Let us calculate the number of sides of this polygon. No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-130^{\\circ})}$ $=7.2$ Hence not possible. $\\\\$ Question 12: Is it possible to have a regular polygon whose interior angles measure measures $1$ $\\frac{3}{4}$ of a right angle. Yes. Let us calculate the number of sides of this polygon. No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-157.5^{\\circ})}$ $=16$ Hence possible $\\\\$ Question 13: Find the number of sides of a regular polygon, if its interior angle is equal to exterior angle. $Interior \\ angles = 180^{\\circ}-Exterior \\ Angle$ This means that each of the interior and the exterior angles $=90^{\\circ}$ No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-90^{\\circ})}$ $=4$ $\\\\$ Question 14: The ratio between the exterior angle and the interior angles is $2:7$. Find the number of sides of the polygon. $Interior \\ angles = 180^{\\circ}-Exterior \\ Angle$ Let the Exterior Angle $=2x$ and the Interior Angle be $7x$ $\\Rightarrow 2x+7x=180 \\ or\\ x=20$ Therefore Interior angle $=140^{\\circ}$ No. of Sides $(n)=$ $\\frac{360^{\\circ}}{180^{\\circ}-140^{\\circ}}$ $= 9$ $\\\\$ Question 15:", "of sides of a new polygon be x So, 180° (x – 2) + 540° = 180° (n – 2) 180x – 360° + 540° = 180n – 360° 180x – 180n = -540° 180 (x – n) = -540° x – n = $$\\frac{540}{180}$$ x – n = -3 n – x = 3 n = x + 3 Hence, from the above, We can conclude that We have to add 3 more sides to the original convex polygon Maintaining Mathematical Proficiency Find the value of x. Question 53. Question 54. The given figure is: From the given figure, We can observe that 113° and x° are the corresponding angles We know that, According to the “Corresponding Angles Theorem”, the corresponding angles are congruent i.e., equal So, x° = 113° Hence, from the above, We can conclude that the value of x is: 113° Question 55. Question 56. The given figure is: From the given figure, We can observe that (3x + 10)° and (6x – 19)° are the corresponding angles We know that, According to the “Corresponding angles Theorem”, the corresponding angles are congruent i.e., equal So, (3x + 10)° = (6x – 19)° 6x – 3x = 19° + 10° 3x° = 29° x° = $$\\frac{29}{3}$$ x° = 9.6° Hence, from the above, We can", "(5 – 2) = 180° (3) = 540° So, 140° + 138° + 59° + x° + 86° = 540° 423° + x° = 540° x° = 540° – 423° x° = 117° Hence, x is 117° Question 17. x = 150° Explanation: Question 18. x = 88.6° Explanation: From the given figure, the polygon has 8 sides. The given angle measures of a polygon with 8 sides are: 143°, 2x°, 152°, 116°, 125°, 140°, 139°, and x° The sum of the angle measures of the interior angles of a polygon with 8 sides = 180° (8 – 2) = 180° (6) = 1080° So, 143° + 2x° + 152° + 116° + 125° + 140° + 139° + x° = 1080° 815° + 3x° = 1080° 3x° = 1080° – 815° 3x° = 265° x° = $$\\frac{265}{3}$$ x° = 88.6° Hence, x is: 88.6° In Exercises 19 – 22, find the measures of ∠X and ∠Y. Question 19. m∠X = m∠Y = 92° Explanation: Question 20. ∠X = ∠Y = 142° Explanation: The given figure is: We can observe that, the above polygon has 5 sides. We know that, if the angles are not mentioned in a polygon, then consider that angles as equal angles. So, the given angle measures of a polygon with 5 sides are:", "# Finding number of sides of a polygon with an unknown A regular polygon has $n$ sides . When the number of sides is doubled, each interior angle increases by $20^{\\circ}$. Find $n$. My workings till I got stuck $1$ int. angle of $n$ sides $=180^{\\circ}n-360^{\\circ}/n$ $1$ int. angle of $2n$ sides $=200^{\\circ}n-360^{\\circ}/n$ $1$ ext. angle of $2n$ sides $=-20^{\\circ}n+360^{\\circ}/n$ $n=360$ divide $1$ ext. angle ... I've done it till $-20n+360=360$ Then I got stuck. Can I get help? Thanks in advance! The total of the angles in an $n$-gon is $180(n-2)$, so each angle is $$a(n) = \\frac{180(n-2)}{n}$$ with the angle measured in degrees. So you want to find $n$ where $a(2n) = a(n) + 20$. This is the same as solving $$\\frac{180(2n-2)}{2n} = \\frac{180(n-2)}{n} + 20.$$ Multiplying through by $n$ gives $$180(n-1) = 180(n-2) + 20n,$$ so $n=9$. Angle sum of $n$-sided polygon $=(n-2) \\times 180^{\\circ}$ Now, \\begin{align*} \\frac{(2n-2)}{2n} \\times 180^{\\circ}- \\frac{(n-2)}{n} \\times 180^{\\circ} &= 20^{\\circ} \\\\ [(n-1)-(n-2)] \\times 180^{\\circ} &= n\\times 20^{\\circ} \\\\ 180 &= 20n \\\\ n &= 9 \\end{align*} The sum of the exterior angles is $360^\\circ$, so each exterior angle on the original polygon is $(360/n)^\\circ$. Doubling the number of sides - now $2n$ - decreases the exterior angle by $20^\\circ$. So: \\begin{align} 360/n -20 &= 360/2n \\\\ 360 -20n &=180\\\\ 20n", "### Home > INT2 > Chapter 9 > Lesson 9.1.1 > Problem9-9 9-9. How many sides does the regular polygon have if each interior angle has the following measure? 1. $60^\\circ$ $60^\\circ = \\frac{180^\\circ(n-2)}{n}$ $n = 3$ sides 1. $156^\\circ$ 1. $90^\\circ$ 1. $140^\\circ$ $9$ sides In addition, for regular polygons: • The measure of each angle in a regular $n$-gon is $\\frac{180^\\circ(n-2)}{n}$ . • The measure of each exterior angle in a regular $n$-gon is $\\frac{360^\\circ}{n}$ .", "measure of (i) 24° (ii) 60° (iii) 72° Solution: (i) Let number of sides of the polygon = n Each exterior angle = 24° ∴ n = $$\\frac{360^{\\circ}}{24^{\\circ}}$$ = 15 sides ∴ Polygon is of 15 sides. (ii) Each interior angle of the polygon = 60° Let number of sides of the polygon = n ∴ n = $$\\frac{360^{\\circ}}{60^{\\circ}}$$ = 6 ∴ Number of sides = 6 (iii) Each interior angle of the polygon = 72° Let number of sides of the polygon = n ∴ n = $$\\frac{360^{\\circ}}{72^{\\circ}}$$ = 5 ∴ Number of sides = 5 Question 5. Find the number of sides of a regular polygon if each of its interior angles is (i) 90° (ii) 108° (iii) 165° Solution: (i) Each interior angle = 90° Let number of sides of the regular polgyon = n ∴ 90° = $$\\frac{2 n-4}{n}$$ × 90° ⇒ $$\\frac{2 n-4}{n}=\\frac{90^{\\circ}}{90^{\\circ}}$$ = 1 ⇒ 2n – 4 = n ⇒ 2n – n = 4 ⇒ n = 4 ⇒ n = 4 ∴ It is a square. (ii) Each interior angle = 108° Let number of sides of the regular polygon = n ∴ 108° = $$\\frac{2 n-4}{n}$$ × 90° ⇒ $$\\frac{2 n-4}{n}=\\frac{108^{\\circ}}{90^{\\circ}}=\\frac{6}{5}$$ ⇒ 10n – 20 = 6n ⇒ 10n – 6n = 20 ⇒ 4n = 20 ⇒", "# Each interior angle of a regular polygon lies between 136 to 142. How do we calculate the sides of the polygon? Jan 7, 2016 $n = 9$ #### Explanation: In a regular polygon each interior angle can be obtained in this way: $\\alpha = {180}^{\\circ} - {360}^{\\circ} / n$ From the conditions of the problem: ${136}^{\\circ} < \\alpha < {142}^{\\circ}$ That's the conjugation of this two inequations: ${136}^{\\circ} < \\alpha$ and $\\alpha < {142}^{\\circ}$ Resolving the first inequation ${136}^{\\circ} < {180}^{\\circ} - {360}^{\\circ} / n$ ${136}^{\\circ} < \\frac{{180}^{\\circ} \\cdot n - {360}^{\\circ}}{n}$ => ${136}^{\\circ} \\cdot n < {180}^{\\circ} . n - {360}^{\\circ}$ => ${44}^{\\circ} . n > {360}^{\\circ}$ => $n > 8.18$ Resolving the second inequation ${180}^{\\circ} - {360}^{\\circ} / n < {142}^{\\circ}$ ${180}^{\\circ} \\cdot n - {360}^{\\circ} < {142}^{\\circ} \\cdot n$ => ${38}^{\\circ} . n < {360}^{\\circ}$ => $n < 9.47$ Conjugating the two inequations $8.18 < n < 9.47$ Since $n \\in \\mathbb{N}$, its only value that satisfies the inequation is $n = 9$ By the way $\\alpha = {180}^{\\circ} - {360}^{\\circ} / 9 = {180}^{\\circ} - {40}^{\\circ}$ => $\\alpha = {140}^{\\circ}$", "polygon. For example, we have $90$ as a divisor of $360$ so $90^{\\circ}$ is a possible exterior angle. Since $90 \\times 4 = 360$, the particular polygon with $90^{\\circ}$ exterior angle will have $4$ sides. Another fact: the sum of angles in a polygon with $n$ sides is $180(n - 2)$ degrees. So all the sums will be a multiple of $180$. Use that formula with $n = 8$ to get the sum of all interior angles in an octagon. Remark: Technically, all the choices you have been provided with can be the measures of an exterior angle. If in the future, any question asks about regular polygons, you can predict that the answers will only be the divisors of $360$. - $1.$ The sum of the measure of the exterior angles of a polygon is $360^\\circ$ If it has uniform $n$ sides, each of the exterior angles will be $\\frac{360^\\circ}n$, so for uniform polygon, the value of each exterior angle must be a divisor of $360$ in degrees. $2.$ The sum of angles of a convex polygon with $n$ sides is $180^\\circ(n-2)$ which is clearly multiple of $180$ in degrees. $3.$ The sum of The measure of angles of a octagon is $180^\\circ(8-2)=1080^\\circ$ Now, if the smaller angles is $\\theta$ each, the larger will be $2\\theta$ So,", "+1 vote For a regular polygon having $10$ sides, the interior angle between the sides of the polygon, in degrees, is: 1. $396$ 2. $324$ 3. $216$ 4. $144$ recategorized +1 vote The interior angle between the sides of the polygon $= \\frac{(n-2) \\times 180^{\\circ}}{n},$ where $n =$ number of sides of the polygon. Here, $n = 10,$ therefore the interior angle between the sides of the polygon $= \\frac{(10-2) \\times 180^{\\circ}}{10}$ $\\qquad = 8 \\times 18^{\\circ} = 144^{\\circ}.$ So, the correct answer is $(D).$ by (3.7k points) selected by +1 vote", "polygon be n. We have, Sum of all interior angles = $$(n – 2) \\times 180^{\\circ}$$ and, measure of its each angle=$$\\frac{(n-2) \\times 180^{\\circ}}{n}$$ here we have each angle as $$165^{\\circ}$$ $$\\Rightarrow \\frac{(n-2) \\times 180^{\\circ}}{n}=165$$ $$\\Rightarrow 180n - 360 =165 n \\quad$$[Cross multiplication and opening the brackets] $$\\Rightarrow 180n -165 n=360$$ $$\\Rightarrow 15n = 360$$ $$\\Rightarrow n = \\frac{360}{15}=24$$ So, we have the number of sides of regular polygon as 24. 5.(a) Is it possible to have a regular polygon with measure of each exterior angle a is $$22^{\\circ}$$? (b) Can it be an interior angle of a regular polygon? Why? (a) We have, Number of sides of regular polygon=$$\\frac{360^{\\circ} }{Exterior \\;angle}$$ As we know number of sides by the formula should be an integer so, it is not possible that a regular polygon have its exterior angle of $$22^{\\circ}$$ because $$22^{\\circ}$$ is not divisible by $$360^{\\circ}$$ hence, we can't get a whole number for calculating the number of side. (b) If interior angle =$$22^{\\circ}$$ And we have: measure of each interior angle=$$\\frac{(n-2)\\times180^o}n$$ $$\\Rightarrow 180n -2 \\times 180=22n$$ $$\\Rightarrow 180n -22n=360$$ $$\\Rightarrow 158n=360$$ $$\\Rightarrow n=\\frac{360}{158}$$ But 158 does not divide 360 exactly.So, the polygon is not possible. 6.(a) What is the minimum interior angle possible for a regular polygon? Why? (b) What is the maximum exterior angle possible", "Browse Questions # The difference between any two consecutive interior angles of a polygon is $5^\\circ$. If the smallest angle is $120^ \\circ$, then find the number of sides of the polygon. Toolbox: • The sum of the interior angles of $n$ sided polygon = $(n-2).180^{\\circ}$ • The sum of $n$ terms of an A.P=$\\large\\frac{n}{2}$$[2a+(n-1)d] Let the no. of sides of the polygon be n The sum of the interior angles of n sided polygon = (n-2).180^{\\circ} It is given that the difference between any two consecutive angles is 5^{\\circ} \\therefore The angles of the polygon are assumed as a,a+5,a+10,...........a+(n-1).5 We know that the the sum of the angles = (n-2).180 \\Rightarrow\\:a+(a+5)+(a+10)+...........a+(n-1).5=(n-2).180 This series is an A.P. with common difference d=5 We know that the sum of n terms of an A.P=\\large\\frac{n}{2}$$[2a+(n-1)d]$ $\\Rightarrow\\:\\large\\frac{n}{2}$$[2a+(n-1)5]=(n-2)180 Also given that the smallest angle is 120^{\\circ} \\therefore\\:a=120 \\Rightarrow\\:\\large\\frac{n}{2}$$[2\\times 120+(n-1)5]=(n-2)180$ $\\Rightarrow\\:n[240+5n-5]=(n-2).360$ $\\Rightarrow\\:5n^2+235n=360n-720$ $\\Rightarrow\\:5n^2-125n+720=0$ Taking $5$ common $\\Rightarrow\\:n^2-25n+144=0$ $\\Rightarrow\\:n^2-16n-9n+144=0$ $\\Rightarrow\\:n(n-16)-9(n-16)=0$ $\\Rightarrow\\:(n-16)(n-9)=0$ $\\Rightarrow\\:n=16\\:\\:or\\:\\:n=9$" ]

[ "47°, 119°, 90°, x°, and x° The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2) = 180° (3) = 540° So, 47° + 119° + 90° + x° + x° = 540° 256° + 2x° = 540° 2x° = 540° – 256° 2x° = 284° x° = $$\\frac{284}{2}$$ x° = 142° Hence, ∠X = ∠Y = 142° Question 21. m∠X = m∠Y = 100.5° Explanation: Question 22. ∠X = ∠Y = 140° Explanation: The given figure is: We can observe that, the polygon has 6 sides. We know that, if the angles are not mentioned in a polygon, the consider that angles as equal angles. The given angle measures of a polygon with 6 sides are: 110°, 149°, 91°, 100°, x°, and x° The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2) = 180° (4) = 720° So, 110° + 149° + 91° + 100 + x° + x° = 720° 440° + 2x° = 720° 2x° = 720° – 440° 2x° = 280° x° = $$\\frac{280}{2}$$ x° = 140° Hence, ∠X = ∠Y = 140° In Exercises 23-26, find the value of x. Question 23. X = 111° Explanation: The sum of the", "Is it possible to have a regular polygon, each of whose interior angles is a. $${45^ \\circ }$$ b. $${60^ \\circ }$$ c. $${75^ \\circ }$$ Ans: a. Given interior angle $$= {45^ \\circ }$$ We know that the interior angle of a polygon of $$n$$ number of sides is given by $$\\frac{{\\left({n – 2} \\right) \\times {{180}^ \\circ }}}{n}.$$ $$\\Rightarrow {45^ \\circ } = \\frac{{\\left({n – 2} \\right){{180}^ \\circ }}}{n}$$ $$\\Rightarrow n \\times {45^ \\circ } = n \\times {180^ \\circ } – {360^ \\circ }$$ $$\\Rightarrow 360 = 180\\,n – 45\\,n$$ $$\\Rightarrow 360 = 135\\,n$$ $$\\Rightarrow n = \\frac{{360}}{{135}}$$ $$\\Rightarrow n = \\frac{8}{3} = 2\\frac{2}{3}$$ which is not a whole number. So, it is impossible to have a regular polygon, each of whose interior angles is $${45^ \\circ }.$$ b. Given interior angle $$= {60^ \\circ }$$ We know that the interior angle of a polygon of $$n$$ number of sides is given by $$\\frac{{\\left({n – 2} \\right) \\times {{180}^ \\circ }}}{n}.$$ $$\\Rightarrow {60^ \\circ } = \\frac{{\\left({n – 2} \\right){{180}^ \\circ }}}{n}$$ $$\\Rightarrow n \\times {60^ \\circ } = n \\times {180^ \\circ } – {360^ \\circ }$$ $$\\Rightarrow 360 = 180\\,n – 60\\,n$$ $$\\Rightarrow 360 = 120\\,n$$ $$\\Rightarrow n = \\frac{{360}}{{120}}$$ $$\\Rightarrow n = 3$$ So, it is possible to have a regular polygon,", "\\frac{180a}{\\pi}$ $\\therefore \\frac{(a - d)\\pi}{180a} = \\frac{1}{120} \\Rightarrow \\frac{(\\frac{\\pi}{3} - d)\\pi}{3\\frac{\\pi}{3}} = \\frac{1}{2}$ $d = \\frac{\\pi}{3} - \\frac{1}{2}$ Thus, angles are $\\frac{1}{2}, \\frac{\\pi}{3}, \\frac{2\\pi}{3}- \\frac{1}{2}$ radians. 48. Let us solve these one by one: 1. We know that if polygon has $n$ sides then sum of angles is $(n - 2)\\pi$ radians or $(n - 2)180^\\circ$ For pentagon sum of angles $= 3\\pi$ or $540^\\circ$ Measure of one interior angle $= \\frac{3\\pi}{5}$ or $108^\\circ$ 2. Measure of one interior angle $= \\frac{5\\pi}{7}$ or $\\frac{900^\\circ}{7}$ 3. Measure of one interior angle $= \\frac{3\\pi}{4}$ or $135^\\circ$ 4. Measure of one interior angle $= \\frac{5\\pi}{6}$ or $150^\\circ$ 5. Measure of one interior angle $= \\frac{15\\pi}{17}$ or $\\frac{2700^\\circ}{17}$ 49. Let there be $n$ sides in one polygon and $2n$ in another. Angles will be $\\frac{(n - 2)\\pi}{n}$ and $\\frac{(2n - 2)\\pi}{2n} - \\frac{(n - 1)\\pi}{n}$ Given ratio of angles $\\frac{3}{2} = \\frac{n -1}{n - 2} \\Rightarrow 3n - 6 = 2n -2 \\Rightarrow n = 4.$ So one polygon is a square while the other is an octagon. 50. Let number of sides in one polygon be $n$ then number of sides in another $\\frac{5n}{4}.$ The angles will be $\\frac{(n - 2)}{n}180^\\circ, \\frac{(\\frac{5n}{4} - 2)}{\\frac{5n}{4}}180^\\circ.$ Given that difference in angles is $9^\\circ$ Solving for difference we find $n = 8$ and thus", "180° (5 – 2) = 180° (3) = 540° So, 47° + 119° + 90° + x° + x° = 540° 256° + 2x° = 540° 2x° = 540° – 256° 2x° = 284° x° = $$\\frac{284}{2}$$ x° = 142° Hence, from the above, We can conclude that ∠X = ∠Y = 142° Question 21. Question 22. The given figure is: From the given figure, We can observe that The polygon has 6 sides We know that, If the angles are not mentioned in a polygon, the consider that angles as equal angles Now, The given angle measures of a polygon with 6 sides are: 110°, 149°, 91°, 100°, x°, and x° So, The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2) = 180° (4) = 720° So, 110° + 149° + 91° + 100 + x° + x° = 720° 440° + 2x° = 720° 2x° = 720° – 440° 2x° = 280° x° = $$\\frac{280}{2}$$ x° = 140° Hence, from the above, We can conclude that ∠X = ∠Y = 140° In Exercises 23-26, find the value of x. Question 23. Question 24. The given figure is: From the given figure, We can observe that The polygon has 7 sides The given angle measures", "interior angle • Pentadecagon - 15 sides - 156° interior angle • Hexakaidecagon - 16 sides - 157.5° interior angle • Heptadecagon - 17 sides - 158.824° interior angle • Octakaidecagon - 18 sides - 160° interior angle • Enneadecagon - 19 sides - 161.053° interior angle • Icosagon - 20 sides - 162° interior angle ### area of a Regular Polygon formula $$\\large{ A_{area} = \\frac{a^2\\;n}{4 \\; tan \\; \\left( \\frac{180}{n} \\right) } }$$ $$\\large{ A_{area} = \\frac{R^2 \\;n \\; sin \\; \\left( \\frac{360}{n} \\right) }{2} }$$ $$\\large{ A_{area} = r^2 \\;n \\; tan \\; \\left( \\frac{180}{n} \\right) }$$ Where: $$\\large{ A_{area} }$$ = area $$\\large{ r }$$ = inside radius (apothem) $$\\large{ R }$$ = outside radius $$\\large{ n }$$ = number of edges $$\\large{ P }$$ = perimeter $$\\large{ a }$$ = edge ### Central Angle of a Regular Polygon formula $$\\large{ CA = \\frac{360}{n} }$$ Where: $$\\large{ CA }$$ = central angle $$\\large{ n }$$ = number of edges ### Circumcircle Radius of a Regular Polygon formula $$\\large{ R = \\frac{a}{2 \\; sin \\; \\left( \\frac{180}{n} \\right) } }$$ Where: $$\\large{ R }$$ = outside radius $$\\large{ a }$$ = edge $$\\large{ n }$$ = number of edges ### Edge of a Regular Polygon formula $$\\large{ a = 2 \\; r \\; tan \\;", "### Home > INT2 > Chapter 9 > Lesson 9.1.1 > Problem9-9 9-9. How many sides does the regular polygon have if each interior angle has the following measure? 1. $60^\\circ$ $60^\\circ = \\frac{180^\\circ(n-2)}{n}$ $n = 3$ sides 1. $156^\\circ$ 1. $90^\\circ$ 1. $140^\\circ$ $9$ sides In addition, for regular polygons: • The measure of each angle in a regular $n$-gon is $\\frac{180^\\circ(n-2)}{n}$ . • The measure of each exterior angle in a regular $n$-gon is $\\frac{360^\\circ}{n}$ .", "(5 – 2) = 180° (3) = 540° So, 140° + 138° + 59° + x° + 86° = 540° 423° + x° = 540° x° = 540° – 423° x° = 117° Hence, x is 117° Question 17. x = 150° Explanation: Question 18. x = 88.6° Explanation: From the given figure, the polygon has 8 sides. The given angle measures of a polygon with 8 sides are: 143°, 2x°, 152°, 116°, 125°, 140°, 139°, and x° The sum of the angle measures of the interior angles of a polygon with 8 sides = 180° (8 – 2) = 180° (6) = 1080° So, 143° + 2x° + 152° + 116° + 125° + 140° + 139° + x° = 1080° 815° + 3x° = 1080° 3x° = 1080° – 815° 3x° = 265° x° = $$\\frac{265}{3}$$ x° = 88.6° Hence, x is: 88.6° In Exercises 19 – 22, find the measures of ∠X and ∠Y. Question 19. m∠X = m∠Y = 92° Explanation: Question 20. ∠X = ∠Y = 142° Explanation: The given figure is: We can observe that, the above polygon has 5 sides. We know that, if the angles are not mentioned in a polygon, then consider that angles as equal angles. So, the given angle measures of a polygon with 5 sides are:", "# Each interior angle of a regular polygon lies between 136 to 142. How do we calculate the sides of the polygon? Jan 7, 2016 $n = 9$ #### Explanation: In a regular polygon each interior angle can be obtained in this way: $\\alpha = {180}^{\\circ} - {360}^{\\circ} / n$ From the conditions of the problem: ${136}^{\\circ} < \\alpha < {142}^{\\circ}$ That's the conjugation of this two inequations: ${136}^{\\circ} < \\alpha$ and $\\alpha < {142}^{\\circ}$ Resolving the first inequation ${136}^{\\circ} < {180}^{\\circ} - {360}^{\\circ} / n$ ${136}^{\\circ} < \\frac{{180}^{\\circ} \\cdot n - {360}^{\\circ}}{n}$ => ${136}^{\\circ} \\cdot n < {180}^{\\circ} . n - {360}^{\\circ}$ => ${44}^{\\circ} . n > {360}^{\\circ}$ => $n > 8.18$ Resolving the second inequation ${180}^{\\circ} - {360}^{\\circ} / n < {142}^{\\circ}$ ${180}^{\\circ} \\cdot n - {360}^{\\circ} < {142}^{\\circ} \\cdot n$ => ${38}^{\\circ} . n < {360}^{\\circ}$ => $n < 9.47$ Conjugating the two inequations $8.18 < n < 9.47$ Since $n \\in \\mathbb{N}$, its only value that satisfies the inequation is $n = 9$ By the way $\\alpha = {180}^{\\circ} - {360}^{\\circ} / 9 = {180}^{\\circ} - {40}^{\\circ}$ => $\\alpha = {140}^{\\circ}$", "+1 vote For a regular polygon having $10$ sides, the interior angle between the sides of the polygon, in degrees, is: 1. $396$ 2. $324$ 3. $216$ 4. $144$ recategorized +1 vote The interior angle between the sides of the polygon $= \\frac{(n-2) \\times 180^{\\circ}}{n},$ where $n =$ number of sides of the polygon. Here, $n = 10,$ therefore the interior angle between the sides of the polygon $= \\frac{(10-2) \\times 180^{\\circ}}{10}$ $\\qquad = 8 \\times 18^{\\circ} = 144^{\\circ}.$ So, the correct answer is $(D).$ by (3.7k points) selected by +1 vote", "# Math Help - radians of a polygon 1. ## radians of a polygon The measure of each interior angle of a regular polygon with n sides is [(n-2)180/n] degrees. What is the measure of each interior angle of a regular polygon with n sides, in radians? How do you convert to radians? 2. Originally Posted by ninobrn99 The measure of each interior angle of a regular polygon with n sides is [(n-2)180/n] degrees. What is the measure of each interior angle of a regular polygon with n sides, in radians? How do you convert to radians? In any circle, there are $360^\\circ$ or $2\\pi^C$. So $1^\\circ = \\frac{2\\pi^C}{360} = \\frac{\\pi^C}{180}$. So to convert degrees to radians, multiply by $\\frac{\\pi^C}{180}$.", "$\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)}$ i) $30$ $180-30=150^{\\circ}$ $n=$ $\\frac{360}{180-150}$ $=12$ ii) $36$ $180-36=144^{\\circ}$ $n=$ $\\frac{360}{180-144}$ $=10$ iii) $40$ $180-40=140^{\\circ}$ $n=$ $\\frac{360}{180-140}$ $=9$ iv) $18$ $180-18=162^{\\circ}$ $n=$ $\\frac{360}{180-162}$ $=20$ $\\\\$ Question 11: Is it possible to have a regular polygon whose interior angles measure $130^{\\circ}$ No. Let us calculate the number of sides of this polygon. No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-130^{\\circ})}$ $=7.2$ Hence not possible. $\\\\$ Question 12: Is it possible to have a regular polygon whose interior angles measure measures $1$ $\\frac{3}{4}$ of a right angle. Yes. Let us calculate the number of sides of this polygon. No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-157.5^{\\circ})}$ $=16$ Hence possible $\\\\$ Question 13: Find the number of sides of a regular polygon, if its interior angle is equal to exterior angle. $Interior \\ angles = 180^{\\circ}-Exterior \\ Angle$ This means that each of the interior and the exterior angles $=90^{\\circ}$ No. of Sides $n=$ $\\frac{360^{\\circ}}{(180^{\\circ}-Interior Angle)} = \\frac{360^{\\circ}}{(180^{\\circ}-90^{\\circ})}$ $=4$ $\\\\$ Question 14: The ratio between the exterior angle and the interior angles is $2:7$. Find the number of sides of the polygon. $Interior \\ angles = 180^{\\circ}-Exterior \\ Angle$ Let the Exterior Angle $=2x$ and the Interior Angle be $7x$ $\\Rightarrow 2x+7x=180 \\ or\\ x=20$ Therefore Interior angle $=140^{\\circ}$ No. of Sides $(n)=$ $\\frac{360^{\\circ}}{180^{\\circ}-140^{\\circ}}$ $= 9$ $\\\\$ Question 15:", "### Find the measure of each interior angle of a regular polygon of $10$ sides. $144 ^ \\circ$ 1. We know that the sum of all exterior angles of a polygon is $360 ^ \\circ.$ 2. Each exterior angle of a regular polygon of $10$ sides $= { \\Bigg ( \\dfrac{ 360 } { 10 } \\Bigg ) } ^ \\circ = 36 ^ \\circ.$ 3. Thus, each interior angle of a regular polygon of $10$ sides $= 180 ^ \\circ - 36 ^ \\circ = 144 ^ \\circ$.", "and get $\\alpha=108^\\circ$. 3) and 4) The sum of all interior angles in an n-sided regular polygon is $n\\alpha = (n-2)\\cdot 180^\\circ$. For a 7-sided regular polygon we set n=7 and get $7\\alpha = (7-2)\\cdot 180^\\circ = 5 \\cdot 180^\\circ = 900^\\circ$. 5) If we make n bigger and bigger, the value for $\\alpha$ approaches 180°. We can express this with the limit: $\\lim_{n\\rightarrow \\infty} \\alpha$ $= \\lim_{n\\rightarrow \\infty} \\frac{(n-2)}{n} \\cdot 180^\\circ$ $= \\lim_{n\\rightarrow \\infty} \\left(1-\\frac{2}{n}\\right) \\cdot 180^\\circ$ $= 180^\\circ$ ## Wallstreet Journal articles Posted by Ed on November 9, 2012 Recently, I’ve been reading the Wallstreet Journal. I particularly enjoy reading the section on life, health and work at the end. Here are two articles that I like: Some Don’t Live to Earn, but Earn to Live By RHEA WESSEL Self-Help for Skeptics By ELIZABETH BERNSTEIN", "polygon be n. We have, Sum of all interior angles = $$(n – 2) \\times 180^{\\circ}$$ and, measure of its each angle=$$\\frac{(n-2) \\times 180^{\\circ}}{n}$$ here we have each angle as $$165^{\\circ}$$ $$\\Rightarrow \\frac{(n-2) \\times 180^{\\circ}}{n}=165$$ $$\\Rightarrow 180n - 360 =165 n \\quad$$[Cross multiplication and opening the brackets] $$\\Rightarrow 180n -165 n=360$$ $$\\Rightarrow 15n = 360$$ $$\\Rightarrow n = \\frac{360}{15}=24$$ So, we have the number of sides of regular polygon as 24. 5.(a) Is it possible to have a regular polygon with measure of each exterior angle a is $$22^{\\circ}$$? (b) Can it be an interior angle of a regular polygon? Why? (a) We have, Number of sides of regular polygon=$$\\frac{360^{\\circ} }{Exterior \\;angle}$$ As we know number of sides by the formula should be an integer so, it is not possible that a regular polygon have its exterior angle of $$22^{\\circ}$$ because $$22^{\\circ}$$ is not divisible by $$360^{\\circ}$$ hence, we can't get a whole number for calculating the number of side. (b) If interior angle =$$22^{\\circ}$$ And we have: measure of each interior angle=$$\\frac{(n-2)\\times180^o}n$$ $$\\Rightarrow 180n -2 \\times 180=22n$$ $$\\Rightarrow 180n -22n=360$$ $$\\Rightarrow 158n=360$$ $$\\Rightarrow n=\\frac{360}{158}$$ But 158 does not divide 360 exactly.So, the polygon is not possible. 6.(a) What is the minimum interior angle possible for a regular polygon? Why? (b) What is the maximum exterior angle possible", "An = nr2 n-+00 Exterior angle of a regular polygon having n sides = $$\\dfrac{360^\\circ}{n}$$ Interior angle of a regular polygon having n sides = $$180^\\circ$$ - Exterior angle; Apothem falls on the midpoint of a side dividing it into two equal parts. For example, consider the polygon shown below: This polygon can be divided into a combination of triangles and trapezium. To get the area of the whole polygon, just add up the areas of all the little triangles (\"n\" of them): Area of Polygon = n × side × apothem / 2. Let’s work out a few example problems about area of a regular polygon. The area of this polygon is n times the area of triangle, since n triangles make up this polygon. 2 π r = n × a. where r = radius of circle, a = side of polygon with n sides. π is a mathematical constant. Given the radius (circumradius) If you know the radius (distance from the center to a vertex, see figure above): where r is the radius (circumradius) n is the number of sides sin is the sine function calculated in degrees (see Trigonometry Overview) . As we know, Area (A) = ½ x p x a, here p = 44 cm and a = 10 cm = ½", "a decagon (10 sides) -- when you find the sum of the interior angles you use (n - 2) * 180 so now you do the reverse. #6: [ (5-2) + (7-2) + (9-2)] x 180 = (3 + 5 + 7) x 180 = 5 x 3 x 180 = 2700 degrees. #7:$$\\frac{720}{180}$$= 4 and 4 + 2 = 6 so this is a regular hexagon and one of its interior is $$\"\\frac{720}{6}$$ = 120 degrees. Or you can also do $$\\frac{720 + 360}{180}$$ = 6 because all the interior angles and their exterior angles are supplementary and the sum of any exterior angle of a convex polygon is 360 degrees. Using this method, you get how many sides instantly. #8: This one is tricky because there are actually two different polygon pairs that the total number of degrees add up to 1980 degrees. One is a regular hexagon and a nonagon. (720 + 1260) Another is a heptagon and an octagon (900 + 1080). Looking at the sum of their diagonals, the latter is the right pair. (14 + 20 = 34) so the answer is 8 - 7 = 1 #9: 360 - 108 - 120 = 132 degrees. #10: To find the central angle B, you do $$\\frac{360}{5}$$ = 72 degrees. BA and BC", "box with the volume of the cube. Volume of rectangular box = L × B × H = 20cm × 6cm × 4cm = 480cm3 Volume of cube = s3 = 2cm3 = 8cm3 => Number of cubes that will fit into the box = $\\frac{480cm^3}{8cm^3}$ = $\\frac{\\cancel{480}^{60}\\cancel{cm^3}}{\\cancel8_1\\cancel{cm^3}}$ = 60 29. The interior angle of a regular polygon is 120°. How many sides has this polygon? A. 3 B. 4 C. 5 D. 6 D. 6 Solution Interior angle of regular polygon = $\\frac{180°(n-2)}{n}$ =>120° = $\\frac{180°(n-2)}{n}$ =>120°× $n$ = 180°$(n-2)$ =>120°$n$ = 180°$n$ – 360° =>360° = 180°$n$ – 120°$n$ =>360° = 60°$n$ =>$\\frac{360°}{60°}$ = $\\frac{60°n}{60°}$ =>$\\frac{\\cancel{360°}^6}{\\cancel{60°}_1}$ = $\\frac{\\cancel{60°}n}{\\cancel{60°}}$ =>n = 6 30. In the diagram above, length of PS = length of SQ and angle SQR = 112°. Find the value of $x$. A. 68° B. 56° C. 46° D. 44° D. 44° Solution ∠PQS + ∠SQR = 180° => ∠PQS + 112° = 180° => ∠PQS = 180° – 112° => ∠PQS = 68° △PSQ is an equilateral triangle because |PS| = |SQ|. This means that ∠PQS = ∠SPQ = 68°. For △PSQ, ∠PQS + ∠QSP + ∠SPQ = 180° => 68° + $x$ + 68° = 180° => $x$ + 136° = 180° => $x$ = 180° – 136° => $x$ = 44° 31.", "of $10$. The sum of interior angles of a pentagon is $$(5-2)\\times 180^o=3\\times 180^o=540^o$$ The interior angle of a pentagon is $$\\frac {540^o}5=108^o$$ The exterior angle of a pentagon is $$\\frac {360^o}5=72^o$$ The perimeter of a pentagon with the side length of $10$ is $$5\\times 10=50$$ The area of a pentagon with the side length of $10$ is $$A=\\frac {50\\times 10}{4\\times\\tan 36^o}\\approx 172.0477$$ The Area & Perimeter of any Polygon work with steps shows the complete step-by-step calculation for finding the sum of interior angles, measures of interior and exterior angles, perimeter and area of a regular pentagon with the side length of $10$ using the regular polygon formulas. For any other number of sides and the value of length of the side of a regular polygon, just supply a positive integer as the number of sides of a regular polygon and a positive real number or parameter as the side length of a regular polygon and click on the GENERATE WORK button. The grade school students may use this Area & Perimeter of any polygon calculator to generate the work, verify the results of geometry problems in the plane or do their homework problems efficiently. ### Real World Problems Using Area & Perimeter of a Regular Polygon Calculating areas and perimeters of regular polygons play an important", "# In a regular polygon each interior angle is 135° greater than each exterior angle. How many sides has the polygon? Dec 21, 2017 The polygon is Hexakaidecagon having $16$ sides. #### Explanation: Exterior angle of regular polygon of $n$ sides is $E = \\frac{360}{n}$ Interior angle of regular polygon of $n$ sides is $I = \\frac{180 \\left(n - 2\\right)}{n}$ Given condition , $I = E + 135 \\therefore \\frac{180 \\left(n - 2\\right)}{n} = \\frac{360}{n} + 135$ or $\\frac{180 \\left(n - 2\\right)}{n} - \\frac{360}{n} = 135$. Multiplying by $n$ on both sides we get , $180 \\left(n - 2\\right) - 360 = 135 n$ or $180 n - 360 - 360 = 135 n \\mathmr{and} 180 n - 135 n = 720$ or $45 n = 720 \\mathmr{and} n = \\frac{720}{45} \\mathmr{and} n = 16$ The polygon is Hexakaidecagon having $16$ sides. [Ans]", "2) \\times 180^\\circ & = 1980^\\circ\\\\180^\\circ n - 360^\\circ & = 1980^\\circ\\\\180^\\circ n & = 2340^\\circ\\\\n & = 13$ The polygon has 13 sides. Example 2: How many degrees does each angle in an equiangular nonagon have? Solution: First we need to find the sum of the interior angles, set $n = 9.$ $(9 - 2) \\times 180^\\circ = 7 \\times 180^\\circ = 1260^\\circ$ “Equiangular” tells us every angle is equal. So, each angle is $\\frac{1260^\\circ}{9} = 140^\\circ$. Equiangular Polygon Formula: For any equiangular $n-$gon, the measure of each angle is $\\frac{(n-2) \\times 180^\\circ}{n}$. If $m \\angle 1 = m \\angle 2 = m \\angle 3 = m \\angle 4 = m \\angle 5 = m \\angle 6 = m \\angle7 = m \\angle 8$, then each angle is $\\frac{(8 - 2) \\times 180^\\circ}{8} = \\frac{6 \\times 180^\\circ}{8} = \\frac{1080^\\circ}{8} = 135^\\circ$ In the Equiangular Polygon Formula, the word equiangular can be switched with regular. Regular Polygon: When a polygon is equilateral and equiangular. Example 3: An interior angle in a regular polygon is $135^\\circ$. How many sides does this polygon have? Solution: Here, we will set the Equiangular Polygon Formula equal to $135^\\circ$ and solve for $n$. $\\frac{(n - 2) \\times 180^\\circ}{n} & = 135^\\circ\\\\180^\\circ n - 360^\\circ & = 135^\\circ n\\\\-360^\\circ & = -45^\\circ n\\\\n &" ]

[ "# How many natural numbers divisible by 7 are there between 3 and 200? How many natural numbers divisible by 7 are there between 3 and 200? [A]27 [B]28 [C]29 [D]36 28 Number just greater than 3 which is divisible by 7 = 7 Number just smaller than 200 which is divisible by 7 = 196 Here, a = 7, an = 196, d = 7, n = 8 ∴ an = a + (n – 1)d => 196 = 7 + (n – 1) $\\times$ 7 $=> n - 1 = \\frac{196 - 7}{7} = 27$ => n = 27 + 1 = 28 Hence option [B] is right answer.", "since 9 does not divide 300. Since the number is divisible by 3 but not 9, it can't be a perfect square/ - 2 years, 5 months ago", "Facebook Twitter Email. Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999. OPtion 1) 312750 2) 2502 Sum 1 ro 250 = 250 … , 249, 250. What is the sum of first 350 odd numbers? What is the sum of first 500 natural numbers? Other. . . There are 3 pairs and they can go in either order, so there are 3 x 2 = 6 ways they can be used to create a three digit number with 1. . Sum of First 50 Odd Numbers; Sum of First 50 Even Numbers; How to Find Sum of First 50 Natural Numbers? Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. That's 250 numbers. Consecutive integers are numbers that follow each other in order. t = the sum of Natural Numbers from 1 to 200 which are divisible by 5. u = the sum of Natural Numbers from 1 to 200. Enter a number: 10 [1] \"The sum is 55\" Here, we ask the user for a number and display the sum of natural numbers upto that number. The number series 1, 2, 3, 4, . That's 500 numbers. In simple words, Consecutive integers are integers that follow in sequence, where the difference", "chance that a number selected at random from the natural numbers 1 to 20 is divisible by either 2 or 3. TheXSquaredFactor Aug 1, 2017 edited by TheXSquaredFactor Aug 1, 2017", "Question: How many three-digit natural numbers are divisible by 9? Solution: The three-digit natural numbers divisible by 9 are 108, 117, 126, ..., 999. Clearly, these number are in AP. Here, = 108 and d = 117 − 108 = 9 $a_{n}=999$ $\\Rightarrow 108+(n-1) \\times 9=999 \\quad\\left[a_{n}=a+(n-1) d\\right]$ $\\Rightarrow 9 n+99=999$ $\\Rightarrow 9 n=999-99=900$ $\\Rightarrow n=100$ Hence, there are 100 three-digit numbers divisible by 9. Let this AP contains n terms. Then,", "want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?", "how many .", "We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?", "- 1) is divisible by 7 for all natural numbers n. Previous", "\\\\$ So, there are total 14 natural numbers between 100 and 200 which are divisible by 7 i.e., n=7 Using formula given by equation (2), we get Sum of all the natural numbers between 100 and 200 which are divisible by 7 is 2107. Natural numbers between 100 and 200 which are divisible by 13 are 104,112,….,195. Clearly, these numbers are forming an arithmetic progression with a common difference of 13. Here ${a_1} = 104$, d=13 and ${a_n} = 195$ Using formula given by equation (1), we will get the value of n. $\\Rightarrow 195 = 104 + 13\\left( {n - 1} \\right) \\\\ \\Rightarrow 13\\left( {n - 1} \\right) = 91 \\\\ \\Rightarrow n - 1 = \\dfrac{{91}}{{13}} = 7 \\\\ \\Rightarrow n = 7 + 1 = 8 \\\\$ So, there are total 8 natural numbers between 100 and 200 which are divisible by 13 i.e., n=8 Using formula given by equation (2), we get Sum of all the natural numbers between 100 and 200 which are divisible by 13 is 1196. There is only one natural number between 100 and 200 which are divisible by both 7 and 13 (i.e., natural number between 100 and 200 which is divisible by 91) is 182. So, the final sum of all natural numbers ‘n’ such that 1001", "the natural numbers ordered by divisibility? • 346,162 hits", "100 and 300 are divisible by 13? A. 16 B. 15 C. 14 D. 13 Prev 1 2 3 4 5", "# Divisibility For how many natural numbers $$x$$, is the expression: $$x ^2 + 2x + 3$$ divisible by $$35$$ ? ×", "9 – Natural numbers, diagrammatically", "the natural number n is divisible by six, then n is divisible by three. From what assumption we started?", "# Count the numbers. The number of natural numbers less than 400 that are not divisible by 17 or 23 is n. Find $\\frac{n}{10}$ ×", "is given by S=2107+1196-182=3121 Hence, option B is correct. Note- In this particular problem, there is only one natural number between 100 and 200 i.e.,182 which is divisible by both 7 and 13. This natural number 182 is counted twice when we are adding the sum of natural numbers between 100 and 200 which are divisible by 7 and the sum of natural numbers between 100 and 200 which are divisible by 13.", "prove We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?", "100 and 200, divisible by 9: ----- Numbers between 100 and 200, divisible by 9: 108 117 126 135 144 153 162 171 180 189 198 The sum : … 2. t = the sum of Natural Numbers from 1 to 200 which are divisible by 5 u = the sum of Natural Numbers from 1 to 200 From the variables (and by common sense), s = u - t Explanation: The sum of a finite arithmetic sequence is equal to the count of the number of terms multiplied by the average of the first and last terms. You just want a way to represent all those numbers and then sum them. The sum of the primes is 1,060. Sum of all numbers from 100 to 200, both inclusive, is a problem in an AP. This Python program allows users to enter Minimum and maximum value. so 100*201 = 20100, Develop An Algorithm And Draw the Flowcharts To Find the Sum of Odd Numbers Between 100 And 200. A prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself. find the sum of the numbers from 1 to 200 inclusive, There is a story that a famous German mathematician named Gauss was given this by his elementary teacher", "root x2. 10. Theorem prove We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started? 11. Algebra X+y=5, find xy (find the product of x and y if x+y = 5)" ]

[ "in the range from $$2$$ to $$31$$: $$2,3,5,7,11,13,17,19,23,29,31;$$ so these are the only primes that \"some number\" might need to be divisible by. But some of these primes (the first three) also appear to higher powers in this range: we have $$2,4,8,16$$; $$3,9,27$$; and $$5,25$$. In each case, the largest one will take care of the rest, so the list of prime-power factors that need to be considered are these: $$7,11,13,16,17,19,23,25,27,29,31.$$ A number is divisible by all the numbers from $$2$$ to $$31$$ if it is divisible by all of these. Conversely, if it is not divisible by some number in the list, it is not divisible by one of these. So we want to find two of these to remove, but only one pair are consecutive numbers: $$16$$ and $$17$$. So the number we are looking for is divisible by all the other nine numbers listed, but not by $$16$$ or $$17$$. The smallest such number is the product of those nine, along with the next smaller power of $$2$$ (namely, $$8$$): $$7\\cdot8\\cdot11\\cdot13\\cdot19\\cdot23\\cdot25\\cdot27\\cdot29\\cdot31 = 2123581660200.$$", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "# Proving the sum of digits is always 9 if number is divisible by 9 • August 5th 2012, 06:02 AM Proving the sum of digits is always 9 if number is divisible by 9 The other day, I was attempting to prove that if the sum of all the digits equal to $9$ then the number is divisible by $9$. E.g. the number $72$— the sum is equal to nine ( $7 + 2 = 9$) and we know for a fact that $8 \\cdot 9 = 72$. Let $N$ be an arbitrary number with an arbitrary number of digits. In expanded form, we can write $N = a \\cdot 10^b + c \\cdot 10^d + e \\cdot 10^f + \\, \\ldots \\, + y \\cdot 10^1 + z \\cdot 10^0 \\, .$ Now, assuming that the sum of all the digits equals to $9$, we can write $9 = a + c + e + \\, \\ldots \\, + y + z$ But now I'm stuck. The proof is complete if we can somehow factor the number $N$ by breaking out a $9$ but I don't really see how I'm supposed to go about to achieve that. The binomial theorem doesn't prove very fruitful (I think) when substituting $10 = 9 +1$. • August 5th 2012,", "# Math Help - Proving the sum of digits is always 9 if number is divisible by 9 1. ## Proving the sum of digits is always 9 if number is divisible by 9 The other day, I was attempting to prove that if the sum of all the digits equal to $9$ then the number is divisible by $9$. E.g. the number $72$— the sum is equal to nine ( $7 + 2 = 9$) and we know for a fact that $8 \\cdot 9 = 72$. Let $N$ be an arbitrary number with an arbitrary number of digits. In expanded form, we can write $N = a \\cdot 10^b + c \\cdot 10^d + e \\cdot 10^f + \\, \\ldots \\, + y \\cdot 10^1 + z \\cdot 10^0 \\, .$ Now, assuming that the sum of all the digits equals to $9$, we can write $9 = a + c + e + \\, \\ldots \\, + y + z$ But now I'm stuck. The proof is complete if we can somehow factor the number $N$ by breaking out a $9$ but I don't really see how I'm supposed to go about to achieve that. The binomial theorem doesn't prove very fruitful (I think) when substituting $10 = 9 +1$. 2. ## Re: Proving the", "# Gap between composite numbers that aren't divisible by 2,3 or 5 It is easy to determine if a number is divisible by 2,3 or 5, so in my experiment i am concentrating on a consecutive list of composite numbers that are not divisible by 2,3 or 5. My question is if there is a maximum gap limit between such numbers, and if so what is it? Please review the image as an example: • The gap is at most $d=2\\cdot3\\cdot5\\cdot7$. To see this look at the arithmetic progression $7+kd$, $k\\in \\Bbb N$. All numbers in that sequence are divisible by $7$, but not by $2,3$ or $5$. Looking at your example this $d$ is probably a very bad upper bound. – leoli1 Jan 26 at 12:00 • The \"eligible\" numbers are the residue classes of $1, 7, 11, 13, 17, 19, 23, 29$ modulo $30$. Although there are infinitely many primes in each of these residue classes (Dirichlet), the density of primes becomes less as the numbers increase in value. So very likely the biggest gaps will occur earliest. – Mark Bennet Jan 26 at 12:09 • $7×(30n+1,7,11,13,17,19,23,29)$ has a maximum gap of 42 – Empy2 Jan 26 at 12:18 • The numbers $2310n+1260+[11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]$ between $2310n+1260+7$ and", "between 10 and 300 , when divisible by 4 leave a remainder 3 are $$11,15,19,23,27, \\ldots .299$$The above numbers are A.P.So, first number $$a=11$$ Common difference $$d=15-11=4$$ Then, the last number is 299 We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$$$\\begin{array}{l}299=11+(n-1) 4 \\\\299=11+4 n-4 \\\\299=7+4 n \\\\299-7=4 n \\\\4 n=292 \\\\n=292 / 4 \\\\n=73\\end{array}\\\\$$The total which are lie between 10 and 300, when divisible by 4 leaves a remainder 3 are 73Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "+ \\ldots + {}^nC_n31^n - 31 - 1$$ $$= 31^2[{}^nC_2 + {}^nC_331 + \\ldots + {}^nC_n31^{n - 2}]$$ The above expression is divisible by $$961.$$ 3. $$3^{2n + 2} - 8n - 9 = 3^2.3^{2n} - 8n - 9 = 9(1 + 8)^n - 8n - 9$$ $$= 9[1 + 8n + {}^nC_28^2 + {}^nC_38^3 + \\ldots + {}^nC_n8^n] - 8n - 9$$ $$= 64n + 9[{}^nC_28^2 + {}^nC_38^3 + \\ldots + {}^nC_n8^n]$$ The above expression is divisible by $$64$$ 4. $$2^{5n + 5} - 31n - 32 = 2^5.(2^5)^n - 31n - 31 = 32(1 + 31)^n - 31n - 32$$ $$= 32[1 + 31n + {}^nC_231^2 + {}^nC_331^3 + \\ldots + {}^nC_n31^n] - 31n - 32$$ $$= 32[{}^nC_231^2 + {}^nC_331^3 + \\ldots + {}^nC_n31^n]$$ The above expression is divisible by $$31^2$$ i.e. $$961$$ 5. $$3^{2n} - 1 + 24n - 32n^2 = (1 + 8)^n - 1 + 24n - 32n^2$$ $$= 1 + 8n + {}nC_28^2 + {}nC_38^3 + \\ldots + {}nC_n8^n - 1 + 24n - 32n^2$$ $$= 32n + \\frac{n(n - 1)}{2}8^2 + {}nC_38^3 + \\ldots + {}nC_n8^n - 32n^2$$ $$= 32n + 64n^2 - 32n + {}nC_38^3 + \\ldots + {}nC_n8^n$$ $$= 64n^2 + {}nC_38^3 + \\ldots + {}nC_n8^n$$ Above expression is divisible by $$256$$ for $$n > 2$$ 21. Let $$r^{th},", "# What will be the least number which when doubled becomes exactly divisible by 9, 15, 21, and 30? Option 1) Option 2) Option 3) Option 4) Option 5) - $LCM \\: of \\: 9, 15, 21\\: and\\: 30=630$ $Number=315$", "The number must be divisible by $$8000$$, because $$15!$$ contains $$8$$ and $$1000$$. $$d_2$$ is non-zero number, because $$15!$$ contains only three $$5$$s. It implies $$1d_10767436d_2$$ must be divisible by $$8$$. It implies $$36d_2$$ is divisible by $$8$$. Hence, $$d_2=8$$. Now you can use the divisibility by $$9$$ ($$d_1+d_2=11$$) and find $$d_1=3$$. $$15!=2^{11}\\cdot 5^3\\cdot 7^2\\cdot 11\\cdot 13=(1000)X$$ where $$X=2^8\\cdot 3^6\\cdot 7^2\\cdot 11\\cdot 13.$$ The last digits of $$2^8(=16^2), 3^6 (=9^3),7^2, 11,13$$ are, respectively $$6,9,9,1,3 .$$ Modulo $$10$$ we have $$6\\cdot 9\\cdot 9 \\cdot 1\\cdot 3\\equiv 6\\cdot(-1)^2\\cdot 3\\equiv 18\\equiv 8$$. So the last digit of $$X$$ is an $$8$$. Therefore the 2nd digit of $$15!$$ must be a $$3$$ in order for the sum of all its digits to be divisible by $$9$$.", "leaves the digits $1,2,6$, to be arranged so that the resulting number is divisible by $8$. The only option for that is $216$, hence we look at $9873216$. This isn't divisible by $7$ either. Next try, start with $9872$. To arrange the digits $1,3,6$ so that the resulting number is divisible by $8$, we have only the option $136$, but $9872136$ isn't divisible by $7$ either. Thus we try to start with $9871$. To get a multiple of $8$, the only candidate is $9871632$, but that is again not divisible by $7$. Thus we move towards numbers starting with $986$, and the largest even such number is $9867312$. Check: $$9867312 = 9\\cdot 1096368 = 8 \\cdot 1233414 = 7\\cdot 1409616.$$", "= 2^{10} \\cdot 2^3 = 1024 \\cdot 8 = 8192$, and $3^8 = (3^4)^2 = 81^2 = 6561$. We have the 2's and 3's out of the way, and then we have $7 \\cdot 17 \\cdot 19 = 2261$. Now if we multiply all the values calculated, we get $1000 \\cdot 1001 \\cdot 8192 \\cdot 6561 \\cdot 2261 = 121,645,100,408,832,000$. Thus $T = 4, M = 8, H = 0$, and the answer $T + M + H = 12$, thus (C). ## Solution 3 We know that $9$ and $11$ are both factors of $19!$. Furthermore, we know that H is 0 because $19!$ ends in three zeroes. We can simply use the divisibility rules for $9$ and $11$ for this problem to find T and M. For $19!$ to be divisible by $9$, the sum of digits must be divisible by $9$. Summing the digits, we get that T + M + $33$ must be divisible by $9$. This leaves either A or C as our answer choice. Now we test for divisibility by $11$. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. $2728$, $2$-$7$+$2$-$8$ = -11 so $2728$ is divisible by $11$). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by", "until we get a number whose units digit is 5 or 0. We have $123 \\ldots 44$ is divisible by $9$, so we can subtract by $9$ to get $123 \\ldots 35$ and we know that this is divisible by 5. So our answer is $\\boxed{\\textbf{(C) } 9}$ ~Arcticturn", "$49$, which is true because $6+8$ is divisible by $7$ and $$6^6-6^5\\cdot8+...+8^6\\equiv0(\\mod7).$$ Observe that $8^7\\equiv1\\pmod{49}$. Since $7|273$, so $8^{273}\\equiv1\\pmod{49}$. Observe also that $6^7\\equiv-1\\pmod{49}$. Since $7|273$, so $6^{273}\\equiv-1\\pmod{49}$. Adding,$8^{273}+ 6^{273}\\equiv0\\pmod{49}$.", "# Maximum terms in a sequence $1, 2, 3, 4 ,\\ldots, 300$, from this series, you need to erase some numbers and create a new series so that any two numbers’ sum is not divisible by $7$. What is the number of maximum terms that could be in the new series? I guess some are confused regarding the wording of this question. It was originally written in another language and I've simply translated it. The questions asks to make a new sequence from the above sequence which is $1,2,3,4, \\ldots,300$. You will remove numbers from this sequence such that if you choose any two numbers from your new sequence, and them add them up, this summation should NOT be divisible by $7$. You'll have to find the maximum numbers of terms of your new sequence. How do I approach such a problem? I've come to a part in this problem where I've found out that numbers that add up to $7$'s multiples should not be included in the sequence. For example, in the case of $13$ and $1$, they add up to $14$, which is divisible by $7$. So both these numbers can't be included in this new sequence. If I consider the number $1$, I can add it to the following sequence: $6, 13, 20, 27,\\ldots$ and", "\\cdot 3 = 28$$ to our total. $28$ is congruent to $1 \\pmod 9$. Thus our sum is congruent to $3 \\pmod 9$, and $k = 1 \\implies \\boxed{B}$. ## Solution 2 As our first trial with 3, We can say that $N\\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\\pmod{3}$. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that $N\\equiv 19+20+21+22+...+91+92\\pmod{3}$, and adding that up using the rainbow strategy, we get $N\\equiv (111\\times37)\\pmod{3}$. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is $\\boxed{B}$ ~mathpro12345", "Does such a value for n even exist (besides the n for our example $$F=527$$)? In Step 3 and 4 we want to examine for which values of n the expression $$(9 \\cdot 10^n - 41)$$ is divisible by 17. ------------------------------------------------------ STEP 3 We already know that $$F=527$$ fulfills the swap condition. We obtain $$F=527$$ for $$n=3$$: $$F = \\frac{1}{17} (9 \\cdot 10^3 -41) = \\frac{1}{17} (9000 - 41) = \\frac{1}{17}(8959) = 527$$ There is a rule for testing whether a number is divisible by 17, see here: http://www.egge.net/~savory/maths1.htm and http://primes.utm.edu/curios/page.php/17.html In order to illustrate the rule we test if 8959 is divisible by 17: Step 1: Split 8959 in 895 and 9 (last digit), multiply 9 (the last digit) by 5: $$9 \\cdot 5 = 40$$. Step 2: Substract this from the ramaining digits 895: $$895-9\\cdot5=850$$. Step 3: If the last two or three remaining digits are divisible by 17 then the original number 8959 is divisible by 17, too. In this case, 850 is divisible by 17, thus 8959 is divisible by 17, too. So, for $$n=3$$, F is an integer. What about $$n=4$$: $$F = \\frac{1}{17}(9 \\cdot 10^4 -41) = \\frac{1}{17}(90000-41) = \\frac{1}{17}(89959)$$ Let's test if 89959 is divisible by 17: Step 1: $$8995-9\\cdot 5 = 8955$$ Step 2: $$895 -5 \\cdot 5 = 870$$", "# Two numbers have sum -37 and product 300. What are the numbers? Dec 8, 2016 $- 12$ and $- 25$ #### Explanation: The prime factorisation of $300$ is: $300 = {2}^{2} \\cdot 3 \\cdot {5}^{2}$ If $m \\cdot n = 300$ and $m + n = 37$ then note that $37$ is divisible by neither $2$ nor $5$. So when splitting the factors of $300$ into $m$ and $n$, one of them must be divisible by ${2}^{2} = 4$ and one (possibly the same one) must be divisible by ${5}^{2} = 25$. We then see that $\\frac{300}{25} = 12$ and $25 + 12 = 37$. So the numbers we are looking for are $- 12$ and $- 25$", "$10^{2n}+4\\cdot10^n-5+9$ is clearly divisible by $9$ so $\\sqrt{10^{2n}+4\\cdot10^n-5+9} = 10^n + 2$ would have to be divisible by $3$. (I guess that's the third and in this case most pertinent argument). Jul 23, 2021 at 18:46 • Why have you downvoted me ? My answer is correct. Jul 24, 2021 at 7:43", "so $17^{200}-1$ must be divisible by $10$.", "# In how many ways can you group $3$ different numbers from $1$ to $12$ wherein their sum is divisible by $3$? In how many ways can you group $3$ different numbers from $1$ to $12$ wherein their sum is divisible by $3$? This question is one of the questions asked in a Math contest for intermediate level, particularly for grade 5 pupils. I know the answer but I do not know how it was solved. I also have tried the counting method but have counted 44 combinations only. Also tried elimination but stuck with 164 combinations, don't know what combinations to subtract from $12C3$. The answer to the question is $76$. Help please? Among our $12$, there are $4$ with remainder $0$ on division by $3$, $4$ with remainder $1$, and $4$ with remainder $2$. We can get a sum divisible by $3$ if we use \"$3$ of a kind\" (same remainder) or if we use $1$ of each kind. There are $3\\cdot \\binom{4}{3}$ ways to choose $3$ of a kind. And there are $4\\cdot 4\\cdot 4$ ways to choose $1$ of each kind. Divide the numbers into $3$ groups by their remainders of division by $3$. $$S_0 = \\{ 3,6,9,12\\}$$ $$S_1 = \\{ 1,4,7,10\\}$$ $$S_2 = \\{ 2,5,8,11\\}$$ In the equation $$x+y+z \\equiv 0 \\mod 3$$" ]

[ "# How many natural numbers divisible by 7 are there between 3 and 200? How many natural numbers divisible by 7 are there between 3 and 200? [A]27 [B]28 [C]29 [D]36 28 Number just greater than 3 which is divisible by 7 = 7 Number just smaller than 200 which is divisible by 7 = 196 Here, a = 7, an = 196, d = 7, n = 8 ∴ an = a + (n – 1)d => 196 = 7 + (n – 1) $\\times$ 7 $=> n - 1 = \\frac{196 - 7}{7} = 27$ => n = 27 + 1 = 28 Hence option [B] is right answer.", "in the range from $$2$$ to $$31$$: $$2,3,5,7,11,13,17,19,23,29,31;$$ so these are the only primes that \"some number\" might need to be divisible by. But some of these primes (the first three) also appear to higher powers in this range: we have $$2,4,8,16$$; $$3,9,27$$; and $$5,25$$. In each case, the largest one will take care of the rest, so the list of prime-power factors that need to be considered are these: $$7,11,13,16,17,19,23,25,27,29,31.$$ A number is divisible by all the numbers from $$2$$ to $$31$$ if it is divisible by all of these. Conversely, if it is not divisible by some number in the list, it is not divisible by one of these. So we want to find two of these to remove, but only one pair are consecutive numbers: $$16$$ and $$17$$. So the number we are looking for is divisible by all the other nine numbers listed, but not by $$16$$ or $$17$$. The smallest such number is the product of those nine, along with the next smaller power of $$2$$ (namely, $$8$$): $$7\\cdot8\\cdot11\\cdot13\\cdot19\\cdot23\\cdot25\\cdot27\\cdot29\\cdot31 = 2123581660200.$$", "# How many numbers $k$ of $200 \\choose k$ are divisible by $3$? $k \\in \\{0,1,2,\\cdots 200\\}$ \"How many of the numbers (200 Choose k), where k is an element of the set {0,1,2,3,4,....,200} are divisible by 3? \" Here is my thinking: (200 Choose 0,1, and 2) are not multiples of 3 but every next combination has at least 1 multiple of 3 in the numerator therefore making the number divisible by 3. Since there are 201 numbers from 0 to 200, and 3 numbers (0,1, and 2) do not work, there are 198 numbers k. Is this answer right and is this the right method? • If $k=9n+r$ , with $r\\in\\{0,1,2\\}$ , then $200\\choose k$ does NOT divide through $3$. – Lucian Dec 7 '13 at 21:17 • You must also consider the factors of $3$ in the denominator. $\\binom{200}9$ is not divisible by $3$. It’s equal to $$\\frac{200\\cdot199\\cdot\\ldots\\cdot192}{9!}\\;,$$ where the multiples of $3$ in the numerator are $198,195$, and $192$, while those in the denominator are $9,6$, and $3$. There are $4$ factors of $3$ in each, so there are no factors of $3$ in the binomial coefficient. – Brian M. Scott Dec 7 '13 at 21:19 • Where did you get $k = 9n+r$ from? – user113027 Dec 7 '13 at 21:20 • Using", "# Two numbers have sum -37 and product 300. What are the numbers? Dec 8, 2016 $- 12$ and $- 25$ #### Explanation: The prime factorisation of $300$ is: $300 = {2}^{2} \\cdot 3 \\cdot {5}^{2}$ If $m \\cdot n = 300$ and $m + n = 37$ then note that $37$ is divisible by neither $2$ nor $5$. So when splitting the factors of $300$ into $m$ and $n$, one of them must be divisible by ${2}^{2} = 4$ and one (possibly the same one) must be divisible by ${5}^{2} = 25$. We then see that $\\frac{300}{25} = 12$ and $25 + 12 = 37$. So the numbers we are looking for are $- 12$ and $- 25$", "Browse Questions # Find the sum of all natural numbers between 45 and 300 that are exactly divisible by 7. $\\begin{array}{1 1} 300 \\\\ 6174 \\\\ 245 \\\\ 36 \\end{array}$ Explanation : $1^{st}\\;term\\;greater\\;than\\;or\\;equal\\;to\\;45\\;that\\;is\\;divisible\\;by\\;7=>\\;49$ $last\\;term\\;<\\;300=>294.$ $a=49\\;,d=7\\;,t_{n}=294\\;find\\;\\;n$ $t_{n}=294=49+7(n-1)$ $7(n-1)=245$ $n-1=35$ $n=36$ $Find\\;sum\\;S_{n}=\\frac{n}{2}\\;[2a+(n-1)d]$ $S_{n}=\\frac{36}{2}\\;[2*49+35*7]$ $=6174.$", "Answer Comment Share Q) Find the sum of all natural numbers between 45 and 300 that are exactly divisible by 7. $\\begin{array}{1 1} 300 \\\\ 6174 \\\\ 245 \\\\ 36 \\end{array}$ 1 Answer Comment A) Answer : (b) 6174 Explanation : $1^{st}\\;term\\;greater\\;than\\;or\\;equal\\;to\\;45\\;that\\;is\\;divisible\\;by\\;7=>\\;49$ $last\\;term\\;<\\;300=>294.$ $a=49\\;,d=7\\;,t_{n}=294\\;find\\;\\;n$ $t_{n}=294=49+7(n-1)$ $7(n-1)=245$ $n-1=35$ $n=36$ $Find\\;sum\\;S_{n}=\\frac{n}{2}\\;[2a+(n-1)d]$ $S_{n}=\\frac{36}{2}\\;[2*49+35*7]$ $=6174.$", "# What will be the least number which when doubled becomes exactly divisible by 9, 15, 21, and 30? Option 1) Option 2) Option 3) Option 4) Option 5) - $LCM \\: of \\: 9, 15, 21\\: and\\: 30=630$ $Number=315$", "5, 7 and 9. 123_ _ _. There can be $10^3=1000$ such numbers. Since every 315th number is divisible by 5,7 and 9; for any range of 1000 numbers, there'll be $\\frac{1000}{315}=3.17=3$ such numbers. So, Option B 1 vote 1 172 views", "# Gap between composite numbers that aren't divisible by 2,3 or 5 It is easy to determine if a number is divisible by 2,3 or 5, so in my experiment i am concentrating on a consecutive list of composite numbers that are not divisible by 2,3 or 5. My question is if there is a maximum gap limit between such numbers, and if so what is it? Please review the image as an example: • The gap is at most $d=2\\cdot3\\cdot5\\cdot7$. To see this look at the arithmetic progression $7+kd$, $k\\in \\Bbb N$. All numbers in that sequence are divisible by $7$, but not by $2,3$ or $5$. Looking at your example this $d$ is probably a very bad upper bound. – leoli1 Jan 26 at 12:00 • The \"eligible\" numbers are the residue classes of $1, 7, 11, 13, 17, 19, 23, 29$ modulo $30$. Although there are infinitely many primes in each of these residue classes (Dirichlet), the density of primes becomes less as the numbers increase in value. So very likely the biggest gaps will occur earliest. – Mark Bennet Jan 26 at 12:09 • $7×(30n+1,7,11,13,17,19,23,29)$ has a maximum gap of 42 – Empy2 Jan 26 at 12:18 • The numbers $2310n+1260+[11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]$ between $2310n+1260+7$ and", "Browse Questions # Find the sum of all natural numbers between 45 and 300 that are exactly divisible by 7. $\\begin{array}{1 1} 300 \\\\ 6174 \\\\ 245 \\\\ 36 \\end{array}$ Can you answer this question? Answer : (b) 6174 Explanation : $1^{st}\\;term\\;greater\\;than\\;or\\;equal\\;to\\;45\\;that\\;is\\;divisible\\;by\\;7=>\\;49$ $last\\;term\\;<\\;300=>294.$ $a=49\\;,d=7\\;,t_{n}=294\\;find\\;\\;n$ $t_{n}=294=49+7(n-1)$ $7(n-1)=245$ $n-1=35$ $n=36$ $Find\\;sum\\;S_{n}=\\frac{n}{2}\\;[2a+(n-1)d]$ $S_{n}=\\frac{36}{2}\\;[2*49+35*7]$ $=6174.$ answered Jan 5, 2014 by", "# How many composite numbers are between 1 and 100? May 27, 2017 $74$ #### Explanation: Note that the following answer assumes that \"between 1 and 100\" is intended to include $1$ and $100$, following common English usage. Start with the numbers $1 , 2 , 3 , \\ldots 100$ • $1$ is not composite, because it's a unit, so that leaves $99$ other numbers. • The numbers $4 = {2}^{2} , 6 , 8 , \\ldots , 100$ are all divisible by $2$, so composite. There are $\\frac{100 - 4}{2} + 1 = 49$ of these, leaving $99 - 49 = 50$ other numbers. • The numbers $9 = {3}^{2} , 15 , 21 , \\ldots , 99$ are all divisible by $3$ and not by $2$. There are $\\frac{99 - 9}{6} + 1 = 16$ of these, leaving $50 - 16 = 34$ other numbers. • The numbers $25 = {5}^{2} , 35 , 55 , 65 , 85 , 95$ are all divisible by $5$ and not by $2$ or $3$. There are $6$ of these, leaving $34 - 6 = 28$ other numbers. • The numbers $49 = {7}^{2} , 77 , 91$ are divisible by $7$ and not by $2$, $3$ or $5$. There are $3$ of these, leaving $28 - 3 =", "# Show that among any consecutive $16$ natural numbers one is coprime to all others Show that among any consecutive $16$ natural numbers one is coprime to all others. Is it useful to use the division algorithm on $16$? $16k,16k+1,16k+2,...16k+15$ • See Solution-O51 in projectpen.files.wordpress.com/2009/06/pen-a9-a37-o511.pdf May 11, 2016 at 14:26 • @HamidRezaEbrahimi modulo $30$ is used because $2\\cdot 3\\cdot 5=30$, so the numbers congruent to one of $1,7,11,13,17,19,23,29$ are automatically not divisible by $2,3,5$. May 11, 2016 at 14:59 • The proof referenced above is wrong. It claims that if we have 8 consecutive odd numbers, then we can always find one which is not divisible by any of 2, 3, 5, 7, 11 or 13. But we have 20573 divisible by 7, 20575 by 5, 20577 by 3, 20579 by 13, 20581 by 11, 20583 by 3, 20585 by 5 and 20587 by 7. May 11, 2016 at 15:40 • A deceptively simple question. May 11, 2016 at 21:16 • I thought someone else would have mentioned by now that \"any consecutive 16 natural numbers\" is not quite the same as $16k, 16k + 1, \\ldots, 16k + 15\". For example, compare 273, ..., 288 to 276, ..., 291. May 12, 2016 at 5:08 ## 2 Answers First off, you should observe that since any two numbers", "# Find the sum Find the sum of all natural numbers from 1 and 100, which are divisible by 2 or 5 Result s = 3050 #### Solution: $2,4,6,8,…,100 \\ \\\\ 5,10,15,…,100 \\ \\\\ \\ \\\\ s_{1}=2 \\cdot \\ (1+2+3+4 ... +50) \\ \\\\ s_{1}=\\dfrac{ n_{1} }{ 2 } \\cdot \\ (a_{50}+a_{1}) \\ \\\\ s_{1}=\\dfrac{ 50 }{ 2 } \\cdot \\ (100+2)=2550 \\ \\\\ \\ \\\\ s_{2}=5(1 + 3 + 5 + … + 19) \\ \\\\ s_{2}=\\dfrac{ n_{2} }{ 2 } \\cdot \\ (b_{19}+b_{1}) \\ \\\\ s_{2}=\\dfrac{ 10 }{ 2 } \\cdot \\ (95+5)=500 \\ \\\\ \\ \\\\ s=s_{1}+s_{2}=2550+500=3050$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! ## Next similar math problems: 1. Digits Show that if x, y, z are 3 consecutive nonzero digits, zyx-xyz = 198, where zyx and xyz are three-digit numbers created from x, y, z. 2. Theorem prove We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we", "\\\\$ So, there are total 14 natural numbers between 100 and 200 which are divisible by 7 i.e., n=7 Using formula given by equation (2), we get Sum of all the natural numbers between 100 and 200 which are divisible by 7 is 2107. Natural numbers between 100 and 200 which are divisible by 13 are 104,112,….,195. Clearly, these numbers are forming an arithmetic progression with a common difference of 13. Here ${a_1} = 104$, d=13 and ${a_n} = 195$ Using formula given by equation (1), we will get the value of n. $\\Rightarrow 195 = 104 + 13\\left( {n - 1} \\right) \\\\ \\Rightarrow 13\\left( {n - 1} \\right) = 91 \\\\ \\Rightarrow n - 1 = \\dfrac{{91}}{{13}} = 7 \\\\ \\Rightarrow n = 7 + 1 = 8 \\\\$ So, there are total 8 natural numbers between 100 and 200 which are divisible by 13 i.e., n=8 Using formula given by equation (2), we get Sum of all the natural numbers between 100 and 200 which are divisible by 13 is 1196. There is only one natural number between 100 and 200 which are divisible by both 7 and 13 (i.e., natural number between 100 and 200 which is divisible by 91) is 182. So, the final sum of all natural numbers ‘n’ such that 1001", "so $17^{200}-1$ must be divisible by $10$.", "# What is the probability that sum of two natural numbers is divisible by 4? [closed] As we can see that the favourable case are those where sum of pair of two natural numbers is divisible by 4, and such pairs may be listed as below: \\begin{align*} &(0,0), (0,4), (0,8),\\\\ &(1,3), (1,7),\\\\ &(2,2), (2,6),\\\\ &(3,1), (3,5), (3,9)\\\\ &(4,0), (4,4), (4,8)\\\\ &(5,3), (5,7)\\\\ &(6,2), (6,6)\\\\ &(7,1), (7,5), (7,9),\\\\ &(8,0), (8,4), (8,8),\\\\ &(9,3), (9,7). \\end{align*} Thus, there are 25 favourable cases where sum of two natural numbers is divisible by 4. Now, for exhaustive cases, we can see that there total 10 natural number viz. 0--9. Therefore, there are 10$\\cdot$10=\\,100. Therefore, the probability comes out to be \\begin{align*} P=\\,\\frac{25}{100}=\\frac{1}{4}. \\end{align*} Is my approach is correct? • First of all, by natural numbers you mean numbers $0,1,2,\\dots,9$? Because $10,11, \\dots$ etc. are also natural numbers. Second, no, you should not ignore $9$ from the possible cases. These are \"all possible cases\" not \"potentially favorable cases\". Apr 4, 2018 at 12:43 • What about $(9,3)$ (and quite a few more)? Apr 4, 2018 at 12:44 • The sum has to be less than 10? Apr 4, 2018 at 12:45 • The sum should be divisible by 4 only Apr 4, 2018 at 12:45 • Should $(9,0)$ really be included? Apr 4, 2018", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "three: 9, 21, 27, and 33 because I have had their factorizations memorized since third grade; 81 because it’s a power of three so I have it memorized too; and 39, 63, 69, 93, and 99 because they consist of two digits each of which is a multiple of three. As you probably know, there is also a simple test for determining divisibility by three: just add the digits and see whether the result is divisible by three. This test identifies a few more “nonobvious” multiples of three which you might otherwise think are prime: $51 = 3 \\cdot 17$, $57 = 3 \\cdot 19$, and $87 = 3 \\cdot 29$. 3. What’s left? There are only three numbers less than 100 which are composite and not divisible by 2, 3, or 5: • $49 = 7 \\cdot 7$ and $77 = 7 \\cdot 11$ are easy to spot. • $91 = 7 \\cdot 13$ is the one I always used to forget about—I call it a “fool’s prime” because it “looks” prime but isn’t. It’s worth just memorizing the fact that it is composite (and its factorization). So, to sum up, faced with a number less than 100 that I want to test for primality, I can quickly rule it out if it is divisible by 2,", "200 and 300 (both inclusive) are no [#permalink] ### Show Tags 03 Feb 2019, 22:52 3 3 Divisible by 2 = $$\\frac{300-200}{2} +1 = 51$$ Divisible by 3 = $$\\frac{300-201}{2} +1 = 34$$ Divisible by 5 = $$\\frac{300-200}{5} +1 = 21$$ Divisible by 2 and 3 =$$\\frac{300-204}{6} +1 = 17$$ Divisible by 3 and 5 = $$\\frac{300-210}{15} +1 = 7$$ Divisible by 5 and 2 = $$\\frac{300-200}{10} +1 = 11$$ Divisible by 2,3 and 5 = $$\\frac{300-210}{30} +1 = 4$$ positive integers that are not divisible by 2, 3 or 5 = $$101 - (51+34+21) + (17+7+11) - 4 = 26$$ IMO D _________________ ##### General Discussion Director Joined: 09 Mar 2018 Posts: 994 Location: India Re: How many positive integers between 200 and 300 (both inclusive) are no [#permalink] ### Show Tags 03 Feb 2019, 23:16 Bunuel wrote: How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5? A. 3 B. 16 C. 24 D. 26 E. 75 IMO D Total numbers will be 101 (300-200+1) Not divisible will be, if we can find the numbers divisible by 2, 3 or 5, we can subtract it from Total number Since the terms will be in an AP, we can use $$a_n$$= a + (n-1)d Divisible by 2 = 300-200/2", "$$252 + 259 + 266 + 273+\\cdots\\cdots+336+343+350$$ Since this is an arithmetic sequence, i.e. This is an A.P. The formula to count the number of integers is below: Count = B - A + 1 Plugging the numbers A = 10 and B = 50 into the formula, we get: 50 - 10 + 1 Count = 41 Therefore, the sum of inclusive integers from 10 to 50 = 30 x 41 = 1230 Watch the Inclusive Number Word Problems Video Then + 499 = 62500. how would it affect the code if i wanted say the sum of the first 250 prime numbers instead of the prime numbers under 250? Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … On each iteration, we add the number num to sum, which gives the total sum in the end. . Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999. Q:-Find the sum of integers from 1 to 100 that are divisible by 2 or 5. getcalc.com's Arithmetic Progression (AP) calculator, formula & workout to find what is the sum of first 250 natural numbers. The below workout" ]

[ "200 is not a multiple of 7. There just isn't enough information.", "# Multiples of 6 or 7 but not both Discrete Mathematics Level 3 How many integers between 1 and 2015 are multiples of 6 OR 7 but not both? ×", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "Multiples of 6 or 7 but not both How many integers between 1 and 2015 are multiples of 6 OR 7 but not both? ×", "How many integers between 1 and 1200 are multiples of 2? How many integers between 1 and 1200 are multiples of 3? How many integers between 1 and 1200 are multiples of 5? Some integers are multiples of both 2 and 3, of both 2 and 5, of both 3 and 5... Some integers are multiples of 2, 3 and 5... Can the Venn diagram in the question help, if you consider that the multiples of 2, 3 and 5 each correspond to one of the sets (circles)?", "# Multiples of 5 and 7 Number Theory Level 3 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# How many multiple? Number Theory Level 1 How many multiples of 7 are between 345 and 563 inclusive? ×", "Number 3 Number Theory Level 2 How many integers that are multiples of 3 are there between 1,000 and 1,000,000? ×", "# Find The Non-multiples Probability Level 1 How many integers between 1 and 1000 (inclusive) are neither multiples of 2 nor multiples of 5? ×", "# Which multiple of 5 is missing? This file contains the multiples of 5 between 5 and 955, inclusive. One of the multiples of 5 is missing though. Which number is missing?", "+0 How many of the first 500 positive integers are divisible by 3, 4 and 5? 0 113 1 +71 How many of the first 500 positive integers are divisible by 3, 4 and 5? Firewolf Jun 17, 2018 #1 +2194 +1 The least common multiple of 3,4, and 5 is the smallest integer that is also divisible by 3,4, and 5. Because 3,4, and 5 do not share any common factors, the least common multiple can be determined by finding the product of 3,4, and 5. $$3*4*5=60$$ All multiples of 60 bounded between 1 and 500 are the only integers that satisfy the original condition. $$500/60$$ or $$8.\\overline{3}$$ represents the number of multiples of 60. Of course, multiples can only be counted as whole numbers, so there are only 8 integers that are divisible by 3,4, and 5. TheXSquaredFactor Jun 17, 2018 edited by TheXSquaredFactor Jun 17, 2018", "the languages is 200 - (E + F + S + ...) Sep 16, 2014 at 7:35", "of a multiple of 10 or 25.", "there are an infinite amount of multiples of 75 that can be obtained by multiplying 75 by integers larger than 10.", "must be chosen to be sure that at least 5 of them are even?(c). How manv different integers between and 300 (inclusive) must be chosen to be sure that at least 2 of them add up to 40?(d)_ How many different integers between and 300 (inclusive must be chosen to be sure th 1. Counting and Pigeonhole Principle (a). A set of three different integers is chosen at random between and 300 (inclusive). How many different outcomes are possible? (6). How many different integers between and 300 (inclusive) must be chosen to be sure that at least 5 of them are even? (c). How ma...", "than 100 are neither multiples of 2 or [#permalink] ### Show Tags 15 Apr 2017, 11:44 devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Set comprises the integers 1-99 inclusive. Number of items in set=99. Number of integers that are a multiple of 2: [(98-2)/2)]+1=49 Number of integers that are a multiple of 3: [(99-3)/2)]+1=33. Of these, 16 are even and are therefore counted in the number of multiples of 20 (49). So there are 17 additional integers to add that are multiples of 3 but not multiples of 2. 99-(49+17)=99-66=33 Agree? Intern Joined: 12 Dec 2016 Posts: 7 Re: How many positive integers less than 100 are neither multiples of 2 or [#permalink] ### Show Tags 18 Apr 2017, 05:49 how i approached this problem - we have to eliminate all the multiples of 2 & 3... Therefore, within 100 it's all about the prime numbers and 1. There are 25 prime numbers - 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 Out of which 2 & 3 are not included so - 23. Also, there is 1. So 24. Next the multiples of the prime numbers", "wrote: How many positive integers less than 1000 are multiples of 5 but NOT of 4 or 7? (A) 114 (B) 121 (C) 122 (D) 129 (E) 136 995/5=all multiples of 5=199 980/20=all multiples of 5 and 4=49 980/35=all multiples of 5 and 7=28 980/140=duplicate multiples of 5, 4 and 7=7 199-49-28+7=129 D. Manager Joined: 20 Jan 2017 Posts: 62 Location: United States (NY) GMAT 1: 610 Q41 V34 Followers: 1 Kudos [?]: 2 [0], given: 14 Re: How many positive integers less than 1000 are multiples of 5 [#permalink] ### Show Tags 20 Jan 2017, 17:37 1)Multiples of 5: (995-5)/5+1=198+1=199 2)Multiples of 5&4: 20: (980-20)/20+1=960/20+1=48+1=49 3)Multiples of 5&7: 35: (980-35)/35+1=945/35+1=27+1=28 4)Multiples of 5&4&7: 140: (980-140)/140+1=840/140+1=6+1=7 5)Multiples of 5 but not 4 and not 7: 199-(49-7)-(28-7)-7=129 Re: How many positive integers less than 1000 are multiples of 5 [#permalink] 20 Jan 2017, 17:37 Similar topics Replies Last post Similar Topics: 2 How many positive integers less than 100 are multiples of a prime 4 15 Oct 2016, 18:16 43 How many positive integers less than 100 are neither multiples of 2 or 20 02 Apr 2016, 14:01 7 How many positive integers less than 50 are multiples of 4 but NOT mul 11 20 Dec 2015, 04:42 18 How many positive integers less than 250 are multiple of", "Tags 21 Aug 2014, 09:11 Doubt: if we change the range from 100 <= n <= 200 to 200 <= n <= 300 and use the same line of reasoning - it leads to erroneous results as illustrated below... number of even integers between 200 and 300 = (300 - 200)/2 + 1 = 51 integers; even multiples of 7 : (294 - 203)/14 + 1 = 7.5 : clearly, this is not possible. and works for even multiples of 9 : (297-207)/18 + 1 = 6. Bunuel (or anyone), could you please explain. Thanks! Math Expert Joined: 02 Sep 2009 Posts: 56269 How many even integers n, where 100 <= n <= 200, are divisib [#permalink] ### Show Tags 22 Aug 2014, 04:52 mrekafandroid wrote: Doubt: if we change the range from 100 <= n <= 200 to 200 <= n <= 300 and use the same line of reasoning - it leads to erroneous results as illustrated below... number of even integers between 200 and 300 = (300 - 200)/2 + 1 = 51 integers; even multiples of 7 : (294 - 203)/14 + 1 = 7.5 : clearly, this is not possible. and works for even multiples of 9 : (297-207)/18 + 1 = 6. Bunuel (or anyone), could you please explain. Thanks! The number", "n that is a multiple of 20.", "integers from 1 to 400." ]

[ "ways Total ways 168+147=315 ways The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41 Since $c<9$, we can have the following viable combinations for $b\\times\\ c\\ =96$ (given our objective is to minimize the sum): $48\\times\\ 2$ ; $32\\times3$ ; $24\\times\\ 4$ ; $16\\times6$ ; $12\\times8$ Similarly, we can factorize $a\\times\\ b\\ = 432$ into its factors. On close observation, we notice that $18\\times24\\ and\\ 24\\ \\times\\ 4\\$ corresponding to $a\\times b\\ and\\ b\\times\\ c\\$ respectively together render us with the least value of the sum of $a+b\\ +\\ c\\ \\ =\\ 18+24+4\\ =46$ Hence, Option D is the correct answer. After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19 Given, The average of the four prime numbers = 35. a +", "count: 5, 10, 15, 20, 25, etc. 3. Multiply the number with 1, 2, 3, 4 and so on: $$2 \\times 1 = 2$$, $$2 \\times 2 = 4$$, $$2 \\times 3 = 6$$, $$2 \\times 4 = 8$$, etc. ## List of a few Multiples First ten multiples of Multiples 3 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 4 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 5 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 10 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 ## Properties of Multiples Every number is a multiple of itself For example, the first multiple of 7 is 7 because: $$7 \\times 1 = 7$$ The multiples of a number are infinite We know that numbers are infinite. Therefore, the multiples of a number are infinite. For example, if we need to list the multiples of 3, we start with: 3, 6, 9, 12, 15, 18….. However, will you be able to list all the multiples here? No, because they are infinite. The multiple of a number is greater than or equal to the number itself. For example, if we take the multiples of 5 5, 10, 15, 20, 25, 30… We can see that: The 1st multiple of 5 is equal", "digit is fixed as zero In remaining 9 places Lets choose 5 places and place 1s there Number of ways are $$\\displaystyle{{C}_{{5}}^{{9}}}$$ 126 Step 3 How many integers in the range 1 through 140 are integer multiples of 2, 5, or 7 There are 70 multiples of 2 23 multiples of 3 left we have 9 multiples of 5 that have not yet been included. Total is 102", "5 and 7? a) 42 b) 41 c) 40 d) 43 Solution: The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41 Question 6: Let N, x and y be positive integers such that $N=x+y,2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible for N? Solution: Possible values of x = 3,4,5,6,7,8,9 When x = 3, there is no possible value of y When x = 4, the possible values of y = 22 When x = 5, the possible values of y=21,22 When x = 6, the possible values of y = 20.21,22 When x = 7, the possible values of y = 19,20,21,22 When x = 8, the possible values of y=18,19,20,21,22 When x = 9, the possible values of y=17,18,19,20,21,22 The unique values of N = 26,27,28,29,30,31 Question 7: How many integers in the set {100, 101, 102, …, 999} have at least", "W Common multiples of 2 and 3 or multiples of 6: first multiple of 6 will be 54 and last multiple will be 96 so the number of multiples of 6 between 54 and 99 inclusive will be 96=54+(n-1)6 or 42=(n-1)6 or 7=n-1 or n=8 So, number of integers in set W will be 42-8=34 Board of Directors Joined: 17 Jul 2014 Posts: 2598 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Re: If W is the set of all the integers between 49 and 99, inclusive, that [#permalink] Show Tags 14 Dec 2016, 15:09 reto wrote: If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers? A. 26 B. 32 C. 33 D. 34 E. 35 my approach between 49 and 99, there are 50 numbers, 25 of which are even. so definitely >25 or 26. A is out. now, multiples of 3: 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 -> 54, 60, 66, 72, 78, 84, 90, 96 are already counted in the 25 from above. remains 51, 57, 63, 69, 75,", "there are two cases possible Case 1: When 7 is at the left extreme In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9) So total ways 3(8)(7)= 168 Case 2: When 7 is not at the extremes Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left) Hence in total 3(7)(7)=147 ways Total ways 168+147=315 ways $ab\\ =\\ 4^{2017}=2^{4034}$ The total number of factors = 4035. out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017. And since the given number is a perfect square we have one set of two equal factors. .’. many pairs(a, b) of positive integers are there such that $a\\leq b$ and $ab=4^{2017}$ = 2018. The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60", "14:02 • It does require that the start and end points are adjusted to exact multiples, the \"discard the remainder\" approach suggested by Jacob is not reliable. Consider: How many multiples of 7 are in the interval [6,8]? Now, how many in the interval [1,13]? Or [8,10]? – Ben Voigt Sep 29 '15 at 18:26 • I should admit I had never heard it described as \"fenceposts\" before either (that I remember). I only used the term to tie into porglezomp's comment on the OP. – Paul Sinclair Sep 29 '15 at 22:12 There are just as many multiples of $3$ between $252$ and $348$ (inclusive) as there are between $0$ and $348-252$, that is, $0$ and $96$. Now what one does is a matter of taste. The number between $0$ and $96$ is the same as the number from $3$ to $99$, which is the same as the number of integers between $1$ and $33$. Or else there are $32$ between $3$ and $96$. But don't forget about $0$. Or else there are just as many as there are integers from $0$ to $32$. Again, don't forget about $0$. There are $33$ of these. Remark: I prefer to figure it out each time. Rules are hard to remember. But I am lucky. I don't have to solve", "of course best done using using computer. We illustrate the process by finding all prime number below $n=169$. ___________________________________________________________________________________________________________________ We start with Figure 1 which is a listing of all integers below 169. Figure 1 – Positive Integers Below 169 Next, in Figure 2, we shade all multiples of 2 that are above 2. In other words we shade all even integers below 169 (excluding 2). Figure 2 – Multiples of 2 are Shaded The first number greater than 2 that has not been shaded is 3. In Figure 3 below, we shade all multiples of 3 that are greater than 3 and that have not been previously shaded (e.g. 9, 15, etc). Figure 3 – Multiples of 3 are Shaded The first number greater than 3 that has not been shaded is 5. In Figure 4 below, we shade all multiples of 5 that are greater than 5 and that have not been previously shaded. Figure 4 – Multiples of 5 are Shaded The first number greater than 5 that has not been shaded is 7. In Figure 5 below, we shade all multiples of 7 that are greater than 7 and that have not been previously shaded (e.g. 49, 77, 91, etc). Figure 5 – Multiples of 7 are Shaded The first number greater than 7", "by two. So for the range $[1,100]$ you get $\\frac{100-0}{2}=50$ odd integers. - The number of odds from $1$ to $100$ is the same as the number of evens from $2$ to $101$ which is the same as the number of evens from $2$ to $100$ which is the same as the number of integers from $1$ to $50$. This is $50$. With a bit of practice this will work flawlessly, and you will retain full control over the logic. It also generalizes nicely. How many numbers of the form $5k+2$ are there from $1$ to $200$? There are just as many as there are numbers divisible by $5$ from $4$ to $203$, which is the same as the number of numbers divisible by $5$ from $5$ to $200$, which is the same as the number of integers from $1$ to $40$, which is $40$.", "7×14=98 7×15=105 7×16=112 7×17=119 7×18=126 7×19=133 7×20=140 So the first 20 multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133 and 140. ## FAQs on Multiples of 7 Q: How do we find multiples of 7? Ans: Multiplying 7 with natural numbers we get the multiples of 7. Q. What are the first 5 multiples of 7? Ans: The first 5 multiples of 7 are 7, 14, 21, 28, and 35. Q. What are the first 10 multiples of 7? Ans: The first 10 multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, and 70 Q. Are factors of 7 and multiples of 7 the same? Ans: Yes, the factors of 7 and multiples of 7 are the same. Q. What is the least common multiple of 5 and 7 Ans: The least common multiple of 5 and 7 is 35.", "sets. Let's list out the possible sets. Since they are in lists of 4, we need multiples of 6 or 7. Let's list them out. 11, 12, 13, 14 12, 13, 14, 15 But if a student has 11 or 15, they would know. So that wouldn't work. The second set would be this: $\\{34,35,36,37\\}$ $\\{47,48,49,50\\}$ $\\{76,77,78,79\\}$ $\\{89,90,91,92\\}$ They couldn't know since the multiples of 7 is 35, and if you list it out, they wouldn't know about it since there are other sets such as 35, 36, 37, 38, and 33, 34, 35, 36. The people with these sets would say no, so the second set would WORK. The other sets are only 2 numbers, and since 2 numbers are sufficiently smaller than our 2nd set, we have concluded the answer is 37+50+79+92 = $\\boxed{258}$. ~Arcticturn ## Solution 2 (Illustrations) Note that $\\mathrm{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$ In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa.", "list of multiples of 3: \\begin{align} 3, 6, 9, 12, 15, 18, {\\bf{\\large 21}}, & \\\\ 24, 27, 30, 33, 36, 39, {\\bf{\\large 42}}, & \\\\ 45, 48, 51, 54, 57, 60, {\\bf{\\large 63}}, & \\\\ 66, 69, 72, 75, 78, 81, {\\bf{\\large 84}} \\end{align} we can see that the first four multiples of 7 that appear in the list of multiples of 3 are 21, 42, 63, and 84. 21 is $3\\times7$. We got 42 as a multiple of 7 because $42=6\\times 7$. We can rewrite it as follows: $$6\\times7 = (2\\times3)\\times7 = 2\\times(3\\times7) = 2\\times 21$$ This is the same as 2 groups of 21. The next one they have in common is 63, which came from $9\\times7$. As before, we can see that this is a multiple of 21: $$9\\times7 = (3\\times3)\\times7 = 3\\times(3\\times7) = 3\\times 21$$ In general, the multiples of 7 that appear in the list of multiples of 3 are also multiples of 21, and these happen each 7th multiple of 3 because each seven groups of 3 make a multiple of 7. Solution: 2 arithmetic 1. The first ten multiples of 3 are listed below: $$3, 6, 9, 12, 15, 18, 21, 24, 27, 30.$$ 2. The multiples of 6 in the list are highlighted in larger, bold face: $$3, {\\bf{\\large", "multiples of 33 are: 33, 66, 99, 132, 165, 198,. . . , 957, 990, 1023 So, we want the number of multiples of 33 from 132 to 990 inclusive Observe: 132 = (33)(4) 165 = (33)(5) 198 = (33)(6) . . . 957 = (33)(29) 990 = (33)(30) We can see that the number of multiples of 33 from 132 to 990 inclusive is the SAME as the number of integers from 4 to 30 inclusive. To determine the above, we can apply the following rule: the number of integers from x to y inclusive equals y - x + 1 So, the number of integers from 4 to 30 inclusive = 30 - 4 + 1 = 27 Answer: B Cheers, Brent _________________ Test confidently with gmatprepnow.com GMAT Club Legend Joined: 11 Sep 2015 Posts: 4542 Location: Canada GMAT 1: 770 Q49 V46 Re: How many multiples of 33 lie between 101 and 1,000, inclusive? [#permalink] ### Show Tags 20 Sep 2018, 05:23 Top Contributor 2 MathRevolution wrote: [Math Revolution GMAT math practice question] How many multiples of $$33$$ lie between $$101$$ and $$1,000,$$ inclusive? $$A. 24$$ $$B. 27$$ $$C. 33$$ $$D. 36$$ $$E. 48$$ Another approach is to apply the following rule: If x and y are multiples of k, then the number of multiples", "counted as multiples of 6. • All multiples of 56 (7 × 8), but don't double-count the ones that are also multiples of 6 or 20. If I did the arithmetic correctly, this brings us up to 22/105. • All multiples of 72 (8 × 9) or 90 (9 × 10), but these are already multiples of 6. • All multiples of 110 (10 × 11), being careful not to double-count multiples of 6, 20, or 56. We're now at 491/2310. Continue the pattern to get a lower bound on the probability. I bet it converges to something, but I haven't bothered to compute what. • Don't forget that 1/2 gives you the other bound, for the same reason as the original answer – Eric Dec 14 '18 at 8:24 • True: The product of two consecutive integers is always even, so no odd number can have the property we're looking for. – Dan Dec 14 '18 at 13:29 • Continuing until 190×191, you get a density of approximately 0.2215. It gets exponentially harder to calculate as you go along, and doesn't converge amazingly fast. (Exact fraction $30692078612850182537626772507400566302461540183482511947993/138565669082349191587117161961128738469412352721071780196786$). – Veedrac Dec 14 '18 at 18:00 • @Dan I doubt it. The 0.22194814 I measured manually has fairly low error (I'd guess 4-5 s.f.). Using an approximate version of", "# Problem of the Week Problem D and Solution Missing the Fives II ## Problem Bobbi lists the positive integers, in order, excluding all multiples of 5. Her resulting list is $1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, \\ldots$ If the $$n$$th integer in Bobbi’s list is $$2023$$, what is the value of $$n$$? ## Solution Solution 1 Note that $$2023$$ is two integers before $$2025$$, which is a multiple of $$5$$. Beginning at $$1$$, $$2025$$ is the $$405$$th multiple of $$5$$, since $$\\frac{2025}{5} = 405$$. That is, the integers from $$1$$ to $$2025$$ contain $$405$$ groups of $$5$$ integers. Each of these $$405$$ groups contain one integer that is a multiple of $$5$$, and so Bobbi leaves out $$406$$ integers (including $$2024$$) in the list of all integers from $$1$$ to $$2025$$. If the $$n$$th integer in Bobbi’s list is $$2023$$, then $$n = 2025 - 406 = 1619$$. Solution 2 Note that $$2023$$ is two integers before $$2025$$, which is a multiple of $$5$$. Beginning at $$1$$, $$2025$$ is the $$405$$th multiple of $$5$$, since $$\\frac{2025}{5} = 405$$. That is, the integers from $$1$$ to $$2025$$ contain $$405$$ groups of $$5$$ integers. In Bobbi’s list of integers, she leaves out the integers that are multiples of $$5$$, and so", "multiple of two or more numbers. ### Worked example 2.7: Common multiples Find three common multiples of 5 and 6. 1. Step 1: List the first ten multiples of each number. The first ten multiples of 5 are: $5;\\space10;\\space 15;\\space 20;\\space 25;\\space 30;\\space 35;\\space 40;\\space 45;\\space 50$ The first ten multiples of 6 are: $6;\\space 12;\\space 18;\\space 24;\\space 30;\\space 36;\\space 42;\\space 48;\\space 54;\\space 60$ 2. Step 2: Identify the numbers that appear in both lists. The number 30 is the only number that appears in both lists. 3. Step 3: Expand the lists until you find another number that appear in both lists. Multiples of 5: $5;\\space 10;\\space 15;\\space 20;\\space 25;\\space 30;\\space 35;\\space 40;\\space 45;\\space 50;\\space 55;\\space 60$ Multiples of 6: $6;\\space 12;\\space 18;\\space 24;\\space 30;\\space 36;\\space 42;\\space 48;\\space 54;\\space 60$ The number 60 also appears in both lists. 4. Step 4: Repeat Step 3 until you find the required number of common multiples. Multiples of 5: $5;\\space 10;\\space 15;\\space 20;\\space 25;\\space 30;\\space 35;\\space 40;\\space 45;\\space 50;\\space 55;\\space 60;\\space 65;\\space 70;\\space 75;\\space 80;\\space 85;\\space 90$ Multiples of 6: $6;\\space 12;\\space 18;\\space 24;\\space 30;\\space 36;\\space 42;\\space 48;\\space 54;\\space 60;\\space 66;\\space 72;\\space 78;\\space 84;\\space 90$ The number 90 also appears in both lists. 5. Step 5: Write down the common multiples. Three common multiples of 5 and 6 are: $30;\\space", "number of integers. If a and b are integers then for every multiplication b = a × n we can find a bigger multiple b = a × (n+1). ## Examples of Multiples In this section we give you a few examples of multiples. We use the 3 and 7 which both have the multiple 21. If we divide 21 by 3 we obtain 7 and when we divide 21 by 7 we get 3. ## Examples of Submultiples We also want to give a few examples of submultiples for which we use 49 and 777, which, despite being a big number has relatively few submultiples. The submultiples of 49 are: 1, 7, 49 The submultiples of 777 are: 1, 3, 7, 21, 37, 111, 259, 777 ## Multiples Exercises As a small exercise try to find the multiples of 9,12, and 15: The multiples of 9 are 0, 9, 18, 27, 36, 45 ... The multiples of 12 are 0, 12, 24, 36, 48, 60 ... The multiples of 15 are 0, 15, 30, 45, 60, 75 ... ## Submultiples Exercises Try to find the submultiples of 113: Answer: As 113 is prime you will only find two submultiples. Use the number 255 to figure out how many submultiples it has. Answer: By counting the entries in", "subtracting 47 once. Similar questions to practice: what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html what-is-the-total-number-of-positive-integers-that-are-less-128104.html how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html how-many-even-integers-n-where-100-n-200-are-divisib-103779.html Hope it helps. Hi Bunuel, Is there a shorter way around this question?This approach took me 5 minutes! Manager Joined: 20 Dec 2013 Posts: 231 Location: India Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 04 Apr 2014, 20:59 Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ Intern Joined: 26 Mar 2014 Posts: 4 Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 16 Apr 2014, 22:53 2 AKG1593 wrote: Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those", "the sets. Let's list out the possible sets. Since they are in lists of 4, we need multiples of 6 or 7. Let's list them out. 11, 12, 13, 14 12, 13, 14, 15 But if a student has 11 or 15, they would know. So that wouldn't work. The second set would be this: $\\{34,35,36,37\\}$ $\\{47,48,49,50\\}$ $\\{76,77,78,79\\}$ $\\{89,90,91,92\\}$ They couldn't know since the multiples of 7 is 35, and if you list it out, they wouldn't know about it since there are other sets such as 35, 36, 37, 38, and 33, 34, 35, 36. The people with these sets would say no, so the second set would WORK. The other sets are only 2 numbers, and since 2 numbers are sufficiently smaller than our 2nd set, we have concluded the answer is 37+50+79+92 = $\\boxed{258}$. ~Arcticturn ## Solution 2 (Illustrations) Note that $\\text{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$ In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S.$ By observations,", "$$3$$, $$N \\times 3 = 9$$, $$9$$ is a multiple of $$3$$, etc. Multiples of 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... Multiples of 2 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... Multiples of 3 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ... Multiples of 4 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ... Multiples of 5 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ... Multiples of 6 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ... Multiples of 7 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ... Multiples of 8 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ... Multiples of 9 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ... Multiples of 10 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ... Multiples of 11 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, ... Multiples of 12 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ... Multiples of 13 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, ... Multiples of 14 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, ... Multiples of 15 15, 30, 45, 60, 75, 90," ]

[ "ways Total ways 168+147=315 ways The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41 Since $c<9$, we can have the following viable combinations for $b\\times\\ c\\ =96$ (given our objective is to minimize the sum): $48\\times\\ 2$ ; $32\\times3$ ; $24\\times\\ 4$ ; $16\\times6$ ; $12\\times8$ Similarly, we can factorize $a\\times\\ b\\ = 432$ into its factors. On close observation, we notice that $18\\times24\\ and\\ 24\\ \\times\\ 4\\$ corresponding to $a\\times b\\ and\\ b\\times\\ c\\$ respectively together render us with the least value of the sum of $a+b\\ +\\ c\\ \\ =\\ 18+24+4\\ =46$ Hence, Option D is the correct answer. After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19 Given, The average of the four prime numbers = 35. a +", "digit is fixed as zero In remaining 9 places Lets choose 5 places and place 1s there Number of ways are $$\\displaystyle{{C}_{{5}}^{{9}}}$$ 126 Step 3 How many integers in the range 1 through 140 are integer multiples of 2, 5, or 7 There are 70 multiples of 2 23 multiples of 3 left we have 9 multiples of 5 that have not yet been included. Total is 102", "there are two cases possible Case 1: When 7 is at the left extreme In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9) So total ways 3(8)(7)= 168 Case 2: When 7 is not at the extremes Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left) Hence in total 3(7)(7)=147 ways Total ways 168+147=315 ways $ab\\ =\\ 4^{2017}=2^{4034}$ The total number of factors = 4035. out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017. And since the given number is a perfect square we have one set of two equal factors. .’. many pairs(a, b) of positive integers are there such that $a\\leq b$ and $ab=4^{2017}$ = 2018. The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60", "and 96, inclusive? 21 22 23 24 25 My answer was 21 and that's incorrect. $$# \\ of \\ multiples \\ of \\ x \\ in \\ the \\ range = \\frac{Last \\ multiple \\ of \\ x \\ in \\ the \\ range \\ - \\ First \\ multiple \\ of \\ x \\ in \\ the \\ range}{x}+1$$. In the original case: $$\\frac{96-12}{4}+1=22$$. If the question were: how many multiples of 5 are there between -7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5; $$\\frac{30-(-5)}{5}+1=8$$. OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21; $$\\frac{-7-(-21)}{7}+1=3$$. Hope it helps. I have one concern regarding this How many multiples and how many integers? How many of the three-digit numbers are divisible by 7? Generally we do--->999-100= 899 so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1 but the answer is only 128,How is that possible????? As per our knowledge we know that ->If the question says both inclusive we add +1 at the last, For eg How many of the three-digit", "than 100 are neither multiples of 2 or [#permalink] ### Show Tags 15 Apr 2017, 11:44 devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Set comprises the integers 1-99 inclusive. Number of items in set=99. Number of integers that are a multiple of 2: [(98-2)/2)]+1=49 Number of integers that are a multiple of 3: [(99-3)/2)]+1=33. Of these, 16 are even and are therefore counted in the number of multiples of 20 (49). So there are 17 additional integers to add that are multiples of 3 but not multiples of 2. 99-(49+17)=99-66=33 Agree? Intern Joined: 12 Dec 2016 Posts: 7 Re: How many positive integers less than 100 are neither multiples of 2 or [#permalink] ### Show Tags 18 Apr 2017, 05:49 how i approached this problem - we have to eliminate all the multiples of 2 & 3... Therefore, within 100 it's all about the prime numbers and 1. There are 25 prime numbers - 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 Out of which 2 & 3 are not included so - 23. Also, there is 1. So 24. Next the multiples of the prime numbers", "14:02 • It does require that the start and end points are adjusted to exact multiples, the \"discard the remainder\" approach suggested by Jacob is not reliable. Consider: How many multiples of 7 are in the interval [6,8]? Now, how many in the interval [1,13]? Or [8,10]? – Ben Voigt Sep 29 '15 at 18:26 • I should admit I had never heard it described as \"fenceposts\" before either (that I remember). I only used the term to tie into porglezomp's comment on the OP. – Paul Sinclair Sep 29 '15 at 22:12 There are just as many multiples of $3$ between $252$ and $348$ (inclusive) as there are between $0$ and $348-252$, that is, $0$ and $96$. Now what one does is a matter of taste. The number between $0$ and $96$ is the same as the number from $3$ to $99$, which is the same as the number of integers between $1$ and $33$. Or else there are $32$ between $3$ and $96$. But don't forget about $0$. Or else there are just as many as there are integers from $0$ to $32$. Again, don't forget about $0$. There are $33$ of these. Remark: I prefer to figure it out each time. Rules are hard to remember. But I am lucky. I don't have to solve", "question, we must count the number of 5's in the product of multiples of 3 between 1 and 100 5 will occur for mulitple of 15 LCM(3,5) between 1 and 100 => 6 multiples of 15 => prod has $$5^6$$ 5 will also occur for multiple of 75 LCM(3,25) between 1 and 100 => 1 multiple of 75 => prod has one more 5 so totally, product has $$5^7$$ = ($$5^6 * 5$$). Ofcourse, number of 2s will be definitely greater than number of 5s, but to form a 10, we need 2 and 5, so number of 10s is restricted by number of 5s so maximum m, for which, prod / $$10^m$$ is an integer is 7 If the question had been, prod of multiples of 3 between 1 and 200 then, 5 will occur for mulitple of 15 LCM(3,5) between 1 and 200 => 13 multiples of 15 => product has $$5^{13}$$ 5 will also occur for multiple of 75 LCM(3,25) between 1 and 200 => 2 multiples of 75 => product has another $$5^2$$ 5 will also occur for multiple of 375 LCM(3,125) between 1 and 200 => has 0 multiples of 375 5 will also occur for multiple of 1875 LCM(3,625) between 1 and 200 => has 0 multiples of 1875 . . goes", "we must count the number of 5's in the product of multiples of 3 between 1 and 100 5 will occur for mulitple of 15 LCM(3,5) between 1 and 100 => 6 multiples of 15 => prod has $$5^6$$ 5 will also occur for multiple of 75 LCM(3,25) between 1 and 100 => 1 multiple of 75 => prod has one more 5 so totally, product has $$5^7$$ = ($$5^6 * 5$$). Ofcourse, number of 2s will be definitely greater than number of 5s, but to form a 10, we need 2 and 5, so number of 10s is restricted by number of 5s so maximum m, for which, prod / $$10^m$$ is an integer is 7 If the question had been, prod of multiples of 3 between 1 and 200 then, 5 will occur for mulitple of 15 LCM(3,5) between 1 and 200 => 13 multiples of 15 => product has $$5^{13}$$ 5 will also occur for multiple of 75 LCM(3,25) between 1 and 200 => 2 multiples of 75 => product has another $$5^2$$ 5 will also occur for multiple of 375 LCM(3,125) between 1 and 200 => has 0 multiples of 375 5 will also occur for multiple of 1875 LCM(3,625) between 1 and 200 => has 0 multiples of 1875 . . goes on", "wrote: How many positive integers less than 1000 are multiples of 5 but NOT of 4 or 7? (A) 114 (B) 121 (C) 122 (D) 129 (E) 136 995/5=all multiples of 5=199 980/20=all multiples of 5 and 4=49 980/35=all multiples of 5 and 7=28 980/140=duplicate multiples of 5, 4 and 7=7 199-49-28+7=129 D. Manager Joined: 20 Jan 2017 Posts: 62 Location: United States (NY) GMAT 1: 610 Q41 V34 Followers: 1 Kudos [?]: 2 [0], given: 14 Re: How many positive integers less than 1000 are multiples of 5 [#permalink] ### Show Tags 20 Jan 2017, 17:37 1)Multiples of 5: (995-5)/5+1=198+1=199 2)Multiples of 5&4: 20: (980-20)/20+1=960/20+1=48+1=49 3)Multiples of 5&7: 35: (980-35)/35+1=945/35+1=27+1=28 4)Multiples of 5&4&7: 140: (980-140)/140+1=840/140+1=6+1=7 5)Multiples of 5 but not 4 and not 7: 199-(49-7)-(28-7)-7=129 Re: How many positive integers less than 1000 are multiples of 5 [#permalink] 20 Jan 2017, 17:37 Similar topics Replies Last post Similar Topics: 2 How many positive integers less than 100 are multiples of a prime 4 15 Oct 2016, 18:16 43 How many positive integers less than 100 are neither multiples of 2 or 20 02 Apr 2016, 14:01 7 How many positive integers less than 50 are multiples of 4 but NOT mul 11 20 Dec 2015, 04:42 18 How many positive integers less than 250 are multiple of", "W Common multiples of 2 and 3 or multiples of 6: first multiple of 6 will be 54 and last multiple will be 96 so the number of multiples of 6 between 54 and 99 inclusive will be 96=54+(n-1)6 or 42=(n-1)6 or 7=n-1 or n=8 So, number of integers in set W will be 42-8=34 Board of Directors Joined: 17 Jul 2014 Posts: 2598 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Re: If W is the set of all the integers between 49 and 99, inclusive, that [#permalink] Show Tags 14 Dec 2016, 15:09 reto wrote: If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers? A. 26 B. 32 C. 33 D. 34 E. 35 my approach between 49 and 99, there are 50 numbers, 25 of which are even. so definitely >25 or 26. A is out. now, multiples of 3: 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 -> 54, 60, 66, 72, 78, 84, 90, 96 are already counted in the 25 from above. remains 51, 57, 63, 69, 75,", "7×14=98 7×15=105 7×16=112 7×17=119 7×18=126 7×19=133 7×20=140 So the first 20 multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133 and 140. ## FAQs on Multiples of 7 Q: How do we find multiples of 7? Ans: Multiplying 7 with natural numbers we get the multiples of 7. Q. What are the first 5 multiples of 7? Ans: The first 5 multiples of 7 are 7, 14, 21, 28, and 35. Q. What are the first 10 multiples of 7? Ans: The first 10 multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, and 70 Q. Are factors of 7 and multiples of 7 the same? Ans: Yes, the factors of 7 and multiples of 7 are the same. Q. What is the least common multiple of 5 and 7 Ans: The least common multiple of 5 and 7 is 35.", "of course best done using using computer. We illustrate the process by finding all prime number below $n=169$. ___________________________________________________________________________________________________________________ We start with Figure 1 which is a listing of all integers below 169. Figure 1 – Positive Integers Below 169 Next, in Figure 2, we shade all multiples of 2 that are above 2. In other words we shade all even integers below 169 (excluding 2). Figure 2 – Multiples of 2 are Shaded The first number greater than 2 that has not been shaded is 3. In Figure 3 below, we shade all multiples of 3 that are greater than 3 and that have not been previously shaded (e.g. 9, 15, etc). Figure 3 – Multiples of 3 are Shaded The first number greater than 3 that has not been shaded is 5. In Figure 4 below, we shade all multiples of 5 that are greater than 5 and that have not been previously shaded. Figure 4 – Multiples of 5 are Shaded The first number greater than 5 that has not been shaded is 7. In Figure 5 below, we shade all multiples of 7 that are greater than 7 and that have not been previously shaded (e.g. 49, 77, 91, etc). Figure 5 – Multiples of 7 are Shaded The first number greater than 7", "of those, the 75 or the 125 would only appear together with the 25. Because the least common multiple would be 75 or 125 (25 divides either). If you wanted all possible pairs to have dual conjunctions, you would need a period that factored into more than two prime factors ($$3^2\\cdot 5^3 = 1125$$). Essentially you'd want each comet's period to include its own prime factor. With 1125, we're basically limited to some combination of threes and fives. $$45 = 3^2\\cdot 5$$ $$75 = 3 \\cdot 5^2$$ $$125 = 5^3$$ And since you already chose 45, we can't use 3, 5, 9, or 15 without 45 always appearing with whichever of those. If you allowed something other than 45, we could have 9, 75, 125 or similar. To get the triple conjunction at 125, we need a multiple of 9 and a multiple of 125. Because 9 and 125 have 1125 as the least common multiple. • This is perfect! The comets are a regular enough occurrence to be a mild problem when they show up individually while the dual conjunctions would pose more of a problem, yet do not occur with as much regularity. Then you get the triple whammy, which is unique in that it is the only time where 45 and 125 get to be", "17:49 1 This post was BOOKMARKED Bunuel wrote: marcusaurelius wrote: How many multiples of 4 are there between 12 and 96, inclusive? 21 22 23 24 25 My answer was 21 and that's incorrect. $$# \\ of \\ multiples \\ of \\ x \\ in \\ the \\ range = \\frac{Last \\ multiple \\ of \\ x \\ in \\ the \\ range \\ - \\ First \\ multiple \\ of \\ x \\ in \\ the \\ range}{x}+1$$. In the original case: $$\\frac{96-12}{4}+1=22$$. If the question were: how many multiples of 5 are there between -7 and 35, not inclusive? Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5; $$\\frac{30-(-5)}{5}+1=8$$. OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21; $$\\frac{-7-(-21)}{7}+1=3$$. Hope it helps. I have one concern regarding this How many multiples and how many integers? How many of the three-digit numbers are divisible by 7? Generally we do--->999-100= 899 so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1 but the answer is only 128,How is that possible????? As per our knowledge we know that", "this purpose. If I understand your question, the number of unstable integers that do not exceed $10^k$ for $k=1,2,3,4,5,6,7$ is $0,1,36,582,6933,74067,761605$. Jun 7, 2019 at 21:00 I'm not seeing a smart way (e.g. generating functions) to count these. So I will suggest a way to estimate (or overcount) them. As a nice example, let's start with 30, which is stable. Certain multiples of 30 are unstable. Let q be coprime to 30. Then (3q)(30)^n is unstable. (I assume n is positive.) Of all the numbers less than N, this gives about 8N/2700 which are multiples of 90, 8N/81000 which are multiples of 2700, and so on. So just from looking at 30, one gets about 8N/2610 unstable numbers whose exponent of three is one more than the exponents of both two and five. There is a similar count (about 3/10 as many) for certain unstable multiples of 300. Note that while this nicely partitions some of the multiples of 90 or 300 which are unstable, it does not capture all unstable multiples of 30. Further, some unstable multiples of 42 are included in this count. Nevertheless, this captures many of the unstable multiples of 30, and can be easily estimated. Now one can do something similar with other prime triplets, and there is a good chance of counting", "by two. So for the range $[1,100]$ you get $\\frac{100-0}{2}=50$ odd integers. - The number of odds from $1$ to $100$ is the same as the number of evens from $2$ to $101$ which is the same as the number of evens from $2$ to $100$ which is the same as the number of integers from $1$ to $50$. This is $50$. With a bit of practice this will work flawlessly, and you will retain full control over the logic. It also generalizes nicely. How many numbers of the form $5k+2$ are there from $1$ to $200$? There are just as many as there are numbers divisible by $5$ from $4$ to $203$, which is the same as the number of numbers divisible by $5$ from $5$ to $200$, which is the same as the number of integers from $1$ to $40$, which is $40$.", "of multiples of 11 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(990-11)}{11}+1=90$$; # of multiples of 35 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(980-35)}{35}+1=28$$; # of multiples of both 11 and 35 is 2 ($$11*35=385$$ and 770); So, # of multiples of 11 or 35 in the given range is $$90+28-2=116$$. Thus numbers which are not divisible by either of them is $$1000-116=884$$. Hi Bunuel, I had come across a similar question \"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?\". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer. But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770). There is one critical aspect of your question (\"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but", "subtracting 47 once. Similar questions to practice: what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html what-is-the-total-number-of-positive-integers-that-are-less-128104.html how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html how-many-even-integers-n-where-100-n-200-are-divisib-103779.html Hope it helps. Hi Bunuel, Is there a shorter way around this question?This approach took me 5 minutes! Manager Joined: 20 Dec 2013 Posts: 231 Location: India Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 04 Apr 2014, 20:59 Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ Intern Joined: 26 Mar 2014 Posts: 4 Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 16 Apr 2014, 22:53 2 AKG1593 wrote: Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those", "wrote: devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Hi, To answer this Q we require to know 1) multiples of 2 till 100 $$= \\frac{100}{2} = 50$$ 2) Multiples of 3 till 100 = $$\\frac{100}{3} = 33.33= 33$$ add the two $$50+33=83$$ ; subtract common terms that are multiple of both 2 and 3.. LCM of 2 and 3 = 6 Multiples of 6 till 100 = $$\\frac{100}{6} = 16.66 = 16$$ so total multiples of 2 and 3 = 83-16 = 67 ans = $$100-67 = 33$$ D Hi Chetan, I searched for the formula of no. of multiples in a range and I got: No. of multiples of x in the range = ((Last multiple of x in the range - First multiple of x in the range)/x)+1 But you have directly applied 100/16..100/3. Thanks. GMAT Club Legend Joined: 11 Sep 2015 Posts: 4605 GMAT 1: 770 Q49 V46 Re: How many positive integers less than 100 are neither multiples of 2 or [#permalink] ### Show Tags 01 Dec 2019, 12:47 Top Contributor devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Multiples of 2: 2, 4, 6, ..., 96,", "7 multiples Multiples of 18 = 108, 126, 144, .... 198 = 6 multiples Multiples of both 14 and 18 = Multiples of 126 in the range 100 to 200 inclusive = 1 only (which is 126) So total numbers in range 100 to 200 inclusive which are divisible by either 14 or 18 or both = 7 + 6 - 1 = 12 Senior Manager Joined: 27 Dec 2016 Posts: 309 Re: If 100<= x <=200, then how many Integer values of x are even multiples [#permalink] ### Show Tags 27 May 2017, 15:50 Not sure if my way is correct, but this is how I approached this problem. Between 100 to 200, we can count how many multiples of 7 and 9 there are. # of multiples of 7 between 100 to 200: 105 to 196- ((196-105)/7)+1)= 14 multiples total # of multiples of 9 between 100 to 200: 108 to 198- ((198-108)/9)+1)= 11 multiples total Now, we can count the even multiples. 14 has 7 even multiples and 11 and 5 even multiples. S0, 7+5= 12, Hence, D. Please let me know if my way is correct. Thanks! Intern Joined: 18 May 2017 Posts: 13 WE: Corporate Finance (Health Care) If 100<= x <=200, then how many Integer values of x are even multiples [#permalink]" ]

[ "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "# Multifactorial and divisors Fill in the blank. $$N$$ has $$3$$ prime factors. $$N^2$$ has $$7!!$$ positive divisors. $$N^3$$ has $$10!!!$$ positive divisors. $$\\text{____}$$ has $$13!!!!$$ positive divisors. Note: You might want to read up the definition of multifactorials first. ×", "Find the number of positive divisors of $10!$. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "divisor $n/d$, which is different from $d$ unless $d=\\sqrt{n}$. Since $n$ ends in $0$ it has $2$ and $5$ and possibly others as prime factors. But $15$ gives very few possibilities: just two $2$'s and four $5$'s, or the other way around. The number of positive factors is 15, which is 1*3*5. So, there are no more than 3 prime factors of $n$. Two of them are 2 and 5. How to prove that there's no third divisor? – mbaitoff Jul 1 '12 at 6:16 Baaw, got it. $1=(a_i+1)$. Thus, $a_i=0$. – mbaitoff Jul 1 '12 at 6:17", "Free Version Moderate # Factors of 9 Million COMBIN-5GZGKK How many positive factors of 9000000 are divisible by 120? A $24$ B $30$ C $36$ D $40$ E $48$ F none of the above", "Let $D(x)$ denote the number of positive divisors of $x$. Determine the number of values of $x$ for which $x = D(x)^4$.", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×", "## Algebra 1 $-4(-11)=44$ Since there are an even number of negative factors, the product is positive.", "# Funky Factorials What is the value of $$3!!$$ where $$!!$$ denotes the double factorial notation. ×", "factor A proper factor of a positive integer is a factor of other than 1 or (Derbyshire 2004, p. 32). For example, 2 and 3 are positive proper factors of 6, but 1 and 6 are not.Compare with the term proper divisor, which means a factor of other than (but including 1). Proper divisor A positive proper divisor is a positive divisor of a number , excluding itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not. The number of proper divisors of is therefore given bywhere is the divisor function. For , 2, ..., is therefore given by 0, 1, 1, 2, 1, 3, 1, 3, 2, 3, ... (OEIS A032741). The largest proper divisors of , 3, ... are 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, ... (OEIS A032742).The term \"proper divisor\" is sometimes used to include negative integer divisors of a number (excluding ). Using this definition, , , , 1, 2, and 3 are the proper divisors of 6, while and 6 are the improper divisors.To make matters even more confusing, the proper divisor is often defined so that and 1 are also excluded. Using this alternative definition, the proper divisors of 6 would then be , , 2, and 3, and the", "# The only positive divisor of both $a$ and $a + 1$ is $1$ Prove that if $a \\in \\mathbb Z$ then the only positive divisor of both $a$ and $a + 1$ is $1$. When I saw this statement I didn't understand it. The only way that I can see it being true is if a is a negative number but since a ∈ Z a could also be positive and thus it could have more than one positive divisor. If that's the case would I have to disprove this instead of proving? And what would be the best way to do that? • What common divisors do 2 and 3 have? Or 3 and 4 or 9 and 10? – Paul Sundheim Sep 22 '14 at 0:18 • I think it's asking for a single number that divides both $a$ and $a+1$. – Trurl Sep 22 '14 at 0:19 • Please choose descriptive titles. The original title, Prove or Disprove??? carried virtually no information of what the problem is about. – user147263 Sep 22 '14 at 1:11 You seem to have misunderstood the question. As far as I can tell you are thinking about divisors of $a$ and $a+1$ separately, but the question is asking you to show that the only positive number which is a", "# Factorial divisors True or false: For any positive integer $$n$$, there exists another positive integer $$m$$, such that $$m!$$ has exactly $$n$$ trailing number of zeros. For example, if $$n=3$$, one value for $$m$$ would be 16 as $$16! = 20922789888\\underline{000}$$. ×", "# Squares-Cubes-Factors Let $$A$$ be the least number such that $$10A$$ is a perfect square and $$35A$$ is a perfect cube. Then the number of positive divisors of $$A$$ is? ×", "of divisors is = (e1+1)(e2+1)(e3+1) ... (ek+1). where ek denotes powers of prime factors of the number. For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors. Manager Joined: 23 Jul 2008 Posts: 203 Followers: 1 Kudos [?]: 76 [0], given: 0 ### Show Tags 18 Nov 2008, 12:33 prasun84 wrote: the number of divisors is = (e1+1)(e2+1)(e3+1) ... (ek+1). where ek denotes powers of prime factors of the number. For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors. Thanks a lot. Re: Prime factors [#permalink] 18 Nov 2008, 12:33 Display posts from previous: Sort by", "## Algebra 2 (1st Edition) $4$ Recall: The quotient of two numbers with the same signs is positive. Thus, $-28 \\div (-7) = 4$", "two of them is at least $2$. The answer $12$ is INVALID because divisors are $[1, 2, 3, 4, 6, 12]$. And the difference between, for example, divisors $2$ and $3$ is less than $d=2$.", "Posted by absvalue The question is how many positive divisors do each of the following numbers have: 4) $2^{n}$ I think the answer is n+1 You should be able to count them, try n = 1,2,3,4, and 5. You'll see the pattern. Try the same for $10^{n}$ • Sep 9th 2009, 05:04 PM absvalue Quote: Originally Posted by Plato For #4, there are $(n+1)$ factors. Why? For #5, there are $(n+1)(n+1)$ factors. Why? For #6, how many positive integers divide zero? Thanks, I think I see the pattern. That's beautiful. For #4 there are n + 1 factors because: $2^{n} = (n + 1)$ $2^{3} = (3 + 1) = 4$ (there are 4 positive divisors of 8) $2^{4} = (4 + 1) = 4$ (there are 5 positive divisors of 16) For #5 there are $(n + 1)^2$ factors because: $10^{n} = (n + 1)^2$ $10^2 = (2 + 1)^2 = 9$ (there are 9 positive divisors of 100) $10^3 = (3 + 1)^2 = 16$ (there are 16 positive divisors of 1000) I understand the pattern above. But what if the question was a number other than $2^{n}$ or $10^{n}$? Is there a way to find the number of positive divisors for, say, $8^{n}$? For #6: Zero divided by any number (except zero, because that is", "# Problem #9 If $$12p+1, p^{2}$$ have 3 (positive) factors, find the sum of all possible positive integer values of $$p$$. Factors are also called divisors, and include 1 and itself. e.g. 7 has 2 factors. This problem is part of the set Easy Problems ×", "# Proportional factors? A positive integer $n$ has $d$ positive divisors. After doubling $n,$ you find that $2n$ has $2d$ positive divisors. Is it possible that after tripling $n,$ you will find that $3n$ has $3d$ positive divisors? ×", "# Factorial factors Number Theory Level 4 How many positive integers are divisors of $$21!$$ but are not divisors of $$20!$$? ×" ]

[ "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "we have in total $3 \\times 2 \\times 3 =18$ such positive divisors. Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "# Multifactorial and divisors Fill in the blank. $$N$$ has $$3$$ prime factors. $$N^2$$ has $$7!!$$ positive divisors. $$N^3$$ has $$10!!!$$ positive divisors. $$\\text{____}$$ has $$13!!!!$$ positive divisors. Note: You might want to read up the definition of multifactorials first. ×", "two of them is at least $2$. The answer $12$ is INVALID because divisors are $[1, 2, 3, 4, 6, 12]$. And the difference between, for example, divisors $2$ and $3$ is less than $d=2$.", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×", "• across Funsie: How many positive divisors does$-340000000000000000000000000000000000000$have? :P This is actually doable. Step 1: Find the prime-power factorization of that number. Step 2: Use the $$\\tau$$ function. The answer is $$2964$$. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "divisible by $8$ is $12$.", "$a$ is 5, and they are all equally likely, so the probability that the divisor is divisible by 4 is $\\boxed{3/5}$. Note that we basically ignored what was happening with the primes other $2$. You can be reassured that this is ok through an alternate solution: By the factorization above, we see there are $(4+1)(2+1)(1+1)=5 \\cdot 3 \\cdot 2$ total divisors. If the divisor is a multiple of $4$, then the value of $a$ above must be $2$, $3$, or $4$. Combining this with the possibilities for the other primes besides 2, we have $3 \\cdot 3 \\cdot 2$ divisors that are multiples of $4$. When we take the ratio, the factors contributed from the primes $3$ and $5$ cancel out: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$.", "want to factor the negative, you have to remember that you are taking out a factor of $-1$. So that means $3(-2x^2 + 5x + 12) = -1\\cdot 3(2x^2 - 5x - 12)$ $= -3(2x^2 - 5x - 12)$. Do you see how when you take out a negative factor, all of the signs change?", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "Find the number of positive divisors of $10!$. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "60 views The ratio of two positive numbers is $2:3.$ If $10$ and the sum of the numbers are added to their product, square of $16$ is obtained. The greater number is : 1. $6$ 2. $9$ 3. $12$ 4. $18$ Assume that numbers are $2x,3x$. Now according to the question: $(2x \\times 3x) + 3x+2x+10=256$ $\\implies6x^2+5x-246=0\\implies x=6$ So the number are: $6\\times 2=12,6\\times 3=18$ Option (D) is correct. 4.7k points 1", "+0 # help 0 80 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 #1 +5088 +2 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$ . Jan 1, 2019", "the original number ($175$). If it was one and six, then we had an exponent of $5$ (since this is $6 - 1$) for our prime. If it was two and three, then we had exponents of $1$ and $2$ for our primes. Keep in mind that we don't know what these primes are yet. But if we take the smallest possible primes, we could have $2^5$ (for the case of one prime to the fifth power), $2^1 \\cdot 3^2$ or $2^2 \\cdot 3^1$ for the case of one to the first power, one to the second power). The smallest of these three numbers is $12$. Now for a more involved example, in hopes that you can generalize from here, incorporating the other answers: What is the least positive integer that has the same number of positive factors as 2048? Since $2048 = 2^{11}$, there are $11 + 1 = 12$ positive divisors/factors of $2048$. So we want the smallest positive integer $n$ that has $12$ positive divisors. But when we factor $12$, it gets a little more complicated! For $12 = 12 \\cdot 1$, we have $n =$ (some prime to the eleventh power, which we already know is) $2^{11}$. For $12 = 2 \\cdot 6$, we have $n = p^1 \\cdot q^5$, the smallest two being", "($2 \\times 2$). So for example in the box $6 \\times 6$, each factor of $6$ is divisble by $2$ and so the product $6 \\times 6$ is divisible by $4$ (even though neither factor of $6$ is divisible by $4$). Since $2$ and $3$ are prime numbers they cannot be written as a product of smaller whole numbers and so these isolated shaded boxes do not occur for multiples of $2$ and $3$.", "Free Version Moderate # Factors of 9 Million COMBIN-5GZGKK How many positive factors of 9000000 are divisible by 120? A $24$ B $30$ C $36$ D $40$ E $48$ F none of the above", "# Definition:Solid Number/Side A example of a solid number is $12 = 2 \\times 2 \\times 3$. Its divisors $2$ and $3$ are its sides.", "the digits of $12$ is $1+2 = \\boxed{\\textbf{(B)}\\ 3}$, which is our answer. Note that this solution is rather fast with this problem because the numbers given in the question are low, this may not always be the case, however, when given higher numbers.", "divisor $n/d$, which is different from $d$ unless $d=\\sqrt{n}$. Since $n$ ends in $0$ it has $2$ and $5$ and possibly others as prime factors. But $15$ gives very few possibilities: just two $2$'s and four $5$'s, or the other way around. The number of positive factors is 15, which is 1*3*5. So, there are no more than 3 prime factors of $n$. Two of them are 2 and 5. How to prove that there's no third divisor? – mbaitoff Jul 1 '12 at 6:16 Baaw, got it. $1=(a_i+1)$. Thus, $a_i=0$. – mbaitoff Jul 1 '12 at 6:17" ]

[ "# Counting Positive Factors 12 has 6 positive divisors: 1, 2, 3, 4, 6, 12. Which of the two numbers has more divisors, $12\\times 6$ or $12\\times 9?$ ×", "we have in total $3 \\times 2 \\times 3 =18$ such positive divisors. Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "$a$ is 5, and they are all equally likely, so the probability that the divisor is divisible by 4 is $\\boxed{3/5}$. Note that we basically ignored what was happening with the primes other $2$. You can be reassured that this is ok through an alternate solution: By the factorization above, we see there are $(4+1)(2+1)(1+1)=5 \\cdot 3 \\cdot 2$ total divisors. If the divisor is a multiple of $4$, then the value of $a$ above must be $2$, $3$, or $4$. Combining this with the possibilities for the other primes besides 2, we have $3 \\cdot 3 \\cdot 2$ divisors that are multiples of $4$. When we take the ratio, the factors contributed from the primes $3$ and $5$ cancel out: $\\dfrac{3 \\cdot 3 \\cdot 2}{5 \\cdot 3 \\cdot 2} = \\boxed{\\dfrac35}$.", "two of them is at least $2$. The answer $12$ is INVALID because divisors are $[1, 2, 3, 4, 6, 12]$. And the difference between, for example, divisors $2$ and $3$ is less than $d=2$.", "# Multifactorial and divisors Fill in the blank. $$N$$ has $$3$$ prime factors. $$N^2$$ has $$7!!$$ positive divisors. $$N^3$$ has $$10!!!$$ positive divisors. $$\\text{____}$$ has $$13!!!!$$ positive divisors. Note: You might want to read up the definition of multifactorials first. ×", "• across Funsie: How many positive divisors does$-340000000000000000000000000000000000000$have? :P This is actually doable. Step 1: Find the prime-power factorization of that number. Step 2: Use the $$\\tau$$ function. The answer is $$2964$$. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# How many divisors? For a positive integer $x$, let $f(x)$ be the function which returns the number of distinct positive factors of $x$. If $p$ is a prime number, what is the minimum possible value of $f(75p^2)?$ This problem is posed by Michael T. Details and assumptions You may read up on Divisors of an Integer. The distinct positive factors of $x$ need not be prime. For example, $f(12) = 6$, since 12 has distinct positive factors of $1, 2, 3, 4, 6, 12$. ×", "## How To Find Divisors Of A Number What is the fastest way to find the divisors of a number Program to find divisor of a number in C++ with output.... For example, I have to find all positive divisors of $372$. The prime factorization of $372$ is $2^2 \\cdot 3 \\cdot 31$ Now, I wonder if there is a fast method to find all positive divisors of $372$. Divisor of a Number Program in C++ Studytonight 1/06/2009 · fastest method to find all divisors. GameOn. i need to find the all the divisors of a num(int) i am applying this approach you shouldn't use it to find primes, since on average it takes more iterations to find all the divisors of a number than to check if the number is prime. EDIT: n=12 (int)sqrt(12)=3 12%1==0, so 1 and 12, because 12/1==12 12%2==0, so 2 and 6, because 12/2==6 …... Definition: a superabundant number is a number that have more divisors than any number smaller than it. Example: $$12$$ is superabundant because it has 6 divisors: 1,2,3,4,6,12 and no other smaller number has at least 6 divisors. Divisors of a Number Calculator List - Online Software Tool For example, I have to find all positive divisors of $372$. The prime factorization of $372$ is $2^2 \\cdot 3 \\cdot", "divisor $n/d$, which is different from $d$ unless $d=\\sqrt{n}$. Since $n$ ends in $0$ it has $2$ and $5$ and possibly others as prime factors. But $15$ gives very few possibilities: just two $2$'s and four $5$'s, or the other way around. The number of positive factors is 15, which is 1*3*5. So, there are no more than 3 prime factors of $n$. Two of them are 2 and 5. How to prove that there's no third divisor? – mbaitoff Jul 1 '12 at 6:16 Baaw, got it. $1=(a_i+1)$. Thus, $a_i=0$. – mbaitoff Jul 1 '12 at 6:17", "Find the number of positive divisors of $10!$. Notation: $!$ denotes the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "is equal to ( 0 \\). Example: $$N = 10$$, $$\\sqrt{10} \\approx 3.1$$, $$1$$ and $$10$$ are always divisors, test $$2$$: $$10/2=5$$, so $$2$$ and $$5$$ are divisors of $$10$$, test $$3$$, $$10/3 = 3 + 1/3$$, so $$3$$ is not a divisor of $$10$$. Another method calculates the prime factors decomposition of $$N$$ and by combination of them, get all divisors. Example: $$10 = 2 \\times 5$$, divisors are then $$1$$, $$2$$, $$5$$, and $$2 \\times 5 = 10$$ Negative divisors also exist, but they are the same as positive divisors (with the sign close), so they are ignored. ### What is the list of divisors from 1 to 100? NumberList of Divisors Divisor of 11 Divisors of 21,2 Divisors of 31,3 Divisors of 41,2,4 Divisors of 51,5 Divisors of 61,2,3,6 Divisors of 71,7 Divisors of 81,2,4,8 Divisors of 91,3,9 Divisors of 101,2,5,10 Divisors of 111,11 Divisors of 121,2,3,4,6,12 Divisors of 131,13 Divisors of 141,2,7,14 Divisors of 151,3,5,15 Divisors of 161,2,4,8,16 Divisors of 171,17 Divisors of 181,2,3,6,9,18 Divisors of 191,19 Divisors of 201,2,4,5,10,20 Divisors of 211,3,7,21 Divisors of 221,2,11,22 Divisors of 231,23 Divisors of 241,2,3,4,6,8,12,24 Divisors of 251,5,25 Divisors of 261,2,13,26 Divisors of 271,3,9,27 Divisors of 281,2,4,7,14,28 Divisors of 291,29 Divisors of 301,2,3,5,6,10,15,30 Divisors of 311,31 Divisors of 321,2,4,8,16,32 Divisors of 331,3,11,33 Divisors of 341,2,17,34 Divisors of", "# Positive divisors • Jan 25th 2010, 06:18 AM davidman Positive divisors There are $x$ number of positive divisors of $72$, $y$ number of positive divisors of $900$. The positive divisors of $900$ that are not divisors of $72$ are $z$. Hoping there's a formula and you're not actually supposed to go over all possible numbers... (Giggle) Any help appreciated. • Jan 25th 2010, 06:40 AM Plato Quote: Originally Posted by davidman There are $x$ number of positive divisors of $72$, $y$ number of positive divisors of $900$. The positive divisors of $900$ that are not divisors of $72$ are $z$. I find this question a bit confusing. Is this what you mean? Suppose that $t$ is the number of divisors of $72~\\&~900$. Then $z=y-t$. Note that $t\\le x$. • Jan 25th 2010, 07:51 AM davidman Not quite sure what your $t$ represents. $z$ is the number of divisors of $900$ minus the divisors that overlap with $72$. • Jan 25th 2010, 08:09 AM Plato Quote: Originally Posted by davidman Not quite sure what your $t$ represents. $z$ is the number of divisors of $900$ minus the divisors that overlap with $72$. What is the point of this reply? You did not read my reply? It says clearly what $t$ equals. Do you have difficulty translating Enghish? •", "of divisors is = (e1+1)(e2+1)(e3+1) ... (ek+1). where ek denotes powers of prime factors of the number. For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors. Manager Joined: 23 Jul 2008 Posts: 203 Followers: 1 Kudos [?]: 76 [0], given: 0 ### Show Tags 18 Nov 2008, 12:33 prasun84 wrote: the number of divisors is = (e1+1)(e2+1)(e3+1) ... (ek+1). where ek denotes powers of prime factors of the number. For example, 4200 is $$2^3*3^1*5^2*7^1$$, so it has (3+1)(1+1)(2+1)(1+1) = 48 positive divisors. Thanks a lot. Re: Prime factors [#permalink] 18 Nov 2008, 12:33 Display posts from previous: Sort by", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "factor A proper factor of a positive integer is a factor of other than 1 or (Derbyshire 2004, p. 32). For example, 2 and 3 are positive proper factors of 6, but 1 and 6 are not.Compare with the term proper divisor, which means a factor of other than (but including 1). Proper divisor A positive proper divisor is a positive divisor of a number , excluding itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not. The number of proper divisors of is therefore given bywhere is the divisor function. For , 2, ..., is therefore given by 0, 1, 1, 2, 1, 3, 1, 3, 2, 3, ... (OEIS A032741). The largest proper divisors of , 3, ... are 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, ... (OEIS A032742).The term \"proper divisor\" is sometimes used to include negative integer divisors of a number (excluding ). Using this definition, , , , 1, 2, and 3 are the proper divisors of 6, while and 6 are the improper divisors.To make matters even more confusing, the proper divisor is often defined so that and 1 are also excluded. Using this alternative definition, the proper divisors of 6 would then be , , 2, and 3, and the", "60 views The ratio of two positive numbers is $2:3.$ If $10$ and the sum of the numbers are added to their product, square of $16$ is obtained. The greater number is : 1. $6$ 2. $9$ 3. $12$ 4. $18$ Assume that numbers are $2x,3x$. Now according to the question: $(2x \\times 3x) + 3x+2x+10=256$ $\\implies6x^2+5x-246=0\\implies x=6$ So the number are: $6\\times 2=12,6\\times 3=18$ Option (D) is correct. 4.7k points 1", "# Q) What is the greatest common divisor of ${3}^{{3}^{333}}+1$ and ${3}^{{3}^{334}}+1?$ ans;4 • 0 3^334 =3^(333+1) = 3^333*3^1 =3^333*3 =3^(333*3)=3^333^3 so 3^3^334 + 1 = 3^3^333^3 + 1 =( 3^3^333 + 1)( (3^3^333)^2 - 3^3^333 + 1) # h^3 +1 = (h+1)(h^2 -h +1) thus 3^3^333 + 1 is a factor of 3^3^334 + 1 • 1", "+0 # help 0 80 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 #1 +5088 +2 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$ . Jan 1, 2019", "Let $D(x)$ denote the number of positive divisors of $x$. Determine the number of values of $x$ for which $x = D(x)^4$.", "and $10$ are always divisors, test $2$: $10/2=5$, so $2$ and $5$ are divisors of $10$, test $3$, $10/3 = 3 + 1/3$, so $3$ is not a divisor of $10$. Another method calculates the prime factors decomposition of $N$ and by combination of them, get all divisors. Example: $10 = 2 \\times 5$, divisors are then $1$, $2$, $5$, and $2 \\times 5 = 10$ Negative divisors also exist, but they are the same as positive divisors (with the sign close), so they are ignored. ### What is the list of divisors from 1 to 100? NumberList of Divisors Divisor of 11 Divisors of 21,2 Divisors of 31,3 Divisors of 41,2,4 Divisors of 51,5 Divisors of 61,2,3,6 Divisors of 71,7 Divisors of 81,2,4,8 Divisors of 91,3,9 Divisors of 101,2,5,10 Divisors of 111,11 Divisors of 121,2,3,4,6,12 Divisors of 131,13 Divisors of 141,2,7,14 Divisors of 151,3,5,15 Divisors of 161,2,4,8,16 Divisors of 171,17 Divisors of 181,2,3,6,9,18 Divisors of 191,19 Divisors of 201,2,4,5,10,20 Divisors of 211,3,7,21 Divisors of 221,2,11,22 Divisors of 231,23 Divisors of 241,2,3,4,6,8,12,24 Divisors of 251,5,25 Divisors of 261,2,13,26 Divisors of 271,3,9,27 Divisors of 281,2,4,7,14,28 Divisors of 291,29 Divisors of 301,2,3,5,6,10,15,30 Divisors of 311,31 Divisors of 321,2,4,8,16,32 Divisors of 331,3,11,33 Divisors of 341,2,17,34 Divisors of 351,5,7,35 Divisors of 361,2,3,4,6,9,12,18,36 Divisors of 371,37 Divisors of 381,2,19,38 Divisors of 391,3,13,39 Divisors" ]

[ "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ Solution Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", "# In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater than the measure of $\\angle guy$: [asy] unitsize(50); pair y=(cos(45*pi/180),sin(45*pi/180)); pair g=(cos(120*pi/180),sin(120*pi/180)); pair b=(cos(225*pi/180),sin(225*pi/180)); pair u=(0,0); dot(y); dot(g); dot(b); dot(u); label(“$y$”,y,se); label(“$g$”,g,sw); label(“$b$”,b,se); label(“$u$”,u,se); draw(b–y); draw(u–g); [/asy] how many degrees are in the measure of $\\angle guy$?.", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "and multiplying these together proves the formula for isogonal lines. Hence, we have $$\\frac{BE}{15-BE}\\cdot \\frac{9}{6}=\\frac{169}{196}\\implies BE=\\frac{2535}{463}$$ so our desired answer is $\\boxed{463}.$ ## Solution 6 (Tangent subtraction formulas) Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle. $[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label(\"G\",G,N); label(\"F\",F,N); [/asy]$ Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\\angle DAG = \\alpha$. Now we know that $\\overline{DG} = \\frac{24}{5}$ and $\\overline{GC} = \\frac{18}{5}$. Therefore, $\\overline{AG} = \\frac{52}{5}$, so $\\tan{(\\alpha)} = \\frac{6}{13}$. Our goal now is to use tangent $\\angle EAG$ in triangle $AEG$. We set $\\overline{BE}$ to $x$, so $\\overline{ED} = 9 - x$ and $\\overline{EC} = 15 - x$, so $\\overline{EG} = \\frac{4}{5}(15-x)$ and $\\overline{GC} = \\frac{3}{5}(15-x)$ so $\\overline{AG} = \\frac{3x+25}{5}$. Now we just need tangent of $\\angle EAG$. We find this using $\\tan{(EAG)} =", "= 0.94280904158 // sqrt((2/3)2 + (2/3)2) var f = 0.7453559925 // sqrt((1/3)2 + (2/3)2) var g = 0.47140452079 // sqrt((1/3)2 + (1/3)2) var feasableLines = [a,b,c,d,e,f,g]; var feasableLabels = [\"a\",\"b\",\"c\",\"d\",\"e\",\"f\",\"g\"]; // Line configurations var aLines = [\"AG\", \"DJ\"]; var bLines = [\"AF\",\"AH\",\"BG\",\"CJ\",\"DI\",\"DK\",\"EJ\",\"GL\"]; var cLines = [\"AE\",\"AI\",\"BH\",\"BJ\",\"CG\",\"CI\",\"DH\",\"DL\",\"EK\",\"FJ\",\"FL\",\"GK\"]; var dLines = [\"BI\",\"CH\",\"EL\",\"FK\"]; var eLines = [\"BF\",\"CK\",\"EI\",\"HL\"]; var fLines = [\"BE\",\"BK\",\"CF\",\"CL\",\"EH\",\"FI\",\"HK\",\"IL\"]; var gLines = [\"BL\",\"CE\",\"FH\",\"IK\"]; var configurations = [aLines,bLines,cLines,dLines,eLines,fLines,gLines]; // joint angle requirements var aAngles = [45,45]; var bAngles = [33.69, 56.31]; var cAngles = [18.43, 71.57]; var dAngles = [90,90]; var eAngles = [45,45]; var fAngles = [26.57, 63.43]; var gAngles = [45,45]; var angles = [aAngles,bAngles,cAngles,dAngles,eAngles,fAngles,gAngles]; // Number of configurations for each feasable length var maxA = 2; var maxB = 8; var maxC = 12; var maxD = 4; var maxE = 4; var maxF = 8; var maxG = 4; var convertToCannonical = function(arr){ var mapR1 = { \"A\":\"D\", \"B\":\"E\", \"C\":\"F\", \"D\":\"G\", \"E\":\"H\", \"F\":\"I\", \"G\":\"J\", \"H\":\"K\", \"I\":\"L\", \"J\":\"A\", \"K\":\"B\", \"L\":\"C\" }; var mapR2 = { \"A\":\"G\", \"B\":\"H\", \"C\":\"I\", \"D\":\"J\", \"E\":\"K\", \"F\":\"L\", \"G\":\"A\", \"H\":\"B\", \"I\":\"C\", \"J\":\"D\", \"K\":\"E\", \"L\":\"F\" } var mapR3 = { \"A\":\"J\", \"B\":\"K\", \"C\":\"L\", \"D\":\"A\", \"E\":\"B\", \"F\":\"C\", \"G\":\"D\", \"H\":\"E\", \"I\":\"F\", \"J\":\"G\", \"K\":\"H\", \"L\":\"I\" } var result1 = sort(map(function(str){ return sort(map(function(x){return mapR1[x];},str.split(''))).join('');}, arr)); var result2 = sort(map(function(str){ return sort(map(function(x){return mapR2[x];},str.split(''))).join('');}, arr)); var result3 = sort(map(function(str){ return sort(map(function(x){return", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "draw(A--B--C--cycle); dotfactor = 3; dot(\"A\",A,dir(135)); dot(\"B\",B,dir(215)); dot(\"C\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"D\",D,dir(270)); dot(\"E\",E,dir(180)); dot(\"F\",F,dir(90)); dot(\"G\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$ First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB =", "C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$", "# Difference between revisions of \"2005 AIME II Problems/Problem 14\" ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively.", "pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < 2; ++i) /* left */ fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */ fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label(\"\\frac 12\",(s*3/4,s/2),sm); label(\"\\cdots\",(s*1/4,s/2),sm); label(\"\\frac 12\",shiftR+(s*3/4,s/2),sm); label(\"\\cdots\",shiftR+(s*1/4,s/2),sm); label(\"\\frac 14\",shiftR+(s*1/4,s*3/4),sm); label(\"\\frac 18\",shiftR+(s*3/8,s/4),sm); htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label(\"1\",(s/2,-1),S,sm); label(\"1\",shiftR+(s/2,-1),S,sm); [/asy]$ The infinite geometric series $\\frac 12 + \\frac {1}{2^2} + \\frac {1}{2^3} + \\cdots = 1$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){ for(int i = 0; i < 4; ++i){ draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype(\"2 2\")); draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype(\"2 2\")); } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); } for(int i = 0; i < n; ++i) { fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]); fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]); fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]); fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]); drawTriGrid(h/3^i); } [/asy]$ The infinite geometric series $\\frac 13 + \\frac {1}{3^2} + \\frac {1}{3^3} + \\cdots = \\frac 12$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0); void drawEquilaterals(pair A, real s){ filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]); filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]); filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); } for(int i = 0; i < n; ++i) drawEquilaterals(shiftR + (0,h-h/(2^i)", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of" ]

[ "= EF$ . At the same time , $\\angle DCF = 60^o$ and $CD = CB = CF$ so $\\Delta DCF$ is equilateral . We thus have $FC = FD = FE$ . Therefore , $F$ is the circumcentre of $\\Delta DCE$ so $\\angle CED = \\frac{1}{2}~ \\angle CFD = \\frac{1}{2}~ 60^o = 30^o$ 7. Nice. I like this solution (# means angle) : Triangle BCE: it is given that #ABC = 80 and #BCE = 60; then #BEC = 40 Triangle CDF: this triangle being equilateral, then #CFD = 60; since #BFC = 80, this leaves #DFE = 180-80-60 = 40 Triangle DEF: #EDF = #DEF, so #CED + 40 = 70, hence #CED = 30", "Finally, the answer is $25+7=\\boxed{032}$. ### Solution 2 No restrictions are set on the lengths of the bases, so for calculational simplicity let $\\angle CAF = 30^{\\circ}$. Since $CAF$ is a $30-60-90$ triangle, $AF=\\frac{CF\\sqrt{3}}2=15\\sqrt{7}$. $EF=EA+AF=10\\sqrt{7}+15\\sqrt{7}=25\\sqrt{7}$ The answer is $25+7=\\boxed{032}$. Note that while this is not rigorous, the above solution shows that $\\angle CAF = 30^{\\circ}$ is indeed the only possibility. ### Solution 3 Extend $\\overline {AB}$ through $B$, to meet $\\overline {DC}$ (extended through $C$) at $G$. $ADG$ is an equilateral triangle because of the angle conditions on the base. If $\\overline {GC} = x$ then $\\overline {CD} = 40\\sqrt{7}-x$, because $\\overline{AD}$ and therefore $\\overline{GD}$ $= 40\\sqrt{7}$. By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\\overline{FD} = \\frac{40\\sqrt{7}-x}{2}$, and $\\overline{CF} = \\frac{40\\sqrt{21} - \\sqrt{3}x}{2}$ Similarly $CAF$ is a 30-60-90 triangle and thus $\\overline{CF} = \\frac{10\\sqrt{21}}{2} = 5\\sqrt{21}$. Equating and solving for $x$, $x = 30\\sqrt{7}$ and thus $\\overline{FD} = \\frac{40\\sqrt{7}-x}{2} = 5\\sqrt{7}$. $\\overline{ED}-\\overline{FD} = \\overline{EF}$ $30\\sqrt{7} - 5\\sqrt{7} = 25\\sqrt{7}$ and $25 + 7 = \\boxed{032}$", "and angle GFC add to make 180, so if we subtract EFC from 180, then we get the size of angle CFG. So, we get $\\text{Angle CFG } = 180 - 32 = 148\\degree$. Now, looking at the diagram to the right (and it helps a lot to draw on your diagrams when doing these questions on paper), we can see that angle CFG and the missing angle $x$ are corresponding angles. Therefore, we get $x = 148\\degree$. As mentioned, there are multiple ways to do this question. How else might you do it? ### Example Questions Firstly, because angle HFG and angle EFC are vertically opposite, we get $\\text{angle EFC } = 48\\degree$ Secondly, because angle EFC and angle BCA (angle $x$) are corresponding angles, we get $\\text{angle BCA } = x = 48\\degree$ There are other possible methods for doing this question. As long as you’ve correctly applied angle facts, explained each step, and got the answer to be $48\\degree$, your answer is correct. Firstly, using the fact that angles FGJ and CDG are corresponding angles, we get $\\text{angle CDG } = 121\\degree$. Secondly, because angles on a straight line add to 180, and angles CDG and CDA are on a straight line, we get $\\text{angle CDA } = 180 - 121 = 59\\degree$. Thirdly, again", "\"$6^{\\circ}$,\" not the length of $\\overline{bf}$, right? In other words, is it true that $H'\\ne H$? – Robert Howard Jul 19 '18 at 17:54 • H' is the length of _bf; H is the overall height of the opposite side (that is, H' plus the radius of the starting arc segment ab, which is equal to W, so H' = H - W. Thank you for asking. – Rory Buszka Jul 19 '18 at 18:03 Your drawing is very... unusual. What's wrong with capital letters? Note that $\\angle dea=90^\\circ-6^\\circ=84^\\circ$. $$\\overline{bc} = dx+R\\sin \\angle dea$$ $$\\overline{ga}=\\overline{df}\\cos \\angle bfd-R\\cos \\angle dea + R\\tag{2}$$ $$\\overline{df} \\sin 6^\\circ+R\\sin 84^\\circ=W\\tag{3}$$ $$\\overline{df}\\cos 6^\\circ-R\\cos 84^\\circ + R=W+H\\tag{4}$$ This linear system of two equations, (3) and (4), has two unknows, $\\overline{df}$ and $R$, and can be easily solved in terms of $W$ and $H$. The rest is easy: $$dx=\\overline{df}\\sin 6^\\circ$$ $$dy=\\overline{df}\\cos 6^\\circ-H$$ • Edit time limits are a POS. So let's see if I get this. Given Eqns (3) and (4) above, $\\overline{df}\\cos 6^\\circ\\ - R\\cos 84^\\circ + R = \\overline{df}\\sin 6^\\circ + R\\sin 84^\\circ + H$ and some algebra later $R - R\\cos 84^\\circ - R\\sin 84^\\circ = \\overline{df}\\sin 6^\\circ - \\overline{df}\\cos 6^\\circ + H$, then distributing out $R$ and $\\overline{df}$ I get $R(1 -\\cos 84^\\circ -\\sin 84^\\circ) = \\overline{df}(\\sin 6^\\circ - \\cos 6^\\circ) +", "unknown angle The side $\\overline{GJ}$ is one of the sides of the $\\Delta JGF$ but its angle is unknown. So, it is not possible to express the length of the side $\\overline{GJ}$ in terms of a trigonometric function. Hence, let’s find the angle of this triangle geometrically. The sides $\\overline{CE}$ and $\\overline{JF}$ are parallel lines and $\\overline{CF}$ is their transversal. $\\angle ECF$ and $\\angle JFC$ are interior alternate angles and they are equal geometrically. $\\angle JFC = \\angle ECF = x$ The side $\\overline{FG}$ is a perpendicular line to side $\\overline{CF}$. So, the $\\angle CFG = 90^°$. but the side $\\overline{JF}$ splits the $\\angle CFG$ as $\\angle JFC$ and $\\angle JFG$. $\\angle JFC + \\angle JFG = 90^°$ It is proved about that $\\angle JFC = x$. $\\implies x + \\angle JFG = 90^°$ $\\implies \\angle JFG = 90^°-x$ $\\Delta JGF$ is a right triangle in which $\\angle GJF = 90^°$ and $\\angle JFG = 90^°-x$. The $\\angle JGF$ can be calculated by the sum of angles rule of a triangle. $\\angle JGF + \\angle GJF + \\angle JFG$ $=$ $180^°$ $\\implies \\angle JGF + 90^° + 90^°-x$ $=$ $180^°$ $\\implies \\angle JGF + 180^°-x$ $=$ $180^°$ $\\implies \\angle JGF$ $=$ $180^°-180^°+x$ $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\, \\angle JGF = x$ It is proved that the $\\angle JGF$ is also", "Triangle $DFC$ is isosceles $(CF=CD)$. Hence $\\angle DFC = \\angle FDC = x^{\\circ}$. Hence $\\angle FCD = (180-2x)^{\\circ}$ (angle sum of triangle). Now $\\angle EBC = \\angle FDC = x^{\\circ}$ (opposite angles of a parallelogram) and triangle EBC is isosceles (CE = CB). Hence $\\angle BEC = x^{\\circ}$ and $\\angle ECB = (180-2x)^{\\circ}$ Lines $AD$ and $BC$ are parallel and hence $$\\angle ADC + \\angle BCD = 180^{\\circ}$$ Therefore: $$x+2(180-2x) + 60= 180$$i.e.$420-3x=180$ i.e. $3x=240$ i.e. $x=80$ This problem is taken from the UKMT Mathematical Challenges.", "Correction: I made an unjustified assumption and hurried to the conclusion here. Please OMIT the part below between the ********* lines, and instead as in comments below work with supplementary angles and an easy chain of equalities: After proving $\\angle AGD= \\angle CFD, \\angle AED=\\angle BFD, \\angle BGD=\\angle CED$ Then as supplementary angles $\\angle AED +\\angle CED=180; \\angle BGD+\\angle AGD=180; \\angle CFD+\\angle BFD=180.$ $\\angle BFD = \\angle AED =180 -\\angle CED = 180 - \\angle BGD = \\angle AGD = \\angle CFD$ and BFC is a straight line so $\\angle BFD = \\angle CFD = 90$ And similarly for the other two sides of the triangle. OMIT below Pairs of these are opposite angles in quadrilaterals so must sum to 180 degrees. $\\angle AED + \\angle AGD = 180; \\ \\ \\angle BGD + \\angle BFD = 180; \\ \\ \\angle CED + \\angle CFD = 180.$ Suppose $\\angle ADE = \\angle BFD < 90$ Then by the above sums $\\angle AGD = \\angle BGD > 90$ which is impossible since AGB is a straight line. Similarly these angles cannot be $>90$. We can work similarly around all three sides of the triangle. OMIT above $\\angle AGD = \\angle CFD = \\angle AED = \\angle BFD = \\angle BGD = \\angle CED = 90$ and $AF, BE, CG$", "Triangle $DFC$ is isosceles $(CF=CD)$. Hence $\\angle DFC = \\angle FDC = x^{\\circ}$. Hence $\\angle FCD = (180-2x)^{\\circ}$ (angle sum of triangle). Now $\\angle EBC = \\angle FDC = x^{\\circ}$ (opposite angles of a parallelogram) and triangle EBC is isosceles (CE = CB). Hence $\\angle BEC = x^{\\circ}$ and $\\angle ECB = (180-2x)^{\\circ}$ Lines $AD$ and $BC$ are parallel and hence $$\\angle ADC + \\angle BCD = 180^{\\circ}$$ Therefore: $$x+2(180-2x) + 60= 180$$i.e.$420-3x=180$ i.e. $3x=240$ i.e. $x=80$ This problem is taken from the UKMT Mathematical Challenges. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "2) $\\angle\\, DFI = 60^{\\circ} = \\angle \\, CFD$; 3) $\\angle \\, FCI = \\angle \\, FCD$; 4) Triangles $DFC$ and $IFC$ are congruent by 2) and 3); 5) By 4) $DF = IF$; 6) Analogous arguments yield that triangles $EBF$ and $IBF$ are congruent; 7) By 6) $EF = IF$; 8) By 5) and 7) $DF = IF = EF$ and moreover $\\angle \\, DFI = 120^{\\circ} = \\angle \\, EFI$ 9) By 8) triangles $DEI$ is equilateral and thus $DI = EI$.", "angle fact you used at each step. Currently we cannot see a rule connecting $\\angle EFC$ with $x$ This means we will need multiple steps. (there are a few ways to do this) Firstly, we will use the fact that angles on a straight line add to 180. Specifically, angle $\\angle EFC$ and angle $\\angle GFC$ add to make $180\\degree$. Which means we can do the following: $\\angle CFG = 180\\degree - 32\\degree = 148\\degree$. Now, looking at the diagram we can see that angle $\\angle CFG$ and the missing angle $x$ are corresponding angles. Therefore, we get $x = 148\\degree$. As mentioned, there are multiple ways to do this question. How else might you do it? ### Example Questions Firstly, because angle HFG and angle EFC are vertically opposite, we get $\\text{angle EFC } = 48\\degree$ Secondly, because angle EFC and angle BCA (angle $x$) are corresponding angles, we get $\\text{angle BCA } = x = 48\\degree$ There are other possible methods for doing this question. As long as you’ve correctly applied angle facts, explained each step, and got the answer to be $48\\degree$, your answer is correct. Firstly, using the fact that angles FGJ and CDG are corresponding angles, we get $\\text{angle CDG } = 121\\degree$. Secondly, because angles on a straight line add to 180,", "to get $f^{m-n+1}=f$. We take $M=m-n+1$ to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \\in A$. QED. Problem 6: Show that there exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4),", "across the foot of perpendicular of $C$ to $AB$ . Then $CF = CB$ , $\\angle CFB = 80^o$ . We find that $\\angle ECF = \\angle CFB - \\angle CEF = 80^o - 40^o = \\angle CEF$ so $CF = EF$ . At the same time , $\\angle DCF = 60^o$ and $CD = CB = CF$ so $\\Delta DCF$ is equilateral . We thus have $FC = FD = FE$ . Therefore , $F$ is the circumcentre of $\\Delta DCE$ so $\\angle CED = \\frac{1}{2}~ \\angle CFD = \\frac{1}{2}~ 60^o = 30^o$ • February 19th 2011, 01:32 AM Wilmer Nice. I like this solution (# means angle) : Triangle BCE: it is given that #ABC = 80 and #BCE = 60; then #BEC = 40 Triangle CDF: this triangle being equilateral, then #CFD = 60; since #BFC = 80, this leaves #DFE = 180-80-60 = 40 Triangle DEF: #EDF = #DEF, so #CED + 40 = 70, hence #CED = 30", "\\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$ Case (iii): Here $(a-b)-(d-e)=2c-2=6$ Case (iv): This is satisfied by $(a,b,d,e)=(6,2,3,5)$. Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values. Aliter: Embed the hexagon", "120 + 120 + 120 = 360 = 60 + 120 + 180. We can also allow for the case when any of the isosceles triangles are constructed internally, in which case the corresponding angles $\\alpha, \\beta$ or $\\gamma$ are between $-180^{\\circ}$ and $0^{\\circ}$. (An angle of $-\\theta$ anticlockwise is simply equivalent to an angle of $360-\\theta$ anticlockwise or $\\theta$ clockwise.) Then the above result almost holds when $\\alpha + \\beta + \\gamma =0^{\\circ}$ or $-180^{\\circ}$ though in these cases one or more of the angles $\\angle EDF, \\angle FED, \\angle DFE$ may need to be adjusted by $180^{\\circ}$ to ensure they sum to $\\pm 180^{\\circ}$. In the following figure we chose $\\alpha = 60^{\\circ}, \\beta = \\gamma = -30^{\\circ}$, leading to $\\angle EDF = 60-180=-120^{\\circ}, \\angle FED = \\angle DFE = -30^{\\circ}$ as shown. #### An example from the IMO Now let us look at an even more general case when the triangles erected on the sides of $ABC$ are not necessarily isosceles. Consider the following example from the 1977 IMO. In the figure the triangle ABC is arbitrary and angles are as shown. We are required to show that $DE=DF$ and $DE \\perp DF$. This time we can proceed by considering spiral similarities (dilations composed with rotations) about $B$ and $C$. Here we consider the composition", "equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$ Case (iii): Here $(a-b)-(d-e)=2c-2=6$ Case (iv): This is satisfied by $(a,b,d,e)=(6,2,3,5)$. Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values. Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides", "equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5), c=4$…(iii) The possibility that $(a,b)=(3,4), (a,e)=(2,5)$ in (i), for instance, need not be considered separately, because we can reflect the figure about $x=\\frac{1}{2}$ and interchange these two sets. Case (i): Here $(a-b)-(d-e)=2c-2=10$. Since $a-b=\\pm 3, d-e=\\pm 1$, this is not possible. Case (ii): Here $(a-b)-(d-e)=2c-2=2$. This is satisfied by $(a,b,d,e)=(6,3,5,4)$ Case (iii): Here $(a-b)-(d-e)=2c-2=6$ Case (iv): This is satisfied by $(a,b,d,e)=(6,2,3,5)$. Hence, we have (essentially) two different solutions: $(1,6,3,2,5,4)$ and $(1,6,2,4,3,5)$. It may be verified that I and II are both satisfied by these sets of values. Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides", "ECF$ Actually, $\\angle ECF = x$ $\\therefore \\,\\,\\,\\,\\,\\, \\angle JFC \\,=\\, x$ It is taken that $\\overline{GF} \\perp \\overline{CF}$. So, $\\angle CFG = 90^°$. The side $\\overline{JF}$ divides the $\\angle CFG$ as two angles $\\angle CFJ$ and $\\angle JFG$. $\\angle CFG \\,=\\, \\angle CFJ + \\angle JFG$ $\\implies 90^° = x+\\angle JFG$ $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\, \\angle JFG = 90^°-x$ In $\\Delta JGF$, the $\\angle GJF$ is $90^°$ and $\\angle JFG$ is calculated as $90^°-x$. The $\\angle JGF$ can be evaluated by them by using sum of angles of a triangle rule. $\\angle JGF + \\angle GFJ + \\angle FJG$ $\\,=\\,$ $180^°$ $\\implies$ $\\angle JGF + 90^°-x + 90^°$ $\\,=\\,$ $180^°$ $\\implies$ $\\angle JGF + 180^°-x$ $\\,=\\,$ $180^°$ $\\implies$ $\\angle JGF$ $\\,=\\,$ $180^°-180^°+x$ $\\implies$ $\\angle JGF$ $\\,=\\,$ $\\require{cancel} \\cancel{180^°}-\\cancel{180^°}+x$ $\\,\\,\\, \\therefore \\,\\,\\,\\,\\,\\,$ $\\angle JGF \\,=\\, x$ It is proved that the $\\angle JGF$ is $x$ but the $\\angle ECF$ is also $x$. It proves that $\\angle ECF$ and $\\angle JGF$ are similar triangles. #### Express lengths of sides as trigonometric functions According to $\\Delta JFG$, try to write the length of side $\\overline{GJ}$ in terms of a trigonometric function. $\\cos{x} \\,=\\, \\dfrac{GJ}{GF}$ $\\implies GJ = {GF}\\cos{x}$ Now, substitute the equivalent value of $GJ$ in the expansion of sin of sum of two angles function. $\\implies$ $\\sin{(x+y)}$ $\\,=\\,$ $\\dfrac{{GF}\\cos{x}}{CG}$ $+$", "EF + \\overparen{FG})$$. Figure 4.2.4 Belt pulleys with radii 5 cm and 8 cm First, at the center $$B$$ of the pulley with radius $$8$$, draw a circle of radius $$3$$, which is the difference in the radii of the two pulleys. Let $$C$$ be the point where this circle intersects $$\\overline{BF}$$. Then we know that the tangent line $$\\overline{AC}$$ to this smaller circle is perpendicular to the line segment $$\\overline{BF}$$. Thus, $$\\angle\\,ACB$$ is a right angle, and so the length of $$\\overline{AC}$$ is $AC ~=~ \\sqrt{AB^2 - BC^2} ~=~ \\sqrt{15^2 - 3^2} ~=~ \\sqrt{216} ~=~ 6\\,\\sqrt{6}\\nonumber$ by the Pythagorean Theorem. Now since $$\\overline{AE} \\perp \\overline{EF}$$ and $$\\overline{EF} \\perp \\overline{CF}$$ and $$\\overline{CF} \\perp \\overline{AC}$$, the quadrilateral $$AEFC$$ must be a rectangle. In particular, $$EF = AC$$, so $$EF = 6\\,\\sqrt{6}$$. By Equation \\ref{4.4} we know that $$\\;\\overparen{DE} = EA \\cdot \\angle\\,DAE\\;$$ and $$\\;\\overparen{FG} = BF \\cdot \\angle\\,GBF$$, where the angles are measured in radians. So thinking of angles in radians (using $$\\pi$$ rad $$= 180^\\circ$$), we see from Figure 4.2.4 that $\\nonumber \\angle\\,DAE ~=~ \\pi \\;-\\; \\angle\\,EAC \\;-\\; \\angle\\,BAC ~=~ \\pi \\;-\\; \\frac{\\pi}{2} \\;-\\; \\angle\\,BAC ~=~ \\frac{\\pi}{2} \\;-\\; \\angle\\,BAC ~,$ where $\\nonumber \\sin\\;\\angle\\,BAC ~=~ \\frac{BC}{AB} ~=~ \\frac{3}{15} ~=~ 0.2 \\quad\\Rightarrow\\quad \\angle\\,BAC ~=~ 0.201~\\text{rad.}$ Thus, $$\\;\\angle\\,DAE = \\frac{\\pi}{2} \\,-\\, 0.201 = 1.37$$ rad. So since $$\\overline{AE}$$ and $$\\overline{BF}$$ are", "$$g(360^{\\circ})$$ from $$f(360^{\\circ})$$. \\begin{align*} f(360^{\\circ}) - g(360^{\\circ}) &= 2 - 0 \\\\ &=2 \\end{align*} Given the following graph. State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = ( 90^{\\circ}; 1),~B = ( 180^{\\circ}; -3),~C = (270 ^{\\circ};-1 ) \\text{ and } D = ( 360^{\\circ}; 3)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$2$$ What is the amplitude of $$g(x)$$. $$3$$ Evaluate: $$f(90^{\\circ}) - g(90^{\\circ})$$ . Read off the value of $$f(90^{\\circ})$$ and $$g(90^{\\circ})$$ from the graph. Then subtract $$g(90^{\\circ})$$ from $$f(90^{\\circ})$$. \\begin{align*} f(90^{\\circ}) - g(90^{\\circ}) &= 1 - 0 \\\\ &=1 \\end{align*} Given the following graph: State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = (90 ^{\\circ};3 ),~B = (90 ^{\\circ};2 ),~C = (180 ^{\\circ}; -4) \\text{ and } D = ( 270^{\\circ};2)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$3$$ What is the amplitude of $$g(x)$$. $$2$$ Evaluate: $$f(270^{\\circ}) - g(270^{\\circ})$$ . Read off the value of $$f(270^{\\circ})$$ and $$g(270^{\\circ})$$ from the graph. Then subtract $$g(270^{\\circ})$$ from $$f(270^{\\circ})$$. \\begin{align*} f(270^{\\circ}) - g(270^{\\circ}) &= -3 - (-2)\\\\ &=-1 \\end{align*}", "to get the relation $f^{M}=f$ with $M>1$. Note that this means $f^{M}(i)=f(i)$ for all $i \\in A$. QED. Problem 6: Show that there exists a convex hexagon in the plane such that : a) all its interior angles are equal b) its sides are 1,2,3,4,5,6 in some order. Solution 6: Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that $AB=1$. Let $BC=a, CD=b, DE=c, EF=d, FA=e$. Since the sum of all angles of a hexagon is equal to $(6-2) \\times 180=720 \\deg$, it follows that each interior angle must be equal to $720/6=120\\deg$. Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by $(x,y)$ we then have: $\\overline{AB}=(1,0)$ $\\overline{BC}=(a\\cos 60\\deg, a\\sin 60\\deg)$ $\\overline{CD}=(b\\cos{120\\deg},b\\sin{120\\deg})$ $\\overline{DE}=(c\\cos{180\\deg},c\\sin{180\\deg})=(-c,0)$ $\\overline{EF}=(d\\cos{240\\deg},d\\sin{240\\deg})$ $\\overline{FA}=(e\\cos{300\\deg},e\\sin{300\\deg})$ This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively. Since the sum of all these six vectors is $\\overline{0}$, it implies that $1+\\frac{a}{2}-\\frac{b}{2}-c-\\frac{d}{2}+\\frac{e}{2}=0$ and $(a+b-d-e)\\frac{\\sqrt{3}}{2}=0$ That is, $a-b-2c-d+e+2=0$….call this I and $a+b-d-e=0$….call this II Since $(a,b,c,d,e)=(2,3,4,5,6)$, in view of (II), we have $(a,b)=(2,5), (a,e)=(3,4), c=6$….(i) $(a,b)=(5,6), (a,e)=(4,5), c=2$…(ii) $(a,b)=(2,6), (a,e)=(5,5)," ]

[ "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\\angle ABD$ exceeds $\\angle AHG$ by $12^\\circ$. Find the degree measure of $\\angle BAG$. $[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label(\"A\",A,dir(0)); label(\"B\",B,dir(75)); label(\"C\",C,dir(90)); label(\"D\",D,dir(105)); label(\"E\",E,dir(180)); label(\"F\",F,dir(225)); label(\"G\",G,dir(260)); label(\"H\",H,dir(280)); label(\"I\",I,dir(315));[/asy]$ ## Problem 7 In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\\overline{AD}$. Points $F$ and $G$ lie on $\\overline{CE}$, and $H$ and $J$ lie on $\\overline{AB}$ and $\\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\\overline{GH}$, and $M$ and $N$ lie on $\\overline{AD}$ and $\\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$. $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); [/asy]$ ## Problem 8 For positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$. ## Problem 9 Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1", "# In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater In the diagram below, point $u$ is on segment $\\overline{by}$ and the measure of $\\angle bug$ is $30$ degrees greater than the measure of $\\angle guy$: [asy] unitsize(50); pair y=(cos(45*pi/180),sin(45*pi/180)); pair g=(cos(120*pi/180),sin(120*pi/180)); pair b=(cos(225*pi/180),sin(225*pi/180)); pair u=(0,0); dot(y); dot(g); dot(b); dot(u); label(“$y$”,y,se); label(“$g$”,g,sw); label(“$b$”,b,se); label(“$u$”,u,se); draw(b–y); draw(u–g); [/asy] how many degrees are in the measure of $\\angle guy$?.", "dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ Solution Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the" ]

[ "started. Proportion of flour most pizza doughs resemble bread doughs, standard doughs and rustic (...", "How many mushrooms found each of them? 17. Dining room The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?", "in Europe to embrace the tomato, since it was deemed poisonous in Europe as a member of the nightshade family. With the rise in popularity of tomato, people started using it more and more. Mozzarella cheese was also slowly gaining ground. Mozzarella had become available in Italy only after water buffalo were imported from India in the 7th century (mozzarella was first made with water buffalo milk). Its popularity grew very slowly until the last half of the 18th century. In fact, cheese and tomatoes did not meet on a pizza until 1889. 4. The most commonly considered pizza (tomato, mozzarella, basil) was supposedly created on June 11, 1889 by a pizza-maker named Raffaele Esposito. This Pizzaiolo (pizza-maker in Italian and spelled Pizzaiuolo in Neapolitan) created a special pizza for the visit of Queen Margherita of Savoia. He made three different pizzas, but the Queen fell in love with one in particular, topped with three ingredients representing the three colours of the Italian flag. The Italian flag was represented by the tomatoes (red), mozzarella (white), and basil (green). Esposito named this pizza “Pizza alia Margherita” in honour of the Queen. Whether Esposito was the first to use those ingredients or not, this is known as the classic Neapolitan pizza or the modern-day tomato-and-cheese pizza. 5. In the latter", "who made the pizza i described) was the son of immigrants from sicily, and worked in his parents' pizza shop until he got married and moved away. he made really good pizza, and we were sorry when he went away. @AlanMunn ;) @barbarabeeton I think it's a bit US oriented but the ingredients seems good :)", "a bit, and a lot of it), and 17 toppings, you have: 5*317 = 645700815 ~ 6e8. This means that if you made every permutation of your pizzas and sliced them up so each pizza had 10 slices, you could give everyone in the world a slice. http://www57.wolframalpha.com/input/?i=645700815+pizza [–] 3 points4 points (7 children) sorry, this has been archived and can no longer be voted on Isn't a mushroom and pepperoni pizza the same as a pepperoni and mushroom pizza etc...? I have a feeling we need to use Binomial coefficients here (nCr). 17c1, 17c2 ...17 c 17, but then complications arise with special cases that probably aren't allowed like, putting 17 pepperoni on a pizza. [–] 14 points15 points * (3 children) sorry, this has been archived and can no longer be voted on Yes, a mushroom and pepperoni pizza is the same as a pepperoni and mushroom pizza. The number of topping choices is the same as the number of subsets of the set of 17 toppings, which is 217. Another way to think of it is suppose I handed you a form to fill out like this: Would you like Pepperoni [ ] Yes [ ] No Mushrooms [ ] Yes [ ] No Green Peppers [ ] Yes [ ] No ... You have", "## COCI '14 Contest 5 #1 Funghi View as PDF Points: 3 (partial) Time limit: 1.0s Memory limit: 32M Problem type After having eaten all the cookies from the wicked witch's house, Hansel and Gretel ordered a jumbo pizza. The pizza arrived shortly, cut into eight pieces. Hansel and Gretel are going to split the pizza in half so that each of them gets a complete pizza \"half-circle\" or, in other words, four consecutive pieces. Gretel really likes mushrooms and wants to get as many as she can. Given the fact that some pizza slices contain less and some more mushrooms, Gretel has asked Hansel to split the pizza so that her pieces contain as many mushrooms as possible. Help Hansel and Gretel! They will tell you how many mushrooms there are on each of the eight pizza slices, and your job is to find the largest total number of mushrooms Gretel can get. The following image depicts the optimal division for the second test sample below (1. denotes the first slice given in the input data): #### Input Each of the eight lines of input contains the integer . These numbers are, respectively, the amount of mushrooms on pizza slices, where the slices are given in clockwise order. #### Output The first and only line of output", "enough, which he suggested to be at least 400 micrograms). Mushrooms vary. Typically, in excess of a few grams, to achieve this same state. San Pedro, though variable, too, requires 12-18 inches or a few (bitter-tasting) dried “stars” (x-section, thin-sliced, in the oven @ 150 degrees, until dry like snack chips).", "out into two 12 inch pizzas. Since the dough was for tonight, I placed the pizza stone in a cold oven to warm up to 400 degrees. While the dough was rising, I chopped up a few toppings. Tonight, I picked tomatoes, jalapeno peppers, basil (picked from our patio garden), parm and mozzarella cheese! Chris added some pepperoni to his pizza too. When the stone is hot, take it out of the oven and add a nice dusting of cornmeal. This is a critical step! It keeps the dough from sticking to the stone. After I rolled out the dough, I placed it on the steaming hot stone. Because I prefer pizzas with no tomato sauce, I first added a light layer of olive oil followed by the toppings. Bake in the oven at 400 degrees for 10 mins and then broil 2 minutes just to give it that perfect golden brown color. Great texture in the dough and great toppings = great pizza!", "half of the 19th century, pizza migrated to America with the Italians. By the turn of the century, the Italian immigrants had begun to open their own bakeries and were selling groceries as well as pizza. Gennaro Lombardi opened the first true US pizzeria in 1905 at 531/3, Spring Street in New York City, a part of town known as “Little Italy”. 6. In India, of late, pizza has become a popular food. It has become a fashion and also a manner of showing that one is part of the famous Western culture. In fact, it is more of a fashion statement. The popularity of the food is rocketing. This is evident from a report by Fortune magazine. The two giants of the pizza industry, Pizza Hut and Dominos, are in hot competition with each other in India. India has 134 Pizza Huts and 149 Dominos locations, with each chain opening 50 stores a year. 7. The popularity of pizza in India, Fortune claims, is because of its similarity to India’s native cuisine. Unlike Chinese and Japanese, Indians eat leavened bread (roti/naan), and a popular traditional version slathers it in butter and garlic- not unlike garlic bread, the most often ordered side dish at both Dominos and Pizza Hut franchises in India. Cheese (paneer) is ubiquitous in India’s", "# Against the Grain Pesto Pizza Consumption Studies ### Higher Against The Grain Pesto Pizza Consumption Predicts Very Slightly Lower Guiltiness This individual’s Guiltiness is generally lowest after a daily total of 32 serving of Against The Grain Pesto Pizza consumption over the previous 7 days.", "bread while making a healthy pizza. Always make sure that your pizza and the toppings are packed with protein and veggies, and for non-vegetarians, a whole wheat chicken pizza is a healthy option for your diet. Step 2 – Make use of a milder cheese. When making cheese, always use a combination of part-skim mozzarella and naturally lower-fat parmesan to reduce the number of calories. • Step 3 – Watch your portion sizes. Pizza is one of those dishes that older folks force us to eat because it is so delicious. As a result, it’s important to pay attention to the portions, such as if you actually need 4-5 slices at one sitting. • One of the reasons why pizza is deemed ‘unhealthy’ is because of the food and beverages that are served with it. First, we used to eat cheesy breadsticks and later cheesy dis, drinks, etc. It is only healthy when you leave the food and beverages that were served with the pizza and concentrate just on slices of pizza rather than the entire", "# A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How many different ways can a pizza be made with 2 toppings?​ ## This Post Has 3 Comments 1. Chartwig4576 says: The answer is 21 as there are 7 toppings and u can only put two toppings on each pizza so in the end u hv to do 7 * 3 to get 21 2. lisaxo says: I am verry bad at imaging :^ 3. silasjob09 says: Yes I think this question is weird", "(pizza-maker in Italian and spelled Pizzaiuolo in Neapolitan) created a special pizza for the visit of Queen Margherita of Savoia. He made three different pizzas, but the Queen fell in love with one in particular, topped with three ingredients representing the three colours of the Italian flag. The Italian flag was represented by the tomatoes (red), mozzarella (white), and basil (green). Esposito named this pizza “Pizza alia Margherita” in honour of the Queen. Whether Esposito was the first to use those ingredients or not, this is known as the classic Neapolitan pizza or the modern-day tomato-and-cheese pizza. 5. In the latter half of the 19th century, pizza migrated to America with the Italians. By the turn of the century, the Italian immigrants had begun to open their own bakeries and were selling groceries as well as pizza. Gennaro Lombardi opened the first true US pizzeria in 1905 at 531/3, Spring Street in New York City, a part of town known as “Little Italy”. 6. In India, of late, pizza has become a popular food. It has become a fashion and also a manner of showing that one is part of the famous Western culture. In fact, it is more of a fashion statement. The popularity of the food is rocketing. This is evident from a report by Fortune", "how many pieces of pizza one may consume while on a diet. • It all starts with the numerous ways in which one might think about pizza. • Pizza has a wide variety of nutrients, which makes it a nutrient-dense meal. If we prepare it using cheap and low-quality ingredients, as is the case with most pizza served in restaurants, it will be a calorie-dense meal.The majority of the dough used outdoors is nutritionally poor, and it has a high concentration of preservatives and hydrogenated oil.If you are eating a pizza with greasy cheese and toppings that are frequently frozen, you will suffer as a result of your choice.Consequently, if you want to buy healthy pizza, always select thin crust since a conventional crust is comprised of refined white wheat, which will raise your insulin levels and cause your desires to increase.However, pizzas produced with fresh and whole ingredients might be a healthier option than those cooked with processed or pre-packaged components.When created with only a few ingredients, the pizzas that result may be extremely nutritious. By using vegetables or healthy protein sources like as grilled chicken as toppings on homemade pizza, you may increase the nutritional value of the dish.Vegetables and healthy protein sources such as grilled chicken are naturally high in nutrients.Healthy topping options, like as", "that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 04:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 04:28 Display posts from previous: Sort by", "yellow So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater. #### Real-World Application: Pizza Toppings Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from? We’ll start by making a table and list first choice, second choice and third choice: \\begin{align*}1^{st}\\end{align*} topping \\begin{align*}2^{nd}\\end{align*} topping \\begin{align*}3^{rd}\\end{align*} topping cheese pepperoni mushroom cheese pepperoni pineapple cheese pepperoni olives cheese mushroom pineapple cheese mushroom olives cheese pineapple olives pepperoni mushroom pineapple pepperoni mushroom olives pepperoni pineapple olives mushroom pineapple olives Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a different order. By counting the table entries we see that there are 10 possibilities for a 3-topping pizza. #### Determining Combinations by Looking at Permutations You can see that there are always fewer combinations than permutations in a given situation – but you should also see that knowing which combinations have already been used is important to avoid counting combinations", "see that there are 17 splits.", "pizza? (1) the pizza requires 33 grams of ingredients (2) the ratio of the number of grams of flour to the total number of grams used in the recipe is 1:3 _________________ Director Joined: 20 Feb 2015 Posts: 795 Concentration: Strategy, General Management Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] ### Show Tags 31 May 2016, 03:06 c:f = 2:1 1. 33 grams of ingredients cheese = (2/3)*33 = 22 grams sufficient !! 2. flour: t-flour = 1:3 insufficient !! A Retired Moderator Joined: 19 Mar 2014 Posts: 931 Location: India Concentration: Finance, Entrepreneurship GPA: 3.5 Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] ### Show Tags 18 Aug 2017, 14:20 Bunuel wrote: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 gram, respectively. If cheese and flour are the only ingredients in the recipe, how many grams of cheese are used when making a pizza? (1) the pizza requires 33 grams of ingredients (2) the ratio of the number of grams of flour to the total number of grams used in the recipe is 1:3 $$\\frac{Chees}{Floor} = \\frac{C}{F} = \\frac{2}{1}$$ (1) the pizza requires 33 grams of ingredients Total", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "we will also write down equivalent pairings at the same time. This will help prevent us from repeating combinations: Pairing Equivalent pairings (do not count) Red & blue blue & red Red & green green & red Red & yellow yellow & red Green & blue blue & green Green & yellow yellow & green Yellow & blue blue & yellow So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater . #### Example B Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from? Solution We’ll start by making a table and list first choice, second choice and third choice: $1^{st}$ topping $2^{nd}$ topping $3^{rd}$ topping cheese pepperoni mushroom cheese pepperoni pineapple cheese pepperoni olives cheese mushroom pineapple cheese mushroom olives cheese pineapple olives pepperoni mushroom pineapple pepperoni mushroom olives pepperoni pineapple olives mushroom pineapple olives Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a" ]

[ "# I am mushroom ^-^ In a mushroom forest, there are $90$ distinct mushrooms and Jimmy goes for picking all of them for his dinner. He took 3 identical bags with him, each of them can hold maximum $30$ mushrooms. If he takes home all of the mushrooms, then in how many ways can he put the mushrooms in the bags ? Write your answer as digit sum of the number. This problem is a part of the set Vegetable combinatorics ×", "How many mushrooms found each of them? 17. Dining room The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?", "a bit, and a lot of it), and 17 toppings, you have: 5*317 = 645700815 ~ 6e8. This means that if you made every permutation of your pizzas and sliced them up so each pizza had 10 slices, you could give everyone in the world a slice. http://www57.wolframalpha.com/input/?i=645700815+pizza [–] 3 points4 points (7 children) sorry, this has been archived and can no longer be voted on Isn't a mushroom and pepperoni pizza the same as a pepperoni and mushroom pizza etc...? I have a feeling we need to use Binomial coefficients here (nCr). 17c1, 17c2 ...17 c 17, but then complications arise with special cases that probably aren't allowed like, putting 17 pepperoni on a pizza. [–] 14 points15 points * (3 children) sorry, this has been archived and can no longer be voted on Yes, a mushroom and pepperoni pizza is the same as a pepperoni and mushroom pizza. The number of topping choices is the same as the number of subsets of the set of 17 toppings, which is 217. Another way to think of it is suppose I handed you a form to fill out like this: Would you like Pepperoni [ ] Yes [ ] No Mushrooms [ ] Yes [ ] No Green Peppers [ ] Yes [ ] No ... You have", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "yellow So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater. #### Real-World Application: Pizza Toppings Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from? We’ll start by making a table and list first choice, second choice and third choice: \\begin{align*}1^{st}\\end{align*} topping \\begin{align*}2^{nd}\\end{align*} topping \\begin{align*}3^{rd}\\end{align*} topping cheese pepperoni mushroom cheese pepperoni pineapple cheese pepperoni olives cheese mushroom pineapple cheese mushroom olives cheese pineapple olives pepperoni mushroom pineapple pepperoni mushroom olives pepperoni pineapple olives mushroom pineapple olives Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a different order. By counting the table entries we see that there are 10 possibilities for a 3-topping pizza. #### Determining Combinations by Looking at Permutations You can see that there are always fewer combinations than permutations in a given situation – but you should also see that knowing which combinations have already been used is important to avoid counting combinations", "If $$n$$ is even the maximum vertical stack of mushrooms left un-stacked would be $$2,4,6,8.....$$ depending on value of $$m$$. And in case of odd $$n$$ the number of mushrooms left unstacked would be $$1,3,5,7.....$$ depending on value of $$m$$. For eg. if $$n=6$$ and $$m=4$$ the maximum number of mushrooms left out would be 2. And we will have to subtract the cases where taken mushrooms are not fully stacked as it would lead to repetition of cases. Let's introduce another variable $$t$$ which represents the maximum number of mushrooms left out which can also be written in other form $$2m-n$$. Now, we can find the number of cases to be subtracted from $$P_{m}$$ to get the desired cases. We have to arrange $$m$$ mushrooms so we will have to subtract those cases where the number of horizontal rows is more than $$n-m$$. In order to achieve this we will first lay the mushrooms in horizontal row and work our way backwards by picking up one at a time and then arranging them too. So the number of discarded cases would be $$P_{0} + P_{1} + P_{2} ...$$ until the point we pick up less mushrooms than we have bases to arrange them on. And we will be faced by the same dilemma because of which we", "even the maximum vertical stack of mushrooms left un-stacked would be $2,4,6,8.....$ depending on value of $m$. And in case of odd $n$ the number of mushrooms left unstacked would be $1,3,5,7.....$ depending on value of $m$. For eg. if $n=6$ and $m=4$ the maximum number of mushrooms left out would be 2. And we will have to subtract the cases where taken mushrooms are not fully stacked as it would lead to repetition of cases. Let's introduce another variable $t$ which represents the maximum number of mushrooms left out which can also be written in other form $2m-n$. Now, we can find the number of cases to be subtracted from $P_{m}$ to get the desired cases. We have to arrange $m$ mushrooms so we will have to subtract those cases where the number of horizontal rows is more than $n-m$. In order to achieve this we will first lay the mushrooms in horizontal row and work our way backwards by picking up one at a time and then arranging them too. So the number of discarded cases would be $P_{0} + P_{1} + P_{2} ...$ until the point we pick up less mushrooms than we have bases to arrange them on. And we will be faced by the same dilemma because of which we divided the case", "## COCI '14 Contest 5 #1 Funghi View as PDF Points: 3 (partial) Time limit: 1.0s Memory limit: 32M Problem type After having eaten all the cookies from the wicked witch's house, Hansel and Gretel ordered a jumbo pizza. The pizza arrived shortly, cut into eight pieces. Hansel and Gretel are going to split the pizza in half so that each of them gets a complete pizza \"half-circle\" or, in other words, four consecutive pieces. Gretel really likes mushrooms and wants to get as many as she can. Given the fact that some pizza slices contain less and some more mushrooms, Gretel has asked Hansel to split the pizza so that her pieces contain as many mushrooms as possible. Help Hansel and Gretel! They will tell you how many mushrooms there are on each of the eight pizza slices, and your job is to find the largest total number of mushrooms Gretel can get. The following image depicts the optimal division for the second test sample below (1. denotes the first slice given in the input data): #### Input Each of the eight lines of input contains the integer . These numbers are, respectively, the amount of mushrooms on pizza slices, where the slices are given in clockwise order. #### Output The first and only line of output", "our minimum speed that she must’ve eaten at least 10. Hence these 5 mushrooms must’ve been newly added by Bartholomew. She had 15 mushrooms at time 20. We know from our minimum speed that she must’ve eaten the 5 mushrooms she previously had. Bartholomew could’ve added some new mushrooms (at least 15) at any point between times 10 and 20. If he added them before time 20, then Kaylin would’ve already eaten some of them, but we want to minimize the amount she eats, so we can assume Bartholomew added exactly 15 new mushrooms exactly at time 20. She had 5 mushrooms at time 30. We know from our minimum speed that she must’ve eaten 10 out of the 15 mushrooms she previously had. She then left the 5 remaining mushrooms on her plate and walked away. So, if Kaylin ate at a constant rate, she must’ve eaten at least 10 + 5 + 10 = 25 mushrooms. • » » » » 6 years ago, # ^ | +2 Thanks. I got it after reading \"\"we'll look at how many pieces of mushroom are on her plate\", but sadly, that was after the contest.You should have been the one in charge of writing the statement, not the one who actually did it. • » » » »", "### Home > A2C > Chapter 10 > Lesson 10.2.6 > Problem10-154 10-154. Another pizza parlor is having a super special. They have a rack of small pizzas, each with three or four different toppings. The price is really low because they forgot to indicate on the boxes what toppings were on each pizza. They had eight toppings available, and they made one pizza for each of the possible combinations of three or four toppings. 1. How many pizzas are on the rack? • Find how many types of pizzas with three toppings can be made. • 8C3 • Find how many pizzas with four toppings can be made. • 8C4 • 8C3 + 8C4 = 56 + 70 = 126 1. What is the probability of getting a pizza that has mushrooms on it? If mushrooms are a known topping, then you choose one fewer topping from only 7 remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "# A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How many different ways can a pizza be made with 2 toppings?​ ## This Post Has 3 Comments 1. Chartwig4576 says: The answer is 21 as there are 7 toppings and u can only put two toppings on each pizza so in the end u hv to do 7 * 3 to get 21 2. lisaxo says: I am verry bad at imaging :^ 3. silasjob09 says: Yes I think this question is weird", "# How tall are the mushrooms? In this challenge you will receive a list of non-negative integers. Each one represents a mushroom with a cap of that radius centered at that location. So a 0 means that it occupies no space at all, a 1 means that its cap only occupies space above it, a 2 means it occupies space above it and one unit to the left and right etc. More generally for size $$\\n\\$$ a total of $$\\2n-1\\$$ spaces are occupied, with the exception of 0 which occupies 0 spaces (i.e. there's no mushroom at all). Mushroom caps can't occupy the same space as each other but mushrooms can have different heights to avoid collisions. So for example here we have two mushrooms. They can't be in the same row since they would occupy the same space but if they are given different heights there is no issue: =-=-= | =-=-=-=-= | | | [ 0,0,3,0,0,2,0 ] (Stems are drawn with | for clarity, but can't collide) Your task is to take as input a list of mushrooms and output a list of heights, one for each input mushroom, such that there is no collisions between mushrooms and they occupy the fewest total rows. For example here we have a worked case: =-=-= =-=-=-=-=-=-= = |", "Home > A2C > Chapter 10 > Lesson 10.2.6 > Problem10-154 10-154. 1. Another pizza parlor is having a super special. They have a rack of small pizzas, each with three or four different toppings. The price is really low because they forgot to indicate on the boxes what toppings were on each pizza. They had eight toppings available, and they made one pizza for each of the possible combinations of three or four toppings. Homework Help ✎ 1. How many pizzas are on the rack? 2. What is the probability of getting a pizza that has mushrooms on it? Find how many types of pizzas with three toppings can be made. 8C3 Find how many pizzas with four toppings can be made. 8C4 8C3 + 8C4 = 56 + 70 = 126 If mushrooms are a known topping, then you choose one fewer topping from only 7 remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "million fragments of garlic arranged into a perfect line! Instead of risking our fingers or sanity, we’re going to use a technique known to professionals and amateurs alike: we will chop the garlic at random, over and over, until we generate one million pieces. How many chops does our simple approach require? The answer: Our rustic chop will mince almost as quickly as the professional’s honed technique. Put mathematically, the number of cells in our random chop also grows quadratically with the number of slices we make. With $$n$$ slices, our technique produces $$n + n(n-1)/ 4\\pi$$ bits of garlic when the garlic is a circle. Taking $$n=3540$$ yields, on average, just over one million pieces of garlic. Neat!", "CasyC77 does this make sense? So with this detail how many could you make? I am sorry, when you said the work \"these\" it looked like \"three\" to me. Thus, I assumed you were asking \"how many can you make if you use 3 different one ... \". I have to leave now, maybe, tomorrow I will answer. 7. Originally Posted by CasyC77 Can someone pleae help me urgently..... I feel really silly at the moment. My daughter came home with 2 maths problems for homework and I really dont know how to help her with these. can someone please explain how to work out the answers to the following problems so I can explain it to her. 1. you have 5 pizza toppings(sausage, onion, peppers, pepperoni and mushroom), how many different varities of pizzas can you make with these toppings? There is 1 pizza with 0 toppings There are 5 pizzas with 1 topping There are 4 pizzas with sausage and one other topping, there are three with onion but not sausage, there two pizzas with peppers and something else other than sausage or onion, and one with peperoni and mushrooms, so there are a total of 4+3+2+1=10 two topping pizzas. Now there are as many three topping pizzas as there are two topping pizzas as each", "doing. What’s the smallest number of mushrooms Kaylin could’ve eaten? It doesn’t matter what the exact numbers are except that she can’t eat more than she has; what really matters is the difference between every two consecutive numbers. In particular, she doesn’t have to eat the mushrooms that are left on her plate at the end.The first example works like this:10 5 15 5 She had 10 mushrooms at time 0. She had 5 mushrooms at time 10. This means she must’ve eaten 5. She had 15 mushrooms at time 20. That’s fine, Bartholomew must’ve added some, but we don’t care. She had 5 mushrooms at time 30. This means she must’ve eaten 10. She left the 5 mushrooms on her plate and walked away. That’s it, she has eaten 5 + 10 = 15.Now, if she eats at a constant rate (in the second case we’re interested in), that rate must be at least 5 per 10 seconds because she ate 5 between times 0 and 10, and it must be at least 10 per 10 seconds because she ate 10 between times 20 and 30. Overall, the rate is at least 10 per 10 seconds. With this, we get: She had 10 mushrooms at time 0. She had 5 mushrooms at time 10. We know from", "we can have a pizza with a slice with a topping next to a slice with the same topping. – Emi987 Nov 18 '14 at 23:16 • @Emi Oh... The question needs rewording... – warspyking Nov 18 '14 at 23:23 Pizza's not pizza without cheese; you don't call up the place, order a pizza \"with just pepperoni\", and expect it to show up without cheese (or sauce). A \"normal\" amount of cheese is included in the cost of the pie, and you usually don't save any money ordering a cheeseless pizza. So, I will interpret Z as the number of slices that have only cheese. Given your other requirements, Z = 1 because no two slices are alike, therefore no two slices can have only cheese. The rest of it is simple; there is one slice with pepperoni, one slice with mushrooms, and one slice with both pepperoni and mushrooms, for four unique equal slices of pizza. Reminds me of school..:P This reminds me of binary digits , combination 3 0's and 3 1's ..... So, in that case from 000 to 111 is valid for 3 binary digits . Hence (111)2 = (7)10. So 7 slices Well, the fact none can be identical means nothing. It limits nothing but that it cannot be split into even sized", "The mincing problem How many cuts it take to mince a clove of garlic into one million pieces? A professional chef notes that the usual method will do: First, she slices the garlic one-thousand times along its length and one-thousand times across its width. This gives $1000\\times 1000 = 1\\,000\\,000$ tiny little garlic pieces with only two-thousand cuts. (Our characters remain intentionally two-dimensional—a three-dimensional chef would rotate the garlic to slice along its height, mincing the garlic into one-million pieces with only three-hundred cuts of her knife.) A naïve chef claims a better way. “First,” he says, “I slice the garlic in two. Then, I cut each of those pieces in half with a single slice.” He continues along, carefully rearranging the garlic so that each run of his blade divides every piece in two again. “See?” he points. “Each of my slices doubles the number of garlic bits.” “Two! Four! Eight!” he shouts. “With only twenty slices, I’ll mince it to a million minuscule morsels.” Then there’s us: sloppy, lazy cooks. Unlike the professional chef, we can’t reliably cut a clove of garlic into one-hundred thin slices without risking our fingers. And, we note wryly, it’s only a matter of time before the cocksure chef discovers a problem with his approach: the final cut requires half a", "have, including having all 50 on one pizza. I said you have 50 choices the first time, 49 the next, then 48 until only 1 is left. Based on this simpler example. If you have 3 toppings you have 3x2x1 = 6 different pizzas. Example: Mushroom Tomatoes Bacon Mushroom Tomatoes Mushroom Bacon Tomatoes Bacon Bacon Mushrooms Tomato That gives 7 different kinds not... >> >>7734870 This is a simple combination problem because mushroom/tomato is the same as tomato/mushroom. There are $C(50, 50)$ ways to choose 50 toppings from 50, there are $C(50, 49)$ ways to choose 49 toppings from 50, etc. >> >>7734874 3x2x1 gives 6 different pizzas with 3 topping choices but in reality the answer is 7 because you can have each topping on it's own too. >> >>7734885 With three topping choices you can either choose three toppings or choose two toppings or choose one topping. There are $C(3,3)$ ways to choose three toppings, $C(3,2)$ ways to choose two toppings, and $C(3,1)$ ways to choose one topping. That means there are [eqn]C(3,3) + C(3,2) + C(3,1) = 7[/eqn] ways to choose toppings. Your factorial answer $3 \\times 2 \\times 1$ is not correct. It's a combination problem. [Boards: 3 / a / aco / adv / an / asp / b / biz / c", "we will also write down equivalent pairings at the same time. This will help prevent us from repeating combinations: Pairing Equivalent pairings (do not count) Red & blue blue & red Red & green green & red Red & yellow yellow & red Green & blue blue & green Green & yellow yellow & green Yellow & blue blue & yellow So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater . #### Example B Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from? Solution We’ll start by making a table and list first choice, second choice and third choice: $1^{st}$ topping $2^{nd}$ topping $3^{rd}$ topping cheese pepperoni mushroom cheese pepperoni pineapple cheese pepperoni olives cheese mushroom pineapple cheese mushroom olives cheese pineapple olives pepperoni mushroom pineapple pepperoni mushroom olives pepperoni pineapple olives mushroom pineapple olives Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a" ]

[ "How many mushrooms found each of them? 17. Dining room The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?", "a bit, and a lot of it), and 17 toppings, you have: 5*317 = 645700815 ~ 6e8. This means that if you made every permutation of your pizzas and sliced them up so each pizza had 10 slices, you could give everyone in the world a slice. http://www57.wolframalpha.com/input/?i=645700815+pizza [–] 3 points4 points (7 children) sorry, this has been archived and can no longer be voted on Isn't a mushroom and pepperoni pizza the same as a pepperoni and mushroom pizza etc...? I have a feeling we need to use Binomial coefficients here (nCr). 17c1, 17c2 ...17 c 17, but then complications arise with special cases that probably aren't allowed like, putting 17 pepperoni on a pizza. [–] 14 points15 points * (3 children) sorry, this has been archived and can no longer be voted on Yes, a mushroom and pepperoni pizza is the same as a pepperoni and mushroom pizza. The number of topping choices is the same as the number of subsets of the set of 17 toppings, which is 217. Another way to think of it is suppose I handed you a form to fill out like this: Would you like Pepperoni [ ] Yes [ ] No Mushrooms [ ] Yes [ ] No Green Peppers [ ] Yes [ ] No ... You have", "## COCI '14 Contest 5 #1 Funghi View as PDF Points: 3 (partial) Time limit: 1.0s Memory limit: 32M Problem type After having eaten all the cookies from the wicked witch's house, Hansel and Gretel ordered a jumbo pizza. The pizza arrived shortly, cut into eight pieces. Hansel and Gretel are going to split the pizza in half so that each of them gets a complete pizza \"half-circle\" or, in other words, four consecutive pieces. Gretel really likes mushrooms and wants to get as many as she can. Given the fact that some pizza slices contain less and some more mushrooms, Gretel has asked Hansel to split the pizza so that her pieces contain as many mushrooms as possible. Help Hansel and Gretel! They will tell you how many mushrooms there are on each of the eight pizza slices, and your job is to find the largest total number of mushrooms Gretel can get. The following image depicts the optimal division for the second test sample below (1. denotes the first slice given in the input data): #### Input Each of the eight lines of input contains the integer . These numbers are, respectively, the amount of mushrooms on pizza slices, where the slices are given in clockwise order. #### Output The first and only line of output", "# I am mushroom ^-^ In a mushroom forest, there are $90$ distinct mushrooms and Jimmy goes for picking all of them for his dinner. He took 3 identical bags with him, each of them can hold maximum $30$ mushrooms. If he takes home all of the mushrooms, then in how many ways can he put the mushrooms in the bags ? Write your answer as digit sum of the number. This problem is a part of the set Vegetable combinatorics ×", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "The mincing problem How many cuts it take to mince a clove of garlic into one million pieces? A professional chef notes that the usual method will do: First, she slices the garlic one-thousand times along its length and one-thousand times across its width. This gives $1000\\times 1000 = 1\\,000\\,000$ tiny little garlic pieces with only two-thousand cuts. (Our characters remain intentionally two-dimensional—a three-dimensional chef would rotate the garlic to slice along its height, mincing the garlic into one-million pieces with only three-hundred cuts of her knife.) A naïve chef claims a better way. “First,” he says, “I slice the garlic in two. Then, I cut each of those pieces in half with a single slice.” He continues along, carefully rearranging the garlic so that each run of his blade divides every piece in two again. “See?” he points. “Each of my slices doubles the number of garlic bits.” “Two! Four! Eight!” he shouts. “With only twenty slices, I’ll mince it to a million minuscule morsels.” Then there’s us: sloppy, lazy cooks. Unlike the professional chef, we can’t reliably cut a clove of garlic into one-hundred thin slices without risking our fingers. And, we note wryly, it’s only a matter of time before the cocksure chef discovers a problem with his approach: the final cut requires half a", "# A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How many different ways can a pizza be made with 2 toppings?​ ## This Post Has 3 Comments 1. Chartwig4576 says: The answer is 21 as there are 7 toppings and u can only put two toppings on each pizza so in the end u hv to do 7 * 3 to get 21 2. lisaxo says: I am verry bad at imaging :^ 3. silasjob09 says: Yes I think this question is weird", "and Grow Question 14. Your hamster weighs $$\\frac{13}{16}$$ ounce. Your friend’s hamster weighs $$\\frac{6}{8}$$ ounce. Do the hamsters weigh the same amount? Use a common denominator to write equivalent fractions for the hamsters weight. Step 1: Use the product of the denominators: 16 $\\dpi{100} \\small \\times$ 8 = 128 Step 2: Write equivalent fractions with denominators of 128 $\\dpi{100} \\small \\frac{13}{16} = \\frac{13 \\times 8}{16 \\times 8} = \\frac{104}{128}$ $\\dpi{100} \\small \\frac{6}{8} = \\frac{6 \\times 16}{8 \\times 16} = \\frac{96}{128}$ Hamsters weigh the different amount. One is $\\dpi{100} \\small \\frac{104}{128}$ ounce and the other one is $\\dpi{100} \\small \\frac{96}{128}$ ounce. Question 15. DIG DEEPER! You have three vegetable pizzas of the same size. One has 4 equal slices. The second has 8 equal slices. The third has 6 equal slices. You cut the pizzas until all of them have the same number of slices. How many slices does each pizza have? Given, You have three vegetable pizzas of the same size. One has 4 equal slices. The second has 8 equal slices. The third has 6 equal slices. You cut the pizzas until all of them have the same number of slices. 4 + 6 + 8 = 18 one has $$\\frac{4}{18}$$ second has $$\\frac{6}{18}$$ Third has $$\\frac{8}{18}$$ Total there are 18 slices. ### Find Common Denominators Homework", "have, including having all 50 on one pizza. I said you have 50 choices the first time, 49 the next, then 48 until only 1 is left. Based on this simpler example. If you have 3 toppings you have 3x2x1 = 6 different pizzas. Example: Mushroom Tomatoes Bacon Mushroom Tomatoes Mushroom Bacon Tomatoes Bacon Bacon Mushrooms Tomato That gives 7 different kinds not... >> >>7734870 This is a simple combination problem because mushroom/tomato is the same as tomato/mushroom. There are $C(50, 50)$ ways to choose 50 toppings from 50, there are $C(50, 49)$ ways to choose 49 toppings from 50, etc. >> >>7734874 3x2x1 gives 6 different pizzas with 3 topping choices but in reality the answer is 7 because you can have each topping on it's own too. >> >>7734885 With three topping choices you can either choose three toppings or choose two toppings or choose one topping. There are $C(3,3)$ ways to choose three toppings, $C(3,2)$ ways to choose two toppings, and $C(3,1)$ ways to choose one topping. That means there are [eqn]C(3,3) + C(3,2) + C(3,1) = 7[/eqn] ways to choose toppings. Your factorial answer $3 \\times 2 \\times 1$ is not correct. It's a combination problem. [Boards: 3 / a / aco / adv / an / asp / b / biz / c", "pizza? (1) the pizza requires 33 grams of ingredients (2) the ratio of the number of grams of flour to the total number of grams used in the recipe is 1:3 _________________ Director Joined: 20 Feb 2015 Posts: 795 Concentration: Strategy, General Management Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] ### Show Tags 31 May 2016, 03:06 c:f = 2:1 1. 33 grams of ingredients cheese = (2/3)*33 = 22 grams sufficient !! 2. flour: t-flour = 1:3 insufficient !! A Retired Moderator Joined: 19 Mar 2014 Posts: 931 Location: India Concentration: Finance, Entrepreneurship GPA: 3.5 Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] ### Show Tags 18 Aug 2017, 14:20 Bunuel wrote: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 gram, respectively. If cheese and flour are the only ingredients in the recipe, how many grams of cheese are used when making a pizza? (1) the pizza requires 33 grams of ingredients (2) the ratio of the number of grams of flour to the total number of grams used in the recipe is 1:3 $$\\frac{Chees}{Floor} = \\frac{C}{F} = \\frac{2}{1}$$ (1) the pizza requires 33 grams of ingredients Total", "yellow So there are 6 distinct combinations. There are also 6 “repeat” pairings – for every pair of colors we choose there is 1 combination but 2 permutations. Anne can choose from six distinct color pairs for her sweater. #### Real-World Application: Pizza Toppings Triominoes Pizza Company specializes in 3-topping pizzas. If the available toppings are cheese, pepperoni, mushroom, pineapple and olives, how many different 3-topping combinations can customers choose from? We’ll start by making a table and list first choice, second choice and third choice: \\begin{align*}1^{st}\\end{align*} topping \\begin{align*}2^{nd}\\end{align*} topping \\begin{align*}3^{rd}\\end{align*} topping cheese pepperoni mushroom cheese pepperoni pineapple cheese pepperoni olives cheese mushroom pineapple cheese mushroom olives cheese pineapple olives pepperoni mushroom pineapple pepperoni mushroom olives pepperoni pineapple olives mushroom pineapple olives Note that as we progress through the choices for first topping, the number of combinations we have for the second and third toppings get fewer. This is because some combinations have already been used, in a different order. By counting the table entries we see that there are 10 possibilities for a 3-topping pizza. #### Determining Combinations by Looking at Permutations You can see that there are always fewer combinations than permutations in a given situation – but you should also see that knowing which combinations have already been used is important to avoid counting combinations", "CasyC77 does this make sense? So with this detail how many could you make? I am sorry, when you said the work \"these\" it looked like \"three\" to me. Thus, I assumed you were asking \"how many can you make if you use 3 different one ... \". I have to leave now, maybe, tomorrow I will answer. 7. Originally Posted by CasyC77 Can someone pleae help me urgently..... I feel really silly at the moment. My daughter came home with 2 maths problems for homework and I really dont know how to help her with these. can someone please explain how to work out the answers to the following problems so I can explain it to her. 1. you have 5 pizza toppings(sausage, onion, peppers, pepperoni and mushroom), how many different varities of pizzas can you make with these toppings? There is 1 pizza with 0 toppings There are 5 pizzas with 1 topping There are 4 pizzas with sausage and one other topping, there are three with onion but not sausage, there two pizzas with peppers and something else other than sausage or onion, and one with peperoni and mushrooms, so there are a total of 4+3+2+1=10 two topping pizzas. Now there are as many three topping pizzas as there are two topping pizzas as each", "Therefore 2*x+1*x= T (total gms of cheese and flour combined in the pizza) (1) T=33 3x=33 Hence sufficient (2) F/(C+F) = 1/3 C/F=2/1. This is given in the question. Hence insufficient Ans : A Kudos [?]: 12 [0], given: 30 Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] 20 Aug 2017, 06:07 Display posts from previous: Sort by", "(2) the ratio of the number of grams of flour to the total number of grams used in the recipe is 1:3 C/F = 2/1 Therefore 2*x+1*x= T (total gms of cheese and flour combined in the pizza) (1) T=33 3x=33 Hence sufficient (2) F/(C+F) = 1/3 C/F=2/1. This is given in the question. Hence insufficient Ans : A Re: A pizza recipe calls for cheese and flour in the ratio of 2 grams to 1 [#permalink] 20 Aug 2017, 06:07 Display posts from previous: Sort by", "### Home > A2C > Chapter 10 > Lesson 10.2.6 > Problem10-154 10-154. Another pizza parlor is having a super special. They have a rack of small pizzas, each with three or four different toppings. The price is really low because they forgot to indicate on the boxes what toppings were on each pizza. They had eight toppings available, and they made one pizza for each of the possible combinations of three or four toppings. 1. How many pizzas are on the rack? • Find how many types of pizzas with three toppings can be made. • 8C3 • Find how many pizzas with four toppings can be made. • 8C4 • 8C3 + 8C4 = 56 + 70 = 126 1. What is the probability of getting a pizza that has mushrooms on it? If mushrooms are a known topping, then you choose one fewer topping from only 7 remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "Home > A2C > Chapter 10 > Lesson 10.2.6 > Problem10-154 10-154. 1. Another pizza parlor is having a super special. They have a rack of small pizzas, each with three or four different toppings. The price is really low because they forgot to indicate on the boxes what toppings were on each pizza. They had eight toppings available, and they made one pizza for each of the possible combinations of three or four toppings. Homework Help ✎ 1. How many pizzas are on the rack? 2. What is the probability of getting a pizza that has mushrooms on it? Find how many types of pizzas with three toppings can be made. 8C3 Find how many pizzas with four toppings can be made. 8C4 8C3 + 8C4 = 56 + 70 = 126 If mushrooms are a known topping, then you choose one fewer topping from only 7 remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "million fragments of garlic arranged into a perfect line! Instead of risking our fingers or sanity, we’re going to use a technique known to professionals and amateurs alike: we will chop the garlic at random, over and over, until we generate one million pieces. How many chops does our simple approach require? The answer: Our rustic chop will mince almost as quickly as the professional’s honed technique. Put mathematically, the number of cells in our random chop also grows quadratically with the number of slices we make. With $$n$$ slices, our technique produces $$n + n(n-1)/ 4\\pi$$ bits of garlic when the garlic is a circle. Taking $$n=3540$$ yields, on average, just over one million pieces of garlic. Neat!", "that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 04:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 04:28 Display posts from previous: Sort by", "If $$n$$ is even the maximum vertical stack of mushrooms left un-stacked would be $$2,4,6,8.....$$ depending on value of $$m$$. And in case of odd $$n$$ the number of mushrooms left unstacked would be $$1,3,5,7.....$$ depending on value of $$m$$. For eg. if $$n=6$$ and $$m=4$$ the maximum number of mushrooms left out would be 2. And we will have to subtract the cases where taken mushrooms are not fully stacked as it would lead to repetition of cases. Let's introduce another variable $$t$$ which represents the maximum number of mushrooms left out which can also be written in other form $$2m-n$$. Now, we can find the number of cases to be subtracted from $$P_{m}$$ to get the desired cases. We have to arrange $$m$$ mushrooms so we will have to subtract those cases where the number of horizontal rows is more than $$n-m$$. In order to achieve this we will first lay the mushrooms in horizontal row and work our way backwards by picking up one at a time and then arranging them too. So the number of discarded cases would be $$P_{0} + P_{1} + P_{2} ...$$ until the point we pick up less mushrooms than we have bases to arrange them on. And we will be faced by the same dilemma because of which we", "our minimum speed that she must’ve eaten at least 10. Hence these 5 mushrooms must’ve been newly added by Bartholomew. She had 15 mushrooms at time 20. We know from our minimum speed that she must’ve eaten the 5 mushrooms she previously had. Bartholomew could’ve added some new mushrooms (at least 15) at any point between times 10 and 20. If he added them before time 20, then Kaylin would’ve already eaten some of them, but we want to minimize the amount she eats, so we can assume Bartholomew added exactly 15 new mushrooms exactly at time 20. She had 5 mushrooms at time 30. We know from our minimum speed that she must’ve eaten 10 out of the 15 mushrooms she previously had. She then left the 5 remaining mushrooms on her plate and walked away. So, if Kaylin ate at a constant rate, she must’ve eaten at least 10 + 5 + 10 = 25 mushrooms. • » » » » 6 years ago, # ^ | +2 Thanks. I got it after reading \"\"we'll look at how many pieces of mushroom are on her plate\", but sadly, that was after the contest.You should have been the one in charge of writing the statement, not the one who actually did it. • » » » »" ]

[ "= 4$$, then determine the value of $$\\Large \\frac{dL}{dt}$$. (A calculator is recommended.)", "The value of the expression $\\frac{1}{1+ \\log_u \\: vw} + \\frac{1}{1+ \\log_v \\: wu} + \\frac{1}{1+\\log_w uv}$ is _______", "is larger than $\\frac{1}{6}$. Therefore, we can assume that $\\frac{log_2{x}}{x}$ equals to $\\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\\frac{(2^t-3t)t}{4^t}$ is $\\textbf{(C)}\\ \\frac{1}{12}$.", "at $Q(\\large\\frac{3}{2},\\frac{15}{4})$ The minimum value is $\\large\\frac{22}{7}$ at $S(\\large\\frac{18}{7},\\frac{2}{7})$", "What is the value of $\\frac{1}{2}$ of $\\frac{2}{3}$ of $\\frac{3}{4}$ of $\\frac{4}{5}$ of $\\frac{5}{6}$ of $\\frac{6}{7}$ of $\\frac{7}{8}$ of $\\frac{8}{9}$ of $\\frac{9}{10}$ of $1000$?", "increases to $2$, the value of $\\frac{1}{x}$ comes down and it is minimum wheen $x = 2$ and it is $1 + \\frac{1}{2} = 1.5$ graph{(x^2+1/x)/(x^2-(1-1/x^2)/(1/x+1/x^2) [-0.983, 4.017, 0.23, 2.73]}", "Browse Questions Find the principal value of $\\cos^{-1} \\bigg( -\\frac{1}{2} \\bigg).$ Ans: $\\frac{2\\pi}{3}$", "# $\\frac{a+b^{22}}{a^2+b^{11}}$ If $a>b>1$ and $\\log_a b+\\frac{6}{11}\\log_b a=\\frac{67}{11},$ what is the value of $\\frac{a+b^{22}}{a^2+b^{11}}?$ ×", "• # question_answer What is the value of $\\frac{\\mathbf{1}}{\\mathbf{2}}\\mathbf{lo}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{36-21o}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{3+lo}{{\\mathbf{g}}_{\\mathbf{10}}}\\mathbf{15?}$ A) 2 B) 3 C) 1D) 0 (c): $\\frac{1}{2}{{\\log }_{10}}36-2lo{{g}_{10}}3+lo{{g}_{10}}15$ $=lo{{g}_{10}}{{36}^{1/2}}-lo{{g}_{10}}{{3}^{2}}+lo{{g}_{10}}15$ $=lo{{g}_{10}}6-lo{{g}_{10}}9+lo{{g}_{10}}15$ $={{\\log }_{10}}\\frac{6\\times 15}{9}={{\\log }_{10}}\\frac{90}{9}={{\\log }_{10}}10=1$", "# What is the value of (1/8)^2? May 2, 2018 $\\frac{1}{64}$ You can multiply $\\frac{1}{8}$ by itself (it's being squared): $\\frac{1}{8} \\cdot \\frac{1}{8} = \\frac{1}{64}$ ${\\left(\\frac{a}{b}\\right)}^{c} = {a}^{c} / {b}^{c} \\rightarrow$ \"Distribute\" out the exponent ${\\left(\\frac{1}{8}\\right)}^{2} = {1}^{2} / {8}^{2} = \\frac{1}{64}$", "well defined. If $x = \\frac{75 - 3 \\sqrt{105}}{26} \\approx 1.702$ then $5 - x > 0$ and $35 - {x}^{3} \\approx 35 - 4.93 > 0$, so both logs are well defined Real valued.", "# How do you find the value of 1/3(4-7^2)? ##### 1 Answer Nov 17, 2016 This is simply simplification. $\\frac{1}{3} \\left(4 - 49\\right)$ $\\frac{1}{3} \\left(- 45\\right)$ $- \\frac{45}{3}$", "Browse Questions # Find the principal value of $\\cos^{-1} \\bigg( -\\frac{1}{2} \\bigg).$ Ans: $\\frac{2\\pi}{3}$", "Find the principal value of $\\cos^{-1} \\bigg( -\\frac{1}{2} \\bigg).$ Ans: $\\frac{2\\pi}{3}$ answered Feb 15, 2013", "Step 4: So, $8 \\frac{1}{2} = 8.5$ converting_mixed_number_with_denominator_2_4_5_decimal.htm", "the value of: (i) $$20_{{c}_{18}}$$ ? (ii) $$5_{{p}_{4}}$$ ? Solution: (i) Given, $$20_{{c}_{3}}$$ = $$\\frac{20 * 19 * 18}{3!}$$ = $$\\frac{20 * 19 * 18}{3 * 2 * 1}$$ = 1140 (ii) Given, $$5_{{p}_{4}}$$ = $$\\frac{5!}{4!}$$ = $$\\frac{5 * 4 * 3 * 2 * 1}{4 * 3 * 2 * 1}$$ = 5", "• question_answer In the given figure what is the value of$\\mathbf{3sin}\\theta +\\mathbf{4cos}\\theta$? A) 3 B) 4 C) 5 D) 1 (c) $AC=\\sqrt{{{3}^{2}}+{{4}^{2}}}=5$ $\\therefore 3sin\\theta +4cos\\theta$ $=3\\times \\frac{3}{5}+4\\times \\frac{4}{5}$ $=\\frac{9}{5}+\\frac{16}{5}=\\frac{25}{5}=5$", "{3}{2^2} \\approx 0.477 - 2 \\times 0.301$. It is useful to know what $\\log 2, \\log 3$ is approximately equal to. After the first year, the value is $\\frac{3}{4}\\times 900$. After another year, it is $\\frac{3}{4}\\times\\frac{3}{4}\\times 900$. Continuing this way, one sees that after eight years, the value is $(\\frac{3}{4})^8\\times 900$.", "## Algebra and Trigonometry 10th Edition $A=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x=4$: $A=8$ $A=-\\frac{x^2}{2}+4x$ (see item (a)) $A=-\\frac{1}{2}(x^2-8x)=-\\frac{1}{2}(x^2-8x+16-16)$ $A=-\\frac{1}{2}[x^2-2(4)x+4^2-16]$ $A=-\\frac{1}{2}[(x-4)^2-16]=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x-4=0$, that is, when $x=4$: $A=-\\frac{(4-4)^2}{2}+8=8$", "= \\frac{1}{3} \\ln 8 = \\ln 8^{1/3}$ using the usual log rule $a \\ln b = \\ln b^a$. And you should know that $8^{1/3} = \\sqrt[3] 8 = 2$. • June 13th 2009, 05:26 PM justin016 Quote: Originally Posted by mr fantastic $\\frac{\\ln 8}{3} = \\frac{1}{3} \\ln 8 = \\ln 8^{1/3}$ using the usual log rule $a \\ln b = \\ln b^a$. And you should know that $8^{1/3} = \\sqrt[3] 8 = 2$. Okay i got it, thanx alot" ]

[ "Step 4: So, $8 \\frac{1}{2} = 8.5$ converting_mixed_number_with_denominator_2_4_5_decimal.htm", "FearTheBeard21 9 # Ratio and Proportions z to 4 = 8/5 z = ? $\\frac{z}{4}=\\frac{8}{5}\\\\ 5z=32\\\\ x=\\frac{32}{5}$ $\\frac{z}{4}=\\frac{8}{5}\\ \\ \\ \\ \\ |multiply\\ both\\ sides\\ by\\ 4\\\\\\\\z=4\\times\\frac{8}{5}\\\\\\\\z=\\frac{32}{5}\\\\\\\\\\boxed{z=6\\frac{2}{5}=6.4}$", "# How do you simplify the ratio 10.4 to 8.4? $\\frac{26}{21}$ $\\frac{10.4 \\times 10}{8.4 \\times 10} = \\frac{104}{84}$ Common factor is 4$\\to$ see prime factor tree diagram $\\frac{104 \\div 4}{84 \\div 4} = \\frac{26}{21}$", "FredrickMaughn203 6 # Is 3/8 larger than 3/4? Multiply $\\frac{3}{4}$ by 2 to get a common denominator. That'll then show you that $\\frac{3}{4}$ is bigger, as: $\\frac{3}{4}$ × 2 = $\\frac{6}{8}$ and, $\\frac{6}{8}$ is bigger than $\\frac{3}{8}$", "$(\\frac{1}{1} + \\frac{2}{2} + \\frac{5}{3}) + (\\frac{2}{2} + \\frac{5}{3}) + (\\frac{5}{3}) = 8$. The brackets denote the inner sum $\\sum_{j=i}^{n}{\\frac{a_j}{j}}$, while the summation of brackets corresponds to the sum over $i$.", "$\\frac{1}{2} + \\frac{3}{2} + c \\:=\\:8\\quad\\Rightarrow\\quad\\boxed{c = 6}$ Therefore: . $f(n) \\;=\\;\\frac{1}{2}n^2 + \\frac{3}{2}n + 6\\quad\\Rightarrow\\quad \\boxed{f(n)\\;=\\;\\frac{1}{2}(n^2 + 3n + 12)}$", "[2] . . and hence: . $8B \\,=\\,9G\\quad\\Rightarrow\\quad G\\,=\\,\\frac{8}{9}B$ [3] Substitute [2] and [3] into [1]: . $\\frac{\\frac{2}{3}B + \\frac{2}{3}B}{B + \\frac{8}{9}B} \\;=\\;\\frac{\\frac{4}{3}B}{\\frac{17}{9}B} \\;= \\;\\boxed{\\frac{12}{17}}$", "\\frac {n \\paren {n - 2} } 8$ $\\ds$ $=$ $\\ds \\frac {n \\paren {n - 2} } {24} \\paren {n - 1 + 3}$ factorizing $\\ds$ $=$ $\\ds \\frac {n \\paren {n - 2} \\paren {n + 2} } {24}$ By doing the substitution $n \\to n - 1$, we obtain the formula for odd $n$: $\\dfrac {\\paren {n - 3} \\paren {n - 1} \\paren {n + 1} } {24}$ $\\blacksquare$", "# How do you simplify the ratio 10.4 to 8.4? Dec 22, 2016 $\\frac{26}{21}$ $\\frac{10.4 \\times 10}{8.4 \\times 10} = \\frac{104}{84}$ Common factor is 4$\\to$ see prime factor tree diagram $\\frac{104 \\div 4}{84 \\div 4} = \\frac{26}{21}$", "# What is the ratio of 8/24? Jun 17, 2018 $\\frac{1}{3}$ #### Explanation: Divide the denominator and numerator on 8, and you get your answer. Jun 19, 2018 $\\frac{1}{3}$ #### Explanation: $\\frac{8}{24}$ $= \\frac{1 \\times \\cancel{8}}{3 \\times \\cancel{8}}$ Jun 20, 2018 $\\frac{1}{3}$ #### Explanation: Since both the numerator and denominator have a common factor of $8$, we can divide them by $8$. This gives us $\\frac{1}{3}$ Hope this helps!", "\\frac{4}{8} + \\frac{6}{8} \\;=\\;\\frac{22}{8} \\;=\\;2\\frac{3}{4}$ Therefore: . $616 + 2\\frac{3}{4} \\;=\\;\\boxed{618\\frac{3}{4}}$ 4. ## thanks!!!! i was just stuck on the part where it said 22/8 thats all but thank u!!", "since $$8= 4(2)$$ and $$\\sqrt{8}= 2\\sqrt{2}$$, the simplest form is $$\\frac{2\\sqrt{2}}{3}= \\frac{2}{3}\\sqrt{2}$$.", "the bottom.. when the exponent isn't negative. $ 8x^{-1/2}+\\frac{4}{7}x^{1/2} $ $\\frac{8}{x^{1/2}} + \\frac{4x^{1/2}}{7}$ common denominator is $7x^{1/2}$ ... • Apr 18th 2010, 06:36 PM desiderius1 $ 8 + 4x^{1/2} $ How do you get 4(14 + x)? • Apr 18th 2010, 06:48 PM desiderius1 oh I see, you multiply 8 and 7 to get 56. 56/14 = 4.", "+ 1 , 4 k , 3 k + 5$ becomes (with a little more effort) $\\frac{8}{13} , - \\frac{20}{13} , \\frac{50}{13}$ with a common ratio of $\\left(- \\frac{5}{2}\\right)$", "b √c I think you can take it to: √1/ √2 + √1/ √4 + √1/8..... but does that actually help me?! The steps of what you did would be greatly appreciated! ajkerr $\\sqrt{\\frac{1}{2}}=\\frac{1}{\\sqrt {2}}$ $=\\frac{1}{\\sqrt {2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$ Now you try to simplify further. for adding, make the denominator same of each term. 5. Originally Posted by Shyam $\\sqrt{\\frac{1}{2}}=\\frac{1}{\\sqrt {2}}$ $=\\frac{1}{\\sqrt {2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$ Now you try to simplify further. soo..... then √1/4 = 1/ √4 and then..... 1/ √4 x √4/ √4 = √4/4 then √1/8 = 1/ √8 and then 1/ √8 x √8/ √8 = √8/8 leaving me with... √2 /2 + √4 /4+ √8 /8 denominator to... 4√2 / 8 + 2√4 / 8 + 8√8 to get (4√2+2√4+8√8) / 8 -- is that right? then what? 6. Originally Posted by ajkerr soo..... then √1/4 = 1/ √4 and then..... 1/ √4 x √4/ √4 = √4/4 then √1/8 = 1/ √8 and then 1/ √8 x √8/ √8 = √8/8 leaving me with... √2 /2 + √4 /4+ √8 /8 and then do I times them up to get a common denominator? you realize $\\sqrt{4} = 2$, right? so you have $\\frac 12 + \\frac {\\sqrt{2}}2 + \\frac {\\sqrt{8}}8$ now add the last two fractions 7. Originally Posted by Jhevon you realize", "# How do you evaluate 7/8 + 8/7 ? May 28, 2016 You would have to find a common denominator. #### Explanation: In this case, I use 56 as the common denominator. $\\frac{7}{8} \\cdot \\frac{7}{7} = \\frac{49}{56}$ $\\frac{8}{7} \\cdot \\frac{8}{8} = \\frac{64}{56}$ $\\frac{49}{56} + \\frac{64}{56}$ $\\frac{113}{56}$ Now, are you able to simplify the value any further?", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "ScottyClermont 5 # what is larger 3/8 or 5/16 $\\frac{3}{8} ... \\frac{5}{16}$ First, find the least common denominator between the two. Since 8 is a factor of 16, the LCD is 16. $\\frac{3*2}{8*2}$ $\\frac{3}{8}= \\frac{6}{16}$ Now, you can compare the numerators. $6>5$ Since 6 is greater than 5, you know that $\\frac{6}{16} > \\frac{5}{16}$. Therefore, we now know that $\\frac{3}{8}$ is larger. $\\frac{3}{8}=\\frac{3\\codt2}{8\\cdot2}=\\frac{6}{16} > \\frac{5}{16}\\\\\\\\Answer:\\frac{3}{8}\\ is\\ larger\\ than\\ \\frac{5}{16}.$", "How do you simplify 1/4 + 6/8? $\\frac{1}{4} + \\frac{6}{8} = \\frac{1}{4} + \\frac{3}{4} = \\frac{4}{4} = 1$ $\\mathmr{and} \\frac{1}{4} + \\frac{6}{8} = \\frac{2}{8} + \\frac{6}{8} = \\frac{8}{8} = 1$", "\\frac{3\\cdot 7}{7\\cdot 8} = \\frac{16}{56} + \\frac{21}{56} = \\frac{37}{56}. \\end{align*} But again, there is nothing magic about using the LCM. In the first example, if you wanted to use a common denominator of $24$ and just cross-multiply to find the numerators, that would work fine: $$\\frac{1}{4} + \\frac{5}{6} = \\frac{1\\cdot 6}{4\\cdot 6} + \\frac{5\\cdot 4}{4\\cdot 6} = \\frac{6}{24} + \\frac{20}{24} = \\frac{26}{24} = \\frac{13}{12}.$$ • You can use any denominator that is a multiple of both of the denominators in question. The LCM is the least of these, but as you point out, it may be simpler just to cross-multiply to avoid having to find the LCM. As you can see above, they both give the same answer. And the reason is that, for example, $\\frac{6}{24} = \\frac{3}{12}$ in the above --- using a larger common denominator simply means you are multiplying numerator and denominator by a larger number, which of course does not change its value. – rogerl Oct 16 '14 at 21:51" ]

[ "What is the value of $\\frac{1}{2}$ of $\\frac{2}{3}$ of $\\frac{3}{4}$ of $\\frac{4}{5}$ of $\\frac{5}{6}$ of $\\frac{6}{7}$ of $\\frac{7}{8}$ of $\\frac{8}{9}$ of $\\frac{9}{10}$ of $1000$?", "Step 4: So, $8 \\frac{1}{2} = 8.5$ converting_mixed_number_with_denominator_2_4_5_decimal.htm", "# What is the value of (1/8)^2? May 2, 2018 $\\frac{1}{64}$ You can multiply $\\frac{1}{8}$ by itself (it's being squared): $\\frac{1}{8} \\cdot \\frac{1}{8} = \\frac{1}{64}$ ${\\left(\\frac{a}{b}\\right)}^{c} = {a}^{c} / {b}^{c} \\rightarrow$ \"Distribute\" out the exponent ${\\left(\\frac{1}{8}\\right)}^{2} = {1}^{2} / {8}^{2} = \\frac{1}{64}$", "FredrickMaughn203 6 # Is 3/8 larger than 3/4? Multiply $\\frac{3}{4}$ by 2 to get a common denominator. That'll then show you that $\\frac{3}{4}$ is bigger, as: $\\frac{3}{4}$ × 2 = $\\frac{6}{8}$ and, $\\frac{6}{8}$ is bigger than $\\frac{3}{8}$", "is larger than $\\frac{1}{6}$. Therefore, we can assume that $\\frac{log_2{x}}{x}$ equals to $\\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\\frac{(2^t-3t)t}{4^t}$ is $\\textbf{(C)}\\ \\frac{1}{12}$.", "problem. When $b=4$, what is $\\frac{b}{c}$? Here is the next step: \\begin{aligned} \\cos A\\,\\frac{dA}{dt}&=\\frac{1.5}{8}\\\\ \\frac{b}{c} \\frac{dA}{dt}&=\\frac{1.5}{8}\\\\ \\frac{c}{b}\\cdot\\frac{b}{c}\\frac{dA}{dt}&=\\frac{c} {b}\\cdot\\frac{1.5}{8}\\\\ \\frac{dA}{dt}&=\\frac{c}{b}\\cdot\\frac{1.5}{8}. \\end{aligned} 9. 4/8 or 1/2 10. If the first answer is (3/8), then the second one is -(3/8) right? 11. Originally Posted by Velvet Love 4/8 or 1/2 Correct. The value of $\\frac{dA}{dt}$ is therefore $\\frac{dA}{dt}=\\frac{2}{1}\\cdot\\frac{1.5}{8}.$", "# Thread: mixed number exponents 1. ## mixed number exponents 8 -1 1/3 what is a value? 2. Originally Posted by shawn9356 8 -1 1/3 what is a value? First change the mixed number $1\\frac{1}{3}$ to an improper fraction, $1\\frac{1}{3}=\\frac{4}{3}$ by multiplying the whole number 1 to the denominator and adding it to the numerator. Now we have $8-\\frac{4}{3}$ In order to get a common denominator on the eight we have to multiply the top and the bottom by 3, giving us, $\\frac{24}{3}-\\frac{4}{3}=\\frac{20}{3}=6\\frac{2}{3}$", "• question_answer The value of$\\cos \\frac{\\pi }{{{2}^{2}}}.\\cos \\frac{\\pi }{{{2}^{3}}}....\\cos \\frac{\\pi }{{{2}^{10}}}.\\sin \\frac{\\pi }{{{2}^{10}}}$ is: A) $\\frac{1}{512}$ B) $\\frac{1}{1024}$ C) $\\frac{1}{256}$ D) $\\frac{1}{2}$ Using formula$\\frac{\\sin {{2}^{n}}A}{{{2}^{n}}\\sin A}$ $=\\cos A\\cos 2A\\cos {{2}^{2}}A...\\cos {{2}^{n-1}}A$", "## Algebra and Trigonometry 10th Edition $A=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x=4$: $A=8$ $A=-\\frac{x^2}{2}+4x$ (see item (a)) $A=-\\frac{1}{2}(x^2-8x)=-\\frac{1}{2}(x^2-8x+16-16)$ $A=-\\frac{1}{2}[x^2-2(4)x+4^2-16]$ $A=-\\frac{1}{2}[(x-4)^2-16]=-\\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x-4=0$, that is, when $x=4$: $A=-\\frac{(4-4)^2}{2}+8=8$", "### Home > PC > Chapter 8 > Lesson 8.2.1 > Problem8-71 8-71. 1. Find the following sums. You can use your SUM program or the sum sequence commands from your calculator. Homework Help ✎ 1. What is the value of ? $=\\frac{1}{0!}+\\frac{1}{1!}+\\frac{1}{2!}+\\frac{1}{3!}+\\frac{1}{4!} =1+1+\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{24} =2.708333$ 2.71828 e", "# What is the value? Algebra Level 2 Find the value of $$\\frac{1}{1+\\sqrt{2}}$$+$$\\frac{1}{\\sqrt{2}+\\sqrt{3}}$$+$$\\frac{1}{\\sqrt{3}+\\sqrt{4}}$$+$$\\frac{1}{\\sqrt{4}+\\sqrt{5}}$$+$$\\frac{1}{\\sqrt{5}+\\sqrt{6}}$$+$$\\frac{1}{\\sqrt{6}+\\sqrt{7}}$$+$$\\frac{1}{\\sqrt{7}+\\sqrt{8}}$$+$$\\frac{1}{\\sqrt{8}+\\sqrt{9}}$$", "$(\\frac{1}{1} + \\frac{2}{2} + \\frac{5}{3}) + (\\frac{2}{2} + \\frac{5}{3}) + (\\frac{5}{3}) = 8$. The brackets denote the inner sum $\\sum_{j=i}^{n}{\\frac{a_j}{j}}$, while the summation of brackets corresponds to the sum over $i$.", "# How do you find the unknown value of 8div2=n? Nov 24, 2016 $n = 4$ #### Explanation: Write as $\\text{ } \\frac{8}{2} = n$ Turn it round the other way and it is still true $n = \\frac{8}{2}$ Divide top and bottom of the fraction by 2 $n = \\frac{8 \\div 2}{2 \\div 2} = \\frac{4}{1} = 4$ $n = 4$", "are: $\\frac 1 2$, $\\frac 0 2$. In the second example the exact values of minimum expected values are: $\\frac{132} 8$, $\\frac{54} 8$, $\\frac{30} 8$, $\\frac{17} 8$, $\\frac{12} 8$, $\\frac 7 8$, $\\frac 3 8$, $\\frac 0 8$.", "= 4$$, then determine the value of $$\\Large \\frac{dL}{dt}$$. (A calculator is recommended.)", "since $$8= 4(2)$$ and $$\\sqrt{8}= 2\\sqrt{2}$$, the simplest form is $$\\frac{2\\sqrt{2}}{3}= \\frac{2}{3}\\sqrt{2}$$.", "at $Q(\\large\\frac{3}{2},\\frac{15}{4})$ The minimum value is $\\large\\frac{22}{7}$ at $S(\\large\\frac{18}{7},\\frac{2}{7})$", "# How do you evaluate 7/8 + 8/7 ? May 28, 2016 You would have to find a common denominator. #### Explanation: In this case, I use 56 as the common denominator. $\\frac{7}{8} \\cdot \\frac{7}{7} = \\frac{49}{56}$ $\\frac{8}{7} \\cdot \\frac{8}{8} = \\frac{64}{56}$ $\\frac{49}{56} + \\frac{64}{56}$ $\\frac{113}{56}$ Now, are you able to simplify the value any further?", "The value of $\\frac{1}{3+\\frac{1}{2-\\frac{1}{\\frac{7}{9}}}}+\\frac{17}{22}$ is : [A] $\\frac{12}{22}$ [B] $\\frac{22}{5}$ [C] 1 [D] $\\frac{5}{22}$", "The value of Question: The value of $\\frac{\\sqrt{32}+\\sqrt{48}}{\\sqrt{8}+\\sqrt{12}}$ is (a) $\\sqrt{2}$ (b) 2 (c) 4 (d) 8 Solution: $\\frac{\\sqrt{32}+\\sqrt{48}}{\\sqrt{8}+\\sqrt{12}}=\\frac{\\sqrt{16 \\times 2}+\\sqrt{16 \\times 3}}{\\sqrt{4 \\times 2}+\\sqrt{4 \\times 3}}$ $=\\frac{4 \\sqrt{2}+4 \\sqrt{3}}{2 \\sqrt{2}+2 \\sqrt{3}}=\\frac{4(\\sqrt{2}+\\sqrt{3})}{2(\\sqrt{2}+\\sqrt{3})}=2$" ]

[ "in the second. How many chairs can be around one table? • Kate shares Kate shares a 64-ounce bottle of apple cider with five friends. Each person's serving will be the same number of ounces. Between what two whole number of ounces will each person's serving to be? Explain using division.", "each row. How many students are in the class?", "How many ways can they sit?", "in a row", "# Problematic little chairs There are $$2016$$ chairs in a row and two available colours, blue and red. On how many ways can we paint all chairs, so that number of blue chairs is even? (e. g. A case in which the first two chairs are blue and other are red, is different from the case in which the last two chairs are blue and other are red) In solution section \"inspiration\" for the problem is mentioned. ×", "... with 50 more rows# }", "between chairs ...", "Room Table With Four Chairs,", "third one. How many boxes are in each row?", "of thread relevance. Four in a row.", "school play, 40 rows of chairs are set up. There are 22 chairs in each row. How many chairs are there in all? Options: a. 800 b. 840 c. 880 d. 8,800 c. 880 Explanation: As per the given data For the school play, 40 rows of chairs are available. 22 chairs are available in each row. Then total chairs in school play are = 22 x 40 By using the place value method You can think of 40 as 4 tens 22 x 40 = 22 x 4 tens = 88 tens = 880 Total chairs in school are = 880 Question 2. At West School, there are 20 classrooms. Each classroom has 20 students. How many students are at West School? Options: a. 40 b. 400 c. 440 d. 4,000 b. 400 Explanation: From the given data, Total classrooms in west school = 20 Number of students per each classroom = 20 Then, total students at West School = 20 x 20 By using the associative property You can think of 20 as 2 x 10 20 x 20 = 20 x (2 x 10) = (20 x 2) x 10 =(40) x 10 =400 Total number of students at West School = 400 Spiral Review Question 3. Alex has 48 stickers. This is 6 times", "each row", "put one on each table.", "a row", "of these in a row.", "# Permutation & Combination Two sets of the CAT paper are to be given to twelve students. In how many ways can the students be placed in two rows of six so that there should be no identical sets side by side and that the students sitting one behind the other should have the same sets ? ×", "How many mushrooms found each of them? 17. Dining room The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?", "is also enough for the third group of 22 students. So we need 32 desks in total.", "students. So we need 32 desks in total.", "day by a quarter, etc. How many times will increase its height after 6 days? • Students In the front row sitting three students and in every other row 11 students more than the previous row. Determine how many students are in the room when the room is 9 lines, and determine how many students are in the seventh row." ]

[ "7\\cdot 6$ ways to seat three people so that the end chairs are empty. For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is $1\\cdot 3\\cdot 5 = 15$ 3-digit numbers. All together, there are $75+15 = 90$ 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated. Edit: For #4 a) Digits may be repeated. So, there are $5^3 = 125$ 3-digit numbers using only odd digits. 3. ## Re: Permutations and Combinations Originally Posted by SlipEternal For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many", "of C , so we must have either A or B stand there. Whichever of the two does so, the other may take any of the remaining four positions on the other side of C, along with the other three people. Those four can then be arranged in any order and still met the requirement that C is between A and B: $$A \\ C \\quad 4! \\ \\ \\ ,$$ $$B \\ C \\quad 4! \\ \\ \\ ;$$ or $$4! \\quad C \\ A \\ \\ \\ ,$$ $$4! \\quad C \\ B \\ \\ \\ .$$ So this case accounts for $\\ 2 \\ \\cdot \\ 2 \\ \\cdot \\ 4! \\ = \\ 96 \\$ possible arrangements. In the second case, there are two positions to one side of C and three positions to the other. Again, if A is on one side of C , B must be on the other. On the side of C with two positions available, we could have A or B standing there along with one of the three other people; they can take up one of two possible arrangements. To the other side, the remaining three people may be arranged in any order. We therefore may have: $$(A \\ \\bullet \\ : \\ \\ 3 \\cdot 2!)", "# $20$ people are sitting around a (circular) table. How many ways can we choose $3$ people, no two of whom are neighbors? 0 choices for the 1st person. 17 choices for the 2nd person (must exclude 1st and his/her two neighbours) For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded). For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. So if order matters, total is $20 \\cdot (2 \\cdot 15 + 15 \\cdot 14)$ but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people. So total = $20 \\cdot \\frac{2 \\cdot 15 + 15 \\cdot 14}{6}$ just redid it; does this make any sense? • What is the answer with a straight table and how does turning the table into a corner change the problem? thats the step that you need to account for Oct 4, 2016 at 16:14 • first person - 20 choices , second person 17 choices , then third person 14 choices - then you could have chosen the 3 people in", "you are about to construct you have only 5-1=4 possibilities for $d_1$. For $d_2$ there is no restriction placed on the digit, except that it must not be 5. So there are 10-1=9 possibilities for your choice of $d_2$. Now for digit $d_3$: again, it must not be 5 but also, since it is the leading digit of your number, you must not make it 0 either. So there are 10-1-1=8 possible choices for $d_3$. Since these choices are independent of each other, the total number of possibilities is the product $8\\cdot 9\\cdot 4$ 2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded? Again, imagine yourself selecting the 3 highest placed persons by yourself. For the first place, you have 8 possibilities. For each of these 8 possibilities you have 8-1=7 possibilities to choose a person for the second place. And, finally, you have 8-2=6 possibilities to choose a person for the third place. This gives $8\\cdot 7\\cdot 6$ different ways in all.", "Distribution of five people in a car where exactly two can drive In how many ways can we distribute five people in a car if exactly two of them are can drive? - • For the next seat (we'll label it as S2), there are $4$ possibilities. • For the next seat (S3), there are $3$ possibilities. • For the next seat (S4), there are $2$ possibilities. Simply multiply the possibilities for S1, S2, S3, S4 and S5 which yields $2\\cdot 4 \\cdot 3\\cdot 2 \\cdot 1$.", "$7!/7=6!$ You can view it from a different perspective. First, you sit the first person. There is one combination to do that since all sits are identical. Then, there are $6!$ combinations to sit the other people. Therefore you have $1 \\cdot 6!$ combinations in total.", "could not be solved with the permutation formula. Note: Here you have only begun to explore counting problems. For more information about combinations, permutations, and other types of counting problems, consult a Probability text. Example A You are going on a road trip with 4 friends in a car that fits 5 people. How many different ways can everyone sit if you have to drive the whole way? Solution: A decision chart is a great way of thinking about this problem. You have to sit in the driver’s seat. There are four options for the first passenger seat. Once that person is seated there are three options for the next passenger seat. This goes on until there is one person left with one seat. $1 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=24$ Example B How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes competing? Solution: Since the order does matter with the three medals, this is a permutation problem. You will start with 12 athletes and then choose and arrange 3 different winners. $_{12}P_3=\\frac{12!}{\\left(12-3\\right)!}=\\frac{12!}{9!}=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot \\ldots}{9 \\cdot \\ldots}=12 \\cdot 11 \\cdot 10=1320$ Note that you could also use a decision chart to decide how many possibilities are there for gold", "the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3! ways. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements Example $$\\PageIndex{6}$$ You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books? Solution We first do the problem using the multiplication axiom. Since the math books go in the first three slots, there are 4 choices for the first slot, 3 choices for the second and 2 choices for the third. The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below. 4 3 2 5 4 Therefore, the number of permutations are $$4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 = 480$$. Alternately, we can see that $$4 \\cdot 3 \\cdot 2$$ is really same as 4P3, and $$5 \\cdot 4$$ is 5P2. So the answer can", "$19$ persons in total. Suppose the number of persons on the longest arc is $k>10$. Then two places of $x,y,z$ are just chosen from the two end-points of the arc, and there are $19-k$ possible places for the third person. Once the three places of $x,y,z$ are chosen, there are three possible ways to put $x,y,z$ into them clockwise. Also, note that for any $k>10$, there are $19$ ways to choose an arc of length $k$. Therefore the total number of ways (of the complement) is $$\\sum_{k=11}^{18} 3\\cdot 19 \\cdot (19-k) = 3\\cdot 19 \\cdot (1+\\cdots+8) = 3\\cdot 19\\cdot 36$$ $$3\\cdot \\binom{19}{3} - 3\\cdot 19\\cdot 36 = 3\\cdot 19 \\cdot (51 - 36) = 855$$ NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of $x,y,z$ corresponds to $3$ possible ways to put them in, and that each arc of length $k>10$ has $19$ equitable positions, it is evident that the answer should be divisible by $3\\cdot 19$, which can only be $855$ from the five choices.", "# Difference between revisions of \"2011 AIME II Problems/Problem 12\" ## Problem 12 Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. ## Solution Use complementary probability and PIE. If we consider the delegates from each country to be indistinguishable and number the chairs, we have $$\\frac{9!}{(3!)^3}$$ total ways to seat the candidates. This comes to: $$\\frac{9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4}{6\\cdot6} = 6\\cdot8\\cdot7\\cdot5 = 30\\cdot56.$$ Of these there are $$3 \\times 9 \\times \\frac{6!}{(3!)^2}$$ ways to have the candidates of at least some one country sit together. This comes to $$\\frac{27\\cdot6\\cdot5\\cdot4}6 = 27\\cdot 20.$$ Among these there are $$3 \\times 9 \\times 4$$ ways for candidates from two countries to each sit together. This comes to $$27\\cdot 4.$$ Finally, there are $$9 \\times 2 = 18.$$ ways for the candidates from all the countries to sit in three blocks (9 clockwise arrangements, and 9 counter-clockwise arrangements). So, by PIE, the total count of unwanted arrangements is $$27\\cdot 20 - 27\\cdot 4 + 18 = 16\\cdot27 + 18 = 18\\cdot25.$$ So the fraction $$\\frac mn = \\frac{30\\cdot 56 - 18\\cdot 25}{30\\cdot 56}", "PDA View Full Version : SAT math counting problems Jack 07-20-2013, 04:27 PM SAT math counting problems I need help with the following question: \"In how many different ways can 4 students be seated in two seats?\" Thanks!!:D:D Oliver 07-20-2013, 04:56 PM SAT maths counting problem permutations \"In how many different ways can 4 students be seated in two seats?\" The number of ways you can change the order of a set of objects is called the number of PERMUTATIONS of that set of objects. The total number of ways of arranging n objects, taking r at a time is gibe by the following formula: {{P}^{n}}_{r}=\\frac{n!}{(n-r)!}, where n!=1\\cdot2\\cdot3..\\cdot n So, in your case, you have 4 students taking 2 at a time {{P}^{4}}_{2}=\\frac{4!}{(4-2)!}=\\frac{4!}{2!}=\\frac{1\\cdot 2\\cdot 3\\cdot 4}{1\\cdot 2}=3\\cdot 4=12 ways. The above calculations could be done easily in your GDC casio fx-9860GII SD for example as follows: OPTN>PROB>nPr and set n=4 and r=2 Another way to solve this problem is that the first seat can be filled with 4 ways and the second with the remaining 3ways(students). So, the total ways are 4x3=12 Hope these help!!:D Jack 07-20-2013, 04:56 PM Thanks Oliver!!", "3=\\frac{6 !}{(6-3) !}=\\frac{6 !}{3 !}=\\frac{6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{3 \\cdot 2 \\cdot 1}=120$ b. 7P2 = $$7 \\cdot 6 = 42$$, or $7 \\mathrm{P} 2=\\frac{7 !}{5 !}=\\frac{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}=42$ Next we consider some more permutation problems to get further insight into these concepts. Example $$\\PageIndex{5}$$ In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other? Solution Let us suppose we have four people A, B, C, and D. Further suppose that A and B want to sit together. For the sake of argument, we tie A and B together and treat them as one person. The four people are $$\\boxed{AB}$$ CD. Since $$\\boxed{AB}$$ is treated as one person, we have the following possible arrangements. $\\boxed{AB} CD, \\boxed{AB} DC, C \\boxed{AB}D, D\\boxed{AB}C, CD \\boxed{AB}, DC\\boxed{AB} \\nonumber$ Note that there are six more such permutations because A and B could also be tied in the order BA. And they are $\\boxed{BA}CD, \\boxed{BA} DC, C\\boxed{BA}D, D\\boxed{BA}C, CD\\boxed{BA}, DC \\boxed{BA} \\nonumber$ So altogether there are 12 different permutations. Let us now do the problem using the multiplication axiom. After we tie two of", "# How many possible ways to vote? ## There are 8 judges. People can vote for 0, 1, 2, 3, or 4 people. How many ways can a person vote? May 21, 2018 163 ways. #### Explanation: There is $1$ way to vote for 0 people. There are $8$ ways to vote for 1 person. There are $\\frac{8 \\cdot 7}{2}$ ways to vote for 2 people. There are $\\frac{8 \\cdot 7 \\cdot 6}{2 \\cdot 3}$ ways to vote for 3 people. There are $\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5}{2 \\cdot 3 \\cdot 4}$ ways to vote for 4 people. This is all because you can choose people but there are ways you can order the people. For example, there are $2 \\cdot 3$ ways to order the same 3 people. Adding everything, we get $1 + 8 + 28 + 56 + 70 = 163$.", "a single $6$-cycle. Case: Two $3$-cycles • The first friend has $5$ possible friends to choose. • The friend to whom the book is given has $4$ possible choices. • The next friend has just one choice - to give their book to the first friend. • The first of the $3$ remaining friends has $2$ choices. • The next friend has $1$ choice, to give their book to the last remaining friend. • The last friend gives their book to the first friend of their $3$-cycle. So there are $5 \\cdot 4 \\cdot 2 = 40$ possible choices of two $3$-cycles.", "choices for the first digit. Once this choice has been made, there are four digits left. These can be arranged in $4!$ ways. Then $3\\times 4!$ gives the correct answer. Added Later: From what I can tell, you made two mistakes. First of all, after the choice of the first digit, you said there were three numbers left when in fact there are ... 2 We have four groups of three arranged around the circle. First choose the four seats of the managing directors. This can be done in $3$ ways. Next, distribute the four companies around the four quarters of the circle (around the four chosen seats). That can be done in $4!=24$ ways. Next, for each company, choose who sits to the left an who sits to the right ... 1 It is known as factorial and denoted as $n!$. The case $10!$ can be reduced: \\begin{eqnarray} 1 \\cdot 2 \\cdot 3 \\cdots \\cdot 8 \\cdot 9 \\cdot 10 &=& 10 \\Big(5-4\\Big) \\Big(5-3\\Big) \\cdots \\Big(5+4\\Big)\\\\ &=& 50 \\Big(25-16\\Big) \\Big(25-9\\Big) \\Big(25-4\\Big) \\Big(25-1\\Big)\\\\ &=& 50 \\Big(15-6\\Big) \\Big(15+6\\Big) \\Big(20-4\\Big) ... 1 There are $24!$ permutations of the letters b through y. For each such permutation, if a is to the left of z, it can appear in any of 15 positions (from before the first letter to before the", "1$ way. . . .The other Four people can sit at their table in $3!$ ways. Hence: . $C(5,1)\\!\\cdot\\!0!\\!\\cdot\\!3!\\text{ ways.}$ [2] Two at one table and Three at the other. . . .There are C(5,2) ways to choose the Two people. . . .They can sit at their table in $1!$ way. . . .The other Three people can sit at their table in $2!$ ways. Hence: . $C(5,2)\\!\\cdot\\!1!\\!\\cdot\\!2!\\text{ ways.}$ Therefore: . $C(5,1)\\!\\cdot\\!0!\\cdot\\!3! + C(5,2)\\!\\cdot\\!1!\\!\\cdot\\!2!\\text{ ways.}$ Many Thanks", "## Precalculus (10th Edition) $24$ The number of ways arranging $n$ distinct objects is $n!=n\\cdot(n-1)\\cdot(n-2)\\cdot...\\cdot3\\cdot2\\cdot1.$ Hence, the number of ways of arranging $4$ people in a line is: $4!=4\\cdot3\\cdot2\\cdot1=24.$", "have the same care, then we have $$\\;5!\\;= \\;5\\cdot 4\\cdot 3\\cdot 2 = 120$$ ways, since no one can have a car the same as anyone else. You logic was correct about person $2$ not having as a choice the car selected by Person $1$. By the same logic, Person $3$ cannot have as a choice either the car selected by Person $1$, nor the car selected by Person $2$, leaving Person $3$ only 3 choices, and so on. Given your clarification, then we have that the only restriction is that Person 1 and Person 2 do not have the same car, then your calculation $5\\cdot 4\\cdot 5\\cdot 5$ is correct.", "pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$. For $n=2$, there are two possibilities. The group of $6$ can be split into two groups of $3$, with each group creating a friendship triangle. The first person has $\\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations. However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality. This person has $\\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated \"across\" from one of the original three people, forming a different configuration. Thus, there are $10 \\cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$. As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\\cdot 2 =", "= |B|\\cdot k$$ so $$1<{|B|\\over |A|} = {n-k+1\\over k} \\iff k\\leq {n\\over 2}$$ You can ask yourself: In how many ways can we choose $$k$$ people, including one president, from a population of $$n$$ people? First answer: You choose $$k-1$$ people from $$n$$ people and then from the rest of them a president, so we have $${n\\choose k-1}\\cdot {n-k+1\\choose 1}$$ Second answer: You choose $$k$$ people from $$n$$ people and then from these selected you choose a president, so we have $${n\\choose k}\\cdot {k\\choose 1}$$ So we have:$${n\\choose k-1}\\cdot (n-k+1)=k{n\\choose k}$$ and from here we get $${n\\choose k-1}<{n\\choose k} \\iff k\\leq {n\\over 2}$$" ]

[ "in the second. How many chairs can be around one table? • Kate shares Kate shares a 64-ounce bottle of apple cider with five friends. Each person's serving will be the same number of ounces. Between what two whole number of ounces will each person's serving to be? Explain using division.", "7\\cdot 6$ ways to seat three people so that the end chairs are empty. For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is $1\\cdot 3\\cdot 5 = 15$ 3-digit numbers. All together, there are $75+15 = 90$ 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated. Edit: For #4 a) Digits may be repeated. So, there are $5^3 = 125$ 3-digit numbers using only odd digits. 3. ## Re: Permutations and Combinations Originally Posted by SlipEternal For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $3\\cdot 5\\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many", "number of permutations of D in chair 1 or 5 minus the permutations of both. So, there are $2\\cdot 1\\cdot 3\\cdot 2\\cdot 1 = 12$ permutations where both occur. $120-(48+24-12) = 60$. You have removed twice the cases where A sits in 3 and D sits in 1 or 5. You need to add them back in once. This is the inclusion-exclusion principle. Alternately you can remove all the cases where A sits in 3, then remove the ones where D sits in 1 or 5 but A doesn't sit in 3.", "could not be solved with the permutation formula. Note: Here you have only begun to explore counting problems. For more information about combinations, permutations, and other types of counting problems, consult a Probability text. Example A You are going on a road trip with 4 friends in a car that fits 5 people. How many different ways can everyone sit if you have to drive the whole way? Solution: A decision chart is a great way of thinking about this problem. You have to sit in the driver’s seat. There are four options for the first passenger seat. Once that person is seated there are three options for the next passenger seat. This goes on until there is one person left with one seat. $1 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=24$ Example B How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes competing? Solution: Since the order does matter with the three medals, this is a permutation problem. You will start with 12 athletes and then choose and arrange 3 different winners. $_{12}P_3=\\frac{12!}{\\left(12-3\\right)!}=\\frac{12!}{9!}=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot \\ldots}{9 \\cdot \\ldots}=12 \\cdot 11 \\cdot 10=1320$ Note that you could also use a decision chart to decide how many possibilities are there for gold", "combinations, 4 have both in the same chair. If this is not intended then these 4 combinations are invalid so 16-4 = 12. – Russell McMahon Mar 29 at 13:45 Choose 2 seats out of 4 for the two people and the 2 people can arrange themselves in $2!$ ways. Thus the answer is $$2! \\times \\binom{4}{2} = 2 \\times 6 = 12$$ For $n$ chairs and $m$ people (assuming $\\binom{n}{m} = 0$ for $m \\ge n$) this reduces to choosing $m$ seats out of $n$ and then permuting the $m$ people which is given by the formula $$m! \\times \\binom{n}{m}$$ Here $\\binom{n}{m}$ is the binomial coefficient which denotes the number of ways to choose $m$ objects from a collection of $n$ distinct objects. The number $m! \\times \\binom{n}{m}$ is also denoted as $^nP_m$. - To add some commentary to this: $\\binom{n}{m}$ tells you the number of ways $m$ people can choose from $n$ things, in order, i.e. first person picks, then the second person picks from remaining, then third, then fourth... up to the $m$th person. This is then multiplied by $m!$ which is the total number of ways that $m$ people can be arranged in order. – Dancrumb Mar 28 at 16:19 Seat A first, then B. A has $4$ choices, leaving $3$ choices for B,", "Distribution of five people in a car where exactly two can drive In how many ways can we distribute five people in a car if exactly two of them are can drive? - • For the next seat (we'll label it as S2), there are $4$ possibilities. • For the next seat (S3), there are $3$ possibilities. • For the next seat (S4), there are $2$ possibilities. Simply multiply the possibilities for S1, S2, S3, S4 and S5 which yields $2\\cdot 4 \\cdot 3\\cdot 2 \\cdot 1$.", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "5 persons and 5 chairs There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5? My attempt Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\\cdot3\\cdot1\\cdot2\\cdot1=24$. And in the case that the person D sits on chair 1: $1\\cdot2\\cdot3\\cdot4\\cdot1$. And if person D sits on chair 5: $4 \\cdot 3 \\cdot 2 \\cdot 1 \\cdot 1=24$. Therefore $$120-3\\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)? • The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45 • You need to remove the permutations that", "can three people be seated? . ${\\color{blue}10\\cdot9\\cdot9 \\,=\\,720}$ b) In how many of these will the two end chairs be occupied? The left chair can be occupied by any of the 3 people. The right chair can be occupied by of the other 2 people. The third person has a choice of the 8 middle chairs. There are: . $3\\cdot2\\cdot8 \\,=\\,48$ ways. c) In how many of these will the two end chairs be empty? The end chairs are to be empty. The 3 people are to sit in the 8 other chairs. There are: . $8\\cdot7\\cdot6 \\,=\\,336$ ways. 4. All possible three-digit numbers are formed from the odd digits {1, 3, 5, 7, 9}, (a) How many such numbers are possible if each digit is used only once? .5 x 4 x 3 = 60 (b) How many of the numbers from part (a) are larger than 350? Digits may NOT be repeated. Suppose the number begins with \"3\". The second digit must be 5, 7, or 9 . . . 3 choices. The third digit can be any of the other 3 digits. There are: . $3cdot 3 \\,=\\,9$ numbers. Suppose the number begins with 5, 7, or 9 . . . 3 choices. The other two digits can be any of $4\\cdot3 \\,=\\,12$ choices. There", "$19$ persons in total. Suppose the number of persons on the longest arc is $k>10$. Then two places of $x,y,z$ are just chosen from the two end-points of the arc, and there are $19-k$ possible places for the third person. Once the three places of $x,y,z$ are chosen, there are three possible ways to put $x,y,z$ into them clockwise. Also, note that for any $k>10$, there are $19$ ways to choose an arc of length $k$. Therefore the total number of ways (of the complement) is $$\\sum_{k=11}^{18} 3\\cdot 19 \\cdot (19-k) = 3\\cdot 19 \\cdot (1+\\cdots+8) = 3\\cdot 19\\cdot 36$$ $$3\\cdot \\binom{19}{3} - 3\\cdot 19\\cdot 36 = 3\\cdot 19 \\cdot (51 - 36) = 855$$ NOTE: this multiple choice problem can be done even faster -- after we realized the fact that each choice of the three places of $x,y,z$ corresponds to $3$ possible ways to put them in, and that each arc of length $k>10$ has $19$ equitable positions, it is evident that the answer should be divisible by $3\\cdot 19$, which can only be $855$ from the five choices.", "# How many possible ways to vote? ## There are 8 judges. People can vote for 0, 1, 2, 3, or 4 people. How many ways can a person vote? May 21, 2018 163 ways. #### Explanation: There is $1$ way to vote for 0 people. There are $8$ ways to vote for 1 person. There are $\\frac{8 \\cdot 7}{2}$ ways to vote for 2 people. There are $\\frac{8 \\cdot 7 \\cdot 6}{2 \\cdot 3}$ ways to vote for 3 people. There are $\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5}{2 \\cdot 3 \\cdot 4}$ ways to vote for 4 people. This is all because you can choose people but there are ways you can order the people. For example, there are $2 \\cdot 3$ ways to order the same 3 people. Adding everything, we get $1 + 8 + 28 + 56 + 70 = 163$.", "\\quad C \\quad 3! \\ \\ \\ ,$$ $$(B \\ \\bullet \\ : \\ \\ 3 \\cdot 2!) \\quad C \\quad 3! \\ \\ \\ ;$$ or $$3! \\quad C \\quad (A \\ \\bullet \\ : \\ \\ 3 \\cdot 2!) \\ \\ \\ ,$$ $$3! \\quad C \\quad (B \\ \\bullet \\ : \\ \\ 3 \\cdot 2!) \\ \\ \\ .$$ This second case permits $\\ 2 \\ \\cdot \\ 2 \\ \\cdot \\ 3 \\ \\cdot \\ 2! \\ \\cdot \\ 3! \\ = \\ 144 \\$ possible arrangments. Hence, there are $\\ 96 \\ + \\ 144 \\ = \\ 240 \\$ ways in which these six people may stand in line, subject to the specified condition. $$\\ \\$$ EDIT (5/6) -- An alternative method would be to solve the complementary problem. There are 6! = 720 possible arrangments of the six people in the line. There are four situations in which person C does not find themselves standing between A and B : I : C is at one end of the line, with the other 5 people arranged in any of 5! = 120 ways, thus $$C \\quad 5! \\quad \\quad \\text{or} \\quad \\quad 5! \\quad C$$ for a total of 240 arrangements; II : C is standing one position in from", "## Precalculus (10th Edition) $24$ The number of ways arranging $n$ distinct objects is $n!=n\\cdot(n-1)\\cdot(n-2)\\cdot...\\cdot3\\cdot2\\cdot1.$ Hence, the number of ways of arranging $4$ people in a line is: $4!=4\\cdot3\\cdot2\\cdot1=24.$", "numbers. For n=12, this gives 101001001001 (5 occupied) for me, while starting at chair 4 gives 100101010101 (6 occupied). This method is equivalent to choosing an end, since after three steps you end up at the same configuration (both ends and middle occupied). – Set Big O Nov 19 '14 at 20:46", "of C , so we must have either A or B stand there. Whichever of the two does so, the other may take any of the remaining four positions on the other side of C, along with the other three people. Those four can then be arranged in any order and still met the requirement that C is between A and B: $$A \\ C \\quad 4! \\ \\ \\ ,$$ $$B \\ C \\quad 4! \\ \\ \\ ;$$ or $$4! \\quad C \\ A \\ \\ \\ ,$$ $$4! \\quad C \\ B \\ \\ \\ .$$ So this case accounts for $\\ 2 \\ \\cdot \\ 2 \\ \\cdot \\ 4! \\ = \\ 96 \\$ possible arrangements. In the second case, there are two positions to one side of C and three positions to the other. Again, if A is on one side of C , B must be on the other. On the side of C with two positions available, we could have A or B standing there along with one of the three other people; they can take up one of two possible arrangements. To the other side, the remaining three people may be arranged in any order. We therefore may have: $$(A \\ \\bullet \\ : \\ \\ 3 \\cdot 2!)", "remaining chairs}\\\\ \\text{That gives us }10\\cdot 6 = 60 \\text{ ways to arrange the 5 people as described}$$ The remaining questions have been answered recently here. Search for them. Apr 27, 2019", "this stage in 1 way. Stage 7: Select someone to sit in the 7th chair There are 4 people remaining. So, we can complete stage 7 in 4 ways Stage 8: Select someone to sit in the 8th chair The person seated in the 8th chair must be the partner of the person in the 7th chair So, we can complete this stage in 1 way. Stage 9: Select someone to sit in the 9th chair There are 2 people remaining. So, we can complete stage 9 in 2 ways Stage 10: Select someone to sit in the 10th chair One 1 person remains. So, we can complete this stage in 1 way. By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus seat all 10 people) in (10)(1)(8)(1)(6)(1)(4)(1)(2)(1) ways ([spoiler]= 3840 ways[/spoiler]) Answer: -------------------------- Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775 You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776 Cheers, Brent _________________ Test confidently with gmatprepnow.com Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7590 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There are 5 couples. If they will sit 10 chairs in a row", "tricky Integrated Reasoning way. The important thing is to not waste too much time on a question that's so straight forward and covers a concept you should know well. If the price of chair P is 60$ and the price of chair Q is 90$, the average must be somewhere in the middle. If the average is 75$, then there is an equal amount of each. If the average is 70$as stated here, then you need two chairs of P brand (down by 10$ each) for every one of Q (up by 20$each). This means you need twice as many P's as Q's, so you just look for the numbers that are in a 2-1 ratio. Always start at the top, as you can see that doubling each number will get you out of the range of possible answers fairly quickly. Multiplying by two is also easier than dividing by two for 99.44% of people (stat may be a little arbitrary) so stick to multiplying rather than dividing. 28 doubles to exactly 56, so you need 56 of the cheaper chair and 28 of the more expensive chair. There's no need to do trial-and-error in a question like this, understanding the concept and applying it should take you about a minute or so to get the right answer.", "the cost of the chair to be x. Then the cost of the table =$ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is$125 and that of each table is \\$165. 10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40. Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems. Equations What is an Equation? What", "# Math Help - Urgent help on probability 1. ## Urgent help on probability Five people are selected at random. What is the probability that none of the people in this group were born in the same month? is it 1-(12/12c5) 2. Originally Posted by Amy Five people are selected at random. What is the probability that none of the people in this group were born in the same month? is it 1-(12/12c5) I got $\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 8}{12^5}$ It's equivalent to placing the five people in twelve chairs. 12 choices for the first, 11 for the second ... etc. Divded of course by the number of ways they can sit if they are allowed to sit on top of each other 3. Originally Posted by DivideBy0 I got $\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 8}{12^5}$ Divded of course by the number of ways they can sit if they are allowed to sit on top of each other it is not clear to me... 4. Ok I'm just using: $Pr(Favourable \\ Outcome) = \\frac{number \\ of \\ ways \\ favourable \\ outcome \\ can \\ occur}{total \\ number \\ of \\ outcomes}$ We'll simplify this to $Pr (X) = \\frac{n(X)}{n(\\varepsilon)}$ where Pr(X) is the probability the 5 people are born in" ]

[ "did she have to start with? Spoiled if solved via algebra. III) There are 100 red buttons and 99 black buttons in a large jar. A kid takes them out two by two, and he has some to spare (enough for the whole game). He cannot see which buttons he takes out, the jar is opaque. When he draws two buttons of the same colour, he puts a red button back in the jar. When he draws two buttons of different colours, he puts a black button back in the jar. When only one buttons is left in the jar, which colour is it? Spoiled by probabilities - and that's overkill. You have to ask why. IV) For kids only starting geometry, ask them to prove the Pythagorean Theorem on the special case where you start with half a square, to prove it the ancient way, using only a straight stick, a pen and a compass. For older kids, ask them the same starting with half a rectangle. Spoiled by the modern mathematical proof • (2)–(4) are very nice examples, but (1) seems incomprehensible (and searching, I can’t find an answer anywhere else). Can you give a ROT13’d hint, or a link to an answer, or some other non-spoiling form of help? Feb 24, 2021 at 10:23 •", "labelled N & C, she gets a N, and from the jar labelled C she draws a R. Show how Emily can correctly relable the jars. The jars look like this: . . . $\\#1\\qquad\\#2\\qquad\\#3\\qquad\\#4$ . . $\\boxed{\\begin{array}{c}R\\\\D\\end{array}}\\quad\\boxed {\\begin{array}{c}D\\\\N\\end{array}}\\quad\\boxed{\\begi n{array}{c}N\\\\C\\end{array}}\\quad\\boxed{\\begin{arra y}{c}C\\\\. \\end{array}}$ She drew R from #4. Since the only jar with an R is R&D, #4 is $R\\&D$ She drew N from #3. The only other jar with an N is D&N; #3 is $D\\&N$ She drew N from #1. The only other jar with an N is N&C; #1 is $N\\&C$ This leave $C$ for jar #2. Quote: B) Matt unwraps one chocolate from one jar and is able to relabel it correctly. Make a list of nine ways he could have done this. (1) He draws D from #1. The only other jar with D is D&N; jar #1 is $D\\&N$ (2) He draws R from #2. The only jar with an R is R&D; jar #2 is $R\\&D$ (3) He draws N from #2. The only other jar with an N is N&C; jar #2 is $N\\&C$ (4) He draws D from #2. The only other jar with a D is R&D; jar #2 is $R\\&d$ (5) He draws R from #3. The only jar with an R is R&D; jar #3 is $R\\&D$", "+0 # Mathe 0 355 2 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether?{pls help me bcz I want to learn}#CrAzyGirl Guest Feb 1, 2016 #2 +18956 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 236 \\qquad | \\qquad -\\text{box}\\\\ 2\\cdot \\text{box}-\\text{box} &=& 236 \\\\ \\text{box}", "+0 Mathe 0 532 2 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether?{pls help me bcz I want to learn}#CrAzyGirl Guest Feb 1, 2016 #2 +20108 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 236 \\qquad | \\qquad -\\text{box}\\\\ 2\\cdot \\text{box}-\\text{box} &=& 236 \\\\ \\text{box} &=&", "# Probability of picking buttons from a bag A bag contains $$30$$ buttons that are colored either blue, red or yellow. There are the same number of each color ($$10$$ each). A total $$4$$ buttons are drawn from the bag. Compute the followings: 1. Find $$n(\\Omega)$$. 2. The probability that at least $$3$$ of them are red? 3. The probability that there is at least one of each color? This seems like a basic problem but my professor and I cannot agree on an answer. I think the probabilities are $$2/21$$ and $$100/203$$ for parts $$2$$ and $$3$$ respectively. I used combinations to calculate the probabilities. My professor said $$n(A)/n(\\Omega)$$ is $$3/15$$ for both so that is the answer for both $$2$$ and $$3$$. A bag contains $$30$$ buttons that are colored either blue, red or yellow. There is the same number of each color ($$10$$ each). A total of $$4$$ buttons are drawn from the bag. Compute the following: $$1$$. Find $$n(\\Omega)$$. $$2$$. The probability that at least $$3$$ of them are red. $$\\frac{{10\\choose3}{20\\choose1}+{10\\choose4}}{{30\\choose4}}=\\frac{2610}{27405}=\\frac{2\\cdot3^2\\cdot5\\cdot29}{3^3\\cdot5\\cdot7\\cdot29}=\\frac 2{21}$$ $$3$$. The probability that there is at least one of each color. $$\\frac 12\\times\\frac{{10\\choose1}{10\\choose1}{10\\choose1}{27\\choose1}}{{30\\choose4}}=\\frac 12\\times\\frac{10\\cdot10\\cdot10\\cdot27}{27405}=\\frac 12\\times\\frac{2^3\\cdot3^3\\cdot5^3}{3^3\\cdot5\\cdot7\\cdot29}=\\frac {100}{203}$$ where the factor of $$\\frac12$$ was to cancel double-counting, in that every combination like $$(\\underbrace{R_i}_{{10\\choose1}}\\underbrace{B_i}_{{10\\choose1}}\\underbrace{Y_i}_{{10\\choose1}}\\underbrace{R_j}_{{27\\choose1}})$$ is also counted in $$(R_jB_iY_iR_i)$$. You are correct. Part B", "the other end of the desert. However, if that would have been the case, I wouldn’t have posted it and wasted your time. Hint: You have to make multiple trips back and forth as you transfer all 3000 bananas. Find the most efficient way. You have a (solid) blue cube and an unlimited number of (solid) red cubes, all of which are of the same size. What is the largest number of red cubes that can touch the blue cube along its sides or parts of its sides? Touching along edges or at corner points does not count. You are trapped in a small room with 4 walls. Each wall has a button that is either in an OFF/ON setting though you have no way of telling what the setting is. When you press a button, you change its setting. If you can get all the buttons to have the same setting i.e. either all four are OFF or all four are ON, you are immediately set free. In each move, you can press either two buttons simultaneously or just one button. As soon as this occurs, if you haven't been set free, the whole room spins around you violently, leaving you completely disoriented so that you can never tell which side is which. The starting position is", "### Home > A2C > Chapter 13 > Lesson 13.1.1 > Problem13-23 13-23. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: 1. Two red and one white. $\\frac{_5C_2 \\cdot _4C_1}{_{12}C_5}=\\frac{2}{11}$ 2. Three white. See part (a). 3. One of each color. See part (a). 4. All the same color. $\\frac{_5C_3 + _4C_3 + _3C_3}{_{12}C_3}=\\frac{3}{44}$ 5. One red and two white. See part (a). 6. Two of one color and one of another. How can the probabilities from parts (c) and (d)", "9 Question 4. ___ ÷ 9 = 2 Question 5. Question 6. 9 ÷ ___ = 1 Complete the table. Question 7. Question 8. Problem Solving Question 9. Daryl collects 38 plastic liter bottles and 43 plastic jugs. He divides them equally into 9 recycling bins. How many plastic containers does he put into each bin? Write the steps to solve the problem. Question 10. Annette has 27 pink buttons, 36 red buttons, and 45 green buttons. How can she sort the buttons into 9 equal piles, so there are the same number of different color buttons in each pile? Explain. Texas Test Prep Lesson Check Question 11. Jamal has 63 marbles. For every 8 blue marbles, there is 1 red marble. How many red marbles does Jamal have? (A) 5 (B) 6 (C) 9 (D) 7 Question 12. Anquan made 45 party favors. He puts an equal number of favors on 9 tables. How many party favors are on each table? (A) 6 (B) 5 (C) 7 (D) 8 Question 13. Carla has a coin collection of 54 coins. She puts 9 coins into each page of a coin book. How many pages does Carla fill in her coin book? (A) 9 (B) 7 (C) 6 (D) 8 Question 14. There are 9 people at the diner.", "objects from jar W to jar B. We then thoroughly mix – in fact, we don’t have to – the contents of jar B, following which we transfer $k$ objects, this time, from jar B to jar W. Which is greater now: the percentage of black objects in jar W or the percentage of white objects in jar B? Solution 1: Let us keep track of the number of black and white objects in both the jars before and after the transfers of $k$ objects from one jar to another. So, initially, in jar W, # of white objects = $n$, and # of black objects = $0 \\,$. Also, in jar B, # of white objects = $0 \\,$, and # of black objects = $n$. Now, we transfer $k$ objects from jar W to jar B. So, in jar W, # of white objects = $n-k$, and # of black objects = $0 \\,$. Also, in jar B, # of white objects = $k$, and # of black objects = $n$. Finally, we transfer $k$ objects from jar B to jar W. Let the number of white objects out of those $k$ objects be $k_1$. Then, the number of black objects transferred equals $k-k_1$. Therefore, now, in jar W, # of white objects = $n-k +", "are simultaneously picked at random from a jar containing 2 red balls, 2 green balls, and 6 yellow balls. What is the probability that exactly two of the selected balls will be red?", "choose 6) + (9 choose 7) + (9 choose 8) + (9 choose 9) = 511 total combinations. 13. Jun 2, 2006 ### rhj23 wouldn't it be easier to consider that in any set arrangement any individual button is either on or off, and each independently. so the total number of possibles is 2^9 = 512. it is clear that all of these are distinct (i have one more than you because my method includes 'all off' as a possible arrangement, which it is not in this case) 14. Jun 3, 2006 ### mattmns Yep that is much easier 15. Jun 9, 2006 ### rcgldr Here's an answer that doesn't require knowledge of the choose function: In order to get 3 good buttons, on the first pick, you have 3 good buttons out of 9 total buttons that are good, then on the second pick you have 2 good buttons out of 8 that are good, and on the third pick you have 1 out of 7 buttons that are good. 3/9 x 2/8 x 1/7 = 6/504 = 1/84 Success happens when there are no bad picks: There are 8 possible outcomes when pushing the buttons: Code (Text): (all good, 0 bad) = 3/9 x 2/8 x 1/7 = 6/504 = 1/84 1st bad = 6/9 x", "be 6 non-faded buttons. 3. May 25, 2007 ### danago The buttons arent 0-9, theyre 1-12. There was a diagram with the question that showed this. 4. May 25, 2007 ### Dick No repeated numbers allowed, right? And \"can contain only 1 of the 4\" means \"contains exactly one of the 4\"? If those are both correct then I can't see anything wrong with your solution. 5. May 25, 2007 ### danago Yep thats right. No repeats allowed, and EXACTLY 1 of the 4. Well thanks for clearing that up. Glad i did it correctly then.", "## anonymous 5 years ago three jars contain colored balls as follows: Jar Red White Blue I 10 1 9 II 2 5 13 III 9 1 10 One jar is chosen at random and a ball is withdrawn. The ball is red. What is the probability that it came from jar I? Round to four decimals 1. anonymous So there's a $$\\frac{1}{3}$$ chance that the jar you chose was jar one, and a $$\\frac{10}{20}$$ chance that you chose a red ball if you the first jar. Now you need to combine these two numbers to get the overall probability you're looking for. Do you know how to combine them? 2. anonymous you do you mean like multiple 1/3 by 10/20 3. anonymous Exactly. Does that make sense? 4. anonymous i got an answer of .1667 but when i put it as my answer it says its wrong 5. anonymous Hm. Odd. Let me think on it. 6. anonymous There's a chance that what it's looking for is--given the fact that you already chose a red ball, what are the chances it was in the first jar. In that case, you realize that there are $$21$$ ways to get a red ball. $$10$$ of those ways are in the first jar, so you can get the probability from", "an equiprobable choice of two colours. Thus there are $8$ equiprobable outcomes. Now there is only one such way the three buttons could all show white. Similarly for the all red even. However, when one button is of a different colour than the other two, there are three buttons that could be that different colour. Hence the weight of each of these events is thrice that of each of the all-same-colour events. Thus we have that: $\\quad$ $\\Pr(\\text{all white}) = \\frac 1 8 \\\\ \\Pr(\\text{2 white, 1 red}) = \\frac 3 8 \\\\ \\Pr(\\text{2 red, 1 white}) = \\frac 3 8 \\\\ \\Pr(\\text{all red}) = \\frac 1 8$", "### Home > A2C > Chapter 13 > Lesson 13.1.1 > Problem13-23 13-23. 1. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: Homework Help ✎ 1. Two red and one white. 2. Three white. 3. One of each color. 4. All the same color. 5. One red and two white. 6. Two of one color and one of another. $\\frac{_5C_2 \\cdot _4C_1}{_{12}C_5}=\\frac{2}{11}$ See part (a). See part (a). $\\frac{_5C_3 + _4C_3 + _3C_3}{_{12}C_3}=\\frac{3}{44}$ See part (a). How can the probabilities from parts (c) and (d)", "easily first. • January 5th 2010, 07:28 AM Mukilab Errr, could someone else reiterate the whole answer? I tried it by drawing various lines and giving them each a probability... • January 5th 2010, 07:37 AM Archie Meade You probably have $Nc_r$ and $Np_r$ buttons on your calculator. $Np_r$ will calculate the number of ways of arranging r items from a total of N. $Nc_r$ will count the number of ways of selecting (without arranging) r items from N. Using those keys is the fastest way to answer. The probability of selecting 2 beads the same is $\\frac{selections\\ of\\ 2\\ same}{selections\\ of\\ 2\\ from\\ 30}$ The probability of the beads being different is that subtracted from 1. • January 5th 2010, 07:42 AM Mukilab Quote: Originally Posted by Archie Meade You probably have $Nc_r$ and $Np_r$ buttons on your calculator. $Np_r$ will calculate the number of ways of arranging r items from a total of N. $Nc_r$ will count the number of ways of selecting (without arranging) r items from N. Using those keys is the fastest way to answer. The probability of selecting 2 beads the same is $\\frac{selections\\ of\\ 2\\ same}{selections\\ of\\ 2\\ from\\ 30}$ The probability of the beads being different is that subtracted from 1. Thank you for telling me those keys, very useful", "### Home > CCG > Chapter 12 > Lesson 12.1.4 > Problem12-55 12-55. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: Use combinations to represent the probabilities. 1. Three white balls. 2. Two red balls and one white ball. 3. Three balls of the same color. $\\frac { _{5} C _ { 3 } + _ { 4 } C _ { 3 } + _{3} C _ { 3 } } { _{12} C _ { 3 } } = \\frac { 10 + 4 + 1 } { 220 } \\approx 6.8$%", "incorrect full solution.) My Kinda Elementary but Long-winded Solution Samyak’s Streamlined Solution Multinomial Digression My Solution We shall use a binary encoding. Suppose we had only $32 (=2^5)$ jars. Then we could use five birds to find the poisoned jar in one go. Like this: Number the jars in binary. So $1 = 00001 = 1 \\times 2^0$ $2 = 00010 = 1 \\times 2^1$ $3 = 00011 = 1 \\times 2^1 + 1 \\times 2^0$ $18 = 10010 = 1 \\times 2^4 + 1 \\times 2^1$ all the way to $32 = 100000$. Now we shall name our birds One, Two, Four, Eight and Sixteen. If a jar has a one in the units place, we feed it to One. If it has a one in the two’s place, we feed it to Two. And so on. So jar number 18 = 10010 will be fed to Sixteen and Two because it has ones in ones place and the sixteens place. Jar no. 32 won’t be fed to any pigeon, because it’s got no ones in places One to Sixteen. Then we wait and see which pigeons turn white. We add up the numbers of the pigeons and that tells us the number of the jar. Say pigeons Two and Sixteen turn white. Then the poisoned jar", "236 \\\\ \\end{array}$$ There are 236 buttons in the box $$\\begin{array}{rcll} \\text{red} &=& \\frac{236}{4} \\\\ &=& 59 \\\\\\\\ \\text{yellow} &=& 32+59 \\\\ &=& 91 \\\\\\\\ \\text{blue} &=& 86 \\\\\\\\ 59+91+86 &=& 236 \\end{array}$$ heureka Feb 1, 2016 #1 +27128 +5 buttons (b) = red (1/4b) + yellow(1/4b +32) + blue(86) b = (1/4)b + (1/4)b+ 32 + 86 b = (1/2)b + 118 (1/2)b = 118 b = 236 Alan Feb 1, 2016 #2 +20108 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box}", "never press button $K$, the only higher button. Thus, any sequence of button presses that Bessie takes, excluding the first and last button, uses buttons with numbers at most $K - 2$. We can then solve this using DP. $dp_{a, b, x}$ will be the number of sequences of rooms and button presses which start at room $a$, end at room $b$, and only involve pressing buttons with numbers at most $x$. We can compute transitions as follows. First, we add all of $dp_{a, b, x - 1}$ to $dp_{a, b, x}$, as a sequence only using buttons with numbers at most $x - 1$ will clearly also only use buttons with numbers at most $x$. We then consider the case where we do press button $x$. Note that as we never press any higher buttons, we must only press button $x$ once. Let’s then fix the room $r$ that Bessie is in when she presses button $x$, and consider in isolation the part of the sequence before pressing button $x$ (the left side) and after pressing button $x$ (the right side). The left side, if it is not empty, starts at some room $a$, ends at some room $b$ such that there is an edge from $b$ to $r$, and only presses buttons with number at most" ]

[ "# Probability of picking buttons from a bag A bag contains $$30$$ buttons that are colored either blue, red or yellow. There are the same number of each color ($$10$$ each). A total $$4$$ buttons are drawn from the bag. Compute the followings: 1. Find $$n(\\Omega)$$. 2. The probability that at least $$3$$ of them are red? 3. The probability that there is at least one of each color? This seems like a basic problem but my professor and I cannot agree on an answer. I think the probabilities are $$2/21$$ and $$100/203$$ for parts $$2$$ and $$3$$ respectively. I used combinations to calculate the probabilities. My professor said $$n(A)/n(\\Omega)$$ is $$3/15$$ for both so that is the answer for both $$2$$ and $$3$$. A bag contains $$30$$ buttons that are colored either blue, red or yellow. There is the same number of each color ($$10$$ each). A total of $$4$$ buttons are drawn from the bag. Compute the following: $$1$$. Find $$n(\\Omega)$$. $$2$$. The probability that at least $$3$$ of them are red. $$\\frac{{10\\choose3}{20\\choose1}+{10\\choose4}}{{30\\choose4}}=\\frac{2610}{27405}=\\frac{2\\cdot3^2\\cdot5\\cdot29}{3^3\\cdot5\\cdot7\\cdot29}=\\frac 2{21}$$ $$3$$. The probability that there is at least one of each color. $$\\frac 12\\times\\frac{{10\\choose1}{10\\choose1}{10\\choose1}{27\\choose1}}{{30\\choose4}}=\\frac 12\\times\\frac{10\\cdot10\\cdot10\\cdot27}{27405}=\\frac 12\\times\\frac{2^3\\cdot3^3\\cdot5^3}{3^3\\cdot5\\cdot7\\cdot29}=\\frac {100}{203}$$ where the factor of $$\\frac12$$ was to cancel double-counting, in that every combination like $$(\\underbrace{R_i}_{{10\\choose1}}\\underbrace{B_i}_{{10\\choose1}}\\underbrace{Y_i}_{{10\\choose1}}\\underbrace{R_j}_{{27\\choose1}})$$ is also counted in $$(R_jB_iY_iR_i)$$. You are correct. Part B", "an equiprobable choice of two colours. Thus there are $8$ equiprobable outcomes. Now there is only one such way the three buttons could all show white. Similarly for the all red even. However, when one button is of a different colour than the other two, there are three buttons that could be that different colour. Hence the weight of each of these events is thrice that of each of the all-same-colour events. Thus we have that: $\\quad$ $\\Pr(\\text{all white}) = \\frac 1 8 \\\\ \\Pr(\\text{2 white, 1 red}) = \\frac 3 8 \\\\ \\Pr(\\text{2 red, 1 white}) = \\frac 3 8 \\\\ \\Pr(\\text{all red}) = \\frac 1 8$", "is chosen is unknown. That means that we can reasonably assume that the probability of choosing Jar 1 is the same as the probability of choosing Jar 2. As above, a marble is drawn – without looking inside of the jar – from one of the jars. Here is the first of a new set of questions: 1. If a blue marble is drawn, what is the probability that the jar that was chosen is Jar 1? This problem is ideally set up for using Bayes’s Theorem.99 The (slightly rephrased) question: What is the probability of Jar 1 given a blue marble? is equivalent to asking what $$p(A|B)$$ is, where event $$A$$ is choosing Jar 1 and event $$B$$ is drawing a blue marble. To use Bayes’s Theorem to solve for $$p(A|B)$$, we are going to need to first find $$p(A)$$, $$p(B|A)$$, and $$p(B)$$. $$\\underline{p(A):}$$ Event $$A$$ is Jar 1, so $$p(A)$$ is the probability of choosing Jar 1. We have stipulated that the probability of choosing Jar 1 is the same as the probability of choosing Jar 2. Because one or the other must be chosen (i.e., the sample space $$\\Omega=\\{Jar~1,~Jar~2\\}$$ and the probability of choosing Jar 1 or Jar 2 is one (i.e., $$p(Jar~1\\cup Jar~2)=1$$, or equivalently, $$\\sum\\Omega=1$$), then: $p(A)=\\frac{p(A)}{p(A)+p(B)}=\\frac{p(A)}{2p(A)}=\\frac{1}{2}=0.5$ $$\\underline{p(B|A):}$$ Event $$B$$ is (drawing a)", "# How much bigger is a 1/4 inch button than a 5/8 inch button? Mar 14, 2018 color(purple)((5/8)”) button is color(green)((3/8)” greater than color(brown)((1/4)” button. #### Explanation: To compare the two fractions, we must have common base so that the numerators can be directly matched. Given : 1/4”, 5/8 Now we have to find the L C M for the denominator of the two fractions to make the denominator common. Factors of 4 = 2 * color(red)(2 Factors of 8 = 2 * 2 * color(red)(2 Taking the color(red)(2 once, L C M $= 2 \\cdot 2 \\cdot 2 = 8$ $\\frac{1}{4} , \\frac{5}{8} = \\frac{1 \\cdot 2}{4 \\cdot 2} , \\frac{5}{8} = \\frac{2}{8} , \\frac{5}{8}$ Difference between the two terms is (5-2) / 8 = 3/8” color(purple)((5/8)”) button is color(green)((3/8)” greater than color(brown)((1/4)” button.", "# conditional probability with vs. without replacement independence? I have $$2$$ jars, one of which is chosen randomly ($$.5$$ chance) jar 1: $$8$$ blue marbles (b), $$4$$ red marbles (r) jar 2: $$15$$ blue marbles, $$3$$ red marbles In a situation WITH replacement I know the probability of the first pick being a blue marble given that the second pick (from the same jar as the first pick) is a red marble = $$P(p_1 = b \\;|\\; p_2 = r) = P(p_1=b \\;|\\; jar=1)P(jar=1 \\;|\\; p_2=r) + P(p_1=b \\;|\\; jar=2)P(jar=2 \\;|\\; p_2=r)$$ But what does the same probability WITHOUT replacement look like? I know this might be an easy question but I have tried a lot and nothing seems to add up. Thank in advance! $$\\begin{multline} P(p_1 = b \\mid p_2 = r) = P(p_1=b \\mid jar=1,\\ p_2 = r)P(jar=1 \\mid p_2=r) + {} \\\\ P(p_1=b \\mid jar=2,\\ p_2 = r)P(jar=2 \\mid p_2=r) \\end{multline}$$ When you assume replacement, you can get away with the formula you wrote because the two draws are independent of each other (once the jar has been selected): $$P(p_1=b \\mid jar=1,\\ p_2 = r) = P(p_1=b \\mid jar=1,\\ p_2 = b) = P(p_1=b \\mid jar=1).$$ If you have trouble conceiving how to compute $$P(p_1=b \\mid jar=1,\\ p_2 = r),$$ remember that for any combination", "## anonymous 5 years ago three jars contain colored balls as follows: Jar Red White Blue I 10 1 9 II 2 5 13 III 9 1 10 One jar is chosen at random and a ball is withdrawn. The ball is red. What is the probability that it came from jar I? Round to four decimals 1. anonymous So there's a $$\\frac{1}{3}$$ chance that the jar you chose was jar one, and a $$\\frac{10}{20}$$ chance that you chose a red ball if you the first jar. Now you need to combine these two numbers to get the overall probability you're looking for. Do you know how to combine them? 2. anonymous you do you mean like multiple 1/3 by 10/20 3. anonymous Exactly. Does that make sense? 4. anonymous i got an answer of .1667 but when i put it as my answer it says its wrong 5. anonymous Hm. Odd. Let me think on it. 6. anonymous There's a chance that what it's looking for is--given the fact that you already chose a red ball, what are the chances it was in the first jar. In that case, you realize that there are $$21$$ ways to get a red ball. $$10$$ of those ways are in the first jar, so you can get the probability from", "# Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn 72 views closed Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II? +1 vote by (44.6k points) selected Let event J1 : Ball drawn from jar I, event J2 : Ball drawn from jar II. P(J1) = P(head) = 1/2 P(J2) = P(tail) = 1/2 Let event W: Ball drawn is white. In Jar I, there are total 12 balls, out of which 5 balls are white. ∴ Probability that the ball drawn is white under the condition that it is drawn from Jar I. P(W/J1) = $\\frac {^5C_1}{^{12}C_1} = \\frac {5} {12}$ Similarly, P(W/J2) = $\\frac {^3C_1}{^{15}C_1} = \\frac {3} {15} = \\frac 1 5$ Required probability = P(J2/W) By Bayes’ theorem", "# Probability of picking balls from jar Suppose a jar contains 8 balls among which 5 are red and 3 are blue. If i pick, 3 balls from this jar, what is the probability that \"exactly\" two of them are blue ball? • If you are picking without replacement (which is the natural interpretation, I think) then neither is correct. – André Nicolas Sep 12 '15 at 20:00 The total number of ways to choose $3$ out of $8$ balls is $\\binom83=56$ The number of ways to choose $2$ out of $3$ blue balls and $1$ out of $5$ red balls is $\\binom32\\cdot\\binom51=15$ Hence the probability is $\\dfrac{15}{56}$ • @karthikts: It doesn't make any difference. There are $56$ (not necessarily different) ways to choose $3$ out of $8$, and $15$ of them contain exactly two blue. – barak manos Sep 12 '15 at 20:29", "### Home > CCG > Chapter 12 > Lesson 12.1.4 > Problem12-55 12-55. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: Use combinations to represent the probabilities. 1. Three white balls. 2. Two red balls and one white ball. 3. Three balls of the same color. $\\frac { _{5} C _ { 3 } + _ { 4 } C _ { 3 } + _{3} C _ { 3 } } { _{12} C _ { 3 } } = \\frac { 10 + 4 + 1 } { 220 } \\approx 6.8$%", "# probability distribution practice problems This can be simplified by dividing both 2 and 6 by 2. If the chance of drawing a red button is now 1/3, then 5 of the 15 buttons remaining must be red. 1. He will throw the die and will pay you in dollars the number that comes up. There are 6 blue marbles and 3 red marbles for a total of 9 desired outcomes. 2. D: Shown below is the sample space of possible outcomes for tossing three coins, one at a time. Since there is a possibility of two outcomes (heads or tails) for each coin, there is a total of 2*2*2=8 possible outcomes for the three coins altogether. If there are 2 winners for every 100 tickets, there is 1 winner for every 50 tickets sold. If 25 buttons are removed, there are 15 buttons remaining in the bag. 8. $f\\left( x \\right) = \\left\\{ {\\begin{array}{*{20}{l}}{c\\left( {8{x^3} - {x^4}} \\right)}&{{\\mbox{if }}0 \\le x \\le 8}\\\\0&{{\\mbox{otherwise}}}\\end{array}} \\right.$. 7. John has a special die that has one side with a six, two sides with twos and three sides with ones. 1) The coin is flipped ten times. This can be simplified by dividing both 2 and 6 by 2. Therefore, the probability of throwing either a 3 or 4 is 1", "= 2 11 – 5 = 6 ($2×3$) 29 – 11 = 18 ($6×3$) 83 – 29 = 54 ($18×3$) 245 – 83 = 162 ($54×3$) 731 – 245 = 486 ($162×3$) 20. A Jar contains 18 balls. 3 blue balls are removed from the jar and not replaced.Now the probability of getting a blue ball is 1/5 then how many blue balls jar contains initially ? Sol: x/15 = 1/5 x = 3 3 + 3 (removed 3 blue balls) = 6", "of a given total, rather than percentages, so let's take 10k buttons. 3k are read, and 7k are blue. In the lower left box, you have red buttons that don't work, which is 12% of 3k, or 360. In the upper left box, you have red buttons that work, so that's 3000-360 = 2640. The we have 1400 in the lower right box, and 5600 in the upper right box. We're told to find the probability given that the button works, so we just have the top row. There are 2640+5600=8040 total buttons in that row, and 2640 of them are red. So we take 2640/8040 and get 32.84%.", "### Home > A2C > Chapter 13 > Lesson 13.1.1 > Problem13-23 13-23. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: 1. Two red and one white. $\\frac{_5C_2 \\cdot _4C_1}{_{12}C_5}=\\frac{2}{11}$ 2. Three white. See part (a). 3. One of each color. See part (a). 4. All the same color. $\\frac{_5C_3 + _4C_3 + _3C_3}{_{12}C_3}=\\frac{3}{44}$ 5. One red and two white. See part (a). 6. Two of one color and one of another. How can the probabilities from parts (c) and (d)", "is three times more probable that we have Jar 1 than that we have Jar 2.101 That lines up with the fact that the proportion of blue marbles is three times greater in Jar 1 than it is in Jar 2. Now let’s try a slightly trickier question: 1. After drawing a blue marble from one of the jars, the marble is replaced in the same jar, the jar is shaken, and a second marble is drawn. The second marble drawn is also blue. What is the probability that the jar is Jar 1? To reduce confusion, let’s add some subscripts to $$A$$ and $$B$$: we can add $$_1$$ to refer to anything that happened on the first draw, and $$_2$$ to refer to what’s happening on the second draw. $$\\underline{p(A_2):}$$ Event $$A$$ is again Jar 1, but this time, $$p(A)$$ is different. Having already drawn a blue marble from that same jar, we must update the probability that we have Jar 1 based on that new information. In the last problem, we found the probability that we have Jar 1 given that a blue marble was drawn. That probability is our new $$p(A)$$. Thus, our new $$p(A)$$ – denoted $$p(A_2)$$102 – is $$p(A_2|B_1)$$: the updated probability of having Jar 1 _in this second draw given that a", "was quite happy Manager Joined: 20 Dec 2013 Posts: 227 Location: India Re: An jar is filled with red, white, and blue tokens that are e [#permalink] ### Show Tags 07 Apr 2014, 01:31 Oh!took me 10 minutes to figure out!! And that too by options. All of these options are are multiples of $$3$$.So dividing them equally will not help because in that case we will never have equal probability of selecting two tokens on the one hand and selecting just one on the other. By same logic,we can never have two same no. of differently colored tokens. To fulfill the condition,we need to have all different nos. of differently colored tokens. Then I checked the first three multiples of $$3:3,6$$ & $$9$$ added up to $$18$$. With all mental capacity exhausted,clicked $$18$$! (Now I realize the question asks for smallest no. of tokens.I know this is shoddy logic but still got it right!) Manager Joined: 20 Dec 2013 Posts: 121 Re: A jar is filled with red, white, and blue tokens that are eq [#permalink] ### Show Tags 07 Apr 2014, 03:42 2 AccipiterQ wrote: A jar is filled with red, white, and blue tokens that are equivalent except for their color. The chance of randomly selecting a red token, replacing it, then randomly selecting a", "jar W, # of white objects = $n-k$, and # of black objects = $0 \\,$. Also, in jar B, # of white objects = $k$, and # of black objects = $n$. Finally, we transfer $k$ objects from jar B to jar W. Let the number of white objects out of those $k$ objects be $k_1$. Then, the number of black objects transferred equals $k-k_1$. Therefore, now, in jar W, # of white objects = $n-k + k_1$, and # of black objects = $k-k_1$. Also, in jar B, # of white objects = $k-k_1$, and # of black objects = $n - (k-k_1)$. From here, it is easy to see that the percentage of black objects in jar W is the same as the percentage of white objects in jar B! And, we are done. Solution 2: (I think this is a slicker one, and I found it after pondering a little over the first solution I wrote above!) This one uses the idea of invariants, and there are, in fact, two of ’em in this problem! Note that at any given time, # of white objects = # of black objects = $n$. The above is the first invariant. Also, note that after we transfer $k$ objects from jar W to jar B and then", "a black got put into Jar 2,then: Probability Jar 3 gets a black is (n+1)/(m+n+1) Probability Jar 3 gets a white is m/(m+n+1). If a white got put into Jar 2, then: Probability Jar 3 gets a black is n/(m+n+1) Probability Jar 3 gets a white is (m+1)/(m+n+1). So the probability of getting a white from Jar 3 is: $\\left( \\frac{n}{m+n} \\right) \\, \\left( \\frac{m}{m+n+1} \\right) + \\left( \\frac{m}{m+n} \\right) \\, \\left( \\frac{m+1}{m+n+1} \\right)$ $= \\frac{n m}{(m+n)(m+n+1)} + \\frac{m (m+1)}{(m+n)(m+n+1)}$ $= \\frac{nm + m (m+1)}{(m+n)(m+n+1)} = \\frac{m (n + m+1)}{(m+n)(m+n+1)} = \\frac{m}{m+n}$. Similarly the probability of getting a black from Jar 3 is $\\frac{n}{m+n}$. A tree diagram makes it all very clear. Obviously exactly the same argment can now be used to get the probaility of getting a white or black black from Jar 4, 5, 6, ...... k. 3. thankyou very much for your help.", "there are. There are $12$ numbers in the original set. $144 ÷ 12 = 12$ The mean is $12$. Question 4 ### What is the range of the following data? $3, 3, 14, 9, 10, 12, 15, 12, 15, 15, 18, 18$ A $3$ B $12$ C $15$ D $18$ Question 4 Explanation: The range can be found by subtracting the smallest value from the largest value. The largest value is $18$ and the smallest is $3$: $18 − 3 = 15$ Question 5 ### There are 3 red t-shirts, 5 gray t-shirts, and 1 blue t-shirt in a clothing drawer. What are the odds of grabbing a red t-shirt if you pick at random? A $3$ B $\\dfrac{1}{3}$ C $\\dfrac{1}{2}$ D $1$ out of $9$ Question 5 Explanation: The probability of grabbing a red shirt has to be compared to the probability of grabbing all shirts. Since there are $9$ total shirts and $3$ red shirts, the probability of picking a red shirt is $\\frac{3}{9}$, or $\\frac{1}{3}$. Question 6 ### In a bag are $20$ chips that are numbered $1$ through $20$. What are the odds of drawing a $\\#3$ or a $\\#7$ chip? A $.1$ B $10$ C $\\dfrac{1}{20}$ D $1$ out of $40$ Question 6 Explanation: A probability problem that says or often indicated that", "drawn from Bag 1 and a red is returned to Bag 1. . . [2] A black is drawn from Bag 1 and a black is returned to Bag 1. [1] A red is drawn from Bag 1: . $P(R_1) = \\frac{3}{8}$ . . .Bag 2 now contains:7 red and 2 black discs: . . .A red is drawn from Bag 2: . $P(R_2) \\:=\\:\\frac{7}{9}$ . . . . . $P(R_1R_2) \\:=\\:\\frac{3}{8}\\cdot \\frac{7}{9}$ [2] A black is drawn from Bag 1: $P(B_1) = \\frac{5}{8}$ . . .Bag 2 now contains: 6 red and 3 black discs. . . .A black is drawn from Bag 2: . $P(B_2) = \\frac{3}{9}$ . . . . . $P(B_1B_2) \\:=\\:\\frac{5}{8}\\cdot\\frac{3}{9}$ Therefore: . $P(R_1R_2\\text{ or }R_1B_2) \\;=\\;\\frac{3}{8}\\cdot\\frac{7}{9} + \\frac{5}{8}\\cdot\\frac{3}{9} \\;=\\;\\frac{7}{24} + \\frac{5}{24} \\;=\\;\\frac{12}{24} \\;=\\;\\dfrac{1}{2}$", "# Jellybean question #### intrakate1 ##### New member If i have a jar with 15 red, 13 yellow, and 7 green jelly beans what is the probability that i will pull out 1 red and 2 yellows (without the replacement of the ones i removed) #### soroban ##### Elite Member Hello, intrakate1! If i have a jar with 15 red, 13 yellow, and 7 green jelly beans, what is the probability that i will pull out 1 red and 2 yellows (without the replacement of the ones i removed)? $$\\displaystyle \\text{If you mean Red-\\!Yellow-\\!Yellow }in\\;that\\;order\\text{, the probability is: } \\;\\frac{15}{35}\\cdot\\frac{13}{34}\\cdot\\frac{12}{33} \\;=\\;\\frac{78}{1309}$$ $$\\displaystyle \\text{If you mean 1 red, 2 yellows }in\\;any\\;order\\text{, the probability is: }\\; \\dfrac{{15\\choose1}{13\\choose2}}{{35\\choose3}} \\;=\\;\\frac{1170}{6545} \\;=\\;\\frac{234}{1309}$$" ]

[ "# Probability of picking buttons from a bag A bag contains $$30$$ buttons that are colored either blue, red or yellow. There are the same number of each color ($$10$$ each). A total $$4$$ buttons are drawn from the bag. Compute the followings: 1. Find $$n(\\Omega)$$. 2. The probability that at least $$3$$ of them are red? 3. The probability that there is at least one of each color? This seems like a basic problem but my professor and I cannot agree on an answer. I think the probabilities are $$2/21$$ and $$100/203$$ for parts $$2$$ and $$3$$ respectively. I used combinations to calculate the probabilities. My professor said $$n(A)/n(\\Omega)$$ is $$3/15$$ for both so that is the answer for both $$2$$ and $$3$$. A bag contains $$30$$ buttons that are colored either blue, red or yellow. There is the same number of each color ($$10$$ each). A total of $$4$$ buttons are drawn from the bag. Compute the following: $$1$$. Find $$n(\\Omega)$$. $$2$$. The probability that at least $$3$$ of them are red. $$\\frac{{10\\choose3}{20\\choose1}+{10\\choose4}}{{30\\choose4}}=\\frac{2610}{27405}=\\frac{2\\cdot3^2\\cdot5\\cdot29}{3^3\\cdot5\\cdot7\\cdot29}=\\frac 2{21}$$ $$3$$. The probability that there is at least one of each color. $$\\frac 12\\times\\frac{{10\\choose1}{10\\choose1}{10\\choose1}{27\\choose1}}{{30\\choose4}}=\\frac 12\\times\\frac{10\\cdot10\\cdot10\\cdot27}{27405}=\\frac 12\\times\\frac{2^3\\cdot3^3\\cdot5^3}{3^3\\cdot5\\cdot7\\cdot29}=\\frac {100}{203}$$ where the factor of $$\\frac12$$ was to cancel double-counting, in that every combination like $$(\\underbrace{R_i}_{{10\\choose1}}\\underbrace{B_i}_{{10\\choose1}}\\underbrace{Y_i}_{{10\\choose1}}\\underbrace{R_j}_{{27\\choose1}})$$ is also counted in $$(R_jB_iY_iR_i)$$. You are correct. Part B", "is chosen is unknown. That means that we can reasonably assume that the probability of choosing Jar 1 is the same as the probability of choosing Jar 2. As above, a marble is drawn – without looking inside of the jar – from one of the jars. Here is the first of a new set of questions: 1. If a blue marble is drawn, what is the probability that the jar that was chosen is Jar 1? This problem is ideally set up for using Bayes’s Theorem.99 The (slightly rephrased) question: What is the probability of Jar 1 given a blue marble? is equivalent to asking what $$p(A|B)$$ is, where event $$A$$ is choosing Jar 1 and event $$B$$ is drawing a blue marble. To use Bayes’s Theorem to solve for $$p(A|B)$$, we are going to need to first find $$p(A)$$, $$p(B|A)$$, and $$p(B)$$. $$\\underline{p(A):}$$ Event $$A$$ is Jar 1, so $$p(A)$$ is the probability of choosing Jar 1. We have stipulated that the probability of choosing Jar 1 is the same as the probability of choosing Jar 2. Because one or the other must be chosen (i.e., the sample space $$\\Omega=\\{Jar~1,~Jar~2\\}$$ and the probability of choosing Jar 1 or Jar 2 is one (i.e., $$p(Jar~1\\cup Jar~2)=1$$, or equivalently, $$\\sum\\Omega=1$$), then: $p(A)=\\frac{p(A)}{p(A)+p(B)}=\\frac{p(A)}{2p(A)}=\\frac{1}{2}=0.5$ $$\\underline{p(B|A):}$$ Event $$B$$ is (drawing a)", "+0 # Mathe 0 355 2 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether?{pls help me bcz I want to learn}#CrAzyGirl Guest Feb 1, 2016 #2 +18956 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 236 \\qquad | \\qquad -\\text{box}\\\\ 2\\cdot \\text{box}-\\text{box} &=& 236 \\\\ \\text{box}", "+0 Mathe 0 532 2 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether?{pls help me bcz I want to learn}#CrAzyGirl Guest Feb 1, 2016 #2 +20108 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 236 \\qquad | \\qquad -\\text{box}\\\\ 2\\cdot \\text{box}-\\text{box} &=& 236 \\\\ \\text{box} &=&", "an equiprobable choice of two colours. Thus there are $8$ equiprobable outcomes. Now there is only one such way the three buttons could all show white. Similarly for the all red even. However, when one button is of a different colour than the other two, there are three buttons that could be that different colour. Hence the weight of each of these events is thrice that of each of the all-same-colour events. Thus we have that: $\\quad$ $\\Pr(\\text{all white}) = \\frac 1 8 \\\\ \\Pr(\\text{2 white, 1 red}) = \\frac 3 8 \\\\ \\Pr(\\text{2 red, 1 white}) = \\frac 3 8 \\\\ \\Pr(\\text{all red}) = \\frac 1 8$", "### Home > A2C > Chapter 13 > Lesson 13.1.1 > Problem13-23 13-23. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: 1. Two red and one white. $\\frac{_5C_2 \\cdot _4C_1}{_{12}C_5}=\\frac{2}{11}$ 2. Three white. See part (a). 3. One of each color. See part (a). 4. All the same color. $\\frac{_5C_3 + _4C_3 + _3C_3}{_{12}C_3}=\\frac{3}{44}$ 5. One red and two white. See part (a). 6. Two of one color and one of another. How can the probabilities from parts (c) and (d)", "## anonymous 5 years ago three jars contain colored balls as follows: Jar Red White Blue I 10 1 9 II 2 5 13 III 9 1 10 One jar is chosen at random and a ball is withdrawn. The ball is red. What is the probability that it came from jar I? Round to four decimals 1. anonymous So there's a $$\\frac{1}{3}$$ chance that the jar you chose was jar one, and a $$\\frac{10}{20}$$ chance that you chose a red ball if you the first jar. Now you need to combine these two numbers to get the overall probability you're looking for. Do you know how to combine them? 2. anonymous you do you mean like multiple 1/3 by 10/20 3. anonymous Exactly. Does that make sense? 4. anonymous i got an answer of .1667 but when i put it as my answer it says its wrong 5. anonymous Hm. Odd. Let me think on it. 6. anonymous There's a chance that what it's looking for is--given the fact that you already chose a red ball, what are the chances it was in the first jar. In that case, you realize that there are $$21$$ ways to get a red ball. $$10$$ of those ways are in the first jar, so you can get the probability from", "# conditional probability with vs. without replacement independence? I have $$2$$ jars, one of which is chosen randomly ($$.5$$ chance) jar 1: $$8$$ blue marbles (b), $$4$$ red marbles (r) jar 2: $$15$$ blue marbles, $$3$$ red marbles In a situation WITH replacement I know the probability of the first pick being a blue marble given that the second pick (from the same jar as the first pick) is a red marble = $$P(p_1 = b \\;|\\; p_2 = r) = P(p_1=b \\;|\\; jar=1)P(jar=1 \\;|\\; p_2=r) + P(p_1=b \\;|\\; jar=2)P(jar=2 \\;|\\; p_2=r)$$ But what does the same probability WITHOUT replacement look like? I know this might be an easy question but I have tried a lot and nothing seems to add up. Thank in advance! $$\\begin{multline} P(p_1 = b \\mid p_2 = r) = P(p_1=b \\mid jar=1,\\ p_2 = r)P(jar=1 \\mid p_2=r) + {} \\\\ P(p_1=b \\mid jar=2,\\ p_2 = r)P(jar=2 \\mid p_2=r) \\end{multline}$$ When you assume replacement, you can get away with the formula you wrote because the two draws are independent of each other (once the jar has been selected): $$P(p_1=b \\mid jar=1,\\ p_2 = r) = P(p_1=b \\mid jar=1,\\ p_2 = b) = P(p_1=b \\mid jar=1).$$ If you have trouble conceiving how to compute $$P(p_1=b \\mid jar=1,\\ p_2 = r),$$ remember that for any combination", "## anonymous one year ago A box has 8 beads of the same size, but all are different colors. Tamy draws a bead randomly from the box, notes its color, and then puts the bead back in the box. She repeats this 3 times. What is the probability that Tamy would pick a red bead on the first draw, then a white bead, and finally a red bead again? Answers are 1/4096 1/512 1/64 and 1/8 its not i1/8 cause I tried it and got it wrong so I though it would be 1/64? :) 1. anonymous Hell o again 2. kropot72 The probability of drawing any one of the colors is 1/8. The results of each draw are independent. Therefore the required probability is $\\large \\frac{1}{8}\\times\\frac{1}{8}\\times\\frac{1}{8}=you\\ can\\ calculate$ 3. anonymous hi. so it's 1/8 for each draw since the beads are being replaced. Then you need to multiply them to get the probability $\\frac{ 1 }{ 8 }\\times \\frac{ 1 }{ 8 }\\times \\frac{ 1 }{ 8 }$ 4. anonymous oh I got it thank you guys so much", "### Home > CCG > Chapter 12 > Lesson 12.1.4 > Problem12-55 12-55. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: Use combinations to represent the probabilities. 1. Three white balls. 2. Two red balls and one white ball. 3. Three balls of the same color. $\\frac { _{5} C _ { 3 } + _ { 4 } C _ { 3 } + _{3} C _ { 3 } } { _{12} C _ { 3 } } = \\frac { 10 + 4 + 1 } { 220 } \\approx 6.8$%", "easily first. • January 5th 2010, 07:28 AM Mukilab Errr, could someone else reiterate the whole answer? I tried it by drawing various lines and giving them each a probability... • January 5th 2010, 07:37 AM Archie Meade You probably have $Nc_r$ and $Np_r$ buttons on your calculator. $Np_r$ will calculate the number of ways of arranging r items from a total of N. $Nc_r$ will count the number of ways of selecting (without arranging) r items from N. Using those keys is the fastest way to answer. The probability of selecting 2 beads the same is $\\frac{selections\\ of\\ 2\\ same}{selections\\ of\\ 2\\ from\\ 30}$ The probability of the beads being different is that subtracted from 1. • January 5th 2010, 07:42 AM Mukilab Quote: Originally Posted by Archie Meade You probably have $Nc_r$ and $Np_r$ buttons on your calculator. $Np_r$ will calculate the number of ways of arranging r items from a total of N. $Nc_r$ will count the number of ways of selecting (without arranging) r items from N. Using those keys is the fastest way to answer. The probability of selecting 2 beads the same is $\\frac{selections\\ of\\ 2\\ same}{selections\\ of\\ 2\\ from\\ 30}$ The probability of the beads being different is that subtracted from 1. Thank you for telling me those keys, very useful", "### Home > A2C > Chapter 13 > Lesson 13.1.1 > Problem13-23 13-23. 1. A jar contains five red, four white, and three blue balls. If three balls are randomly selected, find the probability of choosing: Homework Help ✎ 1. Two red and one white. 2. Three white. 3. One of each color. 4. All the same color. 5. One red and two white. 6. Two of one color and one of another. $\\frac{_5C_2 \\cdot _4C_1}{_{12}C_5}=\\frac{2}{11}$ See part (a). See part (a). $\\frac{_5C_3 + _4C_3 + _3C_3}{_{12}C_3}=\\frac{3}{44}$ See part (a). How can the probabilities from parts (c) and (d)", "# Jellybean question #### intrakate1 ##### New member If i have a jar with 15 red, 13 yellow, and 7 green jelly beans what is the probability that i will pull out 1 red and 2 yellows (without the replacement of the ones i removed) #### soroban ##### Elite Member Hello, intrakate1! If i have a jar with 15 red, 13 yellow, and 7 green jelly beans, what is the probability that i will pull out 1 red and 2 yellows (without the replacement of the ones i removed)? $$\\displaystyle \\text{If you mean Red-\\!Yellow-\\!Yellow }in\\;that\\;order\\text{, the probability is: } \\;\\frac{15}{35}\\cdot\\frac{13}{34}\\cdot\\frac{12}{33} \\;=\\;\\frac{78}{1309}$$ $$\\displaystyle \\text{If you mean 1 red, 2 yellows }in\\;any\\;order\\text{, the probability is: }\\; \\dfrac{{15\\choose1}{13\\choose2}}{{35\\choose3}} \\;=\\;\\frac{1170}{6545} \\;=\\;\\frac{234}{1309}$$", "# Probability of picking balls from jar Suppose a jar contains 8 balls among which 5 are red and 3 are blue. If i pick, 3 balls from this jar, what is the probability that \"exactly\" two of them are blue ball? • If you are picking without replacement (which is the natural interpretation, I think) then neither is correct. – André Nicolas Sep 12 '15 at 20:00 The total number of ways to choose $3$ out of $8$ balls is $\\binom83=56$ The number of ways to choose $2$ out of $3$ blue balls and $1$ out of $5$ red balls is $\\binom32\\cdot\\binom51=15$ Hence the probability is $\\dfrac{15}{56}$ • @karthikts: It doesn't make any difference. There are $56$ (not necessarily different) ways to choose $3$ out of $8$, and $15$ of them contain exactly two blue. – barak manos Sep 12 '15 at 20:29", "a black got put into Jar 2,then: Probability Jar 3 gets a black is (n+1)/(m+n+1) Probability Jar 3 gets a white is m/(m+n+1). If a white got put into Jar 2, then: Probability Jar 3 gets a black is n/(m+n+1) Probability Jar 3 gets a white is (m+1)/(m+n+1). So the probability of getting a white from Jar 3 is: $\\left( \\frac{n}{m+n} \\right) \\, \\left( \\frac{m}{m+n+1} \\right) + \\left( \\frac{m}{m+n} \\right) \\, \\left( \\frac{m+1}{m+n+1} \\right)$ $= \\frac{n m}{(m+n)(m+n+1)} + \\frac{m (m+1)}{(m+n)(m+n+1)}$ $= \\frac{nm + m (m+1)}{(m+n)(m+n+1)} = \\frac{m (n + m+1)}{(m+n)(m+n+1)} = \\frac{m}{m+n}$. Similarly the probability of getting a black from Jar 3 is $\\frac{n}{m+n}$. A tree diagram makes it all very clear. Obviously exactly the same argment can now be used to get the probaility of getting a white or black black from Jar 4, 5, 6, ...... k. 3. thankyou very much for your help.", "236 \\\\ \\end{array}$$ There are 236 buttons in the box $$\\begin{array}{rcll} \\text{red} &=& \\frac{236}{4} \\\\ &=& 59 \\\\\\\\ \\text{yellow} &=& 32+59 \\\\ &=& 91 \\\\\\\\ \\text{blue} &=& 86 \\\\\\\\ 59+91+86 &=& 236 \\end{array}$$ heureka Feb 1, 2016 #1 +27128 +5 buttons (b) = red (1/4b) + yellow(1/4b +32) + blue(86) b = (1/4)b + (1/4)b+ 32 + 86 b = (1/2)b + 118 (1/2)b = 118 b = 236 Alan Feb 1, 2016 #2 +20108 +35 Hey guys I'm Kendall pls help me with this question : in a box 1/4 of the buttons are red. There are 32 more yellow buttons than red buttons. The remaining 86 buttons are blue. How many buttons are there in the box altogether? $$\\begin{array}{rcll} \\text{red} &=& \\frac{ \\text{box} } {4} \\\\\\\\ \\text{yellow} &=& 32 + \\text{red} \\\\ &=& 32 + \\frac{ \\text{box} } {4} \\\\\\\\ \\text{blue} &=& 86 \\\\ \\\\ \\hline \\\\ \\text{box} &=& \\text{red} ~+ ~\\text{yellow} ~ + ~\\text{blue}\\\\ \\text{box} &=& \\frac{ \\text{box} } {4} ~+ ~32 + \\frac{ \\text{box} } {4} ~ + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 32 + 86\\\\ \\text{box} &=& 2\\cdot \\frac{ \\text{box} } {4} ~+ 118\\\\ \\text{box} &=& \\frac{ \\text{box} } {2} ~+ 118 \\qquad | \\qquad \\cdot 2\\\\ 2\\cdot \\text{box} &=& \\text{box} ~+ 118\\cdot 2 \\\\ 2\\cdot \\text{box} &=& \\text{box}", "that both die rolls are 8 is $$\\frac{1}{64}$$. 3. How many positive divisors does 4000 = 25 53 have? a) 49 b) 73 c) 65 d) 15 Explanation: Any positive divisor of 4000 must be of the form 2x5y, where x and y are integers satisfying o<=x<=5 and 0<=y<=3. There are 5 possibilities for x and 3 possibilities for y and hence there are 3*5 = 15(rule of product) positive divisors of 4000. 4. Mina has 6 different skirts, 3 different scarfs and 7 different tops to wear. She has exactly one orange scarf, exactly one blue skirt, and exactly one black top. If Mina randomly selects each item of clothing, find the probability that she will wear those clothings for the outfit. a) $$\\frac{1}{321}$$ b) $$\\frac{1}{126}$$ c) $$\\frac{4}{411}$$ d) $$\\frac{2}{73}$$ Explanation: There is a $$\\frac{1}{3}$$ probability that Mina would randomly select the orange scarf, a $$\\frac{1}{6}$$ probability to select the blue skirt, and a $$\\frac{1}{7}$$ probability to select the black top. These events are independent, that is, the selection of the scarf does not affect the selection of the tops and so on. Hence, the probability that she selects the clothings of her choice is $$\\frac{1}{3} * \\frac{1}{6} * \\frac{1}{7}$$ = 126. 5. There are 6 possible routes (1, 2, 3, 4, 5, 6) from Chennai to", "# Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn 72 views closed Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II? +1 vote by (44.6k points) selected Let event J1 : Ball drawn from jar I, event J2 : Ball drawn from jar II. P(J1) = P(head) = 1/2 P(J2) = P(tail) = 1/2 Let event W: Ball drawn is white. In Jar I, there are total 12 balls, out of which 5 balls are white. ∴ Probability that the ball drawn is white under the condition that it is drawn from Jar I. P(W/J1) = $\\frac {^5C_1}{^{12}C_1} = \\frac {5} {12}$ Similarly, P(W/J2) = $\\frac {^3C_1}{^{15}C_1} = \\frac {3} {15} = \\frac 1 5$ Required probability = P(J2/W) By Bayes’ theorem", ". . carefully. Leah is asked to get a red sleeping bag out of the back seat of her mother's car. There are 7 sleeping bags in the car, only 2 of which are red. All the bags are in black zippered cases, so there is no way to tell their colors. Leah picks $\\,s$ cases, with $s < 5$, opens the cases, and discovers . . that not one of these cases contains a red sleeping bag. She tosses each of these cases into the front seat of the car. In terms of $\\,s$, what is the probability that the next case Leah chooses from the back seat will contain a red sleeping bag? There are 2 Red sleeping bags and 5 Others. And there are four cases to consider . . . $[1]\\;s = 1$: Leah gets 1 Other. . . $P(\\text{1 Other}) \\:=\\:\\frac{5}{7}$ There are 2 Reds and 4 Others. . . $P(\\text{Red}) \\:=\\:\\frac{2}{6} \\:=\\:\\frac{1}{3}$ Hence: . $P(s = 1\\,\\wedge\\,\\text{Red}) \\:=\\:\\frac{5}{7}\\cdot\\frac{1}{3} \\:=\\:\\dfrac{5}{21}$ $[2]\\;s = 2$: Leah gets 2 Others. . . $P(\\text{2 Others}) \\:=\\:\\frac{{5\\choose2}}{{7\\choose2}} \\:=\\:\\frac{10}{21}$ There are 2 Reds and 3 Others. . . $P(\\text{Red}) \\:=\\:\\frac{2}{5}$ Hence: . $P(s = 2\\,\\wedge\\,\\text{Red}) \\:=\\:\\frac{10}{21}\\cdot\\frac{2}{5} \\:=\\:\\dfrac{4}{21}$ $[3]\\;s = 3$: Leah gets 3 Others. . . $P(\\text{3 Others}) \\:=\\:\\frac{{5\\choose3}}{{7\\choose2}} \\:=\\:\\frac{2}{7}$ There are 2 Reds and 2 Others. . .", "drawn from Bag 1 and a red is returned to Bag 1. . . [2] A black is drawn from Bag 1 and a black is returned to Bag 1. [1] A red is drawn from Bag 1: . $P(R_1) = \\frac{3}{8}$ . . .Bag 2 now contains:7 red and 2 black discs: . . .A red is drawn from Bag 2: . $P(R_2) \\:=\\:\\frac{7}{9}$ . . . . . $P(R_1R_2) \\:=\\:\\frac{3}{8}\\cdot \\frac{7}{9}$ [2] A black is drawn from Bag 1: $P(B_1) = \\frac{5}{8}$ . . .Bag 2 now contains: 6 red and 3 black discs. . . .A black is drawn from Bag 2: . $P(B_2) = \\frac{3}{9}$ . . . . . $P(B_1B_2) \\:=\\:\\frac{5}{8}\\cdot\\frac{3}{9}$ Therefore: . $P(R_1R_2\\text{ or }R_1B_2) \\;=\\;\\frac{3}{8}\\cdot\\frac{7}{9} + \\frac{5}{8}\\cdot\\frac{3}{9} \\;=\\;\\frac{7}{24} + \\frac{5}{24} \\;=\\;\\frac{12}{24} \\;=\\;\\dfrac{1}{2}$" ]

[ "Question The ratio of the measures of two complementary angles is 1:2. What is the measure of the larger angle? (Hint: Let x and 2x represent the angle measures.) F 30° G 45° H 60° 120°​", "# The ratio of the measure of two complementary angles is 4:5, what are the measures of the angles? $\\frac{90}{9} = 10$ $4 \\times 10 : 5 \\times 10$ ${40}^{\\circ} : {50}^{\\circ}$", "less than its supplement. The measure of the angle is • A. $37^{o}$ • B. $74^{o}$ • C. $148^{o}$ • D. None of these Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q5 Subjective Medium If the complementary of an angle is $62^o$, then find its supplement. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "Single Correct Medium The difference between the complementary and supplementary angles of $70^o$ is: • A. $60^o$ • B. $70^o$ • C. $80^o$ • D. $90^o$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "# Question - Trigonometric Ratios of Complementary Angles and Direct Application Account Register Share Books Shortlist ConceptTrigonometric Ratios of Complementary Angles and Direct Application #### Question Without using trigonometric tables evaluate the following: (i) sin^2 25º + sin^2 65º #### Solution You need to to view the solution Is there an error in this question or solution? S", "# Complementary Angles Complementary angles are two angles whose measures add up to $90°$ . In the figure below, $\\angle 1$ and $\\angle 2$ are complementary. Two angles need not be adjacent to be complementary. The angles in the next figure are also complementary, since $35°+55°=90°$ .", "the two lines. supplementary angles supplementary angles Two angles that add up to $180^\\circ$. transversal transversal A line that intersects two other lines.", "x }^{ 2 }$, find : The value of ${ x }^{ 2 }$ it its supplementary angle is three times its complementary angle. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q5 Subjective Medium The sum of two vertically opposite angles is $166^o$. Find each of the angles. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "# The measures of two complementary angles are in the ratio 1:5. What is the measure of each angle? Let say that the two angles are $a$ and $b$ .Because they are complementary we have that $a + b = {90}^{o}$ and from the ratio we get $\\frac{a}{b} = \\frac{1}{5} \\implies b = 5 a$ $a = {15}^{o}$ and $b = {75}^{o}$", "# Ratio of two complementary angles is 2 : 3. What is the double of the larger angle? 1. 54° 2. 108° 3. 95° 4. 75° Option 2 : 108° ## Detailed Solution Given: The ratio of two complementary angles is 2 : 3 Concept used: The sum of the two angles of the complementary angle is 90° Calculation: Let the two angles be 2x and 3x Now, ⇒ 2x + 3x = 90 ⇒ 5x = 90 ⇒ x = 18 Two complementary angles are 36 and 54 Larger angle = 54 Double of the larger angle = 54 × 2 = 108° ∴ The double of the larger angle is 108°", "## Prealgebra (7th Edition) Complementary angles total $90^o$. Since angles do not have negative measurements, there is no complement for a $120^o$ angle.", "complementary angles are in the ratio $1:9$. The angles are: • A. $54^\\circ,\\, 36^\\circ$ • B. $10^\\circ,\\, 90^\\circ$ • C. $11^\\circ,\\, 79^\\circ$ • D. $9^\\circ,\\, 81^\\circ$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "Mathematics # The ratio between two complementary angles is $2 : 3$: find the smallest angle. $36$ ##### SOLUTION Two angles whose sum is ${ 90 }^{ o }$ are said to be complementary. Given two angles are in the ratio of $2:3$. Let the two angles be $2x$ and $3x$. So, $\\ 2x+3x={ 90 }^{ o }$ $\\Rightarrow 5x={ 90 }^{ o }$ $\\Rightarrow x=\\dfrac { 90 }{ 5 }$ $\\Rightarrow x=18\\\\$ Therefore, the two angles are $2x={ 36 }^{ o }$ and $3x={ 54 }^{ o }$. You're just one step away Single Correct Medium Published on 09th 09, 2020 Questions 120418 Subjects 10 Chapters 88 Enrolled Students 86 #### Realted Questions Q1 Subjective Medium In the joining figure, name the following pairs of angles. Adjacent angles that do not form a linear pair. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q2 Subjective Medium Find the value of unknown x in the following diagram. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q3 Single Correct Medium If one angle at a point is reflex angle, the other at that point may be : • A. Acute angle • B. Obtuse angle • C. Straight angle • D. Acute or", "4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade? Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade. Solve Geometry Applications In the following exercises, translate to a system of equations and solve. The difference of two complementary angles is 55 degrees. Find the measures of the angles. The difference of two complementary angles is 17 degrees. Find the measures of the angles. $$53.5$$ degrees and $$36.5$$ degree. Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles. Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles. 16 degrees and 74 degrees The difference of two supplementary angles is 8 degrees. Find the measures of the angles. The difference of two supplementary angles is 88 degrees. Find the measures of the angles. 134 degrees and 46 degrees Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both", "$$45^\\circ$$. ## Supplementary angles VS Complementary angles Supplementary and complementary angles are pairs of angles that add up to $$180^\\circ$$ and $$90^\\circ$$, respectively. Let’s take a closer look at their differences.", "# Question - Trigonometric Ratios of Complementary Angles and Direct Application Account Register Share Books Shortlist ConceptTrigonometric Ratios of Complementary Angles and Direct Application #### Question \\text{Evaluate }\\frac{\\tan 65^\\circ }{\\cot 25^\\circ} #### Solution You need to to view the solution Is there an error in this question or solution? S", "The ratio of the angle (in radians) to the total angle of a circle (2pi) must be equal to the ratio of the area of the section and the total area of the circle.", "• anonymous 2. The measure of two complementary angles is in the ratio 1 : 4. What are the degree measures of the two angles? 45° and 135° 23° and 68° 36° and 144° 18° and 72° Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "## Algebra: A Combined Approach (4th Edition) $70.4$ Two angles are complementary if the sum of their measures is 90°. The complement of a $19.6°$ angle is an angle whose measure is $90°-19.6° = 70.4°$ To check, notice that $19.6° + 70.4°= 90°$.", "# What is the ratio of the measure of a 120 degrees-angle to the measure of its supplement? Two angles are said to be supplementar if they sum up to 180°. So, the supplement of an angle of 120° is an angle of 60°. The ratio is easily calculated: $\\frac{120}{60} = 2$" ]

[ "two angles is 90: x+$$\\frac{x-10}{4}$$=90. x+$$\\frac{x-10}{4}$$=90 4(x+$$\\frac{x-10}{4}$$=90) 4x+x-10=360 (4+1)x-10=360 5x-10=360 5x-10+10=360+10 5x=370 x=74 The measure of the second angle is 74°, the measure of the first angle is $$\\frac{74^{\\circ}-10^{\\circ}}{4}$$=$$\\frac{64^{\\circ}}{4}$$=16°, and the measure of the third angle is 90°. Solution Four Let x be the measure of the second angle. Then, the first angle is $$\\frac{x-10}{4}$$. The sum of the three angles is x+$$\\frac{x-10}{4}$$+90=180. x+$$\\frac{x-10}{4}$$+90=180 x+$$\\frac{x-10}{4}$$+90-90=180-90 x+$$\\frac{x-10}{4}$$=90 $$\\frac{4 x}{4}$$+$$\\frac{x-10}{4}$$=90 $$\\frac{4 x+x-10}{4}$$=90 4x+x-10=360 5x-10+10=360+10 5x=370 x=74 The measure of the second angle is 74°, the measure of the first angle is $$\\frac{74^{\\circ}-10^{\\circ}}{4}$$=$$\\frac{64^{\\circ}}{4}$$=16°, and the measure of the third angle is 90°. Make sure students see at least four different methods of solving the problem. Conclude this example with the statements below. → Each method is slightly different either in terms of how the variable is defined or how the properties of equality are used to solve the equation. The way you find the answer may be different from your classmates’ or your teacher’s. → As long as you are accurate and do what is mathematically correct, you will find the correct answer. ### Eureka Math Grade 8 Module 4 Lesson 5 Exercise Answer Key Exercises For each of the following problems, write an equation and solve. Exercise 1. A pair of congruent angles are described as follows: The degree measure", "random angle to measure. Measure as many angles as you need to become confident. 5. Sometimes the angle you need to measure will be greater than $\\scriptsize {{180}^{{}^\\circ }}$. . . How will you measure the angle in this case shown here? Here’s a hint. What if you measured the smaller angle created between the two lines that is less than $\\scriptsize 180{}^\\circ$? What would you subtract this angle from to find the bigger angle that is greater than $\\scriptsize 180{}^\\circ$? Open the online interactive activity called Measuring angles between 0 and 360 degrees with a protractor. Here you will find a protractor with which to measure angles between $\\scriptsize 0{}^\\circ$ and $\\scriptsize 360{}^\\circ$. What did you find? Because the online interactive activity generates random angles to measure, it is not possible to give all the answers. But you can check all your answers with the answer box. Were you able to work out how to measure angles greater than $\\scriptsize {{180}^{{}^\\circ }}$? To do so, measure the size of the angle smaller than $\\scriptsize {{180}^{{}^\\circ }}$ created by the two lines and then subtract this angle from $\\scriptsize {{360}^{{}^\\circ }}$ to find the larger angle. We can do this, because the angles around a point always add up to $\\scriptsize {{360}^{{}^\\circ }}$. ### Exercise 1.3 1. Use", "/6 and /8, and name the reflex angle correctly, and is closer 90^... { G } E and the angle of 130^ { \\circ } and \\hat { }... Use angles to find a missing angle of 50^ { \\circ }. ) ,:. Measures of all the angles marked x is 60^ { \\circ } this information to present correct. Very similar to the right side of the shops in the figure above, the lines cut by shaded. Turn between two straight lines, and make an F shape time and. And to personalise content to better how to find missing angles in intersecting lines the needs of our users say: vertically angles. Of 180^ { \\circ }, it means we 're having trouble external. Shows what happens when tangents and secants intersect on a circle answer seems correct, opposite... And the angle to measure this reflex angle with an ordinary protractor only... = 91^ { \\circ }. ) make alternate angles are supplementary in! Finding the measure of inscribed angle 2 is equal to 30^ { \\circ.! You look carefully, you use a protractor to measure the angle of measure is vertical to the right of!, so use the vertical angles angle it is the inner scale shop, or 90-degree, angle know... Or print icon to worksheet to", "that is not shaded and then subtract the answer from $360^{\\circ}$. 3. Step 3: Place the protractor correctly on the angle to measure. • The centre of the protractor should be on the vertex of the angle. • The base line of the protractor should be on one arm of the angle. 4. Step 4: Choose the scale and read off the size of the angle. The acute angle measures $18^{\\circ}$. 5. Step 5: Subtract the measured angle from $360^{\\circ}$ and give the correct answer. Check against your estimation, and name the reflex angle correctly. \\begin {align} \\text{Reflex } Z\\hat{X}Y &= 360^{\\circ} - 18^{\\circ}\\\\ &= 342^{\\circ} \\end {align} Remember that reflex angles are always less than $360^{\\circ}$, but more than $180^{\\circ}$. A full turn or revolution is equal to $360^{\\circ}$. ### Exercise 15.2: Measure angles 1. Use a protractor and measure the angle shown below. Estimate the size of the angle: The angle is an acute angle, so the measurement should be between $0^{\\circ}$ and $90^{\\circ}$. Measure the angle with a protractor. Answer: $B\\hat{A}C = 53^{\\circ}$ 2. Use a protractor and measure the angle shown below. Estimate: acute angle, so between $0^{\\circ}$ and $90^{\\circ}$ Answer: $B\\hat{A}C = 59^{\\circ}$ 3. Use a protractor and measure the angle shown below. Estimate: obtuse angle, so between $90^{\\circ}$ and $180^{\\circ}$ Answer: $B\\hat{A}C", "accuracy. If you and your partner both selected the same angle, then choose a new one together. ### Vocabulary Acute Angle an angle whose measure is less than $90^\\circ$ . Obtuse Angle an angle whose measure is greater than $90^\\circ$ . Right Angle an angle whose measure is equal to $90^\\circ$ . Straight Angle an angle whose measure is equal to $180^\\circ$ . Degrees how an angle is measured. ### Guided Practice Here is one for you to try on your own. True or false. An acute angle can also be a right angle. False. An acute angle measures 90 degrees while a right angle measures exactly 90 degrees. ### Practice Directions: Label each angle as acute, obtuse, right, or straight. 9. $55^\\circ$ 10. $102^\\circ$ 11. $90^\\circ$ 12. $180^\\circ$ 13. $10^\\circ$ 14. $87^\\circ$ 15. $134^\\circ$ ### Vocabulary Language: English Acute Angle Acute Angle An acute angle is an angle with a measure of less than 90 degrees. Degree Degree A degree is a unit for measuring angles in a circle. There are 360 degrees in a circle. Obtuse angle Obtuse angle An obtuse angle is an angle greater than 90 degrees but less than 180 degrees. Perpendicular Perpendicular Perpendicular lines are lines that intersect at a $90^{\\circ}$ angle. The product of the slopes of two perpendicular lines is", "in the middle of the bottom line, where all the degree lines meet. Example 2: Measure the three angles from Example 1, using a protractor. Solution: Just like in Example 1, it might be easier to measure these three angles if you separate them. With measurement, we put an $m$ in front of the $\\angle$ sign to indicate measure. So, $m \\angle XUY = 84^\\circ, \\ m \\angle YUZ = 42^\\circ$ and $m \\angle XUZ = 126^\\circ$. In the last lesson, we introduced the Ruler Postulate. Here we introduce the Protractor Postulate. Protractor Postulate: For every angle there is a number between $0^\\circ$ and $180^\\circ$ that is the measure of the angle in degrees. The angle's measure is then the absolute value of the difference of the numbers shown on the protractor where the sides of the angle intersect the protractor. In other words, you do not have to start measuring an angle at $0^\\circ$, as long as you subtract one measurement from the other. Example 3: What is the measure of the angle shown below? Solution: This angle is not lined up with $0^\\circ$, so use subtraction to find its measure. It does not matter which scale you use. Using the inner scale, $|140 - 25| = 125^\\circ$ Using the outer scale, $|165 - 40| = 125^\\circ$", "Proof: together the measure up of one inscribed angle is same to half the measure up of its intercepted arc, the inscribed angle is half the measure up of that is intercepted arc, that is a right line. As the arc\"s measure is $180^\\circ$, the enrolled angle\"s measure is $180^\\circ\\cdot\\frac12 = 90^\\circ$.$\\blacksquare$ When I confirm the systems on the web there were a whole bunch of various other more complicated proofs. Is mine valid? If that is invalid, might someone tell me why? Thank you, Paul geometry one proof-verification re-superstructure point out follow edited Oct 29 \"13 in ~ 13:26 amWhy request Oct 29 \"13 in ~ 12:55 Paul FilchPaul Filch $\\endgroup$ 6 | show 1 much more comment 5 $\\begingroup$ Hint Draw the radius from the center of the circle come the suggest that friend think it has actually an angle of $90$ degrees and write down the angles: $2(x+y)=180$ re-superstructure mention monitor answer Oct 29 \"13 in ~ 14:44 $\\endgroup$ 2 Thanks for contributing response to dearteassociazione.orgematics stack Exchange! But avoid Asking for help, clarification, or responding to various other answers.Making statements based on opinion; ago them increase with recommendations or an individual experience. You are watching: An angle inscribed in a semicircle Use dearteassociazione.orgJax to style equations. Dearteassociazione.orgJax reference. See more: How To Get Bots", "the same as $${-30^\\circ}$$? ## An Angle and its Reflex Angle An angle is the \"space\" between two rays meeting at a common endpoint. Let’s take the letters A,B, and O to name the angle. The larger angle is a Reflex Angle, but the smaller angle is an Acute Angle It is to avoid this particular ambiguity that we refer to the angle marked above (with the dotted line) as reflex $${\\angle AOB}$$. Note: any $${\\angle AOB}$$ and its reflex $${\\angle AOB}$$ sum to $${360^\\circ}$$ This property of angle and its reflex can be used to measure (and construct) angles that measure more than $${180^\\circ}$$ and less than $${360^\\circ}$$. ## How to construct Angles? We use a protractor to construct angles. Let’s draw a $${50^\\circ}$$ angle. ### Step 1 First, draw a ray OB and align the protractor with OB as shown. ### Step 2 Place a point above the marking on the protractor that corresponds to $${50^\\circ}$$. ### Step 3 Remove the protractor and draw a ray beginning at O that passes through this point. $${\\angle AOB}$$ is the required angle. That is $${\\angle AOB}$$ = $${50^\\circ}$$ ### Step 4 If the ray extends in the other direction, we measure the angle from the $${0^\\circ}$$ mark on the bottom-left. The above image shows how to draw a", "complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles. 28. Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles. 16 degrees and 74 degrees 29. The difference of two supplementary angles is 8 degrees. Find the measures of the angles. 30. The difference of two supplementary angles is 88 degrees. Find the measures of the angles. 134 degrees and 46 degrees 31. Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles. 32. Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles. 37 degrees and 143 degrees 33. The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles. 34. The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the", "Thread: question x & y values. 1. question x & y values. Say there is a angle b/t 0-90. A what point will the Y value be larger than the X value. I know that Y increases towards 90,to ultimately become 1. Closer to the x axis the X value should be large... anyone see what I'm saying? 2. Originally Posted by Brndo4u Say there is a angle b/t 0-90. A what point will the Y value be larger than the X value. I know that Y increases towards 90,to ultimately become 1. Closer to the x axis the X value should be large... anyone see what I'm saying? yes, x and y are trig values. specifically, $x = \\cos{\\theta}$ and $y = \\sin{\\theta}$ $x = y$ at $\\theta = 45^\\circ$ 3. So, after 45 degrees the Y will be a larger value, 45 and below the X will be a greater value.", "= 3x + 10° ………(1) From ΔCOD $$\\overline{\\mathrm{OC}}=\\overline{\\mathrm{CD}}$$ ⇒ ∠O = ∠D …………………(2) [∵ The angles which are opposite to the equal sides are equal]. from (1) & (2) ∠BOA = ∠COD [∵ Vertically opposite angles are equal.] But ∠COD = 90 – x (∵ 2x + ∠O + ∠D = 180 ⇒ 2x + ∠O + ∠O = 180 (∵∠O = ∠D) ⇒ 2∠O = 180 – 2x ∠O = $$\\frac{180-2 x}{2}$$ = 90 – x] ∴ From ∠BOA = ∠COD ⇒ 3x + 10 = 90 – x ⇒ 3x + x = 90 – 10 ⇒ 4x = 80 ∴ x = 20° Question 2. The difference between two numbers is 8. if 2 is added to the bigger number the result will be three times the smaller number. Find the numbers. Solution: Let the bigger number be x. The difference between two numbers 8 ∴ Smaller number = x – 8 If 2 is added to the bigger number the result will be three times the smaller number. So x + 2 = 3(x – 8) x + 2 = 3x – 24 x – 3x = -24 – 2 – 2x = -26 ∴ x = $$\\frac{26}{2}$$ = 13 ∴ Bigger number = 13 Smaller number = 13 – 8 = 5 Question", "angle” may help remind them that acute angles are the smaller ones. 2) Since protractors like the one shown here display two different numbers for the measure of an angle, students may read off the wrong one. Remind them to think about whether the angle appears to be acute or obtuse, and figure out which number makes sense based on that. (Another way to check is to note that the end of the protractor they placed against one side of the angle either reads $0^\\circ$ on the inside “track” and $180^\\circ$ on the outside track, or vice versa. The track on which it reads $0^\\circ$ is the one from which they should read off the measure of the other side of the angle. For example, in the illustration shown in the text, the end of the protractor placed against the bottom side of the angle reads $0^\\circ$ on the inside track, so the inside track is the one to use for determining the angle measure, and so we know it is $50^\\circ$ and not $130^\\circ$.) 3) Students may get their calculations backwards here, possibly due to a vague notion that the larger wheel should rotate a greater number of times. Explain if necessary that since the wheels both travel the same distance along their circumference and the larger", "(b). The matched pairs are as follows, (i)-(c), (ii)-(d), (iii)-(a), (iv)-(e), and (v)-(b). 2. Classify each one of the following angles as right, straight, acute, obtuse or reflex. (a) Ans: From the given figure we can see that the measure of the given angle is less than $90^\\circ$. Also, we know that the angles having measures less than $90^\\circ$ are acute angles. Therefore, the angle given in the figure is classified as an acute angle. (b) Ans: From the given figure we can see that the measure of the given angle is more than $90^\\circ$. Also, we know that the angles having measures more than $90^\\circ$ but less than $180^\\circ$ are obtuse angles. Therefore, the angle given in the figure is classified as an obtuse angle. (c) Ans: From the given figure we can see that the measure of the given angle is $90^\\circ$. Also, we know that the angles having measures of $90^\\circ$ are right angles. Therefore, the angle given in the figure is classified as right angle. (d) Ans: From the given figure we can see that the measure of the given angle is greater than $180^\\circ$. Also, we know that the angles have measures greater than $180^\\circ$ are reflex angles. Therefore, the angle given in the figure is classified as reflex angle. (e) Ans: From", "Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles. Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for measure of each angle. Step 3. Name what we are looking for. Let x=x= the measure of the first angle. y=y= the measure of the second angle Step 4. Translate into a system of equations. The angles are supplementary. The larger angle is twelve less than five times the smaller angle. The system is shown: Step 5. Solve the system of equations substitution. Substitute 5x − 12 for y in the first equation. Solve for x. Substitute 32 for x in the second equation, then solve for y. Step 6. Check the answer in the problem. Step 7. Answer the question. The angle measures are 148 and 32 degrees. Example $$\\PageIndex{14}$$ Translate to a system of equations and then solve: Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles. The angle measures are 42 and 138. Example $$\\PageIndex{15}$$ Translate to a system of equations and then solve: Two angles are supplementary. The measure of the", "measure is more than $90^{circ}$. (ii) $angle COD$ is a right angle since its measure is $90^{circ}$. (iii) $angle FOE$ is a straight angle since its measure is $180^{circ}$. (iv) $angle POQ$ is a reflex angle since its measure is more than $180^{circ}$ but less than $360^{circ}$. (v) $angle HOG$ is an acute angle since its measure is more than 0 but less than $90^{circ}$. (vi) $angle POP$ is a complete angle since its measure is $360^{circ}$. QUESTION – 2: Classify the angles whose magnitudes are given below: (i) $30^{circ}$ (ii) $91^{circ}$ (iii) $179^{circ}$ (iv) $90^{circ}$ (v) $181^{circ}$ (vi) $360^{circ}$ (vii) $128^{circ}$ (viii) $left ( 90.5 right )^{circ}$ (ix) $left ( 38.3 right )^{circ}$ (x) $80^{circ}$ (xi) $0^{circ}$ (xii) $15^{circ}$ Solution: (i) Acute angle This is because its measures is less than $90^{circ}$ but more than $0^{circ}$. (ii) Obtuse angle This because its measure is more than $90^{circ}$ but less than $180^{circ}$. (iii) Obtuse angle This because its measure is more than $90^{circ}$ but less than $180^{circ}$. (iv) Right angle This is because its measure is $90^{circ}$. (v) Reflex angle This because its measure is more than $180^{circ}$ but less than $360^{circ}$. (vi) Complete angle This is because its measure is $360^{circ}$. (vii) Obtuse angle This because its measure is more than $90^{circ}$ but less than $180^{circ}$. (viii)", "it as follows:$90^\\circ=\\tfrac14\\tau$ This arguably makes it even easier to work with angles than when we had degrees! If you want to go $$5/12$$ of the way around the circle, then your angle is just $$\\tfrac{5}{12}\\tau$$: five-twelfths of a turn. Simple and clean. We're definitely closer to having a natural unit of angle measurement, but we haven't really linked it to the circle yet. We could just as easily apply our fraction concept to a square, or a hexagon: So what makes the circle special? As the angle opens, the arc of the circle grows proportionally with it. With any other shape, the amount marked off on the perimeter will grow faster or slower depending on how close the shape is to the origin, but with a circle, double the angle always means double the arc. You could imagine wrapping a string around the circle, starting at the initial side and ending at the terminal side, and then measuring how long it is. The problem is that would depend on how big the circle's radius is. But we can fix that too, by always basing our measurements on the simplest possible circle: the unit circle, with a radius of $$1$$. If we wrapped our string all the way around the unit circle, you'd notice it would be about", "$PT$. It seems that the marked angle is an acute angle. An acute angle is between $0^{\\circ}$ and $90^{\\circ}$. This angle is nearer to $90^{\\circ}$ than to $0^{\\circ}$. 2. Step 2: Place the protractor correctly on the angle. • The centre of the protractor should be on the vertex of the angle. • The base line of the protractor should be on one arm of the angle. 3. Step 3: Choose the correct scale. The arm on which the base line lies is to the right side of the protractor, so use the inner scale. 4. Step 4: Read off the size of the angle from the protractor, and check against your estimation. $67^{\\circ}$ is an acute angle, and is closer to $90^{\\circ}$ than to $0^{\\circ}$. 5. Step 5: Name the angle, and give the answer. ### Worked example 15.6: Measuring an obtuse angle Use a protractor to measure the marked angle shown below. Use the dotted lines shown to help with your estimation. The dotted lines have a $90^{\\circ}$ angle between them. 1. Step 1: Consider the diagram and estimate the size of the angle. The diagram shows two arms: $AB$ and $AC$. It seems as if they form an obtuse angle, which is shaded red. An obtuse angle is between $90^{\\circ}$ and $180^{\\circ}$. This angle is", "# Recall that two angles are complementary if the sum of their measures is 90 degrees. Find the measures of two complementary angles if one angle is nine times the other angles? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 3 Apr 23, 2017 Ans: ${9}^{\\circ} , {81}^{\\circ}$ #### Explanation: Let the smaller angle be $x$ and the bigger one be $9 x$. Since the angles are complementary, you have $9 x + x = {90}^{\\circ}$ $10 x = {90}^{\\circ}$ $x = {9}^{\\circ} \\mathmr{and} 9 x = {81}^{\\circ}$ • 8 minutes ago • 8 minutes ago • 8 minutes ago • 9 minutes ago • 3 minutes ago • 3 minutes ago • 4 minutes ago • 4 minutes ago • 5 minutes ago • 7 minutes ago • 8 minutes ago • 8 minutes ago • 8 minutes ago • 9 minutes ago", "also see from the figure, that the angle marked definitely is not smaller than $90{}^\\circ$ either. So, it’s not an acute angle either. Next, we have a reflex angle. A reflex angle lies between $180{}^\\circ$ and $360{}^\\circ$ always. We can also see that the angle marked does not look like the major sector of a circle. Hence, we can say that it’s not a reflex angle either. However, till now, we have concluded that neither is it bigger than $90{}^\\circ$, nor is it smaller than it. It, in fact, looks like it is exactly equal to $90{}^\\circ$, and angles which are exclusively equal to $90{}^\\circ$, are known as right angles. Note: Don’t get confused between obtuse and reflex angle. To sum it up, if we consider any angle $x{}^\\circ$, then it is acute when \\$0", "on the protractor itself. Remember you can check that you are using the correct scale by making sure your answer fits your angle. If the angle is acute, the measure of the angle should be less than $90^\\circ$. If it is obtuse, the measure will be greater than $90^\\circ$. In this case, the angle is acute, so its measure is $50^\\circ$. ### Example 6 What is the measure of the angle shown below? This angle is not lined up with the zero mark on the protractor, so you will have to use subtraction to find its measure. Using the inner scale, we get $|140-15|=|125|=125^{\\circ}$. Using the outer scale, $|40-165|=|-125|=125^{\\circ}$. Notice that it does not matter which scale you use. The measure of the angle is $125^\\circ$. ### Example 7 Use a protractor to measure $\\angle{RST}$ below. You can either line it up with zero, or line it up with another number and find the absolute value of the differences of the angle measures at the endpoints. Either way, the result is $100^\\circ$. The angle measures $100^\\circ$. Multimedia Link The following applet gives you practice measuring angles with a protractor Measuring Angles Applet. ## Angle Addition Postulate You have already encountered the ruler postulate and the protractor postulate. There is also a postulate about angles that is similar to" ]

[ "# The ratio of the measure of two complementary angles is 4:5, what are the measures of the angles? $\\frac{90}{9} = 10$ $4 \\times 10 : 5 \\times 10$ ${40}^{\\circ} : {50}^{\\circ}$", "less than its supplement. The measure of the angle is • A. $37^{o}$ • B. $74^{o}$ • C. $148^{o}$ • D. None of these Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q5 Subjective Medium If the complementary of an angle is $62^o$, then find its supplement. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020", "Question The ratio of the measures of two complementary angles is 1:2. What is the measure of the larger angle? (Hint: Let x and 2x represent the angle measures.) F 30° G 45° H 60° 120°​", "Courses Courses for Kids Free study material Free LIVE classes More # The measure of the complementary of the angle of measure $90{}^\\circ$ is[a] $0{}^\\circ$[b] $45{}^\\circ$[c] $90{}^\\circ$[d] $60{}^\\circ$ Last updated date: 27th Mar 2023 Total views: 306.3k Views today: 2.84k Verified 306.3k+ views Hint: The measure of complementary angles add up to $90{}^\\circ$. So let x be the measure of complementary angles. Create a linear equation in x. Solve for x. The value of x gives the measure of the complementary angle. Alternatively, you can directly use the complementary angle of measure x is $90{}^\\circ -x$. Complementary angles: Two angles whose measures add up to $90{}^\\circ$ are called complementary angles, e.g. $60{}^\\circ$ and $30{}^\\circ$ are complementary angles. Supplementary angles: Two angles whose measures add up to $180{}^\\circ$ are called supplementary angles, e.g. $60{}^\\circ$ and $120{}^\\circ$ Reflex angle of an angle: The measure of an angle and its reflex sum up to $360{}^\\circ$. Since $90{}^\\circ$ and x are complementary we have $90{}^\\circ +x=90{}^\\circ$ Subtracting $90{}^\\circ$ from both sides, we get \\begin{align} & 90{}^\\circ +x-90{}^\\circ =90{}^\\circ -90 \\\\ & \\Rightarrow x=0{}^\\circ \\\\ \\end{align} Hence the measure of the complementary angle of $90{}^\\circ$ is $0{}^\\circ$. Note: [1] There is usually a confusion whether supplementary angles sum up to $180{}^\\circ$ or whether complementary angles add up to $180{}^\\circ$. In that case, one", "# CSI project fpr For inequality ###### Question: CSI project fpr For inequality ### What is the sum of the measures of two complementary angles? A. 270° B. 45° C. 180° D. 90° What is the sum of the measures of two complementary angles? A. 270° B. 45° C. 180° D. 90°... ### V to the third power = 12. Whoever answers did correctly with a good step by step explanation will get brainliest V to the third power = 12. Whoever answers did correctly with a good step by step explanation will get brainliest... ### George has $12, which is$7 more than Tom has. How much money does Tom have? George has $12, which is$7 more than Tom has. How much money does Tom have?... ### Simplify the following expression -24 - 3.r) + (5x - 2)​ Simplify the following expression -24 - 3.r) + (5x - 2)​... ### I added a picture : Determine which of the lines, if any, are parallel or perpendicular. Explain. write an equation of the line that passes through the given point and is perpendicular to the given line. i added a picture : Determine which of the lines, if any, are parallel or perpendicular. Explain. write an equation of the line that passes through the given point and is perpendicular", "Comment Share Q) # Two angles form a linear pair.The measure of the smaller angle is one-half the measure of the larger angle.Find the Larger angle ? ( A ) $120^{\\circ}$ ( B ) $130^{\\circ}$ ( C ) $110^{\\circ}$ ( D ) $140^{\\circ}$ • Two Angles form Linear Pair i.e $X + Y = 180^{\\circ}$ $\\large\\frac{1}{2}x$ = the smaller angle smaller angle$\\large\\frac{1}{2}x$ + larger angle (x) = 180 $\\rightarrow$Linear Pair $\\large\\frac{1}{2}x$ + x = 180 $\\large\\frac{1}{2}x$ = 60 -------> smaller angle", "# The measures of two complementary angles are in the ratio 1:5. What is the measure of each angle? Let say that the two angles are $a$ and $b$ .Because they are complementary we have that $a + b = {90}^{o}$ and from the ratio we get $\\frac{a}{b} = \\frac{1}{5} \\implies b = 5 a$ $a = {15}^{o}$ and $b = {75}^{o}$", "# Trigonometric ratios for angles greater than $90^\\circ$? The trigonometric ratios of an angle greater than $90^\\circ$ are equal to the supplementary angle's ratios. I'm just clarifying this, but the ratios don't actually exist for angles greater than $90^\\circ$ right? (since by definition these ratios are of a right angle triangle). Is it just mathematical custom to assume the ratios of an angle greater than $90^\\circ$ to be equivalent to its supplementary angle's ratios? For convenience? - It helps to visualize a circle of radius 1 in the $xy$ plane. The sine of the angle, which is measured counter-clock-wise, with the $x$ axis being zero, is $y$ and the cosine is $x$. As we pass 90 degrees, we cross the $y$ axis ans $x$ (i.e. sine) swings negative. – Kaz Apr 25 '12 at 1:07 Instead of ratios in right triangles (which as you notice make sense only for acute angles), one can consider the cosine and sine defined as the $x$ and $y$ coordinate of a point that moves around a unit circle. This works for all angles -- and for acute angles you can inscribe a right triangle in the first quadrant of the unit circle and see that the unit-circle definition matches the right-triangle one. After 90°, the cosine becomes negative, because the point is", "of one angle in a pair of complementary angles. Let’s see how this works. We can see that together, $C$ and $D$ form a right angle. Therefore they are complementary, and they add up to $90^\\circ$ . We know that $C$ has a measure of $44^\\circ$ . How can we find the measure of angle $D$ ? To find the measurement of angle $D$ , we simply subtract the measure of angle $C$ from 90. $\\angle{C}+ \\angle{D} & = 90^\\circ\\\\44^\\circ + \\angle{D} & = 90^\\circ\\\\\\angle {D} & = 90-44\\\\\\angle {D} & = 46^\\circ$ In order for these two angles to be complementary, as the problem states, they must add up to $90^\\circ$ . Angle $D$ therefore measures $46^\\circ$ . We can check our calculation by adding angles $C$ and $D$ . Their sum must be equal to $90^\\circ$ . $44^\\circ + 46^\\circ = 90^\\circ$ We can follow the same process to find the unknown angle in a pair of supplementary angles. As with complementary angles, if we know the measure of one angle in the pair, we can find the measure of the other. Angles $P$ and $Q$ are supplementary angles. If angle $P$ measures $112^\\circ$ , what is the measure of angle $Q$ ? We know that supplementary angles have a total of $180^\\circ$ Therefore we can", "Answer Comment Share Q) # If the measure of one angle of a linear pair is five times that of the other, then the measure of the larger angle is ( A ) 150^{\\circ} ( B ) None of these ( C ) 120^{\\circ} ( D ) 110^{circ}", "# Sine & cosine of complementary angles Learn about the relationship between the sine & cosine of complementary angles, which are angles who together sum up to 90°. We want to prove that the sine of an angle equals the cosine of its complement. $\\sin(\\theta) = \\cos(90^\\circ-\\theta)$ It can be fun to verify that this property works on your calculator. Try plugging in $\\sin(10^\\circ)$ and $\\cos(80^\\circ)$, and check that both give you the same answer: 0.17364817766. Let's start with a right triangle. Notice how the acute angles are complementary, sum to 90$^\\circ$. The angles in a triangle add up to 180$^\\circ$. Since one angle in a right triangle measures 90$^\\circ$, the remaining two angles must add up to 180$^\\circ$ minus 90$^\\circ$, or 90$^\\circ$. So, if we name one of the angles $\\theta$, then the other must be 90$^\\circ$ minus $\\theta$. This ensures that the two angles sum to 90$^\\circ$. When two angles sum to 90$^\\circ$, we call them complementary angles. Now here's the cool part. See how the sine of one acute angle describes the $\\blueD{\\text{exact same ratio}}$ as the cosine of the other acute angle? Incredible! Both functions, $\\sin(\\theta)$ and $\\cos(90^\\circ-\\theta)$, give the exact same side ratio in a right triangle. And we're done! We've shown that $\\sin(\\theta) = \\cos(90^\\circ-\\theta)$. In other words, the sine of an", "# Ratio of two complementary angles is 2 : 3. What is the double of the larger angle? 1. 54° 2. 108° 3. 95° 4. 75° Option 2 : 108° ## Detailed Solution Given: The ratio of two complementary angles is 2 : 3 Concept used: The sum of the two angles of the complementary angle is 90° Calculation: Let the two angles be 2x and 3x Now, ⇒ 2x + 3x = 90 ⇒ 5x = 90 ⇒ x = 18 Two complementary angles are 36 and 54 Larger angle = 54 Double of the larger angle = 54 × 2 = 108° ∴ The double of the larger angle is 108°", "# T-Ratios of Complementary Angles Two angles are complementary if their sum is 90°. If the sum of two angles A and B is 90°, then ∠A and ∠B are complementary angles and each of them is complement of the other. Let XOX′ and YOY′ be a rectangular system of coordinates. Let A be any point on OX. Let ray OA be rotated in an anti clockwise direction and trace an angle θ from its initial position. Let ∠ POM = θ. Draw PM ⊥ OX. Then ΔPMO is a right angled triangle. ∠POM + ∠OPM + ∠PMO = 180° ∠POM + ∠OPM + 90° = 180° ∠POM + ∠OPM = 90° ∠OPM = 90° - ∠POM = 90° - θ Thus, ∠OPM and ∠POM are complementary angles. $$\\sin (90° - \\theta) = \\cos \\theta$$ $$\\cos (90° - \\theta) = \\sin \\theta$$ $$\\tan (90° - \\theta) = \\cot \\theta$$ $$\\cot (90° - \\theta) = \\tan \\theta$$ $$\\csc (90° - \\theta) = \\sec \\theta$$ $$\\sec (90° - \\theta) = \\csc \\theta$$ The above six results are known as trigonometric ratios of complementary angles.", "Mathematics # The ratio between two complementary angles is $2 : 3$: find the smallest angle. $36$ ##### SOLUTION Two angles whose sum is ${ 90 }^{ o }$ are said to be complementary. Given two angles are in the ratio of $2:3$. Let the two angles be $2x$ and $3x$. So, $\\ 2x+3x={ 90 }^{ o }$ $\\Rightarrow 5x={ 90 }^{ o }$ $\\Rightarrow x=\\dfrac { 90 }{ 5 }$ $\\Rightarrow x=18\\\\$ Therefore, the two angles are $2x={ 36 }^{ o }$ and $3x={ 54 }^{ o }$. You're just one step away Single Correct Medium Published on 09th 09, 2020 Questions 120418 Subjects 10 Chapters 88 Enrolled Students 86 #### Realted Questions Q1 Subjective Medium In the joining figure, name the following pairs of angles. Adjacent angles that do not form a linear pair. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q2 Subjective Medium Find the value of unknown x in the following diagram. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q3 Single Correct Medium If one angle at a point is reflex angle, the other at that point may be : • A. Acute angle • B. Obtuse angle • C. Straight angle • D. Acute or", "two angles is 90: x+$$\\frac{x-10}{4}$$=90. x+$$\\frac{x-10}{4}$$=90 4(x+$$\\frac{x-10}{4}$$=90) 4x+x-10=360 (4+1)x-10=360 5x-10=360 5x-10+10=360+10 5x=370 x=74 The measure of the second angle is 74°, the measure of the first angle is $$\\frac{74^{\\circ}-10^{\\circ}}{4}$$=$$\\frac{64^{\\circ}}{4}$$=16°, and the measure of the third angle is 90°. Solution Four Let x be the measure of the second angle. Then, the first angle is $$\\frac{x-10}{4}$$. The sum of the three angles is x+$$\\frac{x-10}{4}$$+90=180. x+$$\\frac{x-10}{4}$$+90=180 x+$$\\frac{x-10}{4}$$+90-90=180-90 x+$$\\frac{x-10}{4}$$=90 $$\\frac{4 x}{4}$$+$$\\frac{x-10}{4}$$=90 $$\\frac{4 x+x-10}{4}$$=90 4x+x-10=360 5x-10+10=360+10 5x=370 x=74 The measure of the second angle is 74°, the measure of the first angle is $$\\frac{74^{\\circ}-10^{\\circ}}{4}$$=$$\\frac{64^{\\circ}}{4}$$=16°, and the measure of the third angle is 90°. Make sure students see at least four different methods of solving the problem. Conclude this example with the statements below. → Each method is slightly different either in terms of how the variable is defined or how the properties of equality are used to solve the equation. The way you find the answer may be different from your classmates’ or your teacher’s. → As long as you are accurate and do what is mathematically correct, you will find the correct answer. ### Eureka Math Grade 8 Module 4 Lesson 5 Exercise Answer Key Exercises For each of the following problems, write an equation and solve. Exercise 1. A pair of congruent angles are described as follows: The degree measure", "# Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. Question: Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. The measure of larger angle is (a) 72° (b) 54° (c) 63° (d) 36° Solution: (b) 54° Let the measure of the required angle be $x^{\\circ}$ Then, the measure of its complement will be $(90-x)^{\\circ}$. $\\therefore 2 x=3(90-x)$ $\\Rightarrow 2 x=270-3 x$ $\\Rightarrow 5 x=270$ $\\Rightarrow x=54$", "complementary angles is 17 degrees. Find the measures of the angles. Exercise $$\\PageIndex{33}$$ Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles. Answer The measures are 44 degrees and 136 degrees. Exercise $$\\PageIndex{34}$$ Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles. Exercise $$\\PageIndex{35}$$ Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles. Answer The measures are 34 degrees and 56 degrees. Exercise $$\\PageIndex{36}$$ Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles. Exercise $$\\PageIndex{37}$$ Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio. Answer The width is 10 feet and the length is 25 feet. Exercise $$\\PageIndex{38}$$ Darrin is hanging 200", "known as complementary. An angle bisector. For example 90° means 90 degrees. Two angles whose measures add to 90 degrees. A Euclidean construction. Then we'll start getting into obtuse angles, 100, 110, 120, 130, 140, 150. This means that 120º is the supplement of 60º. The degree measure of angle A of isosceles triangle ABC is 30 degrees. One of the most important aspects of geometry is constructing angles. Pro Lite, NEET Example: Construct an angle bisector for the following angle: Positive and Negative Angles. Acute angle. And I want to give credit to the person who built this protractor, because I think it's pretty neat. In this section, you will learn how to construct angles using protractor. 96. 138. So we can apply this knowledge to construct a 30° angle. Sum and Difference of Angles in Trigonometry, Table of 90 - Multiplication Table of 90, Vedantu To be able to speak about angles that measure less than $$1^\\circ$$, we use submultiples of a degree, so we avoid working with expressions like the following: This angle measures half a degree; This angle measures $$0,76$$ degrees; Thus, the sexagesimal degree has … Vedantu academic counsellor will be calling you shortly for your Online Counselling session. 2. The radian measure of an angle whose terminal side is along the negative", "/6 and /8, and name the reflex angle correctly, and is closer 90^... { G } E and the angle of 130^ { \\circ } and \\hat { }... Use angles to find a missing angle of 50^ { \\circ }. ) ,:. Measures of all the angles marked x is 60^ { \\circ } this information to present correct. Very similar to the right side of the shops in the figure above, the lines cut by shaded. Turn between two straight lines, and make an F shape time and. And to personalise content to better how to find missing angles in intersecting lines the needs of our users say: vertically angles. Of 180^ { \\circ }, it means we 're having trouble external. Shows what happens when tangents and secants intersect on a circle answer seems correct, opposite... And the angle to measure this reflex angle with an ordinary protractor only... = 91^ { \\circ }. ) make alternate angles are supplementary in! Finding the measure of inscribed angle 2 is equal to 30^ { \\circ.! You look carefully, you use a protractor to measure the angle of measure is vertical to the right of!, so use the vertical angles angle it is the inner scale shop, or 90-degree, angle know... Or print icon to worksheet to", "Comment Share Q) # The difference in the measures of two complementary angles is $12^{\\circ}$. Find the measures of the angles. $( A ) 32^{\\circ},58^{\\circ} \\\\ ( B ) 39^{\\circ},51^{\\circ} \\\\ ( C ) 31^{\\circ},59^{\\circ} \\\\ ( D ) None\\; of\\; these$ • Two angles are complementary when their sum adds to $90^{\\circ}$ Let $x$ be one angle. Then $90-x$ is another angle. So $x - (90-x) = 12$ $x - 90 +x = 12$ $2x = 12 + 90 = 102$ $x = \\frac{102}{2} = 51$" ]

[ "60 degrees triangle ABC is an isosceles or triangle. ) the length in a scalene triangle stock images in HD and of!", "# Area of Inscribed Isosceles Triangle #### Shared by Best of Geometry $ABC$ is an isosceles triangle with $AB = BC$. The circumcircle of $ABC$ has radius $8$. Given that $AC$ is a diameter of the circumcircle, what is the area of triangle $ABC$? ### Details and assumptions The circumcircle of a triangle $ABC$ passes through $A, B, C$. 3 tries left ·", "## What is the area of the right triangle $$ABC?$$ ##### This topic has expert replies Legendary Member Posts: 1421 Joined: 01 Mar 2018 Followed by:2 members ### What is the area of the right triangle $$ABC?$$ by Gmat_mission » Fri Aug 20, 2021 12:21 pm 00:00 A B C D E ## Global Stats What is the area of the right triangle $$ABC?$$ (1) Side $$AB$$ measures $$5$$ meters and Side $$BC$$ measures $$13$$ meters. (2) Angle $$ABC$$ measures $$90$$ degrees.", "+0 # Angles 0 84 1 Triangle ABC is an isosceles right triangle with a right angle at A. Segments BD and BE trisect angle ABC. What is the degree measure of angle BEA? Jun 2, 2021 #1 +181 +1 Because the triangle is isosceles we know that $\\angle{ABC} = 45$ ($\\frac{180-90}{2}$) Because it trisects angle $ABC$, we know that $\\angle{EBA} = 15$. Thus, we subtract $180 - 15 - 90 = 75$, because $EAB$ is a triangle. Therefore, our answer is $\\boxed{75}$. Jun 2, 2021", "Free Version Moderate # Isosceles Triangle ACTMAT-JE13KE $\\Delta$ABC is an isosceles triangle with vertex A and $AB = AC.$ If the measure of ${\\angle A}$ is $\\frac{1}{4}$ the measure of ${\\angle B}$, what is the measure, in degrees, of ${\\angle A}$? A $20°$ B $30°$ C $45°$ D $60°$ E $80°$", "If an isosceles triangle ABC with sides AB=AC=6cm is inscribed inside a circle of radius 9cm.Find the area of triangle ABC.", "right triangle is a triangle with one 90 degree angle.", "In triangle $$ABC$$, $$\\tan \\angle CAB = 22/7$$, and the altitude from $$A$$ divides $$BC$$ into segments of length 3 and 17. What is the area of triangle $$ABC$$? (第六届AIME1988 第7题)", "$$ABC$$ is the required isosceles right-angles triangle. Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school 0", "How long is perimeter of triangle ABC?", "Are you ready to take on this challenging Geometry problem? # Area of Inscribed Isosceles Triangle #### Shared by Arron Kau $ABC$ is an isosceles triangle with $AB = BC$. The circumcircle of $ABC$ has radius $8$. Given that $AC$ is a diameter of the circumcircle, what is the area of triangle $ABC$? ### Details and assumptions The circumcircle of a triangle $ABC$ passes through $A, B, C$. 3 tries left · This section requires Javascript.", "# Isosceles Right-angled Triangle Geometry Level 1 ABC is a right triangle as shown. Given that $$D$$ is the midpoint of $$AB$$, $$E$$ is the midpoint of $$BC$$, and the side lengths of $$AB$$ and $$AC$$ are both 12, find the area of the shaded region. ×", "## If triangle $$ABC$$ is an isosceles triangle, what is $$\\angle ABC?$$ ##### This topic has expert replies Legendary Member Posts: 1570 Joined: 01 Mar 2018 Followed by:2 members ### If triangle $$ABC$$ is an isosceles triangle, what is $$\\angle ABC?$$ by Gmat_mission » Sat Oct 02, 2021 4:47 am 00:00 A B C D E ## Global Stats If triangle $$ABC$$ is an isosceles triangle, what is $$\\angle ABC?$$ (1) $$\\angle CAB = 45$$ degrees. (2) $$\\angle BCA = 90$$ degrees.", "triangle. Right Triangle Right Triangle A right triangle is a triangle with one 90 degree angle.", "+0 # Need Help! 0 74 1 +16 The area of $\\triangle ABC$ is 6 square centimeters. $\\overline{AB}\\|\\overline{DE}$. $BD=4BC$. What is the number of square centimeters in the area of $\\triangle CDE$? Jul 31, 2021 edited by okhurana Jul 31, 2021", "+0 # Help on Geometry question! 0 214 1 The triangle $\\triangle ABC$ is an isosceles triangle where $AB = 4\\sqrt{2}$ and $\\angle B$ is a right angle. If $I$ is the incenter of $\\triangle ABC,$ then what is $BI$? Express your answer in the form $a + b\\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect square other than $1.$ Jul 26, 2020", "angle in the alternate segment. Examples 1. ED is a tangent to a circle, touching at point A. Triangle ABC is a triangle proscribed within the circle. Angle ∠DAC has a value of 59º. What is the value of angle ∠ABC?", "and $x$ is the lengths of the short sides. The triangle $ABC$ is an isosceles right triangle with the length of the hypotenuse $x$ feet and thus Pythagoras Theorem says that the length of $AC$ is $\\frac{x}{\\sqrt 2}$ feet. Thus $d = 2 \\times |AC| + x = \\sqrt 2 x + x = \\left(1 + \\sqrt 2 \\right) x.$ Hence if $x = 6.67$ feet then $d = 16.10$ feet. Penny", "3 Tutor System Starting just at 265/hour # Q.3 Construct an isosceles right-angled triangle ABC, where m$$\\angle ACB$$= 90° and AC = 6 cm. Step 2 Draw $$\\angle$$C = 90°. Step 5 $$\\triangle$$ ACB is the required right-angled triangle.", "the five orange triangles are congruent, all of the short triangle legs that have dashes across them are the same length. What is the measure of $$\\angle Z?$$ Solve for $$x^\\circ.$$ The three horizontal lines in the figure below are parallel. What is the measure of $$\\angle B?$$ Be careful, lots of people get this one wrong! Note: The diagram is not drawn perfectly to scale, so don't just guess or try to measure it with a protractor. ×" ]

[ "# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 16 and the triangle has an area of 64 . What are the lengths of sides A and B? Oct 8, 2016 $8 \\sqrt{2} = 11.3137$ #### Explanation: Given $A = B , C = 16$, Area ${A}_{r} = 64$, $\\implies {A}_{r} = \\frac{1}{2} \\cdot C \\cdot h$ $\\implies 64 = \\frac{1}{2} \\cdot 16 \\cdot h , \\implies h = 8$ ${A}^{2} = {h}^{2} + {\\left(\\frac{C}{2}\\right)}^{2}$ $\\implies {A}^{2} = {8}^{2} + {\\left(\\frac{16}{2}\\right)}^{2} = {8}^{2} + {8}^{2} = 128$ $\\implies A = \\sqrt{128} = \\sqrt{64 \\cdot 2} = 8 \\sqrt{2} = 11.3137$ Hence, $A = B = 11.3137$", "# If either side of an isosceles triangle measures 24 inches and its base measures 16 inches, then what are its angles? Nov 14, 2015 $A = {\\cos}^{-} 1 \\frac{7}{9} , B = C = \\frac{\\pi}{2} - \\frac{A}{2}$ #### Explanation: $24 \\times 24 \\times 16 = A B \\times A C \\times B C$ we use \"cos rule\": ${16}^{2} = {24}^{2} + {24}^{2} - 2 \\cdot 24 \\cdot 24 \\cdot \\cos A$ $\\frac{{16}^{2} - {24}^{2} - {24}^{2}}{- 2 \\cdot 24 \\cdot 24} = \\cos A = - \\frac{{\\left({2}^{4}\\right)}^{2}}{2 \\cdot {\\left({2}^{3}\\right)}^{2} \\cdot {3}^{2}} + 1 = 1 - \\frac{2}{9}$ A + B + C = pi ; B = C = x $A + 2 x = \\pi R i g h t a r r o w x = \\frac{\\pi - A}{2}$", "# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 4 and the triangle has an area of 42 . What are the lengths of sides A and B? Length of sides A & B are $21.1 \\left(1 \\mathrm{dp}\\right)$ unit Let Altitude (h) be drawn from junction of sides A & B on Side $C$. We know area of the triangle is $\\frac{1}{2} \\cdot C \\cdot h = 42 \\mathmr{and} \\frac{1}{2} \\cdot 4 \\cdot h = 42 \\therefore h = 21$The altitude meets side C at midpoint and perpendicular to C.$\\therefore {h}^{2} + {\\left(\\frac{C}{2}\\right)}^{2} = {A}^{2} \\therefore {A}^{2} = {21}^{2} + {\\left(\\frac{4}{2}\\right)}^{2} = 441 + 4 = 445 \\therefore A = B = \\sqrt{445} = 21.1 \\left(1 \\mathrm{dp}\\right)$unit", "3 known isosceles triangles at the center of the circle. the 4th angle add up to 360°. I've got $\\delta \\approx 73.174^\\circ$ 3. Calculate the length of $\\overline{AD} = 2 \\cdot r \\cdot \\sin\\left(\\frac12 \\delta\\right)$", "be 75 degrees, making CAT an isosceles triangle. AT and AC are congruent, so $AC=24$. We now drop an altitude from C, and call the foot this altitude point D. By 30-60-90 triangles, AD equals twelve and CD equals $12\\sqrt{3}$. We also notice that CDB is an isosceles right triangle. CD is congruent to BD, which makes BD also equal to $12\\sqrt{3}$. The base AB is $12+12\\sqrt{3}$, and the altitude CD is $12\\sqrt{3}$. We can easily find that the area triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=\\boxed{291}$. -youyanli", "# Isosceles Triangle In Circle? Geometry Level 2 In a circle of radius $$5 \\text{ cm}$$, $$AB$$ and $$AC$$ are 2 chords such that $$AB=AC=6 \\text{ cm}$$ . the length of chord $$BC$$ is $$\\text{__________ cm}$$. ×", "# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 12 and the triangle has an area of 36 . What are the lengths of sides A and B? Oct 8, 2016 The length of sides $A \\mathmr{and} B$ are $8.49 \\left(2 \\mathrm{dp}\\right) u n i t$ each . #### Explanation: Side $C$ is the base of the isoceles triangle, $A \\mathmr{and} B$ are legs, and $x$ be the altitude. The area of triangle is ${A}_{t} = \\frac{1}{2} \\cdot C \\cdot x \\mathmr{and} 36 = \\frac{1}{2} \\cdot 12 \\cdot x \\mathmr{and} x = 6$ $= \\left({A}^{2} - {\\left(\\frac{C}{2}\\right)}^{2}\\right) = {x}^{2} \\mathmr{and} \\left({A}^{2} - {\\left(\\frac{12}{2}\\right)}^{2}\\right) = 36 \\mathmr{and} {A}^{2} - 36 = 36 \\mathmr{and} {A}^{2} = 72 \\mathmr{and} A = \\sqrt{72} = \\pm 8.485$. Side can not be negative. So the length of sides $A \\mathmr{and} B$ are $8.49 \\left(2 \\mathrm{dp}\\right) u n i t$ each.[Ans]", "# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 8 and the triangle has an area of 32 . What are the lengths of sides A and B? Dec 10, 2016 $\\sqrt{80}$ #### Explanation: Area of a triangle $A = \\frac{1}{2} \\cdot b \\cdot h$, where $b$ is the base and $h$ is the height. Given that $A = B$, base $C = 8$, and area of the triangle $= 32$, $\\implies 32 = \\frac{1}{2} \\cdot 8 \\cdot h$ $\\implies h = 8$ ${A}^{2} = {h}^{2} + {\\left(\\frac{C}{2}\\right)}^{2}$ $\\implies A = \\sqrt{{8}^{2} + {\\left(\\frac{8}{4}\\right)}^{2}} = \\sqrt{80}$ Hence, $A = B = \\sqrt{80}$", "isosceles triangle. \u0003(C) The length of BC is equal to the length of AD. \u0003(D) The length of BC is 10. \u0003(E) The length of AD is 10. 14. In the figure below, AB=BC=CD. If the area of triangle CDE is 42, what is the area of triangle ADG? 15. What are the lengths of sides NO and OP in triangle NOP below? 1 20 2 C 3 C 4 |m-n|<8 5 B 6 B 7 D 8 A 9 D 10 C 11 D 12 A 13 A and D 14 378 15 NO=30 and OP=10\\sqrt { 34 }", "# Math Help - isosceles triangle 1. ## isosceles triangle In the isosceles triangles ABC and CDE below, BC = 16 in., DE = 11 in. What is CD - AD? Here 2. Hello, sanee66! In the isosceles triangles $ABC$ and $CDE$ below, . . $BC = 16\\text{ in.},\\;DE = 11\\text{ in.}$ What is $CD - AD$? Code: A * * * D 11 * * * * * * E * * * * * * * * * B * * * * * * * C 16 We have: . $AC \\:=\\:BC\\:=\\:16$ . . . .and: . $DC \\:=\\:DE\\:=\\:11$ Hence: . $AD \\;=\\;AC - DC \\;=\\;16 - 11\\;=\\;5$ Therefore: . $CD - AD\\;=\\;11 - 5 \\;=\\;6$", "# Math Help - arof an isosceles triangle 1. ## arof an isosceles triangle how do you derermine the area of an isosceles triangle 2. Originally Posted by seller299 how do you derermine the area of an isosceles triangle Hello, an isosceles triangle is divided into 2 right triangles by an altitude. If the length of a side of the isosceles triangle is a you can calculate the height by the Pythagorean theorem: $a^2 = \\left(\\frac12 \\cdot a\\right)^2 + h^2~\\implies~h=\\frac12 \\cdot a \\cdot \\sqrt{3}$ The area of a triangle is calculated by: $A = \\frac12 \\cdot s \\cdot h_s$ where s is the base side and $h_s$ is the height of the triangle perpendicular to s. Plug in the values you know about the isosceles triangle: $A_{isosceles \\ triangle}=\\frac12 \\cdot a \\cdot \\frac12 \\cdot a \\cdot \\sqrt{3} = \\frac14 \\cdot a^2 \\cdot \\sqrt{3}$", "# A triangle has sides A, B, and C. Sides A and B are of lengths 5 and 3, respectively, and the angle between A and B is pi/3. What is the length of side C? Feb 14, 2017 Length of side $C$ is $4.36 \\left(2 \\mathrm{dp}\\right)$ unit. Sides $A = 5 , B = 3$ and angle $\\angle c = \\frac{\\pi}{3} = \\frac{180}{3} = {60}^{0}$. Applying cosine law we can find $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c}$ or $C = \\sqrt{{5}^{2} + {3}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos 60} = \\sqrt{25 + 9 - \\cancel{2} \\cdot 5 \\cdot 3 \\cdot \\frac{1}{\\cancel{2}}} = \\sqrt{19} = 4.36 \\left(2 \\mathrm{dp}\\right)$ unit. Length of side $C$ is $4.36 \\left(2 \\mathrm{dp}\\right)$ unit. [Ans]", ", $22 c m$ and $24.5 c m$ respectively. (answer) Note : To know more about isosceles triangle, please check into : https://en.wikipedia.org/wiki/Isosceles_triangle.", "+0 # Geometry 0 278 2 In the figure below, isosceles $\\triangle ABC$ with base $\\overline{AB}$ has altitude $CH = 24$ cm. $DE = GF$, $HF = 12$ cm, and $FB = 6$ cm. What is the number of square centimeters in the area of pentagon $CDEFG$? Jun 30, 2020 #1 0 CDEFG is 384 cm Jun 30, 2020 #2 0 Nice figure Jun 30, 2020", "an isosceles triangle is 6 cm. ∴ s = $$\\frac{12+12+6}{2}$$ = 15 By Heron’s formula, we know that Area of triangle Therefore, the area of an isosceles triangle is 9√15 cm2. error: Content is protected !!", "+0 # DUDE HELP ME OUT 0 130 4 In a certain isosceles triangle, the base is $1\\frac 12$ times as long as each leg and the perimeter is $63.$ How long is the base? Mar 8, 2020 #1 +21725 +3 isosceles triangle means two legs are the same length and base is 1/12 as long as leg x = leg length x + x + x/12 =63 solve fo 'x' the leg length then base = x/12 Mar 8, 2020 edited by ElectricPavlov Mar 8, 2020 #2 -3 Mar 8, 2020 #3 -3 its wrong Mar 8, 2020 #4 +4 +2 Let's try setting x equal to the length of each leg. Then, we have two sides of length x and one of length $$\\dfrac 32\\cdot x,$$ so the perimeter is $$x+x+\\dfrac 32\\cdot x,$$which is $$\\dfrac 72\\cdot x.$$We also know the perimeter is 63 so we set our two expressions for the perimeter equal to each other: $$\\frac 72\\cdot x = 63.$$ Multiplying both sides by $$\\frac{2}{7}$$ gives The base is $$1\\frac 12\\cdot 18 = \\boxed{27}$$ Hope it helps! Mar 10, 2020 edited by TechnoWizard Mar 10, 2020 edited by TechnoWizard Mar 17, 2020", "# A triangle has sides A, B, and C. Sides A and B are of lengths 2 and 3, respectively, and the angle between A and B is (pi)/8 . What is the length of side C? The length of side $C$ is $1.38 \\left(2 \\mathrm{dp}\\right) u n i t$ Here the sides A=2 ; B=3 and their included angle $c = \\frac{\\pi}{8} = \\frac{180}{8} = {22.5}^{0}$ are known. Applying cosine law we get $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c} \\mathmr{and} C = \\sqrt{{2}^{2} + {3}^{2} - 2 \\cdot 2 \\cdot 3 \\cdot \\cos 22.5} = 1.38 \\left(2 \\mathrm{dp}\\right) u n i t$[Ans]", "+0 # help geometry 0 56 1 In the figure below, isosceles triangle ABC with base AB has altitude CH = 24 cm, DE = GF, HF = 12 cm, and FB = 8 cm. find area of CDEFG Oct 25, 2021 #1 +17 0 Think about using similiar triangles. Notice how triangle CHB is similiar to GFB therefore we can set up a ratio like so to find the length of GF: $$\\frac{CH}{GF} = \\frac{HB}{FB}, \\frac{24}{20} = \\frac{GF}{8}, GF = 9.6.$$ Therefore, we know that the area of GFB = 9.6 * 8 divided by 2. Therefore, GFB = 38.4. Via symmetry, we know that DAE = 38.4. Therefore, the area of CDEFG = (area of entire triangle) - 38.4 * 2 = 480 - 76.8 = 403.2 Oct 25, 2021", "X,Y are the length for other two sides of triangle B. $\\frac{X}{42} = \\frac{14}{21}$ $X = \\frac{14}{21} \\cdot 42$ $X = 28$ $\\frac{Y}{36} = \\frac{14}{21}$ $Y = \\frac{14}{21} \\cdot 36$ $Y = 24$ The length of sides for triangle B are $\\left\\{14 , 28 , 24\\right\\}$", "# Some triangles are quite special! Consider an isosceles $\\triangle ABC$ with $AB=AC=5, BC=6,$ where $I,O,H$ denote its incenter, circumcenter, orthocenter, respectively. Find the area of $\\triangle IOH$. ×" ]

[ "# Area of Inscribed Isosceles Triangle #### Shared by Best of Geometry $ABC$ is an isosceles triangle with $AB = BC$. The circumcircle of $ABC$ has radius $8$. Given that $AC$ is a diameter of the circumcircle, what is the area of triangle $ABC$? ### Details and assumptions The circumcircle of a triangle $ABC$ passes through $A, B, C$. 3 tries left ·", "If an isosceles triangle ABC with sides AB=AC=6cm is inscribed inside a circle of radius 9cm.Find the area of triangle ABC.", "60 degrees triangle ABC is an isosceles or triangle. ) the length in a scalene triangle stock images in HD and of!", "Free Version Moderate # Isosceles Triangle ACTMAT-JE13KE $\\Delta$ABC is an isosceles triangle with vertex A and $AB = AC.$ If the measure of ${\\angle A}$ is $\\frac{1}{4}$ the measure of ${\\angle B}$, what is the measure, in degrees, of ${\\angle A}$? A $20°$ B $30°$ C $45°$ D $60°$ E $80°$", "Are you ready to take on this challenging Geometry problem? # Area of Inscribed Isosceles Triangle #### Shared by Arron Kau $ABC$ is an isosceles triangle with $AB = BC$. The circumcircle of $ABC$ has radius $8$. Given that $AC$ is a diameter of the circumcircle, what is the area of triangle $ABC$? ### Details and assumptions The circumcircle of a triangle $ABC$ passes through $A, B, C$. 3 tries left · This section requires Javascript.", "+0 # Angles 0 84 1 Triangle ABC is an isosceles right triangle with a right angle at A. Segments BD and BE trisect angle ABC. What is the degree measure of angle BEA? Jun 2, 2021 #1 +181 +1 Because the triangle is isosceles we know that $\\angle{ABC} = 45$ ($\\frac{180-90}{2}$) Because it trisects angle $ABC$, we know that $\\angle{EBA} = 15$. Thus, we subtract $180 - 15 - 90 = 75$, because $EAB$ is a triangle. Therefore, our answer is $\\boxed{75}$. Jun 2, 2021", "parallel. Top is 10, other sides are unknown. Area of ABCD is 9 times larger than EFGH. What is the asked by olav on December 11, 2010 55. ## Geometry Isosceles triangle ABE of area 100 square inches is cut by CD into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle ABE from A is 20 inches, what is the number of inches in asked by Curtis on July 9, 2017 56. ## geometry ABCD is a trapezoid with AB parallel to DC. If AB=25, BC=24, CD=50 and AD=7, what is the area of ABCD? asked by Anonymous on March 12, 2013 57. ## maths ABCD is a trapezoid with AB parallel to DC. If AB = 25, BC = 24, CD = 50 and AD = 7, what is the area of ABCD? asked by abc9837 on March 16, 2013 58. ## Maths Isosceles triangle $ABE$ of area 100 square inches is cut by $\\overline{CD}$ into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle $ABE$ from $A$ is 20 inches, what is the asked by Anonymous on February 15, 2018 The base of a right prism is isosceles trapezoid", "+0 # Need Help! 0 74 1 +16 The area of $\\triangle ABC$ is 6 square centimeters. $\\overline{AB}\\|\\overline{DE}$. $BD=4BC$. What is the number of square centimeters in the area of $\\triangle CDE$? Jul 31, 2021 edited by okhurana Jul 31, 2021", "height and median, it implies the triangle $$ABC$$ is isosceles. Hence, $$AC=3$$. From the right triangle $$ACD$$ we can find $$CD=\\sqrt{AC^2-AD^2}=2\\sqrt{2}$$. Finally, the area of the triangle $$ABC$$ is $$\\frac12 \\cdot AB\\cdot CD=\\frac12 \\cdot 3\\cdot 2\\sqrt{2}=3\\sqrt{2}.$$", "# Q 3. Construct an isosceles right-angled triangle ABC, where $m\\angle ACB=90^{\\circ}$ and AC = 6 cm. P Pankaj Sanodiya 1. Draw CA = 6 cm 2. Draw a perpendicular CX to CA at C 3. From C and with length = 6 cm cut an arc on CX. The arc meet CX at B 4. Join BA Triangle ABC is the required triangle Exams Articles Questions", "In triangle $$ABC$$, $$\\tan \\angle CAB = 22/7$$, and the altitude from $$A$$ divides $$BC$$ into segments of length 3 and 17. What is the area of triangle $$ABC$$? (第六届AIME1988 第7题)", "triangle ABC is 5*A/2, where A is the length of OA (consider A>0). Therefore, the question in fact is \"Is 5A/2>15?\" or\"Is A>6?\" (1) Not sufficient, as $$A_1$$ can be below -6, in which case A>6, but if A is above 0 and below 3, obviously the area is less than 15. (2) If the triangle ABC is a right triangle, then necessarily $$\\angle{BAC}$$ is a right angle and BC is the diameter of the circle in which this triangle can be inscribed. Even if the triangle is isosceles, the height AO cannot be greater than the radius of the circle $$AO_1,$$ which is 5/2. So, A is definitely smaller than 6. Sufficient. Great Explanation EvaJager. I completely get it now. EvaJager wrote: For any triangle the max area is can be got by for a equilateral triangle. - This is not a correct statement. You can have a \"tiny\" equilateral triangle or a \"huge\" right or any other triangle. The correct statement is \"For a given perimeter, the maximum area is that of an equilateral triangle.\" The second statement isn't correct either: For any right angled triangle the max area can be got by forming an isosceles triangle. I think what you meant is \"For a right triangle inscribed in a given circle (which means given constant", "of an isosceles triangle measures 42 degrees, 30'. The base is 14.6 meters long. a. Find the length of a leg of the triangle b. Find the altitude of the triangle c. What is the area of the triangle? 2. ### Geometry in triaangle ABC, AB=20 cm, AC=15 cm the length of the altitude AN is 12 cm prove that ABC is a right triangle so far i got that angle ANC is 90 degrees by definition on altitude i got really confused after so can you show me how 3. ### geometry In the accompanying diagram, triangle ABCis similar to triangle DEF, AC = 6, AB = BC = 12, and DF = 8. Find the perimeter of triangle DEF. it an isoslese triangle 6 and 8 are both base my work 6+2=8 12+2=14 2(14)+6=p 24+6=28 I 4. ### Math the right triangle ABC, the altitude from vertex C divides the hypotenuse into two segments, one of length 2 and the other of length 16. Find the perimeter of triangle ABC.", "A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/4, the angle between sides B and C is (pi)/2, and the length of B is 10, what is the area of the triangle? $\\text{Area\"_(triangle\"ABC\") = 50 \"sq.units}$ From the given information we can deduce that the angle between A and C is $\\frac{\\pi}{4}$ and therefore $\\triangle \\text{ABC}$ is an isosceles, right-angled triangle with $\\left\\mid B \\right\\mid = \\left\\mid C \\right\\mid = 10$ and $B \\bot C$ Using $B$ as the base and $C$ as the height $\\textcolor{w h i t e}{\\text{XXX}} A r e {a}_{\\triangle} = \\frac{1}{2} \\cdot B \\cdot C = \\frac{1}{2} \\cdot 10 \\cdot 10 = 50$", "# The Right Brothers Geometry Level 3 $$ABC$$ is an isosceles triangle with $$AB = AC$$ and $$BC = 60$$. $$D$$ is a point on $$BC$$ such that the perpendicular distances from $$D$$ to $$AB$$ and $$AC$$ are $$16$$ and $$32$$, respectively. What is the length of $$AB$$? ×", "### Show Tags 05 Mar 2019, 05:36 Statement 1: gives me 1 side we don't know if the side contains the 90° or not. Insufficient Statement 2: it does not give us any information to find out area. Insufficient Statement 1 & 2 together: no new information is revealed. Insufficient IMO E Posted from my mobile device SVP Joined: 18 Aug 2017 Posts: 2370 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) What is the area of right triangle ABC? (1) Side B is 7*2^(1/2) units [#permalink] ### Show Tags Updated on: 06 Mar 2019, 00:43 Bunuel wrote: What is the area of right triangle ABC ? (1) Side B is $$7\\sqrt{2}$$ units long. (2) Triangle ABC is isosceles. side of right angled triangle 45:45:90 or 30:60:90 45:45:90 or x:x√3:2x #1 Side B is $$7\\sqrt{2}$$ units long. we dont know angles in sufficient #2 Triangle ABC is isosceles so 45:45:90; 45:45:90 sides from 1& 2 ; we cannot determine which side of the right angled triangle is 90* and side length also not sure.. IMO E _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. Originally posted by Archit3110 on 05 Mar 2019, 11:37. Last edited by Archit3110 on 06 Mar 2019, 00:43, edited 1 time in total. Intern", "# Isosceles Right-angled Triangle Geometry Level 1 ABC is a right triangle as shown. Given that $$D$$ is the midpoint of $$AB$$, $$E$$ is the midpoint of $$BC$$, and the side lengths of $$AB$$ and $$AC$$ are both 12, find the area of the shaded region. ×", "be 75 degrees, making CAT an isosceles triangle. AT and AC are congruent, so $AC=24$. We now drop an altitude from C, and call the foot this altitude point D. By 30-60-90 triangles, AD equals twelve and CD equals $12\\sqrt{3}$. We also notice that CDB is an isosceles right triangle. CD is congruent to BD, which makes BD also equal to $12\\sqrt{3}$. The base AB is $12+12\\sqrt{3}$, and the altitude CD is $12\\sqrt{3}$. We can easily find that the area triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=\\boxed{291}$. -youyanli", "? Free Version Difficult # Isosceles Triangles and Triangle Sum Theorem ACTMAT-JKT4IM Triangle ABC is an isosceles triangle with vertex A and $AB = AC.$ Points D and E are placed on segments AB and AC, respectively, so that $CD = DE = AE.$ Refer to the figure below. If the $m{\\angle }A = x$, what is $m{\\angle }BCD$ (in degrees) in terms of $x$? A ${\\cfrac { 1 }{ 2 } x}$ B $x$ C $90-{\\cfrac { 1 }{ 2 } x}$ D $90-{\\cfrac { 5 }{ 2 } x}$ E $90-{\\cfrac { 7 }{ 2 } x}$", "Are you ready to take on this challenging Geometry problem? # Area of Inscribed Isosceles Triangle #### Shared by Best of Geometry $ABC$ is an isosceles triangle with $AB = BC$. The circumcircle of $ABC$ has radius $8$. Given that $AC$ is a diameter of the circumcircle, what is the area of triangle $ABC$? ### Details and assumptions The circumcircle of a triangle $ABC$ passes through $A, B, C$. 3 tries left · ### This Geometry problem was shared by Calvin Lin. Error We encountered an error while talking to our servers. Refresh the page and try again in a few seconds. If the problem persists, please email us." ]

[ "## The price of a skateboard is reduced by 25% . The original price was $120. Find the new price . A:$95 B: $80 C. Question The price of a skateboard is reduced by 25% . The original price was$ 120. Find the new price . A: $95 B:$80 C. $90 in progress 0 10 hours 2021-07-21T23:01:51+00:00 1 Answers 0 views 0 ## Answers ( ) 1. Answer: C.$90 120 – 25%= 90", "3 Skiboards, Inc. has two divisions. The Boards Division makes the board that is made into Skiboards by the Ski Division, but the Board Division can also sell the boards it makes to outside customers. In 2011, The Boards Division reported the following information: Selling price per board \u0007$ 52\u0007\u0007Variable costs per board\u0007\\$ 22\u0007\u0007\u0007\u0007\u0007Number of boards:\u0007\u0007\u0007Produced in 2011\u000710,000\u0007\u0007Sold to the Ski Division \u00078,000\u0007\u0007Sold to outside customers\u00072,000\u0007\u0007Sales from the Boards Division to the Ski Division were made at the same price that sales were made to outside customers.... Attachments:", "0 like 0 dislike The price of a skateboard is reduced by 25%. The original price was $120. Find the new price. A.$95 B. $80 C.$90", "a friend for 60% more than what you paid. If the selling price is more than the buying price then it is called markup Here , the markup is 60% of $22.4 a = p% × w a = 0.6 × 22.4 a = 13.44 So, the markup is$13.44 We know that selling price = cost of buying + markup = 22.4 + 13.44 = 35.84 So , the selling price is $35.84. Try It Question 1. The original price of a skateboard is$50. The skateboard is on sale for 20% off. What is the sale price? Answer: The sales price of skateboard is $40 Explanation: Given , skateboard is$50 with 20% off We know , The sales price be 100% – 20% = 80% of the original price sales price = 80% of 50 = 0.8 × 50 = 40 So, The sales price is $40 Question 2. The discount on a DVD is 50%. It is on sale for$10. What is the original price of the DVD? Answer: The original price of DVD is $20. Explanation: Given , discount on a DVD is 50%. , It is on sale for$10 We know , The sales price be 100% – 50% = 50% a = p% × w 10 = 0.5 × w w = $$\\frac{10}{0.5}$$ w", "## A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage Question A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage discount for the cost of the surfboard. in progress 0 5 months 2021-08-28T09:02:34+00:00 1 Answers 0 views 0", "population mean number of snowboard daily sales has increased by between 43 per day and 72 per day now that ads are posted on Google. Assumptions: We must assume that the first eight days of December are an unbiased sample. Also, since the sample size of 8 is not greater than 30, we must assume that the distribution of differences in sales by date is approximately normal. This assumption has no reason for being true, hence increasing the sample size to over 30 is highly recommended. Uses: A manager would calculate the increased profit by multiplying 43 and 73 by the profit per snowboard sale and compare these to the daily advertising cost that Google charges to make a decision about whether the advertising was successful. Exercise $$\\PageIndex{4}$$ Nine American voters were asked how old is too old to be a member of Congress and how old is told to be the president of the United States. The table below shows the results of the survey. Voter 1 Voter 2 Voter 3 Voter 4 Voter 5 Voter 6 Voter 7 Voter 8 Voter 9 Congress 80 75 65 80 78 70 75 85 69 President 65 70 70 75 67 65 73 80 75 Find and interpret the 95% confidence interval for the difference. The 95% confidence interval", "markup amount? 16. Sally works for a skateboard shop. The company just purchased a skateboard for$89.00 less discounts of 22%, 15%, and 5%. The company has standard expenses of 37% of cost and desires a profit of 25% of the regular unit selling price. What regular unit selling price should Sally set for the skateboard? 17. If an item has a 75% markup on cost, what is its markup on selling price percentage? 18. A product received discounts of 33%, 25%, and 5%. A markup on cost of 50% was then applied to arrive at the regular unit selling price of $349.50. What was the original list price for the product? 19. Mountain Equipment Co-op (MEC) wants to price a new backpack. The backpack can be purchased for a list price of$59.95 less a trade discount of 25% and a quantity discount of 10%. MEC estimates expenses to be 18% of cost and it must maintain a markup on selling price of 35%. a. What is the cost of backpack? b. What is the markup amount? c. What is the regular unit selling price for the backpack? d. What profit will Mountain Equipment Co-op realize? e. What happens to the profits if it sells the backpack at the MSRP instead? 20. Costco can purchase a bag of Starbucks", "before the discount? • Two shops In two different shops, the same skis had the same price. In the first, however, they first became more expensive by 20% and then cheaper by 5%. In the second, they were first cheaper by 5% and then more expensive by 30%, so after these adjustments in the • Selling price Find the selling price. Cost to store:$50 Markup: 10% • Ice skates Ice skates were raised twice, the first time by 25%, the second time by 10%. After the second price, their cost was 82.5 euros. What was the original price of skates?", "– 100p. Revenue (number of units that will sell)(price per unit, p) (50 000 – 100p)p = 50 000p – 100p2 Total Cost (cost for producing the snowboards) This is the total overhead costs plus the cost per snowboard times the number of snowboards: 500 000 + 100(50 000 – 100p) 500 000 + 5 000 000 – 10 000p 5 500 000 – 10 000p b. Write an expression to represent the profit. Profit = Total Revenue – Total Cost = (50 000p – 100p2 ) – (5 500 000 – 10 000p) Therefore, the profit function is f(p) = – 100p2 + 60 000p – 5 500 000. c. What is the selling price of the snowboard that will give the maximum profit? Solve for the vertex of the quadratic function using completing the square. (f(p) = – 100p2 + 60 000p – 5 500 000 = – 100(p2 – 600p + ) – 5 500 000 = – 100(p2 – 600p + 3002 ) – 5 500 000 + 100(300)2 = – 100(p – 300)2 – 5 500 000 + 9 000 000 = – 100(p – 300)2 + 3 500 000) The maximum point is (300,3 500 000). Therefore, the selling price that will yield the maximum profit is $300. d. What is the", "I help? Recilia: I’ve reviewed our financial results for the past 12 months. It looks like we made a profit in some months, and had losses in other months. From what I can tell, we sell each snowboard for $250, our variable cost is$150 per unit, and our fixed cost is $75 per unit. It seems to me that if we sell just one snowboard each month, we should still show a profit of$25, and any additional units sold should increase total profit. Lisa: Your unit sales price of $250 and unit variable cost of$150 look accurate to me, but I’m not sure about your unit fixed cost of $75. Fixed costs total$50,000 a month regardless of the number of units we produce. Trying to express fixed costs on a per unit basis can be misleading because it depends on the number of units being produced and sold, which changes each month. I can tell you that each snowboard produced and sold provides $100 toward covering fixed costs—that is,$250, the sales price of one snowboard, minus $150 in variable cost. Recilia: The$75 per unit for fixed costs was my estimate based on last year’s sales, but I get your point. As you know, I’d like to avoid having losses. Is it possible to determine how many units we", "1. Other 2. Other 3. 1 sk8 is examining an invoice the gross price of... # Question: 1 sk8 is examining an invoice the gross price of... ###### Question details 1) Sk8 is examining an invoice. The gross price of a skateboard is $109.00, and the invoice states it received a trade discount of 15% and quantity discount of 10% as well as a loyalty discount. However, the amount of the loyalty discount is unspecified. a. If Sk8 paid$80.88 for the skateboard, what is the loyalty discount percent? b. If the loyalty discount is applied after all other discounts, what amount of loyalty dollars does Sk8 save per skateboard? 2) Daisy is trying to figure out how much negotiating room she has in purchasing a new car. The car has an MSRP of $34,995.99. She has learned from an industry insider that most car dealerships have a 20% markup on selling price. What does she estimate the dealership paid for the car? 3) A snowboard has a cost of$79.10, expenses of $22.85, and profit of$18.00. a. What is the regular unit selling price? b. What is the markup amount? c. What is the markup on cost percentage? d. What is the markup on selling price percentage? e. What is the break-even selling price? What is the markup on", "Snowboard Company a$17,500 = $37,500 −$20,000. b 87.5 percent = $17,500 ÷$20,000. Carefully review Figure 6.6 \"Sensitivity Analysis for Snowboard Company\". The column labeled Scenario 1 shows that increasing the price by 10 percent will increase profit 87.5 percent ($17,500). Thus profit is highly sensitive to changes in sales price. Another way to look at this is that for every one percent increase in sales price, profit will increase by 8.75 percent, or for every one percent decrease in sales price, profit will decrease by 8.75 percent. The column labeled Scenario 2 shows that decreasing sales volume 10 percent will decrease profit 35 percent ($7,000). Thus profit is also highly sensitive to changes in sales volume. Stated another way, every one percent decrease in sales volume will decrease profit by 3.5 percent; or every one percent increase in sales volume will increase profit by 3.5 percent. When comparing Scenario 1 with Scenario 2, we see that Snowboard Company’s profit is more sensitive to changes in sales price than to changes in sales volume, although changes in either will significantly affect profit. ## Expanding the Use of Sensitivity Analysis Question: Although the focus of sensitivity analysis is typically on how changes in variables will affect profit (as shown in Figure 6.6 \"Sensitivity Analysis for Snowboard Company\"), accountants also use", "# Calculus: General Differentiation – #28357 Question: A snowmobile manufacturer is planning a new line of ski-doos. The price is dependent on the number of ski-doos sold $\\left( x \\right)$, and is given as $p(x)=3432-11x$. What is the maximum revenue that the manufacturer can expect, using this model for revenue? What will be the price for one ski-doo?", "? Free Version Moderate # Mr. Winter's Budget Constraint for Sports Equipment MICRO-YEZX4E Based on the the following graph of Mr. Winter's budget constraint, if the price of skis is $\\$110$, determine the price of snowboards and Mr. Winter's budget. Do not use a dollar sign in your solution (e.g. input 1, instead of \\$1).", "1024 USD, Monday will be 25% discount. How much USD will save, and what will be the price? 17. Apples 2 James has 13 apples. He has 30 percent more apples than Sam. How many apples has Sam?", "### Условие задания: 3 Б. Рис. $$1$$. Snowboarding Snowboarding is a popular winter sport. Lots of people all over the world are fond of snowboarding. It involves gliding over snow slope while standing on a board. It originated in the United States and became hugely popular at the end of the $$XX$$ century and was introduced at the $$1998$$ Winter Olympics. There are various types of snowboarding: freeriding, freestyle, urban, half pipe, boardercross, jibbing, etc. Snowboarding can be dangerous and challenging and requires proper training and using special gear. However, skiing is more dangerous than snowboarding, though people usually think the opposite. All in all, a person should be in good physical shape before trying snowboarding. Despite all the risks this relatively young kind of sport is great fun and more and more people are becoming keen on racing down the snow-covered slopes. Fill in the gaps: 1. Lots of people are fond of snowboarding. 2. There are of snowboarding: freeriding, freestyle, urban, half pipe, boardercross, jibbing, etc. 3. All in all, a person should be before trying snowboarding. Источники: Рис. 1. Snowboarding. Pixabay License CC0. Free for commercial use No attribution required. https://pixabay.com/images/id-2030851/ (Дата обращения 11.10.2021) Вы должны авторизоваться, чтобы ответить на задание. Пожалуйста, войдите в свой профиль на сайте или зарегистрируйтесь.", "come on a given day. How many skiers paying for Heavenly will produce the maximal profit? ### Contributors • Integrated by Justin Marshall.", "### Home > ACC7 > Chapter cc38 > Lesson cc38.1.2 > Problem8-20 8-20. 1. Jack noticed a sale on his favorite model of All-Terrain Vehicle (ATV). The salesperson said the company was cutting the price for that weekend, so Jack got an 8% discount. Homework Help ✎ 1. If the usual price of the ATV was $4298, estimate how much the discount will be. Explain your reasoning. 2. Calculate the amount Jack did pay with the 8% discount. 3. If the store sold three of the ATVs that weekend, how much money was saved by the three customers combined? Round the ATV price to$4300 in order to make the percent easier to approximate. Since we can find that 10% of $4300 is$430, we can also determine that 5% of $4300 would be$215. 8% is roughly in the middle of 10% and 5%. What is your estimation? Find 8% of $4298, then subtract that value from$4298 to find how much Jack paid. 4298 − 343.84 = 4298(.08) = 343.84 $3954.16 For each ATV, the original price was$4298. Based on part (b), it costs \\$3954.16 after the 8% discount. How much did Jack save on one ATV? Now multiply that number by 3.", "# A Trader marks his goods 20% above cost price but allows his customers a discount of 10%, the cost price of a blackboard, which is sold for 216, is [ A ] 196 [ B ] 180 [ C ] 200 [ D ] 108 Answer : Option C Explanation : Selling price = 216 Discount = 10% Marked price = (216X$100\\over90$) = 240 So, Cost price =(240X$100\\over120$) = `200", "Home > MC2 > Chapter 10 > Lesson 10.1.2 > Problem10-18 10-18. Round the ATV price to $4300 in order to make the percent easier to approximate. Since we can find that 10% of$4300 is $430, we can also deduce that 5% of$4300 would be $215. 8% is roughly in the middle of 10% and 5%. What is your estimation? Find 8% of$4298, then subtract that value from $4298 to find how much Jack paid. 4298 − 343.84 = 4298(.08) = 343.84$3954.16 For each ATV, the original price was $4298. Based on part (b), it costs$3954.16 after the 8% discount. How much did Jack save on one ATV? Now multiply that number by 3." ]

[ "not get just 2 items, as we would have thought before we heard about the $\\frac{1}{3}$ discount: we would actually get $\\frac{180}{60}=3$ items. So whereas we thought we would get 2 items for $180 we will actually get 3 items. This means we are one-half – or 50% – better off than without the discount. So a $\\frac{1}{3}$ discount makes us 50% better off. Because $\\frac{1}{3}$ is close to 30%, we should suspect that a 30% discount is probably going to make us closer to 50% better off, than 30% better off. #### So how much better off are we after a 30% discount? What if an item costs$100 and we are given a 30% discount? This means we only pay $70 for the item. So if we were to spend$700 we would get 10 items after the discount. Bu before the discount $700 would have only bought us 7 items, at$100 each. So we are $\\frac{3}{7} \\approx 0.43 = 43\\%$ better off after the 30% discount.", "is helpful PS Forum Moderator Joined: 25 Feb 2013 Posts: 834 Location: India GPA: 3.82 An item is discounted by 15% from its usual selling price. What is the [#permalink] Show Tags 22 Sep 2017, 23:53 An item is discounted by 15% from its usual selling price. What is the usual selling price? (1) The dollar value of the discount is $45. (2) The discounted price is$255. Statement 1: Discount is $$15$$ , if usual selling price is $$100$$. $$45$$ discount corresponds to usual selling price of $$= \\frac{100*45}{15}$$. Sufficient Statement 2: Discounted price is $$100-15=85$$ , if usual selling price is $$100$$. $$255$$ discounted price corresponds to usual selling price of $$= \\frac{100*255}{85}$$. Sufficient Option D An item is discounted by 15% from its usual selling price. What is the [#permalink] 22 Sep 2017, 23:53 Display posts from previous: Sort by", "of an item by 25% and then [#permalink] ### Show Tags 10 Dec 2017, 11:10 Decreasing the original price of an item by 25% and then decreasing the new price by z% is equivalent to decreasing the original price by (A) 0.25(1+3z/100) (B) 0.25(1+z/100) (C) 0.25(1−3z/100) (D) 0.75(1−z/100) (E) 0.75(1+3z/100) Done by POE. So 75 +z/100(75) = 75 [1 + z/100] approx 1/3rd and a little more So only A fits 0.25 *3z/100 +1 as closest to 1/3rd plus one. Rest do not add up. Kudos if the answer was useful Director Joined: 27 May 2012 Posts: 581 Re: Decreasing the original price of an item by 25% and then [#permalink] ### Show Tags 15 Apr 2018, 08:09 megafan wrote: Decreasing the original price of an item by 25% and then decreasing the new price by z% is equivalent to decreasing the original price by (A) $$0.25(1 + \\frac{3z}{100})$$ (B) $$0.25(1 + \\frac{z}{100})$$ (C) $$0.25(1 - \\frac{3z}{100})$$ (D) $$0.75(1 - \\frac{z}{100})$$ (E) $$0.75(1 + \\frac{3z}{100})$$ Source: Gmat Hacks 1800 can anybody help, are we looking for the percentage decrease or are we looking the amount of decrease ? Let initial amount be 100 after 25% discount we get 75 then z% discount will give $$\\frac{(100-z)}{100}*75$$ = $$\\frac{100-z}{4}*3$$ So equivalent discount is 100 -$$\\frac{300+3z}{4}$$ $$\\frac{100+3z}{4}$$ Well as option A", "that means the remaining percentage is $$40\\%$$. If the goods were reduced to $$60\\%$$, the percentage would be $$60\\%$$.", "# The value of a single discount on some amount which is equivalent to a series of discounts of 10%, 20% and 40% on the same amount, is equal to 1. 43.2% 2. 50% 3. 56.8% 4. 70% Option 3 : 56.8% Free Electric charges and coulomb's law (Basic) 77.4 K Users 10 Questions 10 Marks 10 Mins ## Detailed Solution Given: Series of discounts of 10%, 20% and 40% Calculation: Let the original number be 100. After applying a series of discount ⇒ 100(1 - 10%) (1 - 20%)(1 - 40%) ⇒ $$\\frac{90}{100} \\times \\frac{80}{100} \\times \\frac{60}{100} \\times 100$$ ⇒ 43.2% ⇒ The required discount = 100 - 43.2 ⇒ Discount = 56.8% ∴ 56.8% is a single discount of this series.", "Suppose the marked-up price was $100. They discounted it by 40%. So what was the discounted price? If this was the original price they paid, what would be the percentage mark-up on this price to take the selling price up to$100? That's your answer.", "# The price of an item yesterday was $65. Today, the price fell to$26. How do you find the percentage decrease? Nov 23, 2017 The percentage decrease is 60%. #### Explanation: The original price, which is 100%, is $65. This is reduced to $26 (x%; which is what we are finding). So, you subtract: $65-$26=$39 Now we do the cross-multiplying; 100%=$65 x % =$39 You get $x = 60\\$", "product? $100 markup (x) 1.1 =$110 = 10% $100 markup (/) .9 =$111.11 = 10% $100 discount (/) 1.1 =$90.90 = 10% $100 discount (.9) =$90 = 10% I find it easier to mark up (/) .9 .8, which equal 10% 20% etc., and then mark down (x) .9, .8, .7, which equal 10 to 30%. It seems faster if there is no other explanation. I have yet to find anyone who can explain it to me. Thanks, Gary B. His methods are a mix of the forward and backward methods. I replied, explaining what each calculation really does: If you mark up by 10%, you are adding 0.10 to the price of an item: new price = old price + 0.10 * old price = 1 * old price + 0.10 * old price = old price * (1.10) (I am using \"*\" for multiplication.) To determine the original price given the marked up price, you have to undo this by dividing: old price = new price / 1.10 If you discount by 10%, you are subtracting 0.10 of the price of the item: new price = old price - 0.10 * old price = old price * (0.90) To determine the original price given the discounted price, you have to undo this by dividing: old price", "235 views If a seller gives a discount of $15\\%$ on retail price, she still makes a profit of $2\\%$. Which of the following ensures that she makes a profit of $20\\%$? 1. Give a discount of $5\\%$ on retail price 2. Give a discount of $2\\%$ on retail price 3. Increase the retail price by $2\\%$ 4. Sell at retail price Let the retail price be $x$ and the cost price be $y$. Now, $85 \\%$ of $x = 102 \\%$ of $y$ $\\Rightarrow \\frac{85}{100}\\times x = \\frac{102}{100}\\times y$ $\\Rightarrow 5x = 6y$ $\\Rightarrow \\boxed{ \\frac{x}{y} = \\frac{6}{5} } \\quad \\longrightarrow (1)$ Let, her give a discount of $d\\%$ on retail price to make a profit of $20\\%$. Now, $120 \\%$ of $y = d \\%$ of $x$ $\\Rightarrow \\frac{120}{100}\\times y = \\left(\\frac{100-d}{100}\\right)\\times x$ $\\Rightarrow 120 y = 100 x – dx$ $\\Rightarrow 120 y = (100-d)\\left(\\frac{6}{5}y\\right)$ $\\Rightarrow 600 = 600 – 6d$ $\\Rightarrow 6d = 600-600$ $\\Rightarrow \\boxed {d = 0\\% }$ $\\therefore$ She will sell at retail price to get a $20\\%$ discount. $\\textbf{Short Method:}$ Let , • Cost price $= 100$ • Selling price $= 102$ • Retail price $= x$ Now, $0.85 x = 102$ $\\Rightarrow x = \\frac{102}{85}\\times 100$ $\\Rightarrow \\boxed {x = 120}$ If the cost price is $100,$ to make", "we know a = 72 , p = 36% , w = ? a = p% × w a = 36% × w 72 = 0.36 × w w = $$\\frac{72}{0.36}$$ w = 200 So , the original price of the earrings is $200. 19. Given , sales price of the item is$146.54 , percent of discount is 15% , Find original price ? Answer: The Original price of the is $172.4. Explanation: The sales price be 100% – 15% = 85% we know a = 146.54 , p = 85% , w = ? a = p% × w a = 85% × w 146.54 = 0.85 × w w = $$\\frac{146.54}{0.85}$$ w = 172.4 So , the original price of the earrings is$172.4. 20. Given , original price of the item is $60 , sales price of the item is$45 , Find percent of discount ? Answer: The percent of discount is 25% . Explanation: We have a = 45 , w = 60 , p = ? By using proportion , we have $$\\frac{a}{w}$$ = $$\\frac{p}{100}$$ $$\\frac{45}{60}$$ = $$\\frac{p}{100}$$ p = $$\\frac{45 × 100}{60}$$ p = $$\\frac{4,500}{60}$$ p = 75 So, 45 is 75% of 60. To get the percent of discount we have , The percent of discount = The percent of original price", "# The price of an item yesterday was $120. Today, the price fell to$84. How do you find the percentage decrease? Mar 29, 2017 (120-84)*100/120=30% The discount (in terms of USD) is $120 - 84 = 36$ The discount rate Percent = 36*100/120=30%.", "to $90 x + 2250 = 100 x$ $10 x = 2250 \\implies x = \\frac{2250}{10} = 225$ The initial number of items is thus equal to $225$. This means that the initial price per item was $\\textcolor{p u r p \\le}{\\left(1\\right)} \\implies P = \\frac{45000}{225} = \\textcolor{g r e e n}{200}$ So, initially you had $225$ items at $200$ each. The price decreased by 10%# to $\\frac{90}{100} \\cdot 200 = 180$ which means that you can now buy $250$ items, $25$ more thn you had initially $180 \\cdot \\left(225 + 25\\right) = 45000$", "to Problem 5.8.19 affected if the cost per item for the $x$ items, instead of being simply $2, decreases below$2 in proportion to $x$ (because of economy of scale and volume discounts) by 1 cent for each 25 items produced?", "• anonymous The percent decrease of an item that dropped in price from $50 to$43 is _____. a,7% b,14% c,43% d,86% please help me:( Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# The original price of a DVD is represented by the variable x. If the DVD is discounted 45%, write a subtraction expression and a multiplication expression to represent its sale price? ##### 1 Answer Nov 20, 2016 Let's call the sale price $s$ #### Explanation: Subtraction : Then the discount is 45% of $x$ $s = x - \\frac{45}{100} x = x - \\frac{\\cancel{5} \\times 9}{\\cancel{5} \\times 20} x = x - \\frac{9}{20} x$ The answer could also have been expressed as $s = x - 0.45 x$ Multiplication : Then $s$ is 100%-45%=55% of $x$ or: $s = \\frac{55}{100} x = \\frac{\\cancel{5} \\times 11}{\\cancel{5} \\times 20} x = \\frac{11}{20} x$ Or: $s = 0.55 x$", "Calculate the sale price of the software. Solution: Discount = Discount % $$\\times$$ original price Discount = 40 % × 2000 = $$\\frac{40}{100}$$ × 2000 = 800 Discount = Original price – Sale price 800 = 2000 – Sale price Sale price = 2000 – 800 Sale price = 1200 As a result, the sale price of the game is \\$1200. When the price of a commodity is reduced, the discount is the amount of money saved. To put it another way, a discount is a rebate or a reduction in the price of an object. The difference between the original price and the sale price is referred to as the discount. Discount = Original price – Sale price A markup is a value added to the cost price of goods to cover operating costs and to add profit, whereas a discount is a reduction in the catalog price (or list price). To calculate the discount, multiply the discount percent by the original price. Discount = Discount % $$\\times$$ original price To get the sale price, subtract the discount from the original price. Selling price = original price – discount The price of your product is determined by the markup. The profit margin indicates how much money your company makes after expenses are deducted.", "+0 +1 80 1 An item is regularly priced at \\$65. It is now priced at a discount of 55% off the regular price. Find the price now. Jan 25, 2019 #1 +4787 +1 $$55\\% \\text{ off means you pay }(100\\%-55\\%) = 45\\%\\\\ 45\\% \\text{ of }\\65 = (0.45)65 = \\29.25$$ . Jan 25, 2019", "1254,58}$$ 3. $$\\text{R 9875} + \\text{R 790} = \\text{R 10 665}$$ 4. $$\\text{R 15 995} + \\text{R 799,75}= \\text{R 16 794,75}$$ Calculate the percentage discount on each of these items: 1. $$\\frac{\\text{R 1360}}{\\text{R 1523}} = \\text{89}\\%$$. So discount is $$\\text{100}\\% - \\text{89}\\% = \\text{11}\\%$$ 2. $$\\frac{\\text{R 527,40}}{\\text{R 586}} = \\text{90}\\%$$. So discount is $$\\text{100}\\% - \\text{90}\\% = \\text{10}\\%$$", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "IB MYP 3 2021 Edition 4.08 Consumer percentages Lesson ### Mark ups and discounts When buying products, two things that are important to understand are mark ups and discounts. Mark up A mark up is a percentage increase in the price of a product. Discount A discount is a percentage decrease in the price of a product. A mark up indicates that the price of the product has increased from its original price and the percentage of the mark up indicates by how much. For example: if the price of a jumper has been marked up by $20%$20%, then its price has been increased by $20%$20%. Similarly, the percentage of a discount indicates how much the price of a product has decreased from its original price. Since mark ups and discounts are simply percentage increases and decreases, we can calculate them using the same methods. #### Worked Example An encyclopedia priced at $\\$110$$110 is marked up by 40%40%. What is the marked up price of the encyclopedia? Think: We are given that the original price of the encyclopedia is \\110$$110. Since it has been marked up, we are calculating a percentage increase. The amount that we are increasing by is $40%$40%. Do: To increase the original price by $40%$40%, we multiply it by $140%$140%. This gives us: Marked" ]

[ "## The price of a skateboard is reduced by 25% . The original price was $120. Find the new price . A:$95 B: $80 C. Question The price of a skateboard is reduced by 25% . The original price was$ 120. Find the new price . A: $95 B:$80 C. $90 in progress 0 10 hours 2021-07-21T23:01:51+00:00 1 Answers 0 views 0 ## Answers ( ) 1. Answer: C.$90 120 – 25%= 90", "3 Skiboards, Inc. has two divisions. The Boards Division makes the board that is made into Skiboards by the Ski Division, but the Board Division can also sell the boards it makes to outside customers. In 2011, The Boards Division reported the following information: Selling price per board \u0007$ 52\u0007\u0007Variable costs per board\u0007\\$ 22\u0007\u0007\u0007\u0007\u0007Number of boards:\u0007\u0007\u0007Produced in 2011\u000710,000\u0007\u0007Sold to the Ski Division \u00078,000\u0007\u0007Sold to outside customers\u00072,000\u0007\u0007Sales from the Boards Division to the Ski Division were made at the same price that sales were made to outside customers.... Attachments:", "a friend for 60% more than what you paid. If the selling price is more than the buying price then it is called markup Here , the markup is 60% of $22.4 a = p% × w a = 0.6 × 22.4 a = 13.44 So, the markup is$13.44 We know that selling price = cost of buying + markup = 22.4 + 13.44 = 35.84 So , the selling price is $35.84. Try It Question 1. The original price of a skateboard is$50. The skateboard is on sale for 20% off. What is the sale price? Answer: The sales price of skateboard is $40 Explanation: Given , skateboard is$50 with 20% off We know , The sales price be 100% – 20% = 80% of the original price sales price = 80% of 50 = 0.8 × 50 = 40 So, The sales price is $40 Question 2. The discount on a DVD is 50%. It is on sale for$10. What is the original price of the DVD? Answer: The original price of DVD is $20. Explanation: Given , discount on a DVD is 50%. , It is on sale for$10 We know , The sales price be 100% – 50% = 50% a = p% × w 10 = 0.5 × w w = $$\\frac{10}{0.5}$$ w", "## A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage Question A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage discount for the cost of the surfboard. in progress 0 5 months 2021-08-28T09:02:34+00:00 1 Answers 0 views 0", "0 like 0 dislike The price of a skateboard is reduced by 25%. The original price was $120. Find the new price. A.$95 B. $80 C.$90", "population mean number of snowboard daily sales has increased by between 43 per day and 72 per day now that ads are posted on Google. Assumptions: We must assume that the first eight days of December are an unbiased sample. Also, since the sample size of 8 is not greater than 30, we must assume that the distribution of differences in sales by date is approximately normal. This assumption has no reason for being true, hence increasing the sample size to over 30 is highly recommended. Uses: A manager would calculate the increased profit by multiplying 43 and 73 by the profit per snowboard sale and compare these to the daily advertising cost that Google charges to make a decision about whether the advertising was successful. Exercise $$\\PageIndex{4}$$ Nine American voters were asked how old is too old to be a member of Congress and how old is told to be the president of the United States. The table below shows the results of the survey. Voter 1 Voter 2 Voter 3 Voter 4 Voter 5 Voter 6 Voter 7 Voter 8 Voter 9 Congress 80 75 65 80 78 70 75 85 69 President 65 70 70 75 67 65 73 80 75 Find and interpret the 95% confidence interval for the difference. The 95% confidence interval", "– 100p. Revenue (number of units that will sell)(price per unit, p) (50 000 – 100p)p = 50 000p – 100p2 Total Cost (cost for producing the snowboards) This is the total overhead costs plus the cost per snowboard times the number of snowboards: 500 000 + 100(50 000 – 100p) 500 000 + 5 000 000 – 10 000p 5 500 000 – 10 000p b. Write an expression to represent the profit. Profit = Total Revenue – Total Cost = (50 000p – 100p2 ) – (5 500 000 – 10 000p) Therefore, the profit function is f(p) = – 100p2 + 60 000p – 5 500 000. c. What is the selling price of the snowboard that will give the maximum profit? Solve for the vertex of the quadratic function using completing the square. (f(p) = – 100p2 + 60 000p – 5 500 000 = – 100(p2 – 600p + ) – 5 500 000 = – 100(p2 – 600p + 3002 ) – 5 500 000 + 100(300)2 = – 100(p – 300)2 – 5 500 000 + 9 000 000 = – 100(p – 300)2 + 3 500 000) The maximum point is (300,3 500 000). Therefore, the selling price that will yield the maximum profit is $300. d. What is the", "markup amount? 16. Sally works for a skateboard shop. The company just purchased a skateboard for$89.00 less discounts of 22%, 15%, and 5%. The company has standard expenses of 37% of cost and desires a profit of 25% of the regular unit selling price. What regular unit selling price should Sally set for the skateboard? 17. If an item has a 75% markup on cost, what is its markup on selling price percentage? 18. A product received discounts of 33%, 25%, and 5%. A markup on cost of 50% was then applied to arrive at the regular unit selling price of $349.50. What was the original list price for the product? 19. Mountain Equipment Co-op (MEC) wants to price a new backpack. The backpack can be purchased for a list price of$59.95 less a trade discount of 25% and a quantity discount of 10%. MEC estimates expenses to be 18% of cost and it must maintain a markup on selling price of 35%. a. What is the cost of backpack? b. What is the markup amount? c. What is the regular unit selling price for the backpack? d. What profit will Mountain Equipment Co-op realize? e. What happens to the profits if it sells the backpack at the MSRP instead? 20. Costco can purchase a bag of Starbucks", "I help? Recilia: I’ve reviewed our financial results for the past 12 months. It looks like we made a profit in some months, and had losses in other months. From what I can tell, we sell each snowboard for $250, our variable cost is$150 per unit, and our fixed cost is $75 per unit. It seems to me that if we sell just one snowboard each month, we should still show a profit of$25, and any additional units sold should increase total profit. Lisa: Your unit sales price of $250 and unit variable cost of$150 look accurate to me, but I’m not sure about your unit fixed cost of $75. Fixed costs total$50,000 a month regardless of the number of units we produce. Trying to express fixed costs on a per unit basis can be misleading because it depends on the number of units being produced and sold, which changes each month. I can tell you that each snowboard produced and sold provides $100 toward covering fixed costs—that is,$250, the sales price of one snowboard, minus $150 in variable cost. Recilia: The$75 per unit for fixed costs was my estimate based on last year’s sales, but I get your point. As you know, I’d like to avoid having losses. Is it possible to determine how many units we", "Snowboard Company a$17,500 = $37,500 −$20,000. b 87.5 percent = $17,500 ÷$20,000. Carefully review Figure 6.6 \"Sensitivity Analysis for Snowboard Company\". The column labeled Scenario 1 shows that increasing the price by 10 percent will increase profit 87.5 percent ($17,500). Thus profit is highly sensitive to changes in sales price. Another way to look at this is that for every one percent increase in sales price, profit will increase by 8.75 percent, or for every one percent decrease in sales price, profit will decrease by 8.75 percent. The column labeled Scenario 2 shows that decreasing sales volume 10 percent will decrease profit 35 percent ($7,000). Thus profit is also highly sensitive to changes in sales volume. Stated another way, every one percent decrease in sales volume will decrease profit by 3.5 percent; or every one percent increase in sales volume will increase profit by 3.5 percent. When comparing Scenario 1 with Scenario 2, we see that Snowboard Company’s profit is more sensitive to changes in sales price than to changes in sales volume, although changes in either will significantly affect profit. ## Expanding the Use of Sensitivity Analysis Question: Although the focus of sensitivity analysis is typically on how changes in variables will affect profit (as shown in Figure 6.6 \"Sensitivity Analysis for Snowboard Company\"), accountants also use", "### Home > ACC7 > Chapter cc38 > Lesson cc38.1.2 > Problem8-20 8-20. 1. Jack noticed a sale on his favorite model of All-Terrain Vehicle (ATV). The salesperson said the company was cutting the price for that weekend, so Jack got an 8% discount. Homework Help ✎ 1. If the usual price of the ATV was $4298, estimate how much the discount will be. Explain your reasoning. 2. Calculate the amount Jack did pay with the 8% discount. 3. If the store sold three of the ATVs that weekend, how much money was saved by the three customers combined? Round the ATV price to$4300 in order to make the percent easier to approximate. Since we can find that 10% of $4300 is$430, we can also determine that 5% of $4300 would be$215. 8% is roughly in the middle of 10% and 5%. What is your estimation? Find 8% of $4298, then subtract that value from$4298 to find how much Jack paid. 4298 − 343.84 = 4298(.08) = 343.84 $3954.16 For each ATV, the original price was$4298. Based on part (b), it costs \\$3954.16 after the 8% discount. How much did Jack save on one ATV? Now multiply that number by 3.", "### Условие задания: 3 Б. Рис. $$1$$. Snowboarding Snowboarding is a popular winter sport. Lots of people all over the world are fond of snowboarding. It involves gliding over snow slope while standing on a board. It originated in the United States and became hugely popular at the end of the $$XX$$ century and was introduced at the $$1998$$ Winter Olympics. There are various types of snowboarding: freeriding, freestyle, urban, half pipe, boardercross, jibbing, etc. Snowboarding can be dangerous and challenging and requires proper training and using special gear. However, skiing is more dangerous than snowboarding, though people usually think the opposite. All in all, a person should be in good physical shape before trying snowboarding. Despite all the risks this relatively young kind of sport is great fun and more and more people are becoming keen on racing down the snow-covered slopes. Fill in the gaps: 1. Lots of people are fond of snowboarding. 2. There are of snowboarding: freeriding, freestyle, urban, half pipe, boardercross, jibbing, etc. 3. All in all, a person should be before trying snowboarding. Источники: Рис. 1. Snowboarding. Pixabay License CC0. Free for commercial use No attribution required. https://pixabay.com/images/id-2030851/ (Дата обращения 11.10.2021) Вы должны авторизоваться, чтобы ответить на задание. Пожалуйста, войдите в свой профиль на сайте или зарегистрируйтесь.", "1. Other 2. Other 3. 1 sk8 is examining an invoice the gross price of... # Question: 1 sk8 is examining an invoice the gross price of... ###### Question details 1) Sk8 is examining an invoice. The gross price of a skateboard is $109.00, and the invoice states it received a trade discount of 15% and quantity discount of 10% as well as a loyalty discount. However, the amount of the loyalty discount is unspecified. a. If Sk8 paid$80.88 for the skateboard, what is the loyalty discount percent? b. If the loyalty discount is applied after all other discounts, what amount of loyalty dollars does Sk8 save per skateboard? 2) Daisy is trying to figure out how much negotiating room she has in purchasing a new car. The car has an MSRP of $34,995.99. She has learned from an industry insider that most car dealerships have a 20% markup on selling price. What does she estimate the dealership paid for the car? 3) A snowboard has a cost of$79.10, expenses of $22.85, and profit of$18.00. a. What is the regular unit selling price? b. What is the markup amount? c. What is the markup on cost percentage? d. What is the markup on selling price percentage? e. What is the break-even selling price? What is the markup on", "before the discount? • Two shops In two different shops, the same skis had the same price. In the first, however, they first became more expensive by 20% and then cheaper by 5%. In the second, they were first cheaper by 5% and then more expensive by 30%, so after these adjustments in the • Selling price Find the selling price. Cost to store:$50 Markup: 10% • Ice skates Ice skates were raised twice, the first time by 25%, the second time by 10%. After the second price, their cost was 82.5 euros. What was the original price of skates?", "Home > MC2 > Chapter 10 > Lesson 10.1.2 > Problem10-18 10-18. Round the ATV price to $4300 in order to make the percent easier to approximate. Since we can find that 10% of$4300 is $430, we can also deduce that 5% of$4300 would be $215. 8% is roughly in the middle of 10% and 5%. What is your estimation? Find 8% of$4298, then subtract that value from $4298 to find how much Jack paid. 4298 − 343.84 = 4298(.08) = 343.84$3954.16 For each ATV, the original price was $4298. Based on part (b), it costs$3954.16 after the 8% discount. How much did Jack save on one ATV? Now multiply that number by 3.", "# A Trader marks his goods 20% above cost price but allows his customers a discount of 10%, the cost price of a blackboard, which is sold for 216, is [ A ] 196 [ B ] 180 [ C ] 200 [ D ] 108 Answer : Option C Explanation : Selling price = 216 Discount = 10% Marked price = (216X$100\\over90$) = 240 So, Cost price =(240X$100\\over120$) = `200", "# Calculus: General Differentiation – #28357 Question: A snowmobile manufacturer is planning a new line of ski-doos. The price is dependent on the number of ski-doos sold $\\left( x \\right)$, and is given as $p(x)=3432-11x$. What is the maximum revenue that the manufacturer can expect, using this model for revenue? What will be the price for one ski-doo?", "? Free Version Moderate # Mr. Winter's Budget Constraint for Sports Equipment MICRO-YEZX4E Based on the the following graph of Mr. Winter's budget constraint, if the price of skis is $\\$110$, determine the price of snowboards and Mr. Winter's budget. Do not use a dollar sign in your solution (e.g. input 1, instead of \\$1).", "Â I’m glad you enjoyed the experience and will definitely keep your thoughts in mind going forward. Â We strive to make the shopping experience for skis as pleasant and rewarding as possible and that is made possible with constructive feedback like yours. 34. StartUpLift says:$5 awarded to Sierra12346.", "1024 USD, Monday will be 25% discount. How much USD will save, and what will be the price? 17. Apples 2 James has 13 apples. He has 30 percent more apples than Sam. How many apples has Sam?" ]

[ "product of three positive integers?", "Algebra Level 3 There is a triplet of positive numbers such that their sum and their product are equal, for example, $(1, 2, 3):$ $1+2+3=1\\times 2\\times 3.$ True or False? There are infinitely many triplets of positive numbers (not necessarily integers) such that their sum and their product are equal. ×", "three was equal to the sum of the remaining two. The conduct", "## Introductory Algebra for College Students (7th Edition) $-51$ The product of two integers with different signs is negative. Thus, multiplying the two number gives: $=-51$", "## Elementary Technical Mathematics (a) Sum: $12$ (b) Difference: $-4$ (c) Product: $32$ RECALL: (1) For any real number $a$, $a + 0 = a$. (2) For any real number $a$, $a(0) = 0$. (3) $a-b=a+(-b)$. (4) The product of two integers with the same sign is positive. The sum, difference, and product of the given expressions are: (a) Sum: $(+4) + (+8) = +12=12$ (b) Difference: $(+4) - (+8) = -4$ (c) Product: $(+4)(+8) = +32=32$", "Question Use digits and symbols to write ”The product of three and five is greater than the sum of three and five 1. thnahdat 3*5>3+5 Step-by-step explanation: product of three and five 3*5=15 sum of three and five 3+5=8 So, 3*5 > 3+5", "# Do you want to count to three? I have 3 positive integers. The sum of these 3 integers is $k$. The product of these 3 integers is also $k$. What is $k$? ×", "? Free Version Difficult # Sum and Product of Real Numbers GMAT-Y1PX7A If $x$ and $y$ are integers, for how many possibilities does $xy = x + y$ ? A One B Two C Three D Four E None", "The product $$N$$ of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $$N$$. (第二十一届AIME2 2003 第1题)", "The sum of three numbers is $2009$. The sum of the first two numbers is $1004$ and the sum of the last two is $1005$. What is the product of all three numbers?", "## Elementary Technical Mathematics $-24$ The product of two integers with opposite signs is negative. Thus, $(-4)(+6)=-24$", "# 75th Problem 2016 Algebra Level 1 Find the sum of 3 consecutive positive odd integers such that the product of the first and third is 4 less than 7 times the second. Check out the set: 2016 Problems ×", "# Sum is Product Number Theory Level 1 $\\LARGE{2 + 2 = 2 \\times 2 \\\\ 1 + 2 + 3 = 1 \\times 2 \\times 3}$ In the above equations, there are respectively 2 and 3 positive integers whose sum is equal to their product. Find 4 positive integers whose sum is equal to their product. Enter the answer as their sum ×", "# Sizing up factors Is it possible to write $3^{2016}+4^{2017}$ as the product of two integers, both of which are over $2018^{183}$? ×", "The product of three integers X, Y and Z is $192$. Z is equal to $4$ and P is equal to the average of X and Y. What is the minimum possible value of P? 1. $6$ 2. $7$ 3. $8$ 4. $9.5$", "Theory # Product of a positive and negative number When two factors with different signs are multiplied, the product is negative, e.g. $3(\\text{-}5) = \\text{-}3\\cdot5$ which is equal to $\\text{-}15.$", "12$. This may be the interpretation that was intended. If the question has copied the words of the original problem exactly, it may be that the problem was poorly worded. The question seems really to be asking for how many distinct multisets of positive integers exist that each have a sum of $12$.", "## Algebra and Trigonometry 10th Edition $32$ The product of two negative numbers is a positive number. Thus: $(-8)(-4)=+(8)(4)=32$", "+1 vote The sum of two positive numbers is $100$. After subtracting $5$ from each number, the product of the resulting numbers is $0$. One of the original numbers is ______. 1. $80$ 2. $85$ 3. $90$ 4. $95$ edited", "# One Line Problem Find 3 integers whose product (of all 3 integers) is equal to their sum (of all 3 integers). ×" ]

[ "## Elementary Algebra $49=7\\times7$ 49 is only divisible by the prime number 7. Furthermore, 49 divided by 7 is 7. Therefore, the prime factorization of 49 is: $49=7\\times7$", "\\times 7^2 = 490$, whose sum of divisors is $?$.", "# Make numbers from 1-10 with 5 3s In this puzzle book, I came across a question that goes like this: Make the numbers from 1-10 using 5 3s. Rules: You are only allowed to use the four basic arithmetic operators ($$+, -, \\times, \\div$$). You are allowed to use brackets. Here is what I have so far: $$3+3-3+3-3=3$$ $$3+3+3+3-3=9$$ $$(3\\times3)+(3\\div3)-3=7$$ $$3\\div3+3+3+3=10$$ $$3+3\\div3-3+3=4$$ Can someone help me figure out the rest? Complete answer(thanks to Ross Millikan and BAWS): $$3-3\\div3-3\\div3=1$$ $$3-3\\div3\\times3\\div3=2$$ $$3+3-3+3-3=3$$ $$3+3\\div3-3+3=4$$ $$((3\\times3)+(3+3))\\div3=5$$ $$(3\\times3)-(3-3)-3 = 6$$ $$(3\\times3)+(3\\div3)-3=7$$ $$3+3+((3+3)\\div3)=8$$ $$3+3+3+3-3=9$$ $$3\\div3+3+3+3=10$$ $$((3\\times3)+(3+3))\\div3 = 5$$ $$(3\\times3)-(3-3)-3 = 6$$ I'll leave the rest with this new way of looking at it. • Hi, on this site, for answers, you need to hide them in spoiler blocks(>!) most of the times. Thank you. Dec 13 '20 at 5:37 Here are $$1,2$$ and $$8$$ $$3-\\frac 33 - \\frac 33 =1$$ $$3-\\frac 33 \\cdot \\frac 33=2$$ $$3+3+\\frac{3+3}3=8$$ We obtain the missing numbers from the following. $$(3-3)\\times3+\\frac{3}{3}=1$$ $$3-3+\\frac{3+3}{3}=2$$ $$3+\\frac{3}{3}+\\frac{3}{3}=5$$ $$3+3+3\\times (3-3)=6$$", "not divisible by $3$ We try $M=7$ this does work giving $N=16,~M=7$ and thus $3\\times 16=\\boxed{\\textbf{(B)},~48}$ games in their division.", "digit of $3^3$ which is 7..", "that the number has less than $6$ divisors. Since $12$ has $6$ divisors and $10$ has $4$ divisors, James’s desired sum is $\\boxed{10}$.", "# Comparing Two Gigantic Numbers Find the greatest common divisor of the two numbers, $$886! + 1$$ and $$887!$$. Notation: $$!$$ denotes the factorial notation. For example, $$10! = 1\\times2\\times3\\times\\cdots\\times10$$. ×", "# 2005 AIME I Problems/Problem 3 ## Problem How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50? ## Solution (Basic Casework and Combinations) Suppose $n$ is such an integer. Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$. In the first case, the three proper divisors of $n$ are $1$, $p$ and $q$. Thus, we need to pick two prime numbers less than $50$. There are fifteen of these ($2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$) so there are ${15 \\choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type. In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$. Thus we need to pick a prime number whose square is less than $50$. There are four of these ($2, 3, 5,$ and $7$) and so four numbers of the second type. Thus there are $105+4=\\boxed{109}$ integers that meet the given conditions. ~lpieleanu (Minor editing)", "integer $p+4=3k+6=3(k+2)$ is divisible by 3. So, we have proved that at least one of the integers $p$, $p+2$, $p+4$ is divisible by 3. But this is obviously a contradiction, since each of the integers $p$, $p+2$, and $p+4$ are supposed to be primes strictly greater than 3. Thus, our assumption must have been incorrect. Hence, we have proved that the only prime triple is 3,5,7.", "## Elementary Algebra $7\\times7\\times3\\times3$ Since the sum of the digits of $441$ add up to $9$, it is divisible by $9$. When we divide $441$ by $9$ we get $49$. $49$ is the perfect square of $7$, and can be broken down into prime factors $7\\times7$. $9$ can be broken down into prime factors $3\\times3$. Since $3$ and $7$ are prime numbers, the prime factorization of $441$ is $7\\times7\\times3\\times3$", "= 3 \\pmod4.$ Therefore $7^{7^7} = 7^3 = 49*7 = (-1)*7 = -7 = 3 \\pmod{10}$. That says that the last digit of $7^{7^7}$ is $\\Large \\mathbf 3$. Status Not open for further replies.", "## Elementary Algebra $49=7\\times7$ The number $49$ is the perfect square of $7$. Since $7$ is a prime number, we can write the prime factorization of $49$ as $7\\times7$", "# How many positive integers between 50 and 100 are divisible by 7? How many positive integers between 50 and 100 a) are divisible by 7? Which integers are these? This question is in the basic counting section of my textbook and I'm just studying for finals now. For this question, I was wondering if there is a less primitive way of finding this answer. I did: 56, 63, 70, 77, 84, 91, 98; These answers I found by just going on with my 7's multiplications table. The reason I'm wondering this is in case I get asked the same thing but with a much larger number such as 1000. - If it asks you to list them as this question does then you'll have to do it the hard way anyway.... – Simon Hayward Dec 8 '12 at 21:55 $$50 = 7 \\times 7 +1$$ $$100 = 14 \\times 7 +2$$ and the answer is $14-7=7$. This is effectively what you have done, counting $8 \\times 7$, $9 \\times 7$, $10 \\times 7$, $11 \\times 7$, $12 \\times 7$, $13 \\times 7$, and $14 \\times 7$. You will need to be more careful if either of the remainders is $0$. @Ben: Yes: since $100 = 14 \\times 7 +2$ there are $14$ positive multiples of $7$ below", "is the smallest divisor is $$5\\times 5=25$$. And, the first number, distinct from $$7$$, for which $$7$$ is the smallest divisor is $$7\\times 7=49$$.", "by $3$ and so on. Thus this is a sequence of $n - 1$ consecutive composite integers. $\\Box$ This one-liner works because all numbers from $1 \\ldots n$ are divisors of $n!$ 🤦🏽‍♂️", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "Hence positive integers that have exactly $36$ positive divisors are of the form listed above. • Are we looking for the least positive integer with this property or are we seeking all positive integers with this property? – user328442 Nov 7 '17 at 0:21 • Specifically what I don't understand about the linking question: take the first answer for example. With the case, 2*2*2*3, why is the answer stating \"minimum for this factorization is 2^2⋅3⋅5⋅7\"? Notably, where the heck did the 5 and 7 come from and also, how does 2^2⋅3⋅5⋅7 relate to the original number of divisors 24?? – user424890 Nov 7 '17 at 0:22 • @user328442 the latter, all positive integers with said property. For example, the back of my textbook gives, \"all squares of primes\" as an answer to a similar question. Notably 3 divisors. – user424890 Nov 7 '17 at 0:24 • the first 4 primes are 2, 3, 5 and 7 and so an integer will be relatively small. The important part is that we have 3 choices for how many times 2 shows up in a factorization of a divisor of this integer, we have 2 options for 3, 2 for 5 and 2 for 7. This means that there are 3*2*2*2 =24 divisors of that integer. – user328442 Nov 7 '17", "digits in their representation $= 6 + (5 \\times 22) = 116.$ 4. Hence, there are $116$ three-digit integers less than $601$ that have exactly two different digits in their representation.", "# Divisibility of polynomial; basic number theory [duplicate] Show that $42|(n^7 - n)$ for all integers $n$ The only hint I've been given uses Fermat's Little Theorem but I don't know how to use that in this case since $42$ is not a prime. • you can write $42=2\\cdot 3\\cdot 7$ – Yanko Jan 27 '18 at 14:36 • At least you could have viewed my edit :( – Yash Jain Jan 27 '18 at 14:37 Well, $42=2\\times3\\times7$ and: • $n$ and $n^7$ have the same parity, which means that $2\\mid n^7-n$; • by Fermat's little theorem, $3\\mid n^3-n$, and$$n^7-n=n^7-n^5+n^5-n^3+n^3-n=(n^4+n^2+1)(n^3-n),$$which implies that $3\\mid n^7-n$; • by Fermat's little theorem, $7\\mid n^7-n$. Since $2$, $3$, and $7$ are co-prime, this implies that $42\\mid n^7-n$.", "# Consecutive Integers whose Sums of Squares of Divisors are Equal ## Theorem The only two consecutive positive integers whose sums of the squares of their divisors are equal are $6$ and $7$. ## Proof The divisors of $6$ are $1, 2, 3, 6$ and so the sum of the squares of the divisors of $6$ is: $1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$ The divisors of $7$ are $1, 7$ and so the sum of the squares of the divisors of $7$ is: $1^2 + 7^2 = 1 + 49 = 50$ It remains to be shown that there are no more:" ]

[ "\\times 7^2 = 490$, whose sum of divisors is $?$.", "in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \\boxed{448}$.", "# Product of 3 integers is 72, find the 3 integers that give the smallest sum Product of 3 integers a, b, c equals 72, where every factor is positive integer. Find the integers a, b, c with the smallest sum. It's easy to get the factors of 72 manually and see that the 3 smallest factors that give the product as 72 will be the smallest sum. I checked the factorization like this: 1-1-72: 74 1-2-36: 39 1-4-18: 25 1-8-9: 18 2-4-9: 15 3-4-6: 13 It seems like the smallest possible factors will give the smallest sum. But there must be some trick to it or some kind of algorithm to find the smallest integers for any arbitrary product without laborious factoring. Thanks for your help. - The factors must be close to each other. A heuristic: The AM-GM inequality gives us $(a+b+c)/3 \\geq \\sqrt[3]{abc}$ where equality holds when $a=b=c$ and the difference between the LHS and RHS increases when $a$, $b$ and $c$ are far apart from each other. – user17762 Sep 26 '12 at 5:00 Yes @Marvis is right: And 3-4-6 are the closest number in this case. – Sumit Bhowmick Sep 26 '12 at 6:15 One algorithm might be to find the factor of $72$ closest to $\\sqrt[3]{72}$, which is $4$. Then find the", "example, in the unit interval $[0,1]$ there are actually the same number of numbers as there are people on world B from above! which is $\\infty$ly many! However, the amount of space $[0,1]$ takes up is rather small, its just the length which is 1! Funny isn’t it?! 3. DID YOU KNOW… you can tell if 3 fits in any number without using a calculator?! We all know if 2,5, or 10 fit inside a number (if it’s even, ends in 0 or 5, or ends in 0, respectively). What about the number 3? Well turns out if you add the digits up and this number is divisible by 3, then the number is. For example, 108, 1,000,008, are both divisible by 3 as the sum of their digits is 9 which is divisible by 3! There is also one for 7 and 11! Click here for the 3-Rule, here for the 7-Rule, and here for the 11-Rule!!! And to see even more number tricks, click here! 4. DID YOU KNOW…the product of any two integers is equal to the product of their LCM (least common multiple) and GCF (greatest common factor)! Click here for a video on this!Did you know the product of any two integers is equal to the product of their LCM (least common multiple)", "# Consecutive Integers whose Sums of Squares of Divisors are Equal ## Theorem The only two consecutive positive integers whose sums of the squares of their divisors are equal are $6$ and $7$. ## Proof The divisors of $6$ are $1, 2, 3, 6$ and so the sum of the squares of the divisors of $6$ is: $1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$ The divisors of $7$ are $1, 7$ and so the sum of the squares of the divisors of $7$ is: $1^2 + 7^2 = 1 + 49 = 50$ It remains to be shown that there are no more:", "the unit and tens digit should be divisible by $4$ (iii) Divisibility rule of $7$: We need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by $7$, then the original number will also be divisible by $7$. Note: Different types of numbers are: (i) Natural number: $1,2,3,4,$------ (ii) Whole number:$\\;0,1,2,3,4,$ ------ (iii) Integers: $- 4, - 3, - 2, - 1,0,1,2,3,4,$----- (iv) Positive integers: $1,2,3,$----- (v) Negative integers: $- 4, - 3, - 2, - 1$", "chosen, $n(n+1)(n+2)$ is divisible by 3. We have shown that $n(n+1)(n+2)$ is both even and divisible by 3. Thus, the product of any three consecutive integers is divisible by 6.", "# Sum is Product Number Theory Level 1 $\\LARGE{2 + 2 = 2 \\times 2 \\\\ 1 + 2 + 3 = 1 \\times 2 \\times 3}$ In the above equations, there are respectively 2 and 3 positive integers whose sum is equal to their product. Find 4 positive integers whose sum is equal to their product. Enter the answer as their sum ×", "within 10 that we are still working on. We thought about using sums we know to help us find the value of differences. What sum would be helpful in finding the value of $$7 - 3 = \\boxed{\\phantom{7}}$$ ? How would that sum be helpful?” ( $$3+4 = 7$$ if you know you get 7 when you put together 3 and 4, then you know if you start with 7 and take away 3 you will have 4 left. )", "## Elementary Technical Mathematics (a) Sum: $12$ (b) Difference: $-4$ (c) Product: $32$ RECALL: (1) For any real number $a$, $a + 0 = a$. (2) For any real number $a$, $a(0) = 0$. (3) $a-b=a+(-b)$. (4) The product of two integers with the same sign is positive. The sum, difference, and product of the given expressions are: (a) Sum: $(+4) + (+8) = +12=12$ (b) Difference: $(+4) - (+8) = -4$ (c) Product: $(+4)(+8) = +32=32$", "equal integers. $9 = 3 \\times 3$", "# Digit Sum 7 Probability Level 3 How many positive integers less than 1000000 have the sum of their digits equal to 7? Details and assumptions The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is $1 + 1 + 2 + 3 = 7$. ×", "question. Are the sum of integers a and b divisible by 7? 1) a is not divisible by 7 So a is of one of the following formats: 7n+1, 7n+2, 7n+3, 7n+4, 7n+5 and 7n+6 2) a-b is divisible by 7 The format of b is the same as the format of a i.e. if a = 7n + 3, then b = 7m + 3 since a and b vary by a multiple of 7 e.g. if a = 43 (= 7*6+1), then b could be 22 (= 7*3 + 1). The difference between them has to be a multiple of 7 so the remainder obtained by dividing a by 7 will be the same as the remainder obtained by dividing b by 7. Using both statements, when you sum a and b, you get, a+b = 7n+1 + 7m+1 = 7(n+m) + 2 or a+b = 7n+2 + 7m+2 = 7(n+m) + 4 or a+b = 7n+3 + 7m+3 = 7(n+m) + 6 etc I hope you see that the remainder will be some even number (either 2 or 4 or 6 or 8 or 10 or 12). The remainder will never be 7 so the sum will never be a multiple of 7. What happens if we replace 7 by 6? Same logic. a+b =", "SEARCH HOME Math Central Quandaries & Queries Question from Rick: This is a question on my sons pre-ap practice quiz. I think that there is information missing (?) The product of 3 integers is -24 The sum of 3 integers is -12 What are the 3 integers ? This is exactly how it was written on his quiz paper. I have wasted to much time on the internet, trying to find a formula(s) to help him. Please help me. Hi Rick, Suppose the three integers are $a, b$ and $c,$ then since their product is negative either one of them is negative and the other two are positive or all three are negative. When dealing with the product of integers it is quite often helpful to consider the prime factors so I factored 24 to get $a \\times b \\times c = 24 = 2^3 \\times 3.$ Hence the only primes that divide $a, b$ and $c$ are 2 and 3. From this I can see that the divisors of 24 are 1 and 24, 2 and 12, 3 and 8, and 4 and 6. These are the only possible choices for $a, b$ and $c.$ If exactly one of $a, b$ and $c$ is negative then it must be 12 less than the sum of the other", "of three consecutive positive integers, is t a 5 22 Dec 2015, 03:30 3 What is the value of the sum of a list of n odd integers? (1) n = 8 (2 5 12 Dec 2014, 06:34 5 What is the sum of all 50 numbers in a list? 3 04 Sep 2014, 15:53 5 If the sum of three integers is even, is the product of the 6 08 May 2010, 09:02 20 Is the sum of integers a and b divisible by 7? 15 22 Oct 2009, 19:04 Display posts from previous: Sort by", "get it. 7 is prime so $z−x_0,…,z−x_3$ don't exist. – user286826 Jan 19 '16 at 1:09 • Right! You can have at most three distinct integers whose product divides $7$. For instance, $1, 7, -1$. – TonyK Jan 19 '16 at 1:19", "explain why the answer is (C) in the original question.[/highlight] Are the sum of integers a and b divisible by 7? 1) a is not divisible by 7 So a is of one of the following formats: 7n+1, 7n+2, 7n+3, 7n+4, 7n+5 and 7n+6 2) a-b is divisible by 7 The format of b is the same as the format of a i.e. if a = 7n + 3, then b = 7m + 3 since a and b vary by a multiple of 7 e.g. if a = 43 (= 7*6+1), then b could be 22 (= 7*3 + 1). The difference between them has to be a multiple of 7 so the remainder obtained by dividing a by 7 will be the same as the remainder obtained by dividing b by 7. Using both statements, when you sum a and b, you get, a+b = 7n+1 + 7m+1 = 7(n+m) + 2 or a+b = 7n+2 + 7m+2 = 7(n+m) + 4 or a+b = 7n+3 + 7m+3 = 7(n+m) + 6 etc I hope you see that the remainder will be some even number (either 2 or 4 or 6 or 8 or 10 or 12). The remainder will never be 7 so the sum will never be a multiple of 7. [highlight]What happens if", "# Difference between revisions of \"2003 AIME II Problems/Problem 1\" ## Problem The product $N$ of three positive integers is $6$ times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N$. ## Solution Let the three integers be $a, b, c$. $N = abc = 6(a + b + c)$ and $c = a + b$. Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$. Since $a$ and $b$ are positive, $ab = 12$ so $\\{a, b\\}$ is one of $\\{1, 12\\}, \\{2, 6\\}, \\{3, 4\\}$ so $a + b$ is one of $13, 8, 7$ so $N$ is one of $12\\cdot 13 = 156, 12\\cdot 8 = 96, 12\\cdot 7 = 84$ so the answer is $156 + 96 + 84 = \\boxed{336}$.", "12$. This may be the interpretation that was intended. If the question has copied the words of the original problem exactly, it may be that the problem was poorly worded. The question seems really to be asking for how many distinct multisets of positive integers exist that each have a sum of $12$.", "# Product of $N$ Consecutive Integers Divided by $N$ 1 is divisible by 1. $2\\times 3$ is divisible by 2. $4\\times5\\times6$ is divisible by 3. $7 \\times 8 \\times 9 \\times 10$ is divisible by 4. $11 \\times 12 \\times 13 \\times 14 \\times 15$ is divisible by 5. True or False? The product of any $N$ consecutive integers must be divisible by $N.$ ×" ]

[ "+0 # help pls 0 95 1 At the end of the day, a bakery had $$\\frac{6}{7}$$ of a pie left over. Louie, Duey and Huey split the pie, with each taking home the same amount of leftover pie. How much pie did Louie take home? Dec 25, 2020", "# Jeremy and his friends ate 5/8 of a pie. If the pie was cut into eight pieces, how much pie is left over? Nov 19, 2016 $\\frac{3}{8}$ of the pie is left over. $1$ pie = $\\frac{8}{8}$ of a pie. Jeremy and his friends ate (and thus \"subtracted\") $\\frac{5}{8}$ of the pie. $\\frac{8}{8} - \\frac{5}{8} = \\frac{8 - 5}{8} = \\frac{3}{8}$", "### Still have math questions? (1)) Nori had $$2 \\frac { 1 } { 12 }$$ bags of apples. He used $$1 \\frac { 5 } { 12 }$$ bags of apples to make pies. How many bags of chpples does Nori have left? $$\\frac{2}{3}$$", "Susie had a large bowl of cherries. They were all in pairs. Susie took out a pair, she ate one cherry and put the other one back. She took out another pair and did the same again. Then she helped herself to one of the single cherries in the bowl. Susie continued helping herself to the cherries in this way (pair, pair, single - pair, pair, single - ...) After she had done this lots of times, there were just $14$ single cherries left.", "Grandma had made pies for a bake sale. She had carefully put equal amounts of mixture in each pie tin and was now trying to find the weight of the pies. She had a problem; she only had one $200$ gram weight and one $125$ gram weight. She found that one pie balanced on the scale with both weights and a quarter of a pie. How heavy was each pie?", "Stewart's Sequences 1 of 5 A pie factory is booming in business. On the first day of opening they baked only $$2\\frac { 1 }{ 3 }$$ pies, On Day 2 they baked $$4\\frac { 2 }{ 3 }$$ pies! On Day 3 they made $$7$$ pies! $$...$$ On Day $$n$$ they made $$x$$ pies! Assuming that nothing restricts their pie baking skills, this sequence continues each day and that they can bake up to $$\\infty$$ pies per day, find the sum of $$x$$ when $$n=30$$ and $$n=60$$. ×", "Dewey, while Louie is to receive twice as much as Dewey minus 1 dollar. If the total amount to be divided between them if 107 Million Dollars, calculate how much each has to inherit. Let $x$ = amount Dewey will get. \"Huey gets 3 times as much as Dewey.\" . . Then Huey gets: $3x$ \"Louie gets twice as much as Dewey minus 1 dollar. . . Then Louie gets: $2x - 1$ The total of these amount is \\$107,000,000. There is our equation! . . . . $x + 3x + 2x - 1 \\:=\\:107,000,000$ Go for it!", "a party and $$\\frac{3}{4}$$ of the remaining spinach for a pan for her family. She used the rest of the spinach to make a salad. a. What fraction of the spinach did she use to make the salad? b. If Rose used 3 pounds of spinach to make the pan of spinach pie for the party, how many pounds of spinach did Rose use to make the salad?", "frankhester87 16 # you have three fifths of a pie you divide the remaining pie into 5 equal slices what fraction of the original Post in each slice So you divide $\\frac{3}{5}$ by 5: $\\frac{3}{5}$ ÷ $\\frac{5}{1}$ which is: $\\frac{3}{5}$ × $\\frac{1}{5}$ = $\\frac{3}{25}$ So your answer is: $\\frac{3}{25}$", "pie did you eat? $\\frac{1}{3}$ 4. Describe what the diagram below represents using words and numbers. There are $10$ blocks total, but only $1$ shaded block. One tenth, $10\\%$", "of them. Then we actually ate the same amount of pie.", "3 Therefore, out of 12, we shaded 3 portions. Question 7. Lexi ate some of the apples that her mother brought home from the farmer’s fair. One-half of the apples were left over. Starr ate $$\\frac{1}{6}$$ of the apples that were", "# Math Help - Urgent help needed Please 1. ## Urgent help needed Please How do i work this out ? Uncle Scrooge has asked for a strange bequest in his will. Huey is to receive 3 times as much as Duey, while Louie is to receive twice as much as Duey MINUS 1 Dollar. If the total amount to be divided between them if 107 Million Dollars, calculate how much each has to inherit. Thanks 2. Originally Posted by chhoeuk How do i work this out ? Uncle Scrooge has asked for a strange bequest in his will. Huey is to receive 3 times as much as Duey, while Louie is to receive twice as much as Duey MINUS 1 Dollar. If the total amount to be divided between them if 107 Million Dollars, calculate how much each has to inherit. Thanks Let H, L and D denote how much each of the gets. Then the question specifies: H = 3D L = 2D - 1 H + L + D = 107000000 which you should be abl to solve. RonL 3. Hello, chhoeuk! Another approach . . . Uncle Scrooge has asked for a strange bequest in his will. Huey is to receive 3 times as much as Dewey, while Louie is to receive twice as", "Free Version Moderate Pie Portions Free EMETRC-SB94MY Pete receives a whole apple pie on Monday and eats half of it that evening. However, his appetite falls off after that first day. Each subsequent day, Pete will only eat $\\frac{1}{4}$ of the pie that remains at the beginning of that day. If, beginning Tuesday and for each day thereafter, Pete consumes one-fourth of the remaining pie, then how much of the original pie will remain after he eats pie on Friday? A 0% B 2.5% C 7.7% D 15.8% E 25.0%", "Story 1 A generous little monkey had some peaches. On the first day he gave half his peaches away and ate one. On the second day he gave away half of the rest and ate one. On the third day he gave away half of the rest and ate one. On the fourth day he found there was only one left. How many did he have at the beginning? Story 2 A generous little monkey who liked mathematics had 60 peaches. On the first day he decided to keep $\\frac{3}{4}$ of his peaches. He gave the rest away then he ate one. On the second day he decided to keep $\\frac{7}{11}$ of his peaches. He gave the rest away then he ate one. On the third day he decided to keep $\\frac{5}{9}$ of his peaches. He gave the rest away then he ate one. On the fourth day he decided to keep $\\frac{2}{7}$ of his peaches. He gave the rest away then he ate one. On the fourth day he decided to keep $\\frac{2}{3}$ of his peaches. He gave the rest away then he ate one. How many did he have left at the end? Click here for a pdf of the NRICH poster giving an extension to this problem.", "{ 5 }{ 8 }$$ (d) $$\\frac { 6 }{ 8 }$$ Answer: $$\\frac { 5 }{ 8 }$$ Explanation: Given, Mary Jane has $$\\frac { 3 }{ 8 }$$ of a medium pizza left. Hector has $$\\frac { 2 }{ 8 }$$ of another medium pizza left. To find how much pizza do they have altogether we have to add both the fractions. $$\\frac { 3 }{ 8 }$$ + $$\\frac { 2 }{ 8 }$$ = $$\\frac { 5 }{ 8 }$$ Therefore Mary Jane and Hector has $$\\frac { 5 }{ 8 }$$ pizza altogether. Thus the correct answer is option c. Question 2: Jeannie ate $$\\frac { 1 }{ 4 }$$ of an apple. Kelly ate $$\\frac { 2 }{ 4 }$$ of the apple. How much did they eat in all? (a) $$\\frac { 1 }{ 8 }$$ (b) $$\\frac { 2 }{ 8 }$$ (c) $$\\frac { 3 }{ 8 }$$ (d) $$\\frac { 3 }{ 4 }$$ Answer: $$\\frac { 3 }{ 4 }$$ Explanation: Given, Jeannie ate $$\\frac { 1 }{ 4 }$$ of an apple. Kelly ate $$\\frac { 2 }{ 4 }$$ of the apple. $$\\frac { 1 }{ 4 }$$ + $$\\frac { 2 }{ 4 }$$ = $$\\frac { 3 }{ 4 }$$ Thus the correct", "wee wee wee all the way home. \\end{enumerate} \\begin{description} \\item [opportunistic:] This little piggy went to market. \\item [lazy:] This little piggy stayed home. \\item [fat:] This little piggy had roast beef. \\item [hungry:] This little piggy had none. \\item [scared:] This little piggy went wee wee wee wee all the way home. \\end{description} \\end{document} And the output produced goes something like this: • This little piggy went to market. • This little piggy stayed home. • This little piggy had roast beef. • This little piggy had none. • This little piggy went wee wee wee wee all the way home. 1. This little piggy went to market. And he bought the following items: 1. Bob the tomato 2. Larry the cucumber 3. Junior asparagus 2. This little piggy stayed home. 3. This little piggy had roast beef. 4. This little piggy had none. 5. This little piggy went wee wee wee wee all the way home. opportunistic: This little piggy went to market. lazy: This little piggy stayed home. fat: This little piggy had roast beef. hungry: This little piggy had none. scared: This little piggy went wee wee wee wee all the way home. ## Tables and Arrays A surprising variety of material is nicely rendered using tables. If you have not worked with HTML", "# Christmas Pie gone Algebra Level 1 I had a christmas cherry pie sitting in my fridge. At 1am, my son got hungry and ate $$\\frac{1}{5}$$ of it. At 2am, my daughter got hungry and ate $$\\frac{1}{4}$$ of what was left. At 3am, my wife got hungry and ate $$\\frac{1}{3}$$ of what was left. At 4am, my dog got hungry and ate $$\\frac{1}{2}$$ of what was left. What percentage of the pie (in %) was left for me in the morning? ×", "# Michael Michael had a bar if chocolate. He ate 1/2 of it and gave away 1/3. What fraction had he left? Correct result: x = 0.1667 #### Solution: $x=1-\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}=0.1667$ We would be pleased if you find an error in the word problem or inaccuracies and send it to us. Thank you! Tips to related online calculators Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. ## Related math problems and questions: • Pizzas Billy ate 1 1/4 pizzas and John ate 1 2/3 pizzas. How much more pizza did John eat than Billy? • Cake fractions Thomas ate 1/3 of cake, Bohouš of the rest of the cake ate 2/5. What fraction of cake left over for others? • Two pizzas Jacobs mom bought two whole pizzas. He ate 2/10 of the pizza and his dad ate 1 1/5. How much is left. • Erica Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric, how much will she have left? • Martin Martin is making a model of a Native American canoe. He has 5 1/2 feet of wood. He uses 2 3/4 feet for the hull and 1 1/4 feet for a paddle. How much wood does he have left? Martin has feet of wood left. • Simplify", "total cost of \\\\$63 means 6 pies were ordered. In this unit, we will see many situations like this one, and will learn how to use diagrams and equations to answer questions about unknown amounts." ]

[ "# How do you simplify 3/2 div 2 using a model? Jun 17, 2018 See below #### Explanation: Suppose you had 3 pies (an Apple, a Blueberry, and a Cherry). At the end of the day, at Mom's Lunch Bar, half of each pie had been sold. $3$ half pies were left; that is $\\frac{1}{2}$ of an Apple pie, plus $\\frac{1}{2}$ of a Blueberry pie, plus $\\frac{1}{2}$ of a Cherry pie. The amount of pie left is $\\frac{3}{2}$ of a complete pie. Rather than through the remaining pie away, Mom and Pop divide what is left over between the two of them. Mom and Pop each get $\\textcolor{w h i t e}{\\text{xxx.}} \\frac{1}{2}$ of the $\\frac{1}{2}$ of an Apple pie, plus $\\frac{1}{2}$ of the $\\frac{1}{2}$ of a Blueberry pie, plus $\\frac{1}{2}$ of the $\\frac{1}{2}$ of a Cherry pie. Since $\\frac{1}{2}$ of $\\frac{1}{2}$ is $\\frac{1}{4}$ when the left-over $\\frac{3}{2}$ of a complete pie is divided between $2$ people, each gets $3$ times $\\frac{1}{4}$ (or $\\frac{3}{4}$) of a complete pie. That is $\\left(\\frac{3}{2}\\right) \\div 2 = \\frac{3}{4}$. This might have been easier to see with an image but I haven't found a reasonable paint program for quick images on a chromebook.", "Here is my take1 on the June 19th Huge Pie puzzle in the Guardian: A huge pie is divided among 100 guests. The first guest gets 1% of the pie. The second guest gets 2% of the remaining part. The third guest gets 3% of the rest, etc. The last guest gets 100% of the last part. Who gets the biggest piece?\" Tables are always good to get a quick overview of what we’ve got, the tricky part being the rest of the pie: Guest Percentage Rest of pie (before guest is served) 1 $$\\frac{1}{100}$$ $$\\frac{100}{100}$$ 2 $$\\frac{2}{100}$$ $$\\frac{100}{100}\\cdot\\left(1 - \\frac{1}{100}\\right) = \\frac{100}{100}\\cdot\\frac{99}{100}$$ 3 $$\\frac{3}{100}$$ $$\\frac{100}{100}\\cdot\\left(1 - \\frac{1}{100}\\right)\\cdot\\left(1 - \\frac{2}{100}\\right) = \\frac{100}{100}\\cdot\\frac{99}{100}\\cdot\\frac{98}{100}$$ $$\\vdots$$ $$\\vdots$$ $$\\vdots$$ 99 $$\\frac{99}{100}$$ $$\\frac{100}{100}\\cdot\\frac{99}{100}\\cdot\\ \\cdots\\ \\cdot\\frac{2}{100}$$ 100 $$\\frac{100}{100}$$ $$\\frac{100}{100}\\cdot\\frac{99}{100}\\cdot\\ \\cdots\\ \\cdot\\frac{1}{100}$$ If we number the guests $$i = 1...100$$ then each one of them gets a percentage $$p_i = \\frac{i}{100}$$ of the rest of the pie at the time she is being served, and we can write this rest as $$r_i = \\prod_{j=0}^{i-1} \\frac{100 - j}{100}$$. The actual size of the piece of pie that guest $$i$$ gets is $s_i = p_i\\cdot r_i = \\frac{i}{100}\\cdot\\prod_{j=0}^{i-1} \\frac{100 - j}{100}.$ Of course, for the very next guest $$i + 1$$ we have $s_{i+1} = p_{i+1}\\cdot r_{i+1} = \\frac{i+1}{100}\\cdot\\prod_{j=0}^{i} \\frac{100 - j}{100}.$ The difference in their pieces’", "# 6.3: The Key Fraction Rule We know that $$\\frac{a}{b}$$ is the answer to a division problem: $$\\frac{a}{b}$$represents the amount of pie an individual child receives when pies are shared equally by children. What happens if we double the number of pie and double the number of kids? Nothing! The amount of pie per child is still the same: $\\frac{2a}{2b} = \\frac{a}{b} \\ldotp \\nonumber$ For example, as the picture shows, $$\\frac{6}{3}$$ and $$\\frac{12}{6}$$ both give two pies for each child. And tripling the number of pies and the number of children also does not change the final amount of pies per child, nor does quadrupling each number, or one trillion-billion-tupling the numbers! $\\frac{6}{3} = \\frac{12}{6} = \\frac{18}{9} = \\ldots = \\text{two pies per child} \\ldotp \\nonumber$ This leads us to want to believe: ##### Key Fraction Rule $\\frac{xa}{xb} = \\frac{a}{b}, \\nonumber$ (at least for positive whole numbers ). We say that the fractions $$\\frac{xa}{xb}$$ and $$\\frac{a}{b}$$ are equivalent. ##### Example: Fractions equivalent to 3/5 For example, $\\frac{3}{5}\\; \\text{(sharing three pies among five kids)} \\nonumber$ yields the same result as $\\frac{3 \\cdot 2}{5 \\cdot 2} = \\frac{6}{10}\\; \\text{(sharing six pies among ten kids)} \\nonumber$ and as $\\frac{3 \\cdot 100}{5 \\cdot 100} = \\frac{300}{500}\\; \\text{(sharing 300 pies among 500 kids)} \\ldotp \\nonumber$ ##### Think / Pair / Share Write down", "} = \\frac {■ }{ 6}$$ Answer: Question 3: $$\\frac { 8 }{ 10 } – \\frac { 5 }{ 10 } = \\frac {■ }{ 10}$$ Answer: Subtract. Use models to help. Question 4: $$\\frac { 5 }{ 8 } – \\frac { 2 }{ 8 } = \\frac { }{ }$$ Answer: Question 5: $$\\frac { 7 }{ 12 } – \\frac { 2 }{ 12 } = \\frac { }{ }$$ Answer: Question 6: $$\\frac { 3 }{4 } – \\frac { 2 }{ 4 } = \\frac { }{ }$$ Answer: Question 7: $$\\frac { 2 }{ 3 } – \\frac { 1 }{ 3 } = \\frac { }{ }$$ Answer: Question 8: $$\\frac { 7 }{ 8 } – \\frac { 5 }{ 8 } = \\frac { }{ }$$ Answer: Question 9: Explain how you could find the unknown addend in $$\\frac { 2 }{ 6 }$$ + _____ = 1 without using a model. Answer: ### Add Fractions Using Models – Lesson Check – Page No 406 Question 10: Mrs. Ruiz served a pie for dessert two nights in a row. The drawings below show the pie after her family ate dessert on each night. What fraction of the pie did they eat on the second night? $$\\frac { }{", "# Math Help - simple question 1. ## simple question What is the sum of (2 to the n-1)/(3 to the n) where n goes from 1 to infinity? In other words what is the sum of 1/3 + 2/9 + 4/27 + 8/81 ... 2. Originally Posted by shadow1 What is the sum of (2 to the n-1)/(3 to the n) where n goes from 1 to infinity? In other words what is the sum of 1/3 + 2/9 + 4/27 + 8/81 ... Hello, use the sum formula for infinite geometric series: $\\sum_1^{\\infty}\\left( \\frac{2^{n-1}}{3^n} \\right) = \\sum_1^{\\infty}\\left( \\frac{2^{n-1}}{3^n} \\right) = \\sum_1^{\\infty}\\left( \\frac{1}{2} \\cdot \\left(\\frac{2}{3} \\right)^n \\right) = \\frac{\\frac{1}{3}}{1-\\frac{2}{3}} = 1$ 3. Thanks earboth. So what you are saying is, if I have a pie and cut it into three pieces and ate one of them, I would have two one-third pieces left. Then if I cut each of those one-third pieces into three pieces and ate one, I would have four one-ninth pieces. And if I continue cutting each piece into threes and eating one of them, I will eventually eat the entire pie. Right????? 4. Originally Posted by shadow1 Thanks earboth. So what you are saying is, if I have a pie and cut it into three pieces and ate one of them, I would", "Ticket Answer Key Solve the word problem using the RDW strategy. Show all of your work. Mr. Pham mowed $$\\frac{2}{7}$$ of his lawn. His son mowed $$\\frac{1}{4}$$ of it. Who mowed the most? How much of the lawn still needs to be mowed? Fraction Of Lawn mowed by Mr.Pham = $$\\frac{2}{7}$$ Fraction Of Lawn mowed by his Son = $$\\frac{1}{4}$$ lcm of 7 and 4 is 28 . $$\\frac{2}{7}$$ = $$\\frac{8}{28}$$ $$\\frac{1}{4}$$ = $$\\frac{7}{28}$$ $$\\frac{8}{28}$$ > $$\\frac{7}{28}$$ Mr. Pham mowed most than his son . Fraction of lawn still needed to mowed = x 1= $$\\frac{8}{28}$$ + $$\\frac{7}{28}$$ + x $$\\frac{28}{28}$$ = $$\\frac{15}{28}$$ + x x = $$\\frac{28}{28}$$ – $$\\frac{15}{28}$$ x = $$\\frac{13}{28}$$ Therefore, Fraction of lawn still needed to mowed = x = $$\\frac{13}{28}$$ . ### Eureka Math Grade 5 Module 3 Lesson 7 Homework Answer Key Solve the word problems using the RDW strategy. Show all of your work. Question 1. Christine baked a pumpkin pie. She ate $$\\frac{1}{6}$$ of the pie. Her brother ate $$\\frac{1}{6}$$ of it and gave the leftovers to his friends. What fraction of the pie did he give to his friends? Total pie = 1 . Fraction of pie ate by Christine = $$\\frac{1}{6}$$ Fraction of pie ate by her Brother =$$\\frac{1}{6}$$ Fraction of pie leftover to her friends = x", "# Solutions Try a COMPASS math test prep course with a teacher: 1.) $8-15/3+1=8-5+1=4$ 2.) $105-(-9)=105+9=114^{\\circ}\\text{F}$ 3.) $\\frac{1}{2}+\\frac{2}{3}-\\frac{3}{4}=\\frac{6}{12}+\\frac{8}{12}-\\frac{9}{12}=\\frac{5}{12}$ 4.) After the first day $\\frac{2}{3}\\;$ of the pie is leftover. On the second day $\\frac{3}{4}\\;$ of this is eaten, leaving $\\frac{1}{4}\\;$ of $\\frac{2}{3}\\;$ of the original pie, or: $(\\frac{2}{3})(\\frac{1}{4})=\\frac{1}{6}$. 5.) $\\frac{4}{5}\\div\\frac{2}{3}-\\frac{7}{10}=\\\\\\frac{4}{5}\\cdot\\frac{3}{2}-\\frac{7}{10}=\\\\\\frac{12}{10}-\\frac{7}{10}=\\\\\\frac{5}{10}=\\frac{1}{2}$ 6.) $4.6\\cdot4.75+82.63=\\104.48$ 7.) $1.18\\;\\text{billion}=1,180,000,000=1.18\\cdot10^{9}$ 8.) $\\sqrt{3^{4}}=\\sqrt{(3^{2})^{2}}=3^{2}=9$ 9.) $\\frac{x}{9}=\\frac{5}{15}\\\\\\rightarrow \\frac{x}{9}=\\frac{1}{3}\\\\\\rightarrow 3x=9\\\\\\rightarrow x=3$ 10.) $\\frac{x}{5}=\\frac{6.75}{9}\\\\\\rightarrow 9x=5(6.75)\\\\\\rightarrow x=\\3.75$ 11.) $0.54(22,000)=11,880\\;\\text{votes}$ 12.) If 60% are over 25, 40% are under 25. Since half of this 40% are under 21, $(0.5)(0.4)=0.2=20\\%\\;$ are between 21 and 25. 13.) The average age in the room will be the sum of the ages divided by four. Since the sum of ages divided by 10 is equal 14.8, the sum of the ages must be $10\\cdot14.8=148$, and therefore the average is $148/4=37$. 14.) $\\frac{15(66)+25(62)}{40}=63.5$ 15.) $\\frac{(4)^{3}+1}{(4)+1}=\\frac{65}{5}=13$ 16.) $\\frac{((5)-(1))^{2}-(-2)}{(-2)(5)-2(1)}=\\frac{4^{2}+2}{-10-2}=\\frac{18}{-12}=-\\frac{3}{2}$ 17.) $s=1.15(500-w)$ 18.) $\\frac{1^{2}+2^{2}+3^{2}+4^{2}+5x}{1+2+3+4+x}=3.75\\rightarrow\\\\30+5x=3.75(10+x)\\rightarrow\\\\30+5x=37.5+3.75x\\rightarrow\\\\1.25x=7.5\\rightarrow\\\\x=6$ 19.) $4a^{2}b^{2}-3a^{2}b+ab\\;+\\;(ab+a^{2}b+5ab^{2}-3a^{2}b^{2})=\\\\(4-3)a^{2}b^{2}+(-3+1)a^{2}b+5ab^{2}+(1+1)ab=\\\\a^{2}b^{2}-2a^{2}b+5ab^{2}+2ab$ 20.) $5x+1=2(x-4)\\\\\\rightarrow 5x+1=2x-8\\\\\\rightarrow 3x=-9\\\\\\rightarrow x=-3$ 21.) $x^{2}-6x+8=(x-4)(x-2)$. 22.) Begin by multiplying both sides of the equation by 3 to get: $x-\\frac{3}{2}=\\frac{3}{4}\\rightarrow\\\\x=\\frac{3}{4}+\\frac{3}{2}\\rightarrow\\\\x=\\frac{9}{4}$ 23.) $\\frac{-2x^{3}y^{2}z}{6x(yz)^{6}}= -\\frac{1}{3}x^{3-1}y^{2-6}z^{1-6}=-\\frac{1}{3}x^{2}y^{-4}z^{-5}=\\frac{-x^{2}}{3y^{4}z^{5}}$ 24.) $\\frac{x^{2}+6x+5}{x+1}=\\frac{(x+1)(x+5)}{x+1}=x+5$. 25.) $\\frac{-\\sqrt{x}}{2\\sqrt{x}-\\sqrt{y}}\\cdot\\frac{2\\sqrt{x}+\\sqrt{y}}{2\\sqrt{x}+\\sqrt{y}}=\\frac{-(2x+\\sqrt{xy})}{4x-y}=\\frac{2x+\\sqrt{xy}}{y-4x}$. 26.) $\\frac{16-x^{2}}{x^{2}+9x+20}=\\frac{-(x+4)(x-4)}{(x+4)(x+5)}=\\frac{4-x}{x+5}$. 27.) By converting the equation into slope intercept form: $y=-\\frac{3}{2}x-3\\;$, we can see that the y-intercept occurs at $(0,-3)$. 28.) Let $c\\;$ and $k\\;$ represent the number of years Carlos and Katie have worked at the company. From the problem we can derive", "take the help of an example to understand this. Suppose, you bought two pies from place A and two pies from place B. Each of these pies were divided into four equal pieces. You ate three slices from the second pie bought from place A. You are left with a total of five slices or in other words one whole pie and $$\\frac{1}{4}^{th}$$ of a whole pie. From place B you bought two pies and ate two slices from the second pie. You decided to bring the rest back home. You are left with six slices of pie or in other words, one pie along with $$\\frac{2}{4}^{th}$$ of the other pie. Now, to find the total number of pies, we need to add $$1\\frac{1}{4}$$ and $$1\\frac{2}{4}$$ $$1\\frac{1}{4}+1\\frac{2}{4}$$ $$(1+1)=\\left(1\\frac{1}{4}+1\\frac{2}{4}\\right)$$ [Separate the whole and fractional parts] $$2\\left(\\frac{3}{4}\\right)$$ [Add] After returning home, you remember that a total of $$2\\left(\\frac{3}{4}\\right)$$ slices of pie were taken in total from both places. However, you only remember that you took $$1\\left(\\frac{1}{4}\\right)$$ slices of pie from place-A. Now, you wish to find out how many slices were taken home from place-B. For this, you need to subtract $$2\\left(\\frac{3}{4}\\right)$$ from $$2\\left(\\frac{3}{4}\\right)$$. Method 2: The second way to go about addition and subtraction of mixed numbers is to express them as fractions and then solve as shown below", "of piece of pizza Chris ate + Fraction of piece of pizza Stacy ate + Fraction of piece of pizza Rylan ate + Fraction of piece of pizza Alec ate + Fraction of piece of pizza Kelli ate + 1 = $$\\frac{3}{8}$$ + $$\\frac{2}{8}$$ + $$\\frac{4}{8}$$ + $$\\frac{5}{8}$$ + $$\\frac{3}{8}$$ + $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = 3. Fraction of a pizza was left over = Number of pizza they ordered – $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = 3 – $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = $$\\frac{9}{8}$$", "2 Others): . $(_6C_1)(_6C_2) \\:=\\:6\\cdot15 \\:=\\:90$ 2 Cherries (and 1 Other): . $(_6C_2)(_6C_1) \\:=\\:16\\cdot6 \\:=\\:90$ 3 Cherries (and 0 Others): . $(_6C_3)(_6C_0) \\:=\\: 20\\cdot1 \\:=\\:20$ Hence, there are: . $90 + 90 + 20 \\:=\\:200$ ways to get at least one Cherry. Therefore: . $P(\\text{at least 1 Cherry}) \\;=\\;\\frac{200}{220} \\;=\\;\\frac{10}{11}\\quad\\hdots\\quad \\text{ta-}DAA!$", "parts. . . Hence, we want three fourths. $\\text{Hence, }\\frac{3}{4}\\text{ is read \"three-fourths\".}$ $\\text{Example: }\\:\\frac{3}{\\text{-4}}$ It says \"Cut the pie into -4 equal parts, and take 3 slices.\" And we find that we can't do that. We can cut a pie into two equal parts. We can even cut a pie into one equal part (although it sounds silly). We don't know how to cut a pie into zero equal parts. And we certainly don't know how to cut a pie into negative-four equal parts. [2] Common denominator? $\\text{Example: }\\:\\frac{5}{3} + \\frac{3}{\\text{-}4}$ The LCD seems to be -12. $\\text{We have: }\\:\\left(\\frac{\\text{-}4}{\\text{-}4}\\right)\\left(\\frac{5}{3}\\right) + \\left(\\frac{3}{3}\\right)\\left(\\frac{3}{\\text{-}4}\\right) \\;=\\;\\frac{\\text{-}20}{\\text{-}12} + \\frac{9}{\\text{-}12} \\;=\\; \\frac{\\text{-}11}{\\text{-}12} \\;=\\;\\frac{11}{12}$ This takes more work and includes some added risk. . . $\\text{I would prefer to face: }\\:\\frac{5}{3} - \\frac{3}{4}$ FYI $\\text{A fraction has }three\\text{ signs: }\\:+\\frac{+3}{+4} \\quad\\begin{Bmatrix}\\text{sign of the numerator} \\\\ \\text{sign of the denominator} \\\\ \\text{sign of the fraction} \\end{Bmatrix}$ We may change any two of the signs. . . $+\\frac{+3}{+4} \\;=\\;+\\frac{-3}{-4} \\;=\\;-\\frac{-3}{+4} \\;=\\;-\\frac{+3}{-4}$", "of slices of pie and pizza are the same, $$x$$ represents a whole number greater than or equal to 3. That means Marie gets $$\\frac{2}{x}$$ of the pizza and $$\\frac{3}{x}$$ of the pie. Since both fractions have the same denominator $$x$$, they are like fractions. Example 3: Pick out the like fractions from the following: $$\\frac{3}{8}$$, $$\\frac{5}{10}$$, $$\\frac{11}{15}$$, $$\\frac{8}{8}$$, $$\\frac{9}{15}$$, $$\\frac{13}{14}$$ Solution: There are two pairs of like fractions among the provided fractions. They are $$\\frac{5}{8}$$ and $$\\frac{8}{8}$$ and $$\\frac{11}{15}$$ and $$\\frac{9}{15}$$.", "\\cdot \\frac{12}{13} + \\frac{3}{4} \\cdot \\frac{5}{13}} = - \\frac{46}{9}$ So $cos(A + B) \\cdot tan(A - B) = -\\frac{3}{4} \\cdot -\\frac{46}{9} = \\frac{23}{6}$ ================================================== ======= $cos \\left ( \\frac{B}{2} \\right ) = \\pm \\sqrt{\\frac{1 + cos(B)}{2}}$ $cos \\left ( \\frac{B}{2} \\right ) = \\pm \\sqrt{\\frac{1 + \\frac{12}{13}}{2}} = \\pm \\sqrt{\\frac{25}{26}} = \\pm \\frac{5}{\\sqrt{26}}$ Since B is in the 1st quadrant, so is B/2. Thus $cos \\left ( \\frac{B}{2} \\right ) = \\frac{5 \\sqrt{26}}{26}$ -Dan • December 3rd 2006, 04:11 PM topsquark Quote: Originally Posted by slaying_ice Solve the equation in th interval [0,2pie) 2cos^2x-3cosx+1 By the way, the Greek letter $\\pi$ is spelled \"pi\" not \"pie.\" There's no equation to solve here! I'll assume you meant: $2cos^2(x) - 3cos(x) + 1 = 0$ Let's simplify a bit: Let $y = cos(x)$. The we have: $2y^2 - 3y + 1 = 0$ This factors: $(2y - 1)(y - 1) = 0$ So $y = \\frac{1}{2}$ or $y = 1$. Thus $cos(x) = \\frac{1}{2}$ ==> $x = \\frac{\\pi}{3}, \\frac{5 \\pi}{3}$ rad or $cos(x) = 1$ ==> $x = 0$. So the full solution set on $[0, 2 \\pi)$ is $x = 0, \\frac{\\pi}{3}, \\frac{5 \\pi}{3}$. -Dan • December 3rd 2006, 04:18 PM slaying_ice On the second problem you assumed correct on what the equation was. • December 3rd 2006, 04:20", "fraction strips to add $$\\frac{4}{10}$$ and $$\\frac{4}{5}$$. She uses one whole strip to represent the sum. How many fifths strips does she need to complete the sum? (A) 1 (B) 2 (C) 5 (D) 8 Explanation: one extra strip is needed to complete the sum Question 16. Which fraction correctly completes the equation? $$\\frac{6}{8}$$ + $$\\frac{}{}$$ = 1 (A) $$\\frac{1}{2}$$ (B) $$\\frac{1}{8}$$ (C) $$\\frac{3}{8}$$ (D) $$\\frac{1}{4}$$ Explanation: $$\\frac{6}{8}$$ + $$\\frac{1}{4}$$ = $$\\frac{6}{8}$$ + $$\\frac{2}{8}$$ = $$\\frac{8}{8}$$ = 1 Question 17. An apple was cut into 8 equal-size pieces. Stacy ate $$\\frac{1}{4}$$ of the apple. Tony ate $$\\frac{3}{8}$$ of the apple. What part of the apple did Stacy and Tony eat in all? (A) $$\\frac{1}{2}$$ (B) $$\\frac{5}{8}$$ (C) $$\\frac{3}{4}$$ (D) $$\\frac{1}{4}$$ Explanation: Equal the all denominators: Stacy ate $$\\frac{1}{4}$$ of the apple = $$\\frac{1}{4}$$ = $$\\frac{2}{8}$$ Tony ate $$\\frac{3}{8}$$ of the apple = $$\\frac{3}{8}$$ Stacy and Tony ate = $$\\frac{3}{8}$$ + $$\\frac{2}{8}$$= $$\\frac{5}{8}$$ Question 18. Multi-Step Last weekend, Beatrice walked her poodle $$\\frac{2}{3}$$ mile on Saturday and $$\\frac{5}{6}$$ mile on Sunday. Fiona walked her beagle $$\\frac{1}{3}$$ mile on Saturday and $$\\frac{1}{2}$$ mile on Sunday. How much farther did the poodle walk last weekend than the beagle? (A) $$\\frac{1}{2}$$ mile (B) 1$$\\frac{1}{3}$$ miles (C) $$\\frac{2}{3}$$ mile (D) 1$$\\frac{1}{2}$$ miles Explanation: Last weekend, Beatrice walked her poodle $$\\frac{2}{3}$$ mile on Saturday", "of a pie, this means that each student gets $\\frac{3}{8}$ of an apple pie. 3. If 3 pies are divided into 8 equal portions, then 8 of these portions makes 3 pies, a fact that is clearly illustrated in part (a). We can write this in symbols if we use a question mark to represent the amount of pie in one portion: $$8 \\times ? = 3$$ When we know a factor and the product, we can find the other factor by dividing: $$3 \\div 8 = ?$$ So one person's portion is whatever we get when we divide 3 by 8. In part (b), we saw that one portion is $\\frac38$. So that means that $$3 \\div 8 = \\frac38.$$", "4: $$\\frac { 5 }{ 8 } – \\frac { 2 }{ 8 } = \\frac { }{ }$$ Answer: Question 5: $$\\frac { 7 }{ 12 } – \\frac { 2 }{ 12 } = \\frac { }{ }$$ Answer: Question 6: $$\\frac { 3 }{4 } – \\frac { 2 }{ 4 } = \\frac { }{ }$$ Answer: Question 7: $$\\frac { 2 }{ 3 } – \\frac { 1 }{ 3 } = \\frac { }{ }$$ Answer: Question 8: $$\\frac { 7 }{ 8 } – \\frac { 5 }{ 8 } = \\frac { }{ }$$ Answer: Question 9: Explain how you could find the unknown addend in $$\\frac { 2 }{ 6 }$$ + _____ = 1 without using a model. Answer: ### Add Fractions Using Models – Lesson Check – Page No 406 Question 10: Mrs. Ruiz served a pie for dessert two nights in a row. The drawings below show the pie after her family ate dessert on each night. What fraction of the pie did they eat on the second night? $$\\frac { }{ }$$ a. What do you need to know? b. How can you find the number of pieces eaten on the second night? c. Explain the steps you used to solve the problem. Complete the", "took 4 times half of a pizza. It means we took 2 whole pizzas. $$\\frac{a}{a}=1$$ $$\\require{cancel} \\frac{a \\cdot b}{a \\cdot c} = \\frac{\\cancel{a} \\cdot b}{\\cancel{a} \\cdot c} = \\frac{b}{c}$$ $$\\frac{a}{b} \\cdot c = \\frac{a \\cdot c}{b} = \\frac{c}{b} \\cdot a$$ $$\\frac{-a}{b}=-\\frac{a}{b}$$ $$\\frac{1}{\\frac{b}{c}}=\\frac{c}{b}$$ $$\\frac{a}{1}=a$$ $$\\frac{-a}{-b}=\\frac{a}{b}$$ $$\\frac{a}{-b}=-\\frac{a}{b}$$ $$\\frac{a}{\\frac{b}{c}}=\\frac{a\\cdot c}{b}$$ $$\\frac{\\frac{b}{c}}{a}=\\frac{b}{c\\cdot a}$$ $$\\frac{0}{a}=0\\:,\\:a\\ne 0$$", "So there are columns in the rectangle. • The side segment was divided into equal-sized pieces. So there are rows in the rectangle. • A rectangle with columns and rows has $$b \\cdot d$$ pieces. (The area model for whole-number multiplication!) Think / Pair / Share Stick with the general multiplication rule $\\frac{a}{b} \\cdot \\frac{c}{d} = \\frac{a \\cdot c}{b \\cdot d} \\ldotp$ Write a clear explanation for why $$a \\cdot c$$ of the small rectangles will be shaded. Multiplying Fractions by Whole Numbers Often, elementary students are taught to multiply fractions by whole numbers using the fraction rule. Example: Multiply Fractions For example, to multiply $$2 \\cdot \\frac{3}{7}$$, we think of “2” as $$\\frac{2}{1}$$, and compute this way$$2 \\cdot \\frac{3}{7} = \\frac{2}{1} \\cdot \\frac{3}{7} = \\frac{2 \\cdot 3}{1 \\cdot 7} = \\frac{6}{7} \\ldotp\\] We can also think in terms of our original “Pies Per Child” model to answer questions like this. Example: Pies per Child We know that $$\\frac{3}{7}$$ means the amount of pie each child gets when 7 children evenly share 3 pies. If we compute $$2 \\cdot \\frac{3}{7}$$ that means we double the amount of pie each kid gets. We can do this by doubling the number of pies. So the answer is the same as $$\\frac{6}{7}$$: the amount of pie each child gets when 7", "of the denominators 16 and 8. The LCM of 16 and 8 is 16. $$\\frac{5}{16}$$ = $$\\frac{5 × 1}{16 × 1}$$ = $$\\frac{5}{16}$$ $$\\frac{4}{8}$$ = $$\\frac{4 × 2}{8 × 2}$$ = $$\\frac{8}{16}$$ Therefore, we get the like fractions $$\\frac{5}{16}$$ and $$\\frac{8}{16}$$. Now, $$\\frac{5}{16}$$ + $$\\frac{8}{16}$$ = $$\\frac{5 + 8}{16}$$ = $$\\frac{13}{16}$$ Therefore, Michael read in two days $$\\frac{13}{16}$$ of the book. 2. Sarah ate $$\\frac{1}{3}$$ part of the pizza and her sister ate $$\\frac{1}{2}$$ of the pizza. What fraction of the pizza was eaten by both sisters? Solution: Sarah ate $$\\frac{1}{3}$$ part of the pizza. Her sister ate $$\\frac{1}{2}$$ of the pizza. $$\\frac{1}{3}$$ + $$\\frac{1}{2}$$ Let us find the LCM of the denominators 3 and 2. The LCM of 3 and 2 is 6. $$\\frac{1}{3}$$ = $$\\frac{1 × 2}{3 × 2}$$ = $$\\frac{2}{6}$$ $$\\frac{1}{2}$$ = $$\\frac{1 × 3}{2 × 3}$$ = $$\\frac{3}{6}$$ Therefore, we get the like fractions $$\\frac{2}{6}$$ and $$\\frac{3}{6}$$. Now, $$\\frac{2}{6}$$ + $$\\frac{3}{6}$$ = $$\\frac{2 + 3}{6}$$ = $$\\frac{5}{6}$$ Therefore, $$\\frac{5}{6}$$ of the pizza was eaten by both sisters. 3. Catherine is preparing for her final exam. She study $$\\frac{9}{22}$$ hours on Wednesday and $$\\frac{5}{11}$$ hours on Sunday. How many hours she studied in two days? Solution: Catherine study $$\\frac{9}{22}$$ hours on Wednesday. Again, she study $$\\frac{5}{11}$$ hours on Sunday. $$\\frac{9}{22}$$ + $$\\frac{5}{11}$$ Let us find", "# Maya ate 0.4 of her 8 inch pie. Nicole ate 0.5 of her 8 inch pie. How much more did Nicole eat than Maya? Oct 31, 2017 Nicole ate $\\frac{5}{4}$ times more times than Mya did. #### Explanation: The problem says the Mya ate 0.4 of her 8 inch apple pie. Well 0.4 is equal to $\\frac{40}{100} \\mathmr{and} \\frac{4}{10} \\mathmr{and} \\frac{2}{5.}$ So this means the Mya ate $\\frac{2}{5}$ of her 8 inch apple pie. Since there is an \"of\" after 0.4, this means we need to multiply the fraction with 8. Because \"of\", means to multiply, $\\frac{2}{5} \\cdot 8$ Because 8 is equal to 8/1, the problem would become: $\\frac{2}{5} \\cdot \\frac{8}{1} = \\frac{16}{5}$ So for Mya , she ate $\\frac{16}{5}$ inches of pie. Same with Nicole, she ate 0.5 of her 8 inch pie. Meaning she ate $\\frac{50}{100} \\mathmr{and} \\frac{5}{10} \\mathmr{and} \\frac{1}{2.}$ Just like before, you would do $\\frac{1}{2} \\cdot \\frac{8}{1.}$ $\\frac{1}{2} \\cdot \\frac{8}{1} = \\frac{8}{2} = 4$ From here, we need to figure out how many more times Nicole ate pie than Mya. So we would do : $\\frac{4}{1} \\div \\frac{16}{5}$ Because we are dividing fractions, we need to use the keep, switch, and flip method. To do that, we keep the first number(4), switch the operation to multiplication, and then flip the last number" ]

[ "+0 # help pls 0 95 1 At the end of the day, a bakery had $$\\frac{6}{7}$$ of a pie left over. Louie, Duey and Huey split the pie, with each taking home the same amount of leftover pie. How much pie did Louie take home? Dec 25, 2020", "each pays an amount proportional to what he has eaten so far? Jack takes $\\frac{1}{3}$ of the pizza; there is $\\frac{2}{3}$ left. Jill takes $\\frac{1}{3}$ of that: . $\\frac{1}{3}\\cdot\\frac{2}{3} \\:=\\:\\frac{2}{9}$ . . She takes $\\frac{2}{9}$ of the pizza. .There is: . $\\frac{2}{3} - \\frac{2}{9} \\:=\\:\\frac{4}{9}$ of the pizza left. Tom takes $\\frac{1}{2}$ of that: . $\\frac{1}{2}\\cdot\\frac{4}{9}\\:=\\:\\frac{2}{9}$ . . He takes $\\frac{2}{9}$ of the pizza. .There is: . $\\frac{4}{9} - \\frac{2}{9}$ of the pizza left. Their shares are in the ratio: . $\\frac{1}{3}:\\frac{2}{9}:\\frac{2}{9} \\;=\\;\\frac{3}{9}:\\frac{2}{9}:\\frac{2}{9} \\;=\\;3:2:2$ So far they have eaten: . $\\frac{3}{7}:\\frac{2}{7}:\\frac{2}{7}$ of the consumed pizza, respectively. So Jack pays: . $\\frac{3}{7}\\times12.75 \\:=\\:5.4642... \\:\\approx\\:\\5.46$ And Jill and Tom each pay: . $\\frac{2}{7}\\times12.75 \\:=\\:3.6428... \\:\\approx\\:\\3.64$ (b) Can they split the remaining pizza so that everyone ends up having eaten the same total amount? If so, how? Each must have eaten $\\frac{1}{3}$ of the pizza. Jill and Tom will split the remaining $\\frac{2}{9}$ of the pizza. . . Then both will have: . $\\frac{2}{9} + \\frac{1}{9} \\:=\\:\\frac{3}{9} \\:=\\:\\frac{1}{3}$ of the pizza. 4. 2) A large pizza costs \\$12.75, three friends decide to split one. Jack takes a third, and then Jill takes a third of the remainder and then Tom takes half of what's left. (a) How should they split the total cost so that each pays an amount proportional to", "# How do you simplify 3/2 div 2 using a model? Jun 17, 2018 See below #### Explanation: Suppose you had 3 pies (an Apple, a Blueberry, and a Cherry). At the end of the day, at Mom's Lunch Bar, half of each pie had been sold. $3$ half pies were left; that is $\\frac{1}{2}$ of an Apple pie, plus $\\frac{1}{2}$ of a Blueberry pie, plus $\\frac{1}{2}$ of a Cherry pie. The amount of pie left is $\\frac{3}{2}$ of a complete pie. Rather than through the remaining pie away, Mom and Pop divide what is left over between the two of them. Mom and Pop each get $\\textcolor{w h i t e}{\\text{xxx.}} \\frac{1}{2}$ of the $\\frac{1}{2}$ of an Apple pie, plus $\\frac{1}{2}$ of the $\\frac{1}{2}$ of a Blueberry pie, plus $\\frac{1}{2}$ of the $\\frac{1}{2}$ of a Cherry pie. Since $\\frac{1}{2}$ of $\\frac{1}{2}$ is $\\frac{1}{4}$ when the left-over $\\frac{3}{2}$ of a complete pie is divided between $2$ people, each gets $3$ times $\\frac{1}{4}$ (or $\\frac{3}{4}$) of a complete pie. That is $\\left(\\frac{3}{2}\\right) \\div 2 = \\frac{3}{4}$. This might have been easier to see with an image but I haven't found a reasonable paint program for quick images on a chromebook.", "#10 If Jeremy and Sam each eat one fifth of a pizza, and Betty eats one fifteenth, how much pizza is left? The answer is $$\\frac{8}{15}$$ of a pizza. Solution Find the total amount of pizza consumed, and then determine the remaining amount. Find the amount of pizza consumed: $$\\text{Amount}_\\text{Jeremy} + \\text{amount}_\\text{Sam} + \\text{Amount}_\\text{Betty}$$: $$\\frac{1}{5} + \\frac{1}{5} + \\frac{1}{15} \\rightarrow \\text{Find the sum of the 1st two fractions} \\rightarrow \\frac{2}{5} + \\frac{1}{15}$$ $$\\frac{2}{5} + \\frac{1}{15} = \\left( \\frac{2 \\cdot 3}{5 \\cdot 3} + \\frac{1}{15}\\right) = \\left(\\frac{6 +1}{15}\\right) = \\frac{7}{15}$$ Thus, $$\\frac{7}{15}$$ of the pizza has been eaten. Determine the remaining amount of pizza One whole pizza - $$\\frac{7}{15}$$ = ? $$1 - \\frac{78}{15} = \\left( \\frac{1 \\cdot 15}{1 \\cdot 15} - \\frac{7}{15}\\right) = \\left( \\frac{15-7}{15} \\right) = \\frac{8}{15}$$ Check your answer: Verify by making sure that $$\\frac{7}{15} + \\frac{8}{15}$$ equals one, which represents the whole pizza. Therefore, the remaining amount is $$\\frac{8}{15}$$ths of a pizza", "# Math Help - simple question 1. ## simple question What is the sum of (2 to the n-1)/(3 to the n) where n goes from 1 to infinity? In other words what is the sum of 1/3 + 2/9 + 4/27 + 8/81 ... 2. Originally Posted by shadow1 What is the sum of (2 to the n-1)/(3 to the n) where n goes from 1 to infinity? In other words what is the sum of 1/3 + 2/9 + 4/27 + 8/81 ... Hello, use the sum formula for infinite geometric series: $\\sum_1^{\\infty}\\left( \\frac{2^{n-1}}{3^n} \\right) = \\sum_1^{\\infty}\\left( \\frac{2^{n-1}}{3^n} \\right) = \\sum_1^{\\infty}\\left( \\frac{1}{2} \\cdot \\left(\\frac{2}{3} \\right)^n \\right) = \\frac{\\frac{1}{3}}{1-\\frac{2}{3}} = 1$ 3. Thanks earboth. So what you are saying is, if I have a pie and cut it into three pieces and ate one of them, I would have two one-third pieces left. Then if I cut each of those one-third pieces into three pieces and ate one, I would have four one-ninth pieces. And if I continue cutting each piece into threes and eating one of them, I will eventually eat the entire pie. Right????? 4. Originally Posted by shadow1 Thanks earboth. So what you are saying is, if I have a pie and cut it into three pieces and ate one of them, I would", "a pie was left over? Explain how you decided. I can … use representations to find fraction names for whole numbers. Look Back! Jamie cuts another pie into smaller pieces. Each piece of pie is $$\\frac{1}{8}$$ of the whole. Jamie gives away 8 pieces. Does Jamie have any pie left over? Explain how you know. Answer: Jamie does not have left any pie. Explanation: In the above-given question, given that, Jamie cuts another pie into smaller pieces. Each piece of pie is $$\\frac{1}{8}$$ of the whole. Jamie gives away 8 pieces. 8 – 8 = 0. so Jamie does not have left any pie. Essential Question How Can You Use Fraction Names to Represent Whole Numbers? Visual Learning Bridge What are some equivalent fraction names for 1, 2, and 3? You can write a whole number as a fraction by writing the whole number as the numerator and The number line shows 3 wholes. Each whole is divided into 1 equal part. 1 whole divided into 1 equal part can be written as $$\\frac{1}{1}$$. 2 wholes each divided into 1 equal part can be written as $$\\frac{2}{1}$$. 3 wholes each divided into 1 equal part can be written as $$\\frac{1}{1}$$ 1 = $$\\frac{1}{2}$$ 2 = $$\\frac{2}{1}$$ 3 = $$\\frac{3}{1}$$ You can find other equivalent fraction names for whole", "# Solutions Try a COMPASS math test prep course with a teacher: 1.) $8-15/3+1=8-5+1=4$ 2.) $105-(-9)=105+9=114^{\\circ}\\text{F}$ 3.) $\\frac{1}{2}+\\frac{2}{3}-\\frac{3}{4}=\\frac{6}{12}+\\frac{8}{12}-\\frac{9}{12}=\\frac{5}{12}$ 4.) After the first day $\\frac{2}{3}\\;$ of the pie is leftover. On the second day $\\frac{3}{4}\\;$ of this is eaten, leaving $\\frac{1}{4}\\;$ of $\\frac{2}{3}\\;$ of the original pie, or: $(\\frac{2}{3})(\\frac{1}{4})=\\frac{1}{6}$. 5.) $\\frac{4}{5}\\div\\frac{2}{3}-\\frac{7}{10}=\\\\\\frac{4}{5}\\cdot\\frac{3}{2}-\\frac{7}{10}=\\\\\\frac{12}{10}-\\frac{7}{10}=\\\\\\frac{5}{10}=\\frac{1}{2}$ 6.) $4.6\\cdot4.75+82.63=\\104.48$ 7.) $1.18\\;\\text{billion}=1,180,000,000=1.18\\cdot10^{9}$ 8.) $\\sqrt{3^{4}}=\\sqrt{(3^{2})^{2}}=3^{2}=9$ 9.) $\\frac{x}{9}=\\frac{5}{15}\\\\\\rightarrow \\frac{x}{9}=\\frac{1}{3}\\\\\\rightarrow 3x=9\\\\\\rightarrow x=3$ 10.) $\\frac{x}{5}=\\frac{6.75}{9}\\\\\\rightarrow 9x=5(6.75)\\\\\\rightarrow x=\\3.75$ 11.) $0.54(22,000)=11,880\\;\\text{votes}$ 12.) If 60% are over 25, 40% are under 25. Since half of this 40% are under 21, $(0.5)(0.4)=0.2=20\\%\\;$ are between 21 and 25. 13.) The average age in the room will be the sum of the ages divided by four. Since the sum of ages divided by 10 is equal 14.8, the sum of the ages must be $10\\cdot14.8=148$, and therefore the average is $148/4=37$. 14.) $\\frac{15(66)+25(62)}{40}=63.5$ 15.) $\\frac{(4)^{3}+1}{(4)+1}=\\frac{65}{5}=13$ 16.) $\\frac{((5)-(1))^{2}-(-2)}{(-2)(5)-2(1)}=\\frac{4^{2}+2}{-10-2}=\\frac{18}{-12}=-\\frac{3}{2}$ 17.) $s=1.15(500-w)$ 18.) $\\frac{1^{2}+2^{2}+3^{2}+4^{2}+5x}{1+2+3+4+x}=3.75\\rightarrow\\\\30+5x=3.75(10+x)\\rightarrow\\\\30+5x=37.5+3.75x\\rightarrow\\\\1.25x=7.5\\rightarrow\\\\x=6$ 19.) $4a^{2}b^{2}-3a^{2}b+ab\\;+\\;(ab+a^{2}b+5ab^{2}-3a^{2}b^{2})=\\\\(4-3)a^{2}b^{2}+(-3+1)a^{2}b+5ab^{2}+(1+1)ab=\\\\a^{2}b^{2}-2a^{2}b+5ab^{2}+2ab$ 20.) $5x+1=2(x-4)\\\\\\rightarrow 5x+1=2x-8\\\\\\rightarrow 3x=-9\\\\\\rightarrow x=-3$ 21.) $x^{2}-6x+8=(x-4)(x-2)$. 22.) Begin by multiplying both sides of the equation by 3 to get: $x-\\frac{3}{2}=\\frac{3}{4}\\rightarrow\\\\x=\\frac{3}{4}+\\frac{3}{2}\\rightarrow\\\\x=\\frac{9}{4}$ 23.) $\\frac{-2x^{3}y^{2}z}{6x(yz)^{6}}= -\\frac{1}{3}x^{3-1}y^{2-6}z^{1-6}=-\\frac{1}{3}x^{2}y^{-4}z^{-5}=\\frac{-x^{2}}{3y^{4}z^{5}}$ 24.) $\\frac{x^{2}+6x+5}{x+1}=\\frac{(x+1)(x+5)}{x+1}=x+5$. 25.) $\\frac{-\\sqrt{x}}{2\\sqrt{x}-\\sqrt{y}}\\cdot\\frac{2\\sqrt{x}+\\sqrt{y}}{2\\sqrt{x}+\\sqrt{y}}=\\frac{-(2x+\\sqrt{xy})}{4x-y}=\\frac{2x+\\sqrt{xy}}{y-4x}$. 26.) $\\frac{16-x^{2}}{x^{2}+9x+20}=\\frac{-(x+4)(x-4)}{(x+4)(x+5)}=\\frac{4-x}{x+5}$. 27.) By converting the equation into slope intercept form: $y=-\\frac{3}{2}x-3\\;$, we can see that the y-intercept occurs at $(0,-3)$. 28.) Let $c\\;$ and $k\\;$ represent the number of years Carlos and Katie have worked at the company. From the problem we can derive", "Stewart's Sequences 1 of 5 A pie factory is booming in business. On the first day of opening they baked only $$2\\frac { 1 }{ 3 }$$ pies, On Day 2 they baked $$4\\frac { 2 }{ 3 }$$ pies! On Day 3 they made $$7$$ pies! $$...$$ On Day $$n$$ they made $$x$$ pies! Assuming that nothing restricts their pie baking skills, this sequence continues each day and that they can bake up to $$\\infty$$ pies per day, find the sum of $$x$$ when $$n=30$$ and $$n=60$$. ×", "pieces and we eat all 6 pieces, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. So $\\frac{6}{6}=1.$ This leads us to the property of one that tells us that any number, except zero, divided by itself is 1. ## Property of one $\\begin{array}{cccc}\\frac{a}{a}=1\\hfill & & & \\left(a\\ne 0\\right)\\hfill \\end{array}$ Any number, except zero, divided by itself is one. Doing the Manipulative Mathematics activity “Fractions Equivalent to One” will help you develop a better understanding of fractions that are equivalent to one. If a pie was cut in $6$ pieces and we ate all 6, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. If the pie was cut into 8 pieces and we ate all 8, we ate $\\frac{8}{8}$ pieces, or one whole pie. We ate the same amount—one whole pie. The fractions $\\frac{6}{6}$ and $\\frac{8}{8}$ have the same value, 1, and so they are called equivalent fractions. Equivalent fractions are fractions that have the same value. Let’s think of pizzas this time. [link] shows two images: a single pizza on the left, cut into two equal pieces, and a second pizza of the same size, cut into eight pieces on the right. This is a way to show that $\\frac{1}{2}$ is equivalent to $\\frac{4}{8}.$ In other words, they are equivalent fractions .", "pieces and we eat all 6 pieces, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. So $\\frac{6}{6}=1.$ This leads us to the property of one that tells us that any number, except zero, divided by itself is 1. ## Property of one $\\begin{array}{cccc}\\frac{a}{a}=1\\hfill & & & \\left(a\\ne 0\\right)\\hfill \\end{array}$ Any number, except zero, divided by itself is one. Doing the Manipulative Mathematics activity “Fractions Equivalent to One” will help you develop a better understanding of fractions that are equivalent to one. If a pie was cut in $6$ pieces and we ate all 6, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. If the pie was cut into 8 pieces and we ate all 8, we ate $\\frac{8}{8}$ pieces, or one whole pie. We ate the same amount—one whole pie. The fractions $\\frac{6}{6}$ and $\\frac{8}{8}$ have the same value, 1, and so they are called equivalent fractions. Equivalent fractions are fractions that have the same value. Let’s think of pizzas this time. [link] shows two images: a single pizza on the left, cut into two equal pieces, and a second pizza of the same size, cut into eight pieces on the right. This is a way to show that $\\frac{1}{2}$ is equivalent to $\\frac{4}{8}.$ In other words, they are equivalent fractions .", "pieces and we eat all 6 pieces, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. So $\\frac{6}{6}=1.$ This leads us to the property of one that tells us that any number, except zero, divided by itself is 1. ## Property of one $\\begin{array}{cccc}\\frac{a}{a}=1\\hfill & & & \\left(a\\ne 0\\right)\\hfill \\end{array}$ Any number, except zero, divided by itself is one. Doing the Manipulative Mathematics activity “Fractions Equivalent to One” will help you develop a better understanding of fractions that are equivalent to one. If a pie was cut in $6$ pieces and we ate all 6, we ate $\\frac{6}{6}$ pieces, or, in other words, one whole pie. If the pie was cut into 8 pieces and we ate all 8, we ate $\\frac{8}{8}$ pieces, or one whole pie. We ate the same amount—one whole pie. The fractions $\\frac{6}{6}$ and $\\frac{8}{8}$ have the same value, 1, and so they are called equivalent fractions. Equivalent fractions are fractions that have the same value. Let’s think of pizzas this time. [link] shows two images: a single pizza on the left, cut into two equal pieces, and a second pizza of the same size, cut into eight pieces on the right. This is a way to show that $\\frac{1}{2}$ is equivalent to $\\frac{4}{8}.$ In other words, they are equivalent fractions .", "5, 10, 15, 20, 25, 30, 35… So, the LCM of 3 and 5 is 15. Now, we need to multiply the numerator and denominator of the two fractions such that the denominators become 15. $$\\frac{1}{3}~=~\\frac{1~\\times~5}{3~\\times~5}~=~\\frac{5}{15}$$ Similarly, $$\\frac{2}{5}~=~\\frac{2~\\times~3}{5~\\times~3}~=~\\frac{6}{15}$$ We know that $$\\frac{5}{15}$$ and $$\\frac{6}{15}$$ are similar fractions. We can add them directly to get $$\\frac{11}{15}$$ as the sum. ## Solved Like & Unlike Fractions Examples Example 1: Determine whether $$\\frac{12}{25}$$ and $$\\frac{11}{25}$$ are like fractions. If yes, find the sum of the two fractions. Solution: Like fractions are fractions that have the same denominator. That means $$\\frac{12}{25}$$ and $$\\frac{11}{25}$$ are like fractions as they have the same denominator, 25. So, the sum of the two fractions is $$\\frac{12}{25}~+~\\frac{11}{25}~=~\\frac{23}{25}$$. Example 2: Sam gave 2 slices of a pizza and 3 slices of pie to his friend Marie. If the pizza and the pie had the same number of slices, determine whether the fractions that represent the slices of pizza and pie given to Marie are like fractions or not. Solution: Sam gave 2 slices of a pizza and 3 slices of a pie to Marie, so the minimum number of slices that the pie or pizza can be split into is 3. But the number of slices is unknown. Let the number of slices be $$x$$. Since the number", "frankhester87 16 # you have three fifths of a pie you divide the remaining pie into 5 equal slices what fraction of the original Post in each slice So you divide $\\frac{3}{5}$ by 5: $\\frac{3}{5}$ ÷ $\\frac{5}{1}$ which is: $\\frac{3}{5}$ × $\\frac{1}{5}$ = $\\frac{3}{25}$ So your answer is: $\\frac{3}{25}$", "7 kids share 3 pies} \\\\ \\hline ????? \\qquad \\qquad \\qquad \\qquad & \\end{split} \\nonumber$ Since in both cases we have 7 kids sharing the pies, we can imagine that it is the same 7 kids in both cases. First, they share 2 pies. Then they share 3 more pies. The total each child gets by the time all the pie-sharing is done is the same as if the 7 kids had just shared 5 pies to begin with. That is: $\\begin{split} \\text{amount of pie each kids gets when 7 kids share 2 pies} & \\\\ +\\; \\text{amount of pie each kids gets when 7 kids share 3 pies} \\\\ \\hline \\text{amount of pie each kids gets when 7 kids share 5 pies} \\ldotp & \\end{split} \\nonumber$ $\\frac{2}{7} + \\frac{3}{7} = \\frac{5}{7} \\ldotp \\nonumber$ Now let us think about the general case. Our claim is that $\\frac{a}{d} + \\frac{b}{d} = \\frac{a + b}{d} \\ldotp \\nonumber$ Translating into our model, we have kids. First, they share pies between them, and $$\\frac{a}{d}$$ represents the amount each child gets. Then they share more pies, so the additional amount of pie each child gets is $$\\frac{b}{d}$$. The total each kid gets is $$\\frac{a}{d} + \\frac{b}{d}$$. But it does not really matter that the kids first share pies and then share pies. The", "of piece of pizza Chris ate + Fraction of piece of pizza Stacy ate + Fraction of piece of pizza Rylan ate + Fraction of piece of pizza Alec ate + Fraction of piece of pizza Kelli ate + 1 = $$\\frac{3}{8}$$ + $$\\frac{2}{8}$$ + $$\\frac{4}{8}$$ + $$\\frac{5}{8}$$ + $$\\frac{3}{8}$$ + $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = 3. Fraction of a pizza was left over = Number of pizza they ordered – $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = 3 – $$\\frac{8}{8}$$ + $$\\frac{1}{8}$$ = $$\\frac{9}{8}$$", "Ticket Answer Key Solve the word problem using the RDW strategy. Show all of your work. Mr. Pham mowed $$\\frac{2}{7}$$ of his lawn. His son mowed $$\\frac{1}{4}$$ of it. Who mowed the most? How much of the lawn still needs to be mowed? Fraction Of Lawn mowed by Mr.Pham = $$\\frac{2}{7}$$ Fraction Of Lawn mowed by his Son = $$\\frac{1}{4}$$ lcm of 7 and 4 is 28 . $$\\frac{2}{7}$$ = $$\\frac{8}{28}$$ $$\\frac{1}{4}$$ = $$\\frac{7}{28}$$ $$\\frac{8}{28}$$ > $$\\frac{7}{28}$$ Mr. Pham mowed most than his son . Fraction of lawn still needed to mowed = x 1= $$\\frac{8}{28}$$ + $$\\frac{7}{28}$$ + x $$\\frac{28}{28}$$ = $$\\frac{15}{28}$$ + x x = $$\\frac{28}{28}$$ – $$\\frac{15}{28}$$ x = $$\\frac{13}{28}$$ Therefore, Fraction of lawn still needed to mowed = x = $$\\frac{13}{28}$$ . ### Eureka Math Grade 5 Module 3 Lesson 7 Homework Answer Key Solve the word problems using the RDW strategy. Show all of your work. Question 1. Christine baked a pumpkin pie. She ate $$\\frac{1}{6}$$ of the pie. Her brother ate $$\\frac{1}{6}$$ of it and gave the leftovers to his friends. What fraction of the pie did he give to his friends? Total pie = 1 . Fraction of pie ate by Christine = $$\\frac{1}{6}$$ Fraction of pie ate by her Brother =$$\\frac{1}{6}$$ Fraction of pie leftover to her friends = x", "Update all PDFs # How Much Pie? Alignments to Content Standards: 5.NF.B.3 After a class potluck, Emily has three equally sized apple pies left and she wants to divide them into eight equal portions to give to eight students who want to take some pie home. 1. Draw a picture showing how Emily might divide the pies into eight equal portions. Explain how your picture shows eight equal portions. 2. What fraction of a pie will each of the eight students get? 3. Explain how the answer to (b) is related the division problem $3 \\div 8$. ## IM Commentary The purpose of this task is to help students see the connection between $a \\div b$ and $\\frac{a}{b}$ in a particular concrete example. The relationship between the division problem $3 \\div 8$ and the fraction $\\frac{3}{8}$ is actually very subtle. $3 \\div 8$ is the number you multiply 8 by to get 3. $\\frac{3}{8}$ is the number you get by taking 3 copies of the unit fraction $\\frac18$. So $3 \\div 8$ is defined in terms of multiplication, and $\\frac{3}{8}$ is defined in terms of unit fractions. It is not obvious that these two numbers are the same, so students need opportunities to see that they will necessarily always be the same. Note that if $b$ people share", "### Still have math questions? (1)) Nori had $$2 \\frac { 1 } { 12 }$$ bags of apples. He used $$1 \\frac { 5 } { 12 }$$ bags of apples to make pies. How many bags of chpples does Nori have left? $$\\frac{2}{3}$$", "{ 5 }{ 8 }$$ (d) $$\\frac { 6 }{ 8 }$$ Answer: $$\\frac { 5 }{ 8 }$$ Explanation: Given, Mary Jane has $$\\frac { 3 }{ 8 }$$ of a medium pizza left. Hector has $$\\frac { 2 }{ 8 }$$ of another medium pizza left. To find how much pizza do they have altogether we have to add both the fractions. $$\\frac { 3 }{ 8 }$$ + $$\\frac { 2 }{ 8 }$$ = $$\\frac { 5 }{ 8 }$$ Therefore Mary Jane and Hector has $$\\frac { 5 }{ 8 }$$ pizza altogether. Thus the correct answer is option c. Question 2: Jeannie ate $$\\frac { 1 }{ 4 }$$ of an apple. Kelly ate $$\\frac { 2 }{ 4 }$$ of the apple. How much did they eat in all? (a) $$\\frac { 1 }{ 8 }$$ (b) $$\\frac { 2 }{ 8 }$$ (c) $$\\frac { 3 }{ 8 }$$ (d) $$\\frac { 3 }{ 4 }$$ Answer: $$\\frac { 3 }{ 4 }$$ Explanation: Given, Jeannie ate $$\\frac { 1 }{ 4 }$$ of an apple. Kelly ate $$\\frac { 2 }{ 4 }$$ of the apple. $$\\frac { 1 }{ 4 }$$ + $$\\frac { 2 }{ 4 }$$ = $$\\frac { 3 }{ 4 }$$ Thus the correct", "of the whole pie Explanation: Gīven that, An apple pie left from the dinner is $$\\frac{5}{6}$$ Victor plans to eat pie which was left is $$\\frac{1}{6}$$ The whole pie will be left after Victor eats tomorrow =? Pie left from dinner = $$\\frac{5}{6}$$ Victor plans to eat pie which was left = $$\\frac{1}{6}$$ $$\\frac{5}{6}$$ × $$\\frac{1}{6}$$ = $$\\frac{5}{36}$$ To find the whole pie will be left after he eats tomorrow: $$\\frac{5}{6}$$ – $$\\frac{5}{36}$$ LCD = 36 $$\\frac{5}{6}$$ × $$\\frac{6}{6}$$ – $$\\frac{5}{36}$$ $$\\frac{30}{36}$$ – $$\\frac{5}{36}$$ = $$\\frac{25}{36}$$ Therefore, whole pie left after Victor eats tomorrow is $$\\frac{25}{36}$$ Question 15. Everett and Marie are going to make fruit bars for their family reunion. They want to make 4 times the amount the recipe makes. If the recipe calls for $$\\frac{2}{3}$$ cup of oil, how much oil will they need? ______ $$\\frac{□}{□}$$ cup of oil Answer: 2 $$\\frac{2}{3}$$ Explanation: Everett and Marie are going to make fruit bars for their family reunion. They want to make 4 times the amount the recipe makes. 4 × $$\\frac{2}{3}$$ = $$\\frac{8}{3}$$ The mixed fraction of $$\\frac{8}{3}$$ is 2 $$\\frac{2}{3}$$ Thus Everett and Marie need Everett and Marie of oil. Question 16. Matt made the model below to help him solve his math problem. Write an expression that matches Matt’s model. Type below: __________ Answer:" ]

[ "do this because more people will buy their product. ## pineapple on top of pizza Ask: pineapple on top of pizza Yummy sounds delicious makes me wanna order some ## What the solution for this problem At enzos pizza parlor, Ask: What the solution for this problem At enzos pizza parlor, there are seven different toppings, where a costumer can order any number of these toppings. If you dine at the said pizza parlor wiht how many possible toppings can you order your ouzza If there are 7 topping and also 7 choices is needed. we'll be having =(7-7)! =0! =1 ## there are eight different toppings available how many different pizzas Ask: there are eight different toppings available how many different pizzas can be ordered with exactly three toppings? As arrangement don't count, this is a combination problem thus # of different pizzas that can be ordered = 8C3 = 8 .7 .6/3! = 56 different pizzas ## is pizza with toppings heterogeneous??​ Ask: is pizza with toppings heterogeneous??​ YES Explanation: The heterogeneous mixture has a non-uniform composition and has two or more phases. It can be separated out physically. The pizza is obviously physically separated from its toppings. #CarryOnLearning ## what makes food delicious and unique? ​ Ask: what makes food delicious and unique? ​ Flavor", "# 6 ways to sunday The sundae hut offers $$6$$ different toppings for their sundaes. If you can get a sundae with any number of toppings you want (including no toppings), how many different sundaes can you get? Details and assumptions You cannot get the same topping twice. The order of toppings doesn't matter. ×", "# A pizza parlor offers a selection of 3 different cheeses and 9 different toppings. In how many ways can a pizza be made with the following ingredients? $4096$ different ways Therefore there could be ${2}^{12}$ different pizzas $\\textcolor{w h i t e}{\\text{XXX}} {2}^{12} = 4096$", "cherries, and sprinkles). How many different two-scoop cones can be ordered that have two d...", "pair of another). 6. The hand is a straight (consecutive ranks, as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise $$\\PageIndex{14}\\label{ex:combin-14}$$ A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order?", "### Home > CCA2 > Chapter 12 > Lesson 12.1.3 > Problem12-49 12-49. A pizza parlor has $12$ toppings other than cheese. How many different pizzas can be created with five or fewer toppings? List all subproblems and calculate the solution. Homework Help ✎", "is the blend of taste, aroma, and feeling factor sensations. These three sensations occur when food stimulates receptors in our mouth and nose. Let's go back to the chemicals. It is because of the chemical nature of food that the senses are considered chemical sensors. Explanation: ## a restaurant offers four sizes of pizza,two types of crust,and Ask: a restaurant offers four sizes of pizza,two types of crust,and eight toppings.how many possible combinations of pizza with one topping are there? So, 4 sizes ex. (S, M, L, XL) times two types of crust (A, B), there's 8 combinations ( sa sb ma mb la lb xla xlb) then 8 toppings and only one toppings 8×8=64. There's 64 combinations. ## At Enzo's pizza parlor , there are seven different topings Ask: At Enzo's pizza parlor , there are seven different topings , where a costumer can order any number of these toppings.If you dine at the said pizza parlor,with how many possible toppings can you actually order your pizza? 127 Step-by-step explanation: The sum rule of the fundamental principle of counting states that if we have X ways of doing something and Y things of doing another thing, and we cannot do both at the same time, then there are X+Y ways to choose one of the actions. In", "How many mushrooms found each of them? 17. Dining room The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "divide 16 players into two teams of 8 member? • A pizza A pizza place offers 14 different toppings. How many different three topping pizzas can you order? • Combinations How many elements can form six times more combinations fourth class than combination of the second class? • Medals In how many ways can be divided gold, silver, and bronze medal among 21 contestants? • The six The six boys are to be led up the hill by a two-seater lift. How many options are there? • 2nd class combinations From how many elements you can create 4560 combinations of the second class?", "cookies, and also no two bars can be adjacent (so that kids B and C are not skipped). One way to assure this is to only place bars in the spaces between the stars. With 7 stars, there are 6 spots between the stars, so we must choose 3 of those 6 spots to fill with bars. Thus there are $${6 \\choose 3}$$ ways to distribute 7 cookies to 4 kids giving at least one cookie to each kid. Another (and more general) way to approach this modified problem is to first give each kid one cookie. Now the remaining 3 cookies can be distributed to the 4 kids without restrictions. So we have 3 stars and 3 bars for a total of 6 symbols, 3 of which must be bars. So again we see that there are $${6 \\choose 3}$$ ways to distribute the cookies. Stars and bars can be used in counting problems other than kids and cookies. Here are a few examples: Example $$\\PageIndex{1}$$ Your favorite mathematical pizza chain offers 10 toppings. How many pizzas can you make if you are allowed 6 toppings? The order of toppings does not matter but now you are allowed repeats. So one possible pizza is triple sausage, double pineapple, and onions. Solution We get 6 toppings (counting possible", "# Wide Variety Of Flavors An ice cream parlor has 6 flavors of ice creams. If you can have 3 scoops on a cone (where the flavors are not necessarily distinct and the order matters: bottom, middle, top), the number of all possible variations of your ice cream will be $$6\\times6\\times6=216$$. Now, what would the number of variations if, this time, the three scoops must all be distinct flavors and the order of the three flavors still matters? ×", "These questions are from a grade 12, Data Management class. Any help would be great. I am completely stuck. Simple terminolgy in explaining would be useful The first two are more about being able to explain reasoning in words while the last two will involve calculations. Thanks to everyone. 1. Until 1997, most licence plates for passenger cars in Ontario had three numbers followed by three letters. Explain why the goverment began to increase the number of letters to four. 2. A hockey team consists of 17 players (9 forwards, 6 defensemen and 2 goalies). The starting line-up consists of 3 forwards, 2 defensemen and 1 goalie. Explain way the number of ways the players can be selected only is less than if they are selected to specific positions. 3. Six students are asked to secretly choose a number from 1 to 15. Determine the probability that at least two students choose the same number to the nearest thousandth. 4. A pizzeria offers 10 different toppings. A group of people plan to order six pizzas, with up to three toppings on each. They decide to order each topping exactly once and to have at least on topping on each pizza. Determine the different cases possible when distributing the toppings in this way and the number of ways that", "# Dorian is making an ice cream sundae. He has three flavors of ice cream, two flavors of syrup, and three types of toppings. Dorian is making an ice cream sundae. He has three flavors of ice cream, two flavors of syrup, and three types of toppings. If he chooses one flavor, one syrup, and one topping, how many different combinations can he make? ## This Post Has 4 Comments 1. Expert says: That’s is personal so don’t tell you is is about you think sorry 2. Expert says: -9 step-by-step explanation: 3. Expert says: the answer is 21%. 4. Expert says:", "to three sig figs.", "to three sig figs.", "to three sig figs.", "# A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How A pizza shop has available toppings of bacon, pepperoni, sausage, olives, peppers, anchovies, onions and mushrooms. How many different ways can a pizza be made with 2 toppings?​ ## This Post Has 3 Comments 1. Chartwig4576 says: The answer is 21 as there are 7 toppings and u can only put two toppings on each pizza so in the end u hv to do 7 * 3 to get 21 2. lisaxo says: I am verry bad at imaging :^ 3. silasjob09 says: Yes I think this question is weird", "for 4 red napkins. How many yellow napkins will Viera receive for 24 red napkins?", "many wafers did both toppings have?" ]

[ "= \\dfrac{n!}{(n-r)!r!}$ For our example, $_4C_2 = \\dfrac{4!}{(4-2)!2!} = \\dfrac{4 \\times 3 \\times 3 \\times 2 \\times 1}{(2 \\times 1)(2 \\times 1)} = 6$ which is consistent with Table $$\\PageIndex{3}$$. As an example application, suppose there were six kinds of toppings that one could order for a pizza. How many combinations of exactly $$3$$ toppings could be ordered? Here $$n = 6$$ since there are $$6$$ toppings and $$r = 3$$ since we are taking $$3$$ at a time. The formula is then: $_6C_3 = \\dfrac{6!}{(6-3)!3!} = \\dfrac{6\\times 5 \\times 4 \\times 3 \\times 3 \\times 2 \\times 1}{(3 \\times 2 \\times 1)(3 \\times 2 \\times 1)} = 30$ ### Contributor • Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.", "# How many ways can you choose four toppings for a pizza if there are eight toppings to choose from? Jul 6, 2016 $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$ #### Explanation: $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$ Since there are four toppings on the pizza and eight possible toppings to choose from we can determine the varieties available by multiplying the available options. There are 8 possible toppings for the 1st choice. Since we have chosen one topping already There are 7 possible toppings for the 2nd choice There are 6 possible toppings for the 3rd choice and There are 5 possible toppings for the 4th choice $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$", "choices for cheese (none, first type, second type), and $\\,7\\,$ choices for the single topping. By the Multiplication Counting Principle, there are $\\,3\\cdot 3\\cdot 7 = 63\\,$ possible single-topping pizzas.", "+0 # Ice cream store two topping on ice cream cone. 106 ways choose topping how many toppings(including no toppings) 0 455 2 +4 At her favorite ice cream store, Saga can put up to two toppings on an ice cream cone. If there are 106 different ways she can choose toppings (including the choice of no toppings), how many different toppings are there? May 12, 2018 #1 +4247 0 106P2=106!/104!=106*105=11130 Or, is it 106C2=106!/2!*104!=106*105/2*1=5565? Which one? May 12, 2018 #2 +982 +2 We can imagine the problem like this: There are \"x\" toppings to choose from. There are $$\\binom{x}{2}$$ ways to choose the toppings. Therefore $$\\binom{x}{2}$$ = 106 We can rewrite this as: $$\\frac{x\\cdot(x-1)}{2}=106$$ Then we have to add the choice of \"no\" toppings. $$\\frac{x\\cdot(x-1)}{2}+1=106 \\\\ x^2-x-210=0 \\\\ (x+14)(x−15)=0 \\\\ x_1=-14,\\ x_2=15$$ Since there cannot be negative toppings, the answer is 15. We can double check this. If there are 15 toppings and you can have no toppings, how many ways can we \"top\" our ice-cream. $$\\binom{15}{2}+1=106$$, it works! I hope this helped, gavin May 12, 2018", "you have 2 choices with each topping, so you multiply $2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2$. 13. Originally Posted by Mathnasium It should be 256. $2^8=256$. With each topping, you can choose to include it or not include it. Thus, you have 2 choices with each topping, so you multiply $2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2$. thank you. i was debating over that for an hour now. hehe 14. You should get 256 with either way. They're the same. But 2^8 is certainly easier than counting up the combinations from 0 to 8. I was just showing the identity.", "first card, $4$ available choices for the second card, $3$ for the third card, $5$ for the fourth card, and $4$ choices for the last card. This leads to a total of $5\\cdot 4\\cdot 3\\cdot 5\\cdot 4=1200$ configurations for this case. There are $4$ cases of this type. Adding the cases up gives $2\\cdot 120+4\\cdot 600+4\\cdot 1200=7440$ \"happy\" configurations in total. This means that the probability that Kathy is happy will be $\\dfrac{7440}{10\\cdot 9\\cdot 8\\cdot 7\\cdot 6},$ which simplifies to $\\dfrac{31}{126}.$ So the answer is $31+126=\\boxed{157.}$", "# Thread: Combinations - Pizza Toppings 1. ## Combinations - Pizza Toppings I am just wondering if I did these questions correctly as I do not have an answer key for them. A second opinion would be greatly appreciated. Thanks. A pizza company has 15 toppings to choose from, of which 6 are vegetables. 1. How many different three-topping pizzas have exactly one vegetable topping? = (all possible combinations of 1 veggie topping) x (all possible combinations of 2 non-veggie toppings) = $_{6} C _1$ x $_{9} C _2$ = 216 2. How many different three-topping pizzas have at least one vegetable topping? = (all possible combinations of 3-topping pizzas) - (pizzas without any veggies) = $_{15} C _3$ - $_{9} C _3$ = 371 3. How many different three-topping pizzas have exactly two vegetable toppings? = (all possible combinations of 2 veggie toppings) x (all possible combinations of 1 non-veggie topping) = $_{6} C _2$ x $_{9} C _1$ = 135 4. How many different three-topping pizzas have at least two vegetable toppings? = (all possible combinations of 3-topping pizzas) - (pizzas without any veggies) - (pizzas with exactly 1 veggie) = $_{15} C _3$ - $_{9} C _3$ - ( $_{6} C _1$ x $_{9} C _2$ ) = 455 - 84 - 216 = 155", "have, including having all 50 on one pizza. I said you have 50 choices the first time, 49 the next, then 48 until only 1 is left. Based on this simpler example. If you have 3 toppings you have 3x2x1 = 6 different pizzas. Example: Mushroom Tomatoes Bacon Mushroom Tomatoes Mushroom Bacon Tomatoes Bacon Bacon Mushrooms Tomato That gives 7 different kinds not... >> >>7734870 This is a simple combination problem because mushroom/tomato is the same as tomato/mushroom. There are $C(50, 50)$ ways to choose 50 toppings from 50, there are $C(50, 49)$ ways to choose 49 toppings from 50, etc. >> >>7734874 3x2x1 gives 6 different pizzas with 3 topping choices but in reality the answer is 7 because you can have each topping on it's own too. >> >>7734885 With three topping choices you can either choose three toppings or choose two toppings or choose one topping. There are $C(3,3)$ ways to choose three toppings, $C(3,2)$ ways to choose two toppings, and $C(3,1)$ ways to choose one topping. That means there are [eqn]C(3,3) + C(3,2) + C(3,1) = 7[/eqn] ways to choose toppings. Your factorial answer $3 \\times 2 \\times 1$ is not correct. It's a combination problem. [Boards: 3 / a / aco / adv / an / asp / b / biz / c", "# 6 ways to sunday The sundae hut offers $$6$$ different toppings for their sundaes. If you can get a sundae with any number of toppings you want (including no toppings), how many different sundaes can you get? Details and assumptions You cannot get the same topping twice. The order of toppings doesn't matter. ×", "that every one of these would certainly count as just one outcome). Over there are precisely $$6!$$ means to arrange 6 guests, for this reason the exactly answer come the very first question is \\beginequation*\\frac14 \\cdot 13 \\cdot 12 \\cdot 11\\cdot 10 \\cdot 96!.\\endequation* Note that another way to write this is \\beginequation*\\frac14!8!\\cdot 6!.\\endequation* which is what we had actually originally. ### SubsectionExercises ¶1 A pizza parlor supplies 10 toppings. How numerous 3-topping pizzas could they placed on their menu? Assume dual toppings are not allowed. How many total pizzas space possible, with between zero and ten toppings (but not double toppings) allowed? The pizza parlor will certainly list the 10 toppings in 2 equal-sized columns on their menu. How countless ways have the right to they arrange the toppings in the left column? $$10 \\choose 3 = 120$$ pizzas. We must select (in no specific order) 3 the end of the 10 toppings.$$2^10 = 1024$$ pizzas. To speak yes or no to each topping.$$P(10,5) = 30240$$ ways. Assign every of the 5 spots in the left shaft to a distinctive pizza topping. 2 A mix lock consists of a dial through 40 numbers on it. To open the lock, you revolve the dial to the ideal until you with a first number, climate to the left until", "1. ## Counting and Probability Suppose You are ordering two pizzas.A pizza can be small,medium,large or extra large,with any combination of 8 possible toppings( getting no toppings is allowed, as is getting all 8) How many possibilities are there for your two pizzas? Answer: My answer is $\\binom{4}{2}+\\binom{4}{2}*2*(8+28+56+70+56+28+8+1 )=3066$ possibilities are there for ordering my two pizzas. 2. ## Re: Counting and Probability You can get 4 sizes. You can get 0 to 8 toppings . This can be considered an 8 bit field. 1 for the topping is included, 0 for it's not. This is $2^8=256$ different topping combos. So for a single pizza there are $4 \\cdot 256 = 1024$ different combinations. Thus for two pizzas there are $1024^2 = 1,048,576$ different two pizza combos. 3. ## Re: Counting and Probability Originally Posted by romsek You can get 4 sizes. You can get 0 to 8 toppings . This can be considered an 8 bit field. 1 for the topping is included, 0 for it's not. This is $2^8=256$ different topping combos. So for a single pizza there are $4 \\cdot 256 = 1024$ different combinations. Thus for two pizzas there are $1024^2 = 1,048,576$ different two pizza combos. Hello,", "$B \\cap C$ consists of $8$ elements, the set $A\\cap B$ consists of $7$ elements, and the set $C\\cap A$ consists of $7$ elements. The minimum number of elements in the set $A\\cup B\\cup C$ is $8$ $14$ $15$ $22$ 1 vote 8 A Pizza Shop offers $6$ different toppings, and they do not take an order without any topping. I can afford to have one pizza with a maximum of $3$ toppings. In how many ways can I order my pizza? $20$ $35$ $41$ $21$ 1 vote 9 Let $f(x)=1+x+\\dfrac{x^2}{2}+\\dfrac{x^3}{3}...+\\dfrac{x^{2018}}{2018}.$ Then $f’(1)$ is equal to $0$ $2017$ $2018$ $2019$ 10 Let $f’(x)=4x^3-3x^2+2x+k,$ $f(0)=1$ and $f(1)=4.$ Then $f(x)$ is equal to $4x^4-3x^3+2x^2+x+1$ $x^4-x^3+x^2+2x+1$ $x^4-x^3+x^2+2(x+1)$ none of these 1 vote 11 The sum of $99^{th}$ power of all the roots of $x^7-1=0$ is equal to $1$ $2$ $-1$ $0$ 1 vote 12 Let $A=\\{10,11,12,13, \\dots ,99\\}$. How many pairs of numbers $x$ and $y$ are possible so that $x+y\\geq 100$ and $x$ and $y$ belong to $A$? $2405$ $2455$ $1200$ $1230$ 13 In a certain town, $20\\%$ families own a car, $90\\%$ own a phone, $5 \\%$ own neither a car nor a phone and $30, 000$ families own both a car and a phone. Consider the following statements in this regard: $10 \\%$ families own both a car", "Sprinkles a) How many sundaes are possible using 1 flavor of ice cream and 3 toppings? There are ${\\color{blue}4}$ choices of ice cream flavors. Select three toppings from the available six: . ${6\\choose3} \\:=\\:{\\color{blue}20}$ ways. Therefore, there are: . $4\\cdot20 \\:=\\:\\boxed{80}$ possible sundaes. b) How many sundaes are possible using 1 flavor of ice cream and from 0 to 6 toppings? There are ${\\color{blue}4}$ choices of ice cream flavors. For each of the six toppings, there are two choices: .(1) use it, or (2) don't use it. . . Hence, there are: . $2^6 \\:=\\:{\\color{blue}64}$ choices. Therefore, there are: . $4\\cdot64 \\:=\\:\\boxed{256}$ possible sundaes.", "# Combinations • Dec 6th 2010, 06:30 PM tmas Combinations I can't get the last question..... A pizza company advertises that it has 15 toppings from which to choose (There are six vegetable toppings) e) how many different 3 topping pizzas have at least one vegetable? This is what i did: 6nCr3 = 20 15nCr3 = 455 455-19 = 436 The answer in the back of the book is:371 • Dec 6th 2010, 06:54 PM Soroban Hello, tmas! Quote: A pizza company advertises that it has 15 toppings from which to choose. There are six vegetable toppings. e) How many different three-topping pizzas have at least one vegetable? There are 6 Vegetable toppings and 6 Others. There are: . $_{15}C_3 \\:=\\:\\dfrac{15!}{3!\\,12!} \\:=\\:455$ possible three-topping pizzas. The opposite of \"at least one vegetable\" is \"no vegetables.\" There are: . $_9C_3 \\:=\\:\\dfrac{9!}{3!\\,6!} \\:=\\:84$ three-topiing pizzas with no vegetables. Therefore, there are: . $4565 - 84 \\:=\\:371$ pizzas with at least one vegetable. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ If you want to list the pizzas with at least one vegetable, . . be sure you make a complete list. One vegetable", "on, and then adding all of these subset numbers to get the total number. Counting the number of pizzas To illustrate the combinations rule, consider a situation where ine is interested in ordering a pizza and there are six possible toppings. How many toppings can there be in the pizza? Since there are six possible toppings, one can either have 0, 1, 2, 3, 4, 5, or 6 toppings on our pizza. Using combinations rule formula, • There are $${ 6 \\choose 0}$$ pizzas that have no toppings. • There are $${ 6 \\choose 1}$$ pizzas that have exactly one topping. • There are $${ 6 \\choose 2}$$ pizzas that have two toppings. To compute the total number of different pizzas, one continues in this fashion and the total number of possible pizzas is $N = { 6 \\choose 0} + { 6 \\choose 1} + { 6 \\choose 2} + { 6 \\choose 3} + { 6 \\choose 4} + { 6 \\choose 5} + { 6 \\choose 6}.$ The reader can confirm that $$N = 2 ^ 6 = 64$$. 2.6 Arrangements of Non-Distinct Objects First let’s use a simple example to review the two basic counting rules that we have discussed. Suppose one is making up silly words from the letters “a”, “b”, “c”, “d”,", "+1 vote 25 views A Pizza Shop offers $6$ different toppings, and they do not take an order without any topping. I can afford to have one pizza with a maximum of $3$ toppings. In how many ways can I order my pizza? 1. $20$ 2. $35$ 3. $41$ 4. $21$ recategorized | 25 views +1 $^6C_1 + {^6}C_2 + {^6}C_3 = 6 + 15 +20 = 41$ Number of ways = No of pizza with 1 topping + No of pizza with 2 topping + No of pizza with 3 topping = $6C1 + 6C2 + 6C3$ = $6+15+20= 41$ Option C) is correct by Boss (15.6k points) +1 vote Answer: $\\mathbf C$ Explanation: $\\because$ pizza cannot be ordered without any topping and pizza can be ordered with at most 3 toppings. So, only $3$ possibilities are there $$= 6C_1 + 6C_2 + 6C_3$$ $$= 6 + 15 + 20$$ $$= 41$$ $\\therefore \\mathbf C$ is the correct option. by Boss (13.2k points) edited by +1 vote", "soda [3 * 2 * 1... for all 15 days, though again I wonder if it would be 3 * 2 * 3) or maybe I should do a permutation that combines the two? I don't even know how to begin to think about this question. I added up the permutations above but I'm shy of the stated answer. Any insights would be most welcome. • The \"caveat\" implies that whatever lunch she had today doesn't restrict her choices tomorrow, since it came from a different pizzeria. – Micapps Jun 15 '16 at 14:43 Mary has $777 \\cdot 3=2331$ choices for the first day and $2330$ choices each day after that because this day has to be different from the last. This would give $2331 \\cdot 2330^{14} \\approx 3.24\\cdot 10^{50}$ choices. The point of the other pizzeria is to say that she didn't eat here today and has the $2331$ choices tomorrow. Notice that there are $777\\times 3 = 2331$ possible lunches. So Mary must select 15 of them, and then decide in which order to have them. So there are ${2331 \\choose 15} \\times 15! = \\frac{2331!}{2316!}$ ways to plan the lunches. Plugging this in to a computer, this is exactly: $721554522956870749367491575069940819296530824826880000\\approx 7.2\\cdot 10^{53}$ • This is not correct. She is allowed to have the same lunch", "Topping 2: 5 choices (6-1 toppings as one topping is already selected) Topping 3: 4 choices (6-2 toppings as two toppings are already selected) Total= $$6 \\times 5 \\times 4 = 120$$ Now say the three toppings are jelly, jam and nuts. So 120 will contain {jelly, jam, nut},{nut,jelly, jam} ..... and 4 other combinations. Numbers of ways of arranging 3 things $$3!=6$$ So the correct number of options for pancakes =$$\\frac{120}{6}=20$$. For repeating allowed you need to you need to add two more cases 2 toppings are same one different So you have to choose 2 toppings and the order does not matter= $$\\frac{6 \\times 5}{2!}=15$$ ($$C^6_{2}$$) Here you can repeat the first topping 2 times or the second topping twice so Total choices= $$15 \\times 2$$ All 3 toppings same You have to choose 1 topping = $$6$$ ($$C^6_{1}$$) Add them all up $$20 +15 + 6 =41$$. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Supreme Moderator Joined: 01 Nov 2017 Posts: 370 Followers: 5 Kudos [?]: 111 [0], given: 4 Re: At Flo’s Pancake House, pancakes can be ordered with any of [#permalink] 29 Oct 2018, 21:41 Expert's post msawicka wrote: How would this be different if you were", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "item from 9 495 possible four-item combos 45+165+495 = 705, which is a far cry from 20,000. ### So let me get this straight. 05:18 pm So our friends at Dairy Queen have this new \"Sweet Deals\" package, where it's any two of nine items for$3, any three for $4, any four for$5, and they advertise over 20,000 different combinations. Sounds like false advertising to me, there's only 705 by my count: 36 possible: 2 different items from 9 9 possible: 2 of the same item from 9 45 possible two-item combos 84 possible: 3 different items from 9 72 possible: 2 of one item, 1 of a second from 9 9 possible: 3 of one item from 9 165 possible three-item combos 126 possible: 4 different items from 9 252 possible: 2 of one item, 1 of a second, 1 of a third from 9 36 possible: 2 of one item, 2 of a second from 9 72 possible: 3 of one item, 1 of a second from 9 9 possible: 4 of one item from 9 495 possible four-item combos 45+165+495 = 705, which is a far cry from 20,000. ### Looks like it's on. 01:34 am And it's going to be Flyers-Penguins in the first round. Hey, Justin? You up for a gentleman's wager? ###" ]

[ "# 6 ways to sunday The sundae hut offers $$6$$ different toppings for their sundaes. If you can get a sundae with any number of toppings you want (including no toppings), how many different sundaes can you get? Details and assumptions You cannot get the same topping twice. The order of toppings doesn't matter. ×", "+0 # Ice cream store two topping on ice cream cone. 106 ways choose topping how many toppings(including no toppings) 0 455 2 +4 At her favorite ice cream store, Saga can put up to two toppings on an ice cream cone. If there are 106 different ways she can choose toppings (including the choice of no toppings), how many different toppings are there? May 12, 2018 #1 +4247 0 106P2=106!/104!=106*105=11130 Or, is it 106C2=106!/2!*104!=106*105/2*1=5565? Which one? May 12, 2018 #2 +982 +2 We can imagine the problem like this: There are \"x\" toppings to choose from. There are $$\\binom{x}{2}$$ ways to choose the toppings. Therefore $$\\binom{x}{2}$$ = 106 We can rewrite this as: $$\\frac{x\\cdot(x-1)}{2}=106$$ Then we have to add the choice of \"no\" toppings. $$\\frac{x\\cdot(x-1)}{2}+1=106 \\\\ x^2-x-210=0 \\\\ (x+14)(x−15)=0 \\\\ x_1=-14,\\ x_2=15$$ Since there cannot be negative toppings, the answer is 15. We can double check this. If there are 15 toppings and you can have no toppings, how many ways can we \"top\" our ice-cream. $$\\binom{15}{2}+1=106$$, it works! I hope this helped, gavin May 12, 2018", "that every one of these would certainly count as just one outcome). Over there are precisely $$6!$$ means to arrange 6 guests, for this reason the exactly answer come the very first question is \\beginequation*\\frac14 \\cdot 13 \\cdot 12 \\cdot 11\\cdot 10 \\cdot 96!.\\endequation* Note that another way to write this is \\beginequation*\\frac14!8!\\cdot 6!.\\endequation* which is what we had actually originally. ### SubsectionExercises ¶1 A pizza parlor supplies 10 toppings. How numerous 3-topping pizzas could they placed on their menu? Assume dual toppings are not allowed. How many total pizzas space possible, with between zero and ten toppings (but not double toppings) allowed? The pizza parlor will certainly list the 10 toppings in 2 equal-sized columns on their menu. How countless ways have the right to they arrange the toppings in the left column? $$10 \\choose 3 = 120$$ pizzas. We must select (in no specific order) 3 the end of the 10 toppings.$$2^10 = 1024$$ pizzas. To speak yes or no to each topping.$$P(10,5) = 30240$$ ways. Assign every of the 5 spots in the left shaft to a distinctive pizza topping. 2 A mix lock consists of a dial through 40 numbers on it. To open the lock, you revolve the dial to the ideal until you with a first number, climate to the left until", "### Home > CCA2 > Chapter 12 > Lesson 12.1.3 > Problem12-49 12-49. A pizza parlor has $12$ toppings other than cheese. How many different pizzas can be created with five or fewer toppings? List all subproblems and calculate the solution. Homework Help ✎", "# How many ways can you choose four toppings for a pizza if there are eight toppings to choose from? Jul 6, 2016 $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$ #### Explanation: $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$ Since there are four toppings on the pizza and eight possible toppings to choose from we can determine the varieties available by multiplying the available options. There are 8 possible toppings for the 1st choice. Since we have chosen one topping already There are 7 possible toppings for the 2nd choice There are 6 possible toppings for the 3rd choice and There are 5 possible toppings for the 4th choice $8 \\cdot 7 \\cdot 6 \\cdot 5 = 1680$", "# A pizza parlor offers a selection of 3 different cheeses and 9 different toppings. In how many ways can a pizza be made with the following ingredients? $4096$ different ways Therefore there could be ${2}^{12}$ different pizzas $\\textcolor{w h i t e}{\\text{XXX}} {2}^{12} = 4096$", "as in 5, 6, 7, 8, 9, but not all from the same suit). 7. The hand is a flush (all the same suit, but not a straight). 8. The hand is a straight flush (both straight and flush). Exercise (PageIndex{14}label{ex:combin-14}) A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies. 1. How many kinds of pizzas can one order? 2. How many kinds of pizzas can one order with exactly three toppings? 3. How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order? ## Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows: Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins. Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers. In how many different ways can we select 3 items from a set of 8? #### Number", "### Home > CCA2 > Chapter 12 > Lesson 12.1.3 > Problem12-60 12-60. Another pizza parlor had a super special. The parlor had a rack of small pizzas, each with three or four different toppings. The price was really low because an employee forgot to indicate on the boxes what toppings were on each pizza. Eight toppings were available, and one pizza was made for each of the possible combinations of three or four toppings. Homework Help ✎ 1. How many pizzas were on the rack? Find how many types of pizzas with three toppings can be made. 8C3 Find how many pizzas with four toppings can be made. 8C4 8C3 $+$ 8C4$= 56 + 70 = 126$ 2. What was the probability of getting a pizza that had mushrooms on it? If mushrooms are a known topping, then you choose one fewer topping from only $7$ remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "# How many coins? - The Sequel Discrete Mathematics Level 5 My nearest doughnut stall sells donuts in packs of $$1$$, $$2$$, or $$5$$. I keep on ordering such doughnut packs uniformly at random until I have a total of more than or equal to $$400$$ doughnuts. If the expected number of $$5$$ packs I ordered is $$n$$, what is the value of $$\\left\\lfloor 10n\\right\\rfloor$$? Thanks to @bobbym none for checking the answers ×", "Sprinkles a) How many sundaes are possible using 1 flavor of ice cream and 3 toppings? There are ${\\color{blue}4}$ choices of ice cream flavors. Select three toppings from the available six: . ${6\\choose3} \\:=\\:{\\color{blue}20}$ ways. Therefore, there are: . $4\\cdot20 \\:=\\:\\boxed{80}$ possible sundaes. b) How many sundaes are possible using 1 flavor of ice cream and from 0 to 6 toppings? There are ${\\color{blue}4}$ choices of ice cream flavors. For each of the six toppings, there are two choices: .(1) use it, or (2) don't use it. . . Hence, there are: . $2^6 \\:=\\:{\\color{blue}64}$ choices. Therefore, there are: . $4\\cdot64 \\:=\\:\\boxed{256}$ possible sundaes.", "on, and then adding all of these subset numbers to get the total number. Counting the number of pizzas To illustrate the combinations rule, consider a situation where ine is interested in ordering a pizza and there are six possible toppings. How many toppings can there be in the pizza? Since there are six possible toppings, one can either have 0, 1, 2, 3, 4, 5, or 6 toppings on our pizza. Using combinations rule formula, • There are $${ 6 \\choose 0}$$ pizzas that have no toppings. • There are $${ 6 \\choose 1}$$ pizzas that have exactly one topping. • There are $${ 6 \\choose 2}$$ pizzas that have two toppings. To compute the total number of different pizzas, one continues in this fashion and the total number of possible pizzas is $N = { 6 \\choose 0} + { 6 \\choose 1} + { 6 \\choose 2} + { 6 \\choose 3} + { 6 \\choose 4} + { 6 \\choose 5} + { 6 \\choose 6}.$ The reader can confirm that $$N = 2 ^ 6 = 64$$. 2.6 Arrangements of Non-Distinct Objects First let’s use a simple example to review the two basic counting rules that we have discussed. Suppose one is making up silly words from the letters “a”, “b”, “c”, “d”,", "### Home > CCA2 > Chapter 12 > Lesson 12.1.3 > Problem12-60 12-60. 1. Another pizza parlor had a super special. The parlor had a rack of small pizzas, each with three or four different toppings. The price was really low because an employee forgot to indicate on the boxes what toppings were on each pizza. Eight toppings were available, and one pizza was made for each of the possible combinations of three or four toppings. Homework Help ✎ 1. How many pizzas were on the rack? 2. What was the probability of getting a pizza that had mushrooms on it? Find how many types of pizzas with three toppings can be made. 8C3 Find how many pizzas with four toppings can be made. 8C4 8C3 + 8C4 = 56 + 70 = 126 If mushrooms are a known topping, then you choose one fewer topping from only 7 remaining toppings. $\\frac{_{7}C_{3}+_{7}C_{2}}{126}=\\frac{56}{126}=\\frac{4}{9}$", "# Thread: Combinations - Pizza Toppings 1. ## Combinations - Pizza Toppings I am just wondering if I did these questions correctly as I do not have an answer key for them. A second opinion would be greatly appreciated. Thanks. A pizza company has 15 toppings to choose from, of which 6 are vegetables. 1. How many different three-topping pizzas have exactly one vegetable topping? = (all possible combinations of 1 veggie topping) x (all possible combinations of 2 non-veggie toppings) = $_{6} C _1$ x $_{9} C _2$ = 216 2. How many different three-topping pizzas have at least one vegetable topping? = (all possible combinations of 3-topping pizzas) - (pizzas without any veggies) = $_{15} C _3$ - $_{9} C _3$ = 371 3. How many different three-topping pizzas have exactly two vegetable toppings? = (all possible combinations of 2 veggie toppings) x (all possible combinations of 1 non-veggie topping) = $_{6} C _2$ x $_{9} C _1$ = 135 4. How many different three-topping pizzas have at least two vegetable toppings? = (all possible combinations of 3-topping pizzas) - (pizzas without any veggies) - (pizzas with exactly 1 veggie) = $_{15} C _3$ - $_{9} C _3$ - ( $_{6} C _1$ x $_{9} C _2$ ) = 455 - 84 - 216 = 155", "# Combinations • Dec 6th 2010, 06:30 PM tmas Combinations I can't get the last question..... A pizza company advertises that it has 15 toppings from which to choose (There are six vegetable toppings) e) how many different 3 topping pizzas have at least one vegetable? This is what i did: 6nCr3 = 20 15nCr3 = 455 455-19 = 436 The answer in the back of the book is:371 • Dec 6th 2010, 06:54 PM Soroban Hello, tmas! Quote: A pizza company advertises that it has 15 toppings from which to choose. There are six vegetable toppings. e) How many different three-topping pizzas have at least one vegetable? There are 6 Vegetable toppings and 6 Others. There are: . $_{15}C_3 \\:=\\:\\dfrac{15!}{3!\\,12!} \\:=\\:455$ possible three-topping pizzas. The opposite of \"at least one vegetable\" is \"no vegetables.\" There are: . $_9C_3 \\:=\\:\\dfrac{9!}{3!\\,6!} \\:=\\:84$ three-topiing pizzas with no vegetables. Therefore, there are: . $4565 - 84 \\:=\\:371$ pizzas with at least one vegetable. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ If you want to list the pizzas with at least one vegetable, . . be sure you make a complete list. One vegetable", "choices for cheese (none, first type, second type), and $\\,7\\,$ choices for the single topping. By the Multiplication Counting Principle, there are $\\,3\\cdot 3\\cdot 7 = 63\\,$ possible single-topping pizzas.", "+1 vote 25 views A Pizza Shop offers $6$ different toppings, and they do not take an order without any topping. I can afford to have one pizza with a maximum of $3$ toppings. In how many ways can I order my pizza? 1. $20$ 2. $35$ 3. $41$ 4. $21$ recategorized | 25 views +1 $^6C_1 + {^6}C_2 + {^6}C_3 = 6 + 15 +20 = 41$ Number of ways = No of pizza with 1 topping + No of pizza with 2 topping + No of pizza with 3 topping = $6C1 + 6C2 + 6C3$ = $6+15+20= 41$ Option C) is correct by Boss (15.6k points) +1 vote Answer: $\\mathbf C$ Explanation: $\\because$ pizza cannot be ordered without any topping and pizza can be ordered with at most 3 toppings. So, only $3$ possibilities are there $$= 6C_1 + 6C_2 + 6C_3$$ $$= 6 + 15 + 20$$ $$= 41$$ $\\therefore \\mathbf C$ is the correct option. by Boss (13.2k points) edited by +1 vote", "have, including having all 50 on one pizza. I said you have 50 choices the first time, 49 the next, then 48 until only 1 is left. Based on this simpler example. If you have 3 toppings you have 3x2x1 = 6 different pizzas. Example: Mushroom Tomatoes Bacon Mushroom Tomatoes Mushroom Bacon Tomatoes Bacon Bacon Mushrooms Tomato That gives 7 different kinds not... >> >>7734870 This is a simple combination problem because mushroom/tomato is the same as tomato/mushroom. There are $C(50, 50)$ ways to choose 50 toppings from 50, there are $C(50, 49)$ ways to choose 49 toppings from 50, etc. >> >>7734874 3x2x1 gives 6 different pizzas with 3 topping choices but in reality the answer is 7 because you can have each topping on it's own too. >> >>7734885 With three topping choices you can either choose three toppings or choose two toppings or choose one topping. There are $C(3,3)$ ways to choose three toppings, $C(3,2)$ ways to choose two toppings, and $C(3,1)$ ways to choose one topping. That means there are [eqn]C(3,3) + C(3,2) + C(3,1) = 7[/eqn] ways to choose toppings. Your factorial answer $3 \\times 2 \\times 1$ is not correct. It's a combination problem. [Boards: 3 / a / aco / adv / an / asp / b / biz / c", "1. ## Counting and Probability Suppose You are ordering two pizzas.A pizza can be small,medium,large or extra large,with any combination of 8 possible toppings( getting no toppings is allowed, as is getting all 8) How many possibilities are there for your two pizzas? Answer: My answer is $\\binom{4}{2}+\\binom{4}{2}*2*(8+28+56+70+56+28+8+1 )=3066$ possibilities are there for ordering my two pizzas. 2. ## Re: Counting and Probability You can get 4 sizes. You can get 0 to 8 toppings . This can be considered an 8 bit field. 1 for the topping is included, 0 for it's not. This is $2^8=256$ different topping combos. So for a single pizza there are $4 \\cdot 256 = 1024$ different combinations. Thus for two pizzas there are $1024^2 = 1,048,576$ different two pizza combos. 3. ## Re: Counting and Probability Originally Posted by romsek You can get 4 sizes. You can get 0 to 8 toppings . This can be considered an 8 bit field. 1 for the topping is included, 0 for it's not. This is $2^8=256$ different topping combos. So for a single pizza there are $4 \\cdot 256 = 1024$ different combinations. Thus for two pizzas there are $1024^2 = 1,048,576$ different two pizza combos. Hello,", "do this because more people will buy their product. ## pineapple on top of pizza Ask: pineapple on top of pizza Yummy sounds delicious makes me wanna order some ## What the solution for this problem At enzos pizza parlor, Ask: What the solution for this problem At enzos pizza parlor, there are seven different toppings, where a costumer can order any number of these toppings. If you dine at the said pizza parlor wiht how many possible toppings can you order your ouzza If there are 7 topping and also 7 choices is needed. we'll be having =(7-7)! =0! =1 ## there are eight different toppings available how many different pizzas Ask: there are eight different toppings available how many different pizzas can be ordered with exactly three toppings? As arrangement don't count, this is a combination problem thus # of different pizzas that can be ordered = 8C3 = 8 .7 .6/3! = 56 different pizzas ## is pizza with toppings heterogeneous??​ Ask: is pizza with toppings heterogeneous??​ YES Explanation: The heterogeneous mixture has a non-uniform composition and has two or more phases. It can be separated out physically. The pizza is obviously physically separated from its toppings. #CarryOnLearning ## what makes food delicious and unique? ​ Ask: what makes food delicious and unique? ​ Flavor", "# A pizza place offers 4 different cheeses and 10 different toppings. In how many ways can a pizza be made with 1 cheese and 4 toppings? May 6, 2016 $840$ #### Explanation: Picking 1 cheese from 4 = $4 C 1$ Picking 4 toppings from 10 = $10 C 4$ Combining the two: $4 C 1 \\cdot 10 C 4$ $= 4 \\cdot 210$ $= 840$" ]

[ "# Find the following:$\\large\\frac{15}{16}$ of $7\\large\\frac{3}{11}$ ( A ) $\\large\\frac{7}{1}$ ( B )$\\large\\frac{76}{11}$ ( C ) $\\large\\frac{75}{13}$ ( D ) $\\large\\frac{75}{11}$", "\\frac{30}{2} + 30 \\left[1 - {\\left(\\frac{2}{3}\\right)}^{50}\\right]$ $= 30 \\left[\\frac{1}{2} + 1 - {\\left(\\frac{2}{3}\\right)}^{50}\\right]$ $= 30 \\left[\\frac{3}{2} + {\\left(\\frac{2}{3}\\right)}^{50}\\right]$", "get $\\frac{29}{154}$. You can divide the top number by the bottom to get the decimal.", "{4}^{-} 6 \\times {10}^{30}$ $= {10}^{30} / {4}^{6}$ You could also answer as $\\frac{1}{4 \\times {10}^{-} 5} ^ 6$ Which leads to $\\frac{1}{{4}^{6} \\times {10}^{-} 30}$", "= - 7$ $x - \\frac{30}{4} = - 7$ $x - \\frac{15}{2} = - \\frac{14}{2}$ $x = - \\frac{14}{2} + \\frac{15}{2} = \\frac{1}{2}$ Solution: $\\left(\\frac{1}{2} , - \\frac{15}{4}\\right)$", "{\\frac{1} {{24}} - \\frac{1} {{33}}} \\right) = \\left( {\\frac{1} {6} - \\frac{1} {{33}}} \\right)$", "}^{2}}\\left( 30{}^\\circ \\right)$ $=\\frac{{{I}_{0}}}{8}{{\\left( \\frac{\\sqrt{3}}{2} \\right)}^{2}}=\\frac{3{{I}_{0}}}{32}$", "\\;=\\;-\\frac{12}{35}$", "cos30^{o}={\\frac {\\sqrt {3}}{2}}=sin60^{o}}$", "}^4}\\theta d\\theta }$$$$= \\frac{{12}}{{35}}$$", "+ 30^2}$$ Fr = 50 N", "hence, for $n \\ge 30$, $T_{n} \\approx N(5(n-1) + 10\\cdot\\frac{n-1}{2}, 10^2\\cdot\\frac{n-1}{12})$ $\\Rightarrow T_{n} \\approx N(10(n-1), \\frac{25(n-1)}{3})$", "12 3^k+(-1)(-1)^{k-1}+ \\frac 32(-1)^{k-1}=\\frac 32 3^k+ \\frac 12 (-1)^{k-1}=$$ $$=\\frac 12 3^{k+1}+ \\frac 12 (-1)^{k-1}=\\frac 12 3^{k+1}+ \\frac 12 (-1)^{k-1}(-1)^2=$$ $$=\\frac 12 3^{k+1}+ \\frac 12 (-1)^{k+1}$$", "2}{x + 2} > \\frac{2x - 3}{4x - 1}$$ 49. Find the set of real values of $$x$$ for which $$x^2 + 8|x + 4| - 48 > 0$$ 50. Find the values of $$x$$ such that $$\\log_{10}(x^2 - 2x - 2) \\le 0$$", "= - \\frac{27}{30}$ $\\frac{- 59 + 50}{10} = - \\frac{9}{10}$ $- \\frac{9}{10} = - \\frac{9}{10}$", "10. $\\frac{15}{30} = \\frac{x}{2}$ 11. $\\frac{2}{9} = \\frac{n}{63}$ 12. $\\frac{z}{7} = \\frac{12}{21}$ 13. $\\frac{3}{5} = \\frac{t}{60}$ 14. $\\frac{k}{72} = \\frac{5}{12}$ 15. $\\frac{x}{32} = \\frac{4}{8}$", "$\\left(- \\frac{114}{13} , \\frac{206}{13}\\right)$ and $\\left(\\frac{270}{13} , - \\frac{50}{15}\\right)$", "$t = \\frac{1}{0.0333} = 30.0$", "Find $a_{20}$ if $a_n=\\large\\frac{n(n-2)}{n+3}$ $\\begin{array}{1 1}\\frac{360}{23} \\\\ \\frac{240}{23} \\\\ \\frac{380}{23} \\\\ \\frac{420}{23} \\end{array}$", "- \\frac{7}{36}$" ]

[ "# What's the common ratio in 5/12, 1/3, 4/15, 16/75, 64/375? Dec 3, 2015 $\\frac{4}{5}$ #### Explanation: Divide any number in this sequence by the preceding number and you get $\\frac{4}{5}$. That's the common ratio. For example: $\\frac{1}{3} \\div \\frac{5}{12} = \\frac{1}{3} \\cdot \\frac{12}{5} = \\frac{4}{5}$, $\\frac{4}{15} \\div \\frac{1}{3} = \\frac{4}{15} \\cdot 3 = \\frac{12}{15} = \\frac{4}{5}$, $\\frac{16}{75} \\div \\frac{4}{15} = \\frac{16}{75} \\cdot \\frac{15}{4} = \\frac{4}{5}$, and $\\frac{64}{375} \\div \\frac{16}{75} = \\frac{64}{375} \\cdot \\frac{75}{16} = \\frac{4}{5}$. A cool thing related to this is the corresponding geometric series (infinite \"sum\") $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots$. If $a$ is the first term of such a series, and $r$ is the common ratio, with $| r | < 1$, the series converges to $\\frac{a}{1 - r}$. For this example, $a = \\frac{5}{12}$ and $r = \\frac{4}{5}$ so that the series converges to $\\frac{\\frac{5}{12}}{1 - \\frac{4}{5}} = \\frac{\\frac{5}{12}}{\\frac{1}{5}} = \\frac{5}{12} \\cdot 5 = \\frac{25}{12}$ and we write $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots = \\frac{25}{12}$.", "& = \\frac { 1 } { ( 2 - 3 i ) } \\cdot \\color{Cerulean}{\\frac { ( 2 + 3 i ) } { ( 2 + 3 i ) }} \\\\ & = \\frac { ( 2 + 3 i ) } { 2 ^ { 2 } + 3 ^ { 2 } } \\\\ & = \\frac { 2 + 3 i } { 4 + 9 } \\\\ & = \\frac { 2 + 3 i } { 13 } \\end{aligned} To write this complex number in standard form, we make use of the fact that 13 is a common denominator. \\begin{aligned} \\frac { 2 + 3 i } { 13 } & = \\frac { 2 } { 13 } + \\frac { 3 i } { 13 } \\\\ & = \\frac { 2 } { 13 } + \\frac { 3 } { 13 } i \\end{aligned} $$\\frac { 2 } { 13 } + \\frac { 3 } { 13 } i$$ Example $$\\PageIndex{9}$$: Divide: $$\\frac { 1 - 5 i } { 4 + i }$$. Solution \\begin{aligned} \\frac { 1 - 5 i } { 4 + i } & = \\frac { ( 1 - 5 i ) } { ( 4 + i ) }", "5/12, $\\frac{5}{12}=\\frac{5}{12} \\cdot \\color{blue}{1}=\\frac{5}{12} \\cdot \\color{blue}{\\frac{3}{3}}=\\frac{15}{36}$ In the case of 5/18, $\\frac{5}{18}=\\frac{5}{18} \\cdot \\color{blue}{1}=\\frac{5}{18} \\cdot \\color{blue}{\\frac{2}{2}}=\\frac{10}{36}$ If we replace the fractions in equation (8) with their equivalent fractions, we can then add the numerators and divide by the common denominator, as in $\\frac{5}{12}+\\frac{5}{18}=\\frac{15}{36}+\\frac{10}{36}=\\frac{15+10}{36}=\\frac{25}{36}$ Let’s examine a method of organizing the work that is more compact. Consider the following arrangement, where we’ve used color to highlight the form of 1 required to convert the fractions to equivalent fractions with a common denominator of 36. \\begin{aligned} \\frac{5}{12}+\\frac{5}{18} &=\\frac{5}{12} \\cdot \\color{blue}{\\frac{3}{3}}+\\frac{5}{18} \\cdot \\color{blue}{\\frac{2}{2}} \\\\ &=\\frac{15}{36}+\\frac{10}{36} \\\\ &=\\frac{25}{36} \\end{aligned} Let’s look at a more complicated example. Example $$\\PageIndex{3}$$ Simplify the expression $\\frac{x+3}{x+2}-\\frac{x+2}{x+3}$ State all restrictions. Solution The denominators are already factored. If we take each factor that appears to the highest exponential power that appears, our least common denominator is (x+2)(x+3). Our first task is to make equivalent fractions having this common denominator. \\begin{aligned} \\frac{x+3}{x+2}-\\frac{x+2}{x+3} &=\\frac{x+3}{x+2} \\cdot \\color{blue}{\\frac{x+3}{x+3}}-\\frac{x+2}{x+3} \\cdot \\color{blue}{\\frac{x+2}{x+2}} \\\\ &=\\frac{x^{2}+6 x+9}{(x+2)(x+3)}-\\frac{x^{2}+4 x+4}{(x+2)(x+3)} \\end{aligned} Now, subtract the numerators and divide by the common denominator. \\begin{aligned} \\frac{x+3}{x+2}-\\frac{x+2}{x+3} &=\\frac{\\left(x^{2}+6 x+9\\right)-\\left(x^{2}+4 x+4\\right)}{(x+2)(x+3)} \\\\ &=\\frac{x^{2}+6 x+9-x^{2}-4 x-4}{(x+2)(x+3)} \\\\ &=\\frac{2 x+5}{(x+2)(x+3)} \\end{aligned} Note the use of parentheses when we subtracted the numerators. Note further how the minus sign negates each term in the parenthetical expression that follows the minus sign. Tip Always use grouping symbols when", "example: $\\frac 3 5 \\cdot \\frac 5 6 = \\frac {3 \\cdot 5} {5 \\cdot 6} = \\frac {5 \\cdot 3} {5 \\cdot 6} = \\frac 5 5 \\cdot \\frac 3 6 = 1 \\cdot \\frac 3 {2 \\cdot 3} = 1 \\cdot \\frac 3 3 \\cdot \\frac 1 2 = 1 \\cdot 1 \\cdot \\frac 1 2 = \\frac 1 2$ 8. Originally Posted by tkdvipers Cheers I am getting there, What confuses me is the way that. $ \\frac{3}{7} * \\frac{5}{6} $ becomes $ \\frac{1}{7} * \\frac{5}{2} $ I can see that 6/3 = 2 and 3/3 = 1, however I thought the fractions were unrelated and you could only work within the same fraction. for example 3/7 and 5/6 I can do the sums now but want to understand why it is legal. Look at the cancelation that took place in red. The 3 and the 6 have a common factor of 3. $ \\frac{3}{7} * \\frac{5}{6} = \\color{red}\\frac{\\rlap{---}{3}^1}{7} * \\color{red}\\frac{5}{\\rlap{---}{6}^2} =\\color{black}\\frac{1}{7}\\cdot\\frac{5}{2}= \\frac{1\\cdot5}{7\\cdot2}=\\frac{5}{14} $ Another approach: $\\frac{3}{7} \\cdot \\frac{5}{6}=\\frac{3}{6} \\cdot \\frac{5}{7}=\\color{red} \\frac{\\rlap{---}{3}^1}{\\rlap{---}{6}^2}\\cdot\\frac{5}{7}= \\color{black}\\frac{1\\cdot5}{2\\cdot7}=\\frac{5}{14} $", "\\frac{a(r^{n}-1)}{r-1} for r > 1 OR S_{n}= \\frac{a(1-r^{n})}{1-r} for r < 1 a: first term. r: common ration. Example For the sequence: 4, 12, 36, 108 a = 4. r = \\frac{T_{2}}{T_{1}} = \\frac{ 12 }{ 4 } = 3. r is greater than 1, so we use S_{n}= \\frac{a(r^{n}-1)}{r-1} S_{n}= \\frac{a(r^{n}-1)}{r-1} S_{n}= \\frac{4(3^{n}-1)}{3-1} S_{n}= \\frac{4(3^{n}-1)}{2} S_{n}= 2 \\cdot 3^{n}-2 Example For the sequence: 81, 27, 9, 3 a = 81. r = \\frac{T_{2}}{T_{1}} = \\frac{ 27 }{ 81 } = \\frac{1}{3}. r is less than 1, so we use S_{n}= \\frac{a(1-r^{n})}{1-r} S_{n}= \\frac{a(1-r^{n})}{1-r} S_{n}= \\frac{81(1- \\left( \\frac{1}{3} \\right)^{n})}{1- \\frac{1}{3}} S_{n}= \\frac{81- 81 \\cdot \\left( \\frac{1}{3} \\right)^{n}}{\\frac{2}{3}} \\displaystyle S_{n}= \\frac{243}{2} - \\frac{243 \\cdot 3^{-n}}{2} A quadratic sequence is a sequence where the the second difference is common. Below is an example. General Term The general term of a quadratic sequence is T_{n} = an^{2}+bn+c then 2a = second difference. 3a+b = T2-T1 a+b+c = first term. Example For the sequence: (The example above) 6, 12, 22, 36 Second Difference = 4. General term: T_{n} = an^{2}+bn+c. We need to get a, b and c. 2a = Second Difference 2a = 4 \\bullet a = 2 3a+b = T_{2}-T_{1} 3(2)+b = 12-6 \\bullet b = 0 a+b+c = T_{1} (2)+(0)+c = 6 \\bullet c = 4 \\displaystyle T_{n} =", "can actually calculate the probability of all of the stacks containing all colors as in the post above, I will reply in reverse order to the posts. I think that the probability is: \\displaystyle \\large { \\begin{align} & \\left (\\frac{56}{56} \\cdot \\frac{48}{55} \\cdot \\frac{40}{54} \\cdot \\frac{32}{53} \\cdot \\frac{24}{52} \\cdot \\frac{16}{51} \\cdot \\frac{8}{50} \\right ) \\\\ \\cdot & \\left (\\frac{49}{49} \\cdot \\frac{42}{48} \\cdot \\frac{35}{47} \\cdot \\frac{28}{46} \\cdot \\frac{21}{45} \\cdot \\frac{14}{44} \\cdot \\frac{7}{43} \\right ) \\\\ \\cdot & \\left (\\frac{42}{42} \\cdot \\frac{36}{41} \\cdot \\frac{30}{40} \\cdot \\frac{24}{39} \\cdot \\frac{18}{38} \\cdot \\frac{12}{37} \\cdot \\frac{6}{36} \\right ) \\\\ \\cdot & \\left (\\frac{35}{35} \\cdot \\frac{30}{34} \\cdot \\frac{25}{33} \\cdot \\frac{20}{32} \\cdot \\frac{15}{31} \\cdot \\frac{10}{30} \\cdot \\frac{5}{29} \\right ) \\\\ \\cdot & \\left (\\frac{28}{28} \\cdot \\frac{24}{27} \\cdot \\frac{20}{26} \\cdot \\frac{16}{25} \\cdot \\frac{12}{24} \\cdot \\frac{8}{23} \\cdot \\frac{4}{22} \\right ) \\\\ \\cdot & \\left (\\frac{21}{21} \\cdot \\frac{18}{20} \\cdot \\frac{15}{19} \\cdot \\frac{12}{18} \\cdot \\frac{9}{17} \\cdot \\frac{6}{16} \\cdot \\frac{3}{15} \\right ) \\\\ \\cdot & \\left (\\frac{14}{14} \\cdot \\frac{12}{13} \\cdot \\frac{10}{12} \\cdot \\frac{8}{11} \\cdot \\frac{6}{10} \\cdot \\frac{4}{9} \\cdot \\frac{2}{8} \\right ) \\\\ \\cdot & \\left (\\frac{7}{7} \\cdot \\frac{6}{6} \\cdot \\frac{5}{5} \\cdot \\frac{4}{4} \\cdot \\frac{3}{3} \\cdot \\frac{2}{2} \\cdot \\frac{1}{1} \\right ) \\\\ & = \\\\ 8^7 & \\cdot (7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1) \\\\ \\cdot 7^7 & \\cdot (7 \\cdot 6 \\cdot 5 \\cdot", "# How do you rationalize the denominator and simplify sqrt22/sqrt33? $\\frac{\\sqrt{22}}{\\sqrt{33}} = \\frac{\\sqrt{6}}{3}$ #### Explanation: From the given $\\frac{\\sqrt{22}}{\\sqrt{33}} = \\frac{\\sqrt{22}}{\\sqrt{33}} \\cdot \\frac{\\sqrt{33}}{\\sqrt{33}}$ $= \\frac{\\sqrt{22 \\cdot 33}}{\\sqrt{33}} ^ 2 = \\frac{\\sqrt{2 \\cdot 11 \\cdot 3 \\cdot 11}}{33} = \\frac{\\sqrt{6 \\cdot {11}^{2}}}{33} = \\frac{11 \\sqrt{6}}{33} = \\frac{\\sqrt{6}}{3}$ God bless...I hope the explanation is useful.", "avoid using [homework] as the only tag for the question. You should add at least one more suitable tag hinting on the content of the question itself (e.g. [algebra-precalculus] might be fitting here). – Asaf Karagila Sep 25 '11 at 19:12 The new version of the question makes it much more clear what is intended: very likely you should multiply numerator and denominator by $\\sqrt{7} + \\sqrt{12}$ and simplify. – Pete L. Clark Sep 25 '11 at 19:47 ## 1 Answer Here's a similar problem worked out (assuming that \"in surd form\" means something like \"as a sum of roots of square-free integers multiplied by rational numbers\"): \\begin{align} \\frac{\\sqrt{875}}{\\sqrt{11}-\\sqrt{45}} &=\\frac{\\sqrt{875}(\\sqrt{11}+\\sqrt{45})}{(\\sqrt{11}-\\sqrt{45})(\\sqrt{11}+\\sqrt{45})}\\\\ &=\\frac{\\sqrt{875}(\\sqrt{11}+\\sqrt{45})}{\\sqrt{11}^2-\\sqrt{45}^2}\\\\ &=\\frac{\\sqrt{875}(\\sqrt{11}+\\sqrt{45})}{11-45}\\\\ &=\\frac{\\sqrt{5^3\\cdot7}(\\sqrt{11}+\\sqrt{3^2\\cdot5})}{-34}\\\\ &=\\frac{5\\sqrt{5\\cdot7}(\\sqrt{11}+3\\sqrt{5})}{-34}\\\\ &=\\frac{5\\sqrt{5\\cdot7\\cdot11}+15\\sqrt{5\\cdot5\\cdot7}}{-34}\\\\ &=\\frac{5\\sqrt{5\\cdot7\\cdot11}+75\\sqrt{7}}{-34}\\\\ &=-\\frac5{34}\\sqrt{385}-\\frac{75}{34}\\sqrt{7}\\\\ \\end{align} - Could the downvoter please explain the downvote? – joriki Sep 25 '11 at 21:37 $@$joriki: I upvoted your answer. My only guess is that the downvoter felt somehow that you were doing the OP's homework. Of course you didn't answer the problem asked but a similar problem, so in order for the OP to adapt your answer to his question it seems necessary and sufficient for him to learn the techniques behind this type of algebra problem. In my opinion, what you are doing to the OP could best be described as \"teaching\", so I find the downvote very", "# How do you simplify 7/12 + 4/5? Jun 6, 2016 First of all you have to evaluate the common denominator between 12 and 5 searching the minimum comune multiple. Those numbers do not have any common factor so the MCM is $12 \\cdot 5 = 60$ Then you have for the first fraction $\\frac{60}{12} = 5$ to multiply for the numerator. For the second fraction you have $\\frac{60}{5} = 12$ to multiply for the numerator. $\\frac{7}{12} + \\frac{4}{5} = \\left(7 \\cdot \\frac{5}{60}\\right) + \\frac{12 \\cdot 4}{60}$ $= \\frac{35}{60} + \\frac{48}{60}$ now you can sum the numerators $\\frac{35}{60} + \\frac{48}{60}$ $= \\frac{35 + 48}{60}$ $= \\frac{83}{60}$.", "# How do you simplify 2/(3sqrt63) ? Mar 18, 2018 Shown below... #### Explanation: We need to rationalise the denominator, in this case multiply both numerator and denominator by $\\sqrt{63}$ $\\implies \\frac{2}{3 \\sqrt{63}} \\cdot \\frac{\\sqrt{63}}{\\sqrt{63}}$ $\\implies \\frac{2 \\sqrt{63}}{3 \\cdot 63}$ $\\implies \\frac{2 \\cdot \\sqrt{3 \\cdot 3 \\cdot 7}}{189}$ $\\implies \\frac{6 \\sqrt{7}}{189}$ $\\implies \\frac{2 \\sqrt{7}}{63}$ Mar 18, 2018 $\\frac{2 \\sqrt{7}}{63}$ #### Explanation: $\\frac{2}{3 \\sqrt{63}}$ $\\therefore = \\frac{2}{3 \\sqrt{3 \\cdot 3 \\cdot 7}}$ $\\therefore \\sqrt{3} \\cdot \\sqrt{3} = 3$ $\\therefore = \\frac{2}{3 \\cdot 3 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$ $\\therefore \\sqrt{7} \\times \\sqrt{7} = 7$ $\\therefore = \\frac{2 \\sqrt{7}}{9 \\times 7}$ $\\therefore = \\frac{2 \\sqrt{7}}{63}$", "How do you rationalize the denominator and simplify 12/root3(9)? May 16, 2018 $\\frac{12}{\\sqrt[3]{9}} = 4 \\sqrt[3]{3}$ Explanation: $\\frac{12}{\\sqrt[3]{9}}$ Multiplying by ${9}^{\\frac{2}{3}}$ on both numerator and denominator we get $\\frac{12 \\cdot {9}^{\\frac{2}{3}}}{\\sqrt[3]{9} \\cdot {9}^{\\frac{2}{3}}}$ $= \\frac{12 \\cdot {9}^{\\frac{2}{3}}}{{9}^{\\frac{1}{3} + \\frac{2}{3}}} = \\frac{12 \\cdot \\sqrt[3]{81}}{9}$ $\\frac{12 \\cdot 3 \\cdot \\sqrt[3]{3}}{9} = 4 \\sqrt[3]{3}$ [Ans]", "+0 Help! 0 104 1 Simplify and rationalize the denominator: $\\sqrt[3]{\\frac{8}{\\sqrt{27}}}.$ If the simplified expression can be expressed in the form $\\frac{a\\sqrt{b}}{c}$, what is $a + b + c$? Jun 12, 2019 #1 +8759 +3 $$\\sqrt[3]{\\frac{8}{\\sqrt{27}}}\\ =\\ \\Bigg(\\frac{8}{27^{\\frac12}}\\Bigg)^{\\frac13}\\ =\\ \\frac{8^{\\frac13}}{27^{\\frac12\\cdot\\frac13}}\\ =\\ \\frac{8^{\\frac13}}{27^{\\frac13\\cdot\\frac12}}\\ =\\ \\frac{8^{\\frac13}}{(27^{\\frac13})^{\\frac12}}\\ =\\ \\frac{2}{3^{\\frac12}}\\ =\\ \\frac{2}{\\sqrt3}\\ =\\ \\frac{2}{\\sqrt3}\\cdot\\frac{\\sqrt3}{\\sqrt3}\\ =\\ \\frac{2\\sqrt3}{3}$$ Now it is simplified in the form $$\\frac{a\\sqrt{b}}{c}$$ . a + b + c = 2 + 3 + 3 = 8 Jun 12, 2019 #1 +8759 +3 $$\\sqrt[3]{\\frac{8}{\\sqrt{27}}}\\ =\\ \\Bigg(\\frac{8}{27^{\\frac12}}\\Bigg)^{\\frac13}\\ =\\ \\frac{8^{\\frac13}}{27^{\\frac12\\cdot\\frac13}}\\ =\\ \\frac{8^{\\frac13}}{27^{\\frac13\\cdot\\frac12}}\\ =\\ \\frac{8^{\\frac13}}{(27^{\\frac13})^{\\frac12}}\\ =\\ \\frac{2}{3^{\\frac12}}\\ =\\ \\frac{2}{\\sqrt3}\\ =\\ \\frac{2}{\\sqrt3}\\cdot\\frac{\\sqrt3}{\\sqrt3}\\ =\\ \\frac{2\\sqrt3}{3}$$ Now it is simplified in the form $$\\frac{a\\sqrt{b}}{c}$$ . a + b + c = 2 + 3 + 3 = 8 hectictar Jun 12, 2019", "ScottyClermont 5 # what is larger 3/8 or 5/16 $\\frac{3}{8} ... \\frac{5}{16}$ First, find the least common denominator between the two. Since 8 is a factor of 16, the LCD is 16. $\\frac{3*2}{8*2}$ $\\frac{3}{8}= \\frac{6}{16}$ Now, you can compare the numerators. $6>5$ Since 6 is greater than 5, you know that $\\frac{6}{16} > \\frac{5}{16}$. Therefore, we now know that $\\frac{3}{8}$ is larger. $\\frac{3}{8}=\\frac{3\\codt2}{8\\cdot2}=\\frac{6}{16} > \\frac{5}{16}\\\\\\\\Answer:\\frac{3}{8}\\ is\\ larger\\ than\\ \\frac{5}{16}.$", "equivalent fractions. $\\frac{{5}}{{12}}=\\frac{{{5}\\cdot{\\color{green}{{{18}}}}}}{{{12}\\cdot{\\color{green}{{{18}}}}}}=\\frac{{90}}{{216}}$. $\\frac{{7}}{{18}}=\\frac{{{7}\\cdot{\\color{red}{{{12}}}}}}{{{18}\\cdot{\\color{red}{{{12}}}}}}=\\frac{{84}}{{216}}$. Now, subtract fractions $\\frac{{90}}{{216}}-\\frac{{84}}{{216}}=\\frac{{6}}{{216}}$. Reduce if possible: $\\frac{{6}}{{216}}=\\frac{{1}}{{36}}$. Answer: $\\frac{{1}}{{36}}$. Now, let's try to do above example using second way. Example 3. Find $\\frac{{5}}{{12}}-\\frac{{7}}{{18}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({12},{18}\\right)}={36}$. Find equivalent fractions. We need to multiply numerator and denominator of the first fraction by $\\frac{{36}}{{12}}={3}$ to get 36 in denominator: $\\frac{{5}}{{12}}=\\frac{{{5}\\cdot{\\color{green}{{{3}}}}}}{{{12}\\cdot{\\color{green}{{{3}}}}}}=\\frac{{15}}{{36}}$. We need to multiply numerator and denominator of the second fraction by $\\frac{{36}}{{18}}={2}$ to get 36 in denominator: $\\frac{{7}}{{18}}=\\frac{{{7}\\cdot{\\color{red}{{{2}}}}}}{{{18}\\cdot{\\color{red}{{{2}}}}}}=\\frac{{14}}{{36}}$. Now, subtract fractions $\\frac{{15}}{{36}}-\\frac{{14}}{{36}}=\\frac{{1}}{{36}}$. Reduce if possible: $\\frac{{1}}{{36}}$ is irreducible. Answer: $\\frac{{1}}{{36}}$. Note, that using second way we obtained answer without reducing fraction and calculations were simpler. Example 4. Find $-\\frac{{19}}{{8}}-\\frac{{13}}{{16}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({8},{16}\\right)}={16}$. Find equivalent fractions. We need to multiply numerator and denominator of the first fraction by $\\frac{{16}}{{8}}={2}$ to get 16 in denominator: $-\\frac{{19}}{{8}}=-\\frac{{{19}\\cdot{\\color{green}{{{2}}}}}}{{{8}\\cdot{\\color{green}{{{2}}}}}}=-\\frac{{38}}{{16}}$. Second fraction already has required denominator, so we don't need to find equivalent fraction. Now, subtract fractions $-\\frac{{38}}{{16}}-\\frac{{13}}{{16}}=\\frac{{-{38}-{13}}}{{16}}=-\\frac{{51}}{{16}}$. Reduce if possible: $-\\frac{{51}}{{16}}$ is irreducible. Answer: $-\\frac{{51}}{{16}}$. Next example. Example 5. Find $\\frac{{13}}{{6}}-\\frac{{1}}{{2}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({6},{2}\\right)}={6}$. Find equivalent fractions: First fraction already has required denominator so we don't need to find equivalent fraction. We need to multiply second fraction by $\\frac{{6}}{{2}}={3}$ to get 6 in denominator: $\\frac{{1}}{{2}}=\\frac{{{1}\\cdot{\\color{red}{{{3}}}}}}{{{2}\\cdot{\\color{red}{{{3}}}}}}=\\frac{{3}}{{6}}$. Now, add fractions $\\frac{{13}}{{6}}-\\frac{{3}}{{6}}=\\frac{{10}}{{6}}$. Reduce if possible: $\\frac{{10}}{{6}}=\\frac{{5}}{{3}}$. Answer: $\\frac{{5}}{{3}}$. Now, it", "=5+\\frac1{1+\\frac1{6+\\frac1{\\frac53}}}=5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac23}}} =5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{\\frac32}}}}=5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{1+\\frac12}}}}$$ So, the last but one convergent is $$5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{1}}}}=\\frac{88}{15}$$ So, using the convergent property of continued fraction, $223\\cdot15-38\\cdot88=1\\implies (38)^{-1}\\equiv -88\\pmod{223}\\equiv (223-88)=135$ So, $c\\equiv (38)^{-1}a\\pmod{223}\\equiv 135a$ $(-41)^{-1}$ can be calculated similarly. -", "1, 5, 25, 125, 625,... b) We need to find the common ratio first, but how do we do that when we have no terms next to each other that we can divide? Well, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice : once to get to the second term in the sequence, and again to get to five. So we can say $20 \\cdot r \\cdot r = 5$ , or $20 \\cdot r^2 = 5$ . Dividing both sides by 20, we get $r^2 = \\frac{5}{20} = \\frac{1}{4}$ , or $r = \\frac{1}{2}$ (because $\\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}$ ). To get the first missing term, we multiply 20 by $\\frac{1}{2}$ and get 10. To get the second missing term, we multiply 5 by $\\frac{1}{2}$ and get 2.5. Sequence (b) becomes: 20, 10, 5, 2.5, 1.25,... You can see that if we keep multiplying by the common ratio, we can find any term in the sequence that we want—the tenth term, the fiftieth term, the thousandth term.... However, it would be awfully tedious to keep multiplying over and over again in order to find a term that is a long way from the start. What could we do instead of just multiplying repeatedly?", "Rationalize the denominator of the fraction and enter the new denominator below. 8/2-√12 $$\\frac{8}{2-\\sqrt{12}}=\\frac{8}{2-\\sqrt{4\\cdot3}}=\\frac{8}{2-2\\sqrt3}=\\frac{8}{2(1-\\sqrt3)}=\\frac{4}{1-\\sqrt3}\\\\\\\\\\frac{4}{1-\\sqrt3}\\cdot\\frac{1+\\sqrt3}{1+\\sqrt3}=\\frac{4(1+\\sqrt3)}{1^2-(\\sqrt3)^2}=\\frac{4(1+\\sqrt3)}{1-3}=\\frac{4(1+\\sqrt3)}{-2}=-2(1+\\sqrt3) =-2-2\\sqrt3$$ $$if\\ \\frac{8}{2}-\\sqrt{12}=4-\\sqrt{4\\cdot3}=4-2\\sqrt3$$ $$\\frac{8}{2-\\sqrt{12}}\\\\ \\frac{8(2+\\sqrt{12})}{4-12}=\\\\ \\frac{8(2+\\sqrt{12})}{-8}=\\\\ -(2+\\sqrt{12})=\\\\ -2-\\sqrt{12}=\\\\-2-2\\sqrt3$$ RELATED:", "on the recto (front side) of the document; all later content in this summary is present on the verso (back side) of the papyrus. The transition from 60 to 61 is thus both a thematic and physical shift in the papyrus. 61 Seventeen multiplications are to have their products expressed as Egyptian fractions. The whole is to be given as a table. ${\\displaystyle {\\begin{bmatrix}{\\frac {2}{3}}\\cdot {\\frac {2}{3}}={\\frac {1}{3}}+{\\frac {1}{9}}&;&{\\frac {1}{3}}\\cdot {\\frac {2}{3}}={\\frac {1}{6}}+{\\frac {1}{18}}\\\\{\\frac {2}{3}}\\cdot {\\frac {1}{3}}={\\frac {1}{6}}+{\\frac {1}{18}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{6}}={\\frac {1}{12}}+{\\frac {1}{36}}\\\\{\\frac {2}{3}}\\cdot {\\frac {1}{2}}={\\frac {1}{3}}&;&{\\frac {1}{3}}\\cdot {\\frac {1}{2}}={\\frac {1}{6}}\\\\{\\frac {1}{6}}\\cdot {\\frac {1}{2}}={\\frac {1}{12}}&;&{\\frac {1}{12}}\\cdot {\\frac {1}{2}}={\\frac {1}{24}}\\\\{\\frac {1}{9}}\\cdot {\\frac {2}{3}}={\\frac {1}{18}}+{\\frac {1}{54}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{9}}={\\frac {1}{18}}+{\\frac {1}{54}}\\\\{\\frac {1}{4}}\\cdot {\\frac {1}{5}}={\\frac {1}{20}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{7}}={\\frac {1}{14}}+{\\frac {1}{42}}\\\\{\\frac {1}{2}}\\cdot {\\frac {1}{7}}={\\frac {1}{14}}&;&{\\frac {2}{3}}\\cdot {\\frac {1}{11}}={\\frac {1}{22}}+{\\frac {1}{66}}\\\\{\\frac {1}{3}}\\cdot {\\frac {1}{11}}={\\frac {1}{33}}&;&{\\frac {1}{2}}\\cdot {\\frac {1}{11}}={\\frac {1}{22}}\\\\{\\frac {1}{4}}\\cdot {\\frac {1}{11}}={\\frac {1}{44}}&&\\\\\\end{bmatrix}}}$ The syntax of the original document and its repeated multiplications indicate a rudimentary understanding that multiplication is commutative. 61B Give a general procedure for converting the product of 2/3 and the reciprocal of any (positive) odd number 2n+1 into an Egyptian fraction of two terms, e.g. ${\\displaystyle {\\frac {2}{3}}\\cdot {\\frac {1}{2n+1}}={\\frac {1}{p}}+{\\frac {1}{q}}}$ with natural p and q. In other words, find p and q in terms of n. ${\\displaystyle p=2(2n+1)}$ ${\\displaystyle q=6(2n+1)}$ Problem 61B, and the method", "# Math Help - Radical Expressions 2. Originally Posted by reiward Hi reiward, This is truly some industrial strength algebra. I got the first one down fairly simple, but couldn't manipulate the last term to very much. Maybe it's stated incorrectly or something. $\\sqrt[6]{32 \\sqrt{4n^{12}}} - \\frac{n\\sqrt{4n^2+16n+16}}{n+2}+\\frac{n-1}{\\sqrt[3]{n}-1}$ $\\sqrt[6]{32 \\cdot 2n^6} - \\frac{n\\sqrt{(2n+4)^2}}{n+2}+\\frac{n-1}{\\sqrt[3]{n}-1}$ $\\sqrt[6]{2^6 \\cdot n^6} - \\frac{n(2n+4)}{n+2}+\\frac{n-1}{\\sqrt[3]{n}-1}$ $2n - \\frac{2n(n+2}{n+2}+\\frac{n-1}{\\sqrt[3]{n}-1}$ $2n - 2n +\\frac{n-1}{\\sqrt[3]{n}-1}$ $\\frac{n-1}{\\sqrt[3]{n}-1}$ I messed around with this last term a while, but could never get it to anything simpler than it already is. Now, for the second one. $\\frac{\\sqrt{z^6}\\cdot \\sqrt[5]{2x^3y^2}}{\\sqrt[4]{4x^2y^4z^8}} \\div \\frac{\\sqrt[5]{64x^8y^{-3}}}{\\sqrt[6]{8x^3}}$ $\\frac{z^3 \\cdot 2^{\\frac{1}{5}}x^{\\frac{3}{5}}y^{\\frac{2}{5}}}{2^{ \\frac{1}{2}}x^{\\frac{1}{2}}yz^2}\\cdot \\frac{2^{\\frac{1}{2}}x^{\\frac{1}{2}}y^{\\frac{3}{5} }}{2\\cdot 2^{\\frac{1}{5}}x\\cdot x^{\\frac{3}{5}}}=\\frac{zy}{2xy}=\\frac{z}{2x}$ 3. ## thanks! thank you! maybe thats the final answer for number 1. oh well, thank you very much 4. hi, we discussed the answers already, just wanna share. for the last term this is what we did. $\\frac{n-1}{\\sqrt[3]{n}-1} * \\frac{\\sqrt[3]{n}^2 + \\sqrt[3]{n} + 1}{\\sqrt[3]{n}^2 + \\sqrt[3]{n} + 1}$ then $\\frac{n-1} {n-1} \\sqrt[3]{n}^2 + \\sqrt[3]{n} + 1 $ $\\sqrt[3]{n}^2 + \\sqrt[3]{n} + 1", "# How do you find the missing terms in the geometric sequence 32, __, 72, __, 162? Nov 12, 2015 Here is the sequence: $32 , 48 , 72 , 108 , 162$ #### Explanation: In a geometric series, we go from one term to another by multiplying by some number $r$. So if we call our first missing term $x$, we know the following: $32 \\cdot r = x$ $x \\cdot r = 72$ substitute to find the following... $32 \\cdot r \\cdot r = 72$ ${r}^{2} = \\frac{72}{32}$ ${r}^{2} = \\frac{9}{4}$ $r = \\frac{3}{2}$ now we can find the two missing terms... $32 \\cdot \\frac{3}{2} = 48$ $72 \\cdot \\frac{3}{2} = 108$" ]

[ "# What's the common ratio in 5/12, 1/3, 4/15, 16/75, 64/375? Dec 3, 2015 $\\frac{4}{5}$ #### Explanation: Divide any number in this sequence by the preceding number and you get $\\frac{4}{5}$. That's the common ratio. For example: $\\frac{1}{3} \\div \\frac{5}{12} = \\frac{1}{3} \\cdot \\frac{12}{5} = \\frac{4}{5}$, $\\frac{4}{15} \\div \\frac{1}{3} = \\frac{4}{15} \\cdot 3 = \\frac{12}{15} = \\frac{4}{5}$, $\\frac{16}{75} \\div \\frac{4}{15} = \\frac{16}{75} \\cdot \\frac{15}{4} = \\frac{4}{5}$, and $\\frac{64}{375} \\div \\frac{16}{75} = \\frac{64}{375} \\cdot \\frac{75}{16} = \\frac{4}{5}$. A cool thing related to this is the corresponding geometric series (infinite \"sum\") $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots$. If $a$ is the first term of such a series, and $r$ is the common ratio, with $| r | < 1$, the series converges to $\\frac{a}{1 - r}$. For this example, $a = \\frac{5}{12}$ and $r = \\frac{4}{5}$ so that the series converges to $\\frac{\\frac{5}{12}}{1 - \\frac{4}{5}} = \\frac{\\frac{5}{12}}{\\frac{1}{5}} = \\frac{5}{12} \\cdot 5 = \\frac{25}{12}$ and we write $\\frac{5}{12} + \\frac{1}{3} + \\frac{4}{15} + \\frac{16}{75} + \\frac{64}{375} + \\cdots = \\frac{25}{12}$.", "example: $\\frac 3 5 \\cdot \\frac 5 6 = \\frac {3 \\cdot 5} {5 \\cdot 6} = \\frac {5 \\cdot 3} {5 \\cdot 6} = \\frac 5 5 \\cdot \\frac 3 6 = 1 \\cdot \\frac 3 {2 \\cdot 3} = 1 \\cdot \\frac 3 3 \\cdot \\frac 1 2 = 1 \\cdot 1 \\cdot \\frac 1 2 = \\frac 1 2$ 8. Originally Posted by tkdvipers Cheers I am getting there, What confuses me is the way that. $ \\frac{3}{7} * \\frac{5}{6} $ becomes $ \\frac{1}{7} * \\frac{5}{2} $ I can see that 6/3 = 2 and 3/3 = 1, however I thought the fractions were unrelated and you could only work within the same fraction. for example 3/7 and 5/6 I can do the sums now but want to understand why it is legal. Look at the cancelation that took place in red. The 3 and the 6 have a common factor of 3. $ \\frac{3}{7} * \\frac{5}{6} = \\color{red}\\frac{\\rlap{---}{3}^1}{7} * \\color{red}\\frac{5}{\\rlap{---}{6}^2} =\\color{black}\\frac{1}{7}\\cdot\\frac{5}{2}= \\frac{1\\cdot5}{7\\cdot2}=\\frac{5}{14} $ Another approach: $\\frac{3}{7} \\cdot \\frac{5}{6}=\\frac{3}{6} \\cdot \\frac{5}{7}=\\color{red} \\frac{\\rlap{---}{3}^1}{\\rlap{---}{6}^2}\\cdot\\frac{5}{7}= \\color{black}\\frac{1\\cdot5}{2\\cdot7}=\\frac{5}{14} $", "1, 5, 25, 125, 625,... b) We need to find the common ratio first, but how do we do that when we have no terms next to each other that we can divide? Well, we know that to get from 20 to 5 in the sequence we must multiply 20 by the common ratio twice : once to get to the second term in the sequence, and again to get to five. So we can say $20 \\cdot r \\cdot r = 5$ , or $20 \\cdot r^2 = 5$ . Dividing both sides by 20, we get $r^2 = \\frac{5}{20} = \\frac{1}{4}$ , or $r = \\frac{1}{2}$ (because $\\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}$ ). To get the first missing term, we multiply 20 by $\\frac{1}{2}$ and get 10. To get the second missing term, we multiply 5 by $\\frac{1}{2}$ and get 2.5. Sequence (b) becomes: 20, 10, 5, 2.5, 1.25,... You can see that if we keep multiplying by the common ratio, we can find any term in the sequence that we want—the tenth term, the fiftieth term, the thousandth term.... However, it would be awfully tedious to keep multiplying over and over again in order to find a term that is a long way from the start. What could we do instead of just multiplying repeatedly?", "5/12, $\\frac{5}{12}=\\frac{5}{12} \\cdot \\color{blue}{1}=\\frac{5}{12} \\cdot \\color{blue}{\\frac{3}{3}}=\\frac{15}{36}$ In the case of 5/18, $\\frac{5}{18}=\\frac{5}{18} \\cdot \\color{blue}{1}=\\frac{5}{18} \\cdot \\color{blue}{\\frac{2}{2}}=\\frac{10}{36}$ If we replace the fractions in equation (8) with their equivalent fractions, we can then add the numerators and divide by the common denominator, as in $\\frac{5}{12}+\\frac{5}{18}=\\frac{15}{36}+\\frac{10}{36}=\\frac{15+10}{36}=\\frac{25}{36}$ Let’s examine a method of organizing the work that is more compact. Consider the following arrangement, where we’ve used color to highlight the form of 1 required to convert the fractions to equivalent fractions with a common denominator of 36. \\begin{aligned} \\frac{5}{12}+\\frac{5}{18} &=\\frac{5}{12} \\cdot \\color{blue}{\\frac{3}{3}}+\\frac{5}{18} \\cdot \\color{blue}{\\frac{2}{2}} \\\\ &=\\frac{15}{36}+\\frac{10}{36} \\\\ &=\\frac{25}{36} \\end{aligned} Let’s look at a more complicated example. Example $$\\PageIndex{3}$$ Simplify the expression $\\frac{x+3}{x+2}-\\frac{x+2}{x+3}$ State all restrictions. Solution The denominators are already factored. If we take each factor that appears to the highest exponential power that appears, our least common denominator is (x+2)(x+3). Our first task is to make equivalent fractions having this common denominator. \\begin{aligned} \\frac{x+3}{x+2}-\\frac{x+2}{x+3} &=\\frac{x+3}{x+2} \\cdot \\color{blue}{\\frac{x+3}{x+3}}-\\frac{x+2}{x+3} \\cdot \\color{blue}{\\frac{x+2}{x+2}} \\\\ &=\\frac{x^{2}+6 x+9}{(x+2)(x+3)}-\\frac{x^{2}+4 x+4}{(x+2)(x+3)} \\end{aligned} Now, subtract the numerators and divide by the common denominator. \\begin{aligned} \\frac{x+3}{x+2}-\\frac{x+2}{x+3} &=\\frac{\\left(x^{2}+6 x+9\\right)-\\left(x^{2}+4 x+4\\right)}{(x+2)(x+3)} \\\\ &=\\frac{x^{2}+6 x+9-x^{2}-4 x-4}{(x+2)(x+3)} \\\\ &=\\frac{2 x+5}{(x+2)(x+3)} \\end{aligned} Note the use of parentheses when we subtracted the numerators. Note further how the minus sign negates each term in the parenthetical expression that follows the minus sign. Tip Always use grouping symbols when", "and work inside-out. $\\sin \\frac{\\pi}{14} \\sin \\frac{6\\pi}{14} = \\frac{1}{2}\\cos \\frac{5\\pi}{14}$ $\\sin \\frac{2\\pi}{14} \\sin\\frac{5\\pi}{14} = \\frac{1}{2}\\cos \\frac{3\\pi}{14}$ $\\sin \\frac{3\\pi}{14}\\sin \\frac{4\\pi}{13} = \\frac{1}{2}\\cos \\frac{\\pi}{14}$. Thus, $\\cos \\frac{\\pi}{14} \\cos \\frac{3\\pi}{14} \\cos \\frac{5\\pi}{14} = \\frac{\\sqrt{7}}{2^3}$. Generalize! If $n$ is odd then: $\\boxed{ \\cos \\frac{\\pi}{2n} \\cos \\frac{3\\pi}{2n} \\cdot ... \\cdot \\cos \\frac{(n-2)\\pi}{2n} = \\frac{\\sqrt{n}}{2^{(n-1)/2}} }$.", "equivalent fractions. $\\frac{{5}}{{12}}=\\frac{{{5}\\cdot{\\color{green}{{{18}}}}}}{{{12}\\cdot{\\color{green}{{{18}}}}}}=\\frac{{90}}{{216}}$. $\\frac{{7}}{{18}}=\\frac{{{7}\\cdot{\\color{red}{{{12}}}}}}{{{18}\\cdot{\\color{red}{{{12}}}}}}=\\frac{{84}}{{216}}$. Now, subtract fractions $\\frac{{90}}{{216}}-\\frac{{84}}{{216}}=\\frac{{6}}{{216}}$. Reduce if possible: $\\frac{{6}}{{216}}=\\frac{{1}}{{36}}$. Answer: $\\frac{{1}}{{36}}$. Now, let's try to do above example using second way. Example 3. Find $\\frac{{5}}{{12}}-\\frac{{7}}{{18}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({12},{18}\\right)}={36}$. Find equivalent fractions. We need to multiply numerator and denominator of the first fraction by $\\frac{{36}}{{12}}={3}$ to get 36 in denominator: $\\frac{{5}}{{12}}=\\frac{{{5}\\cdot{\\color{green}{{{3}}}}}}{{{12}\\cdot{\\color{green}{{{3}}}}}}=\\frac{{15}}{{36}}$. We need to multiply numerator and denominator of the second fraction by $\\frac{{36}}{{18}}={2}$ to get 36 in denominator: $\\frac{{7}}{{18}}=\\frac{{{7}\\cdot{\\color{red}{{{2}}}}}}{{{18}\\cdot{\\color{red}{{{2}}}}}}=\\frac{{14}}{{36}}$. Now, subtract fractions $\\frac{{15}}{{36}}-\\frac{{14}}{{36}}=\\frac{{1}}{{36}}$. Reduce if possible: $\\frac{{1}}{{36}}$ is irreducible. Answer: $\\frac{{1}}{{36}}$. Note, that using second way we obtained answer without reducing fraction and calculations were simpler. Example 4. Find $-\\frac{{19}}{{8}}-\\frac{{13}}{{16}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({8},{16}\\right)}={16}$. Find equivalent fractions. We need to multiply numerator and denominator of the first fraction by $\\frac{{16}}{{8}}={2}$ to get 16 in denominator: $-\\frac{{19}}{{8}}=-\\frac{{{19}\\cdot{\\color{green}{{{2}}}}}}{{{8}\\cdot{\\color{green}{{{2}}}}}}=-\\frac{{38}}{{16}}$. Second fraction already has required denominator, so we don't need to find equivalent fraction. Now, subtract fractions $-\\frac{{38}}{{16}}-\\frac{{13}}{{16}}=\\frac{{-{38}-{13}}}{{16}}=-\\frac{{51}}{{16}}$. Reduce if possible: $-\\frac{{51}}{{16}}$ is irreducible. Answer: $-\\frac{{51}}{{16}}$. Next example. Example 5. Find $\\frac{{13}}{{6}}-\\frac{{1}}{{2}}$. Find least common multiple of denominators: ${L}{C}{M}{\\left({6},{2}\\right)}={6}$. Find equivalent fractions: First fraction already has required denominator so we don't need to find equivalent fraction. We need to multiply second fraction by $\\frac{{6}}{{2}}={3}$ to get 6 in denominator: $\\frac{{1}}{{2}}=\\frac{{{1}\\cdot{\\color{red}{{{3}}}}}}{{{2}\\cdot{\\color{red}{{{3}}}}}}=\\frac{{3}}{{6}}$. Now, add fractions $\\frac{{13}}{{6}}-\\frac{{3}}{{6}}=\\frac{{10}}{{6}}$. Reduce if possible: $\\frac{{10}}{{6}}=\\frac{{5}}{{3}}$. Answer: $\\frac{{5}}{{3}}$. Now, it", "& = \\frac { 1 } { ( 2 - 3 i ) } \\cdot \\color{Cerulean}{\\frac { ( 2 + 3 i ) } { ( 2 + 3 i ) }} \\\\ & = \\frac { ( 2 + 3 i ) } { 2 ^ { 2 } + 3 ^ { 2 } } \\\\ & = \\frac { 2 + 3 i } { 4 + 9 } \\\\ & = \\frac { 2 + 3 i } { 13 } \\end{aligned} To write this complex number in standard form, we make use of the fact that 13 is a common denominator. \\begin{aligned} \\frac { 2 + 3 i } { 13 } & = \\frac { 2 } { 13 } + \\frac { 3 i } { 13 } \\\\ & = \\frac { 2 } { 13 } + \\frac { 3 } { 13 } i \\end{aligned} $$\\frac { 2 } { 13 } + \\frac { 3 } { 13 } i$$ Example $$\\PageIndex{9}$$: Divide: $$\\frac { 1 - 5 i } { 4 + i }$$. Solution \\begin{aligned} \\frac { 1 - 5 i } { 4 + i } & = \\frac { ( 1 - 5 i ) } { ( 4 + i ) }", "\\textcolor{red}{3}} & = & 5 \\frac{9}{12} \\\\ -2 \\frac{1}{3} & = & -3 \\frac{1 \\cdot \\textcolor{red}{4}}{3 \\cdot \\textcolor{red}{4}} & = & -2 \\frac{4}{12} \\\\ \\hline & & & & 3 \\frac{5}{12} \\end{array}\\nonumber$ Note that the answer is identical to the answer found in Example 6 and Example 7. That is, $5 \\frac{3}{4} - 2 \\frac{1}{3} = 3 \\frac{5}{12}.\\nonumber$ Borrowing in Vertical Format Consider the following example. Example 8 Simplify: $$8 \\frac{1}{4} - 5 \\frac{5}{6}$$. Solution Create equivalent fractions with a common denominator. $\\begin{array}{r r r r r} 8 \\frac{1}{4} & = & 8 \\frac{1 \\cdot \\textcolor{red}{3}}{4 \\cdot \\textcolor{red}{3}} & = & 8 \\frac{3}{12} \\\\ -5 \\frac{5}{6} & = & -5 \\frac{5 \\cdot \\textcolor{red}{2}}{6 \\cdot \\textcolor{red}{2}} & = & -5 \\frac{10}{12} \\\\ \\hline \\end{array}\\nonumber$ You can see the difficulty. On the far right, we cannot subtract 10/12 from 3/12. The fix is to borrow 1 from 8 in the form of 12/12 and add it to the 3/12. $\\begin{array}{r r r r r} 8 \\frac{3}{12} & = & 7 + \\frac{12}{12} + \\frac{3}{12} & = & 7 \\frac{15}{12} \\\\ -5 \\frac{10}{12} & = & -5 \\frac{10}{12} & = & -5 \\frac{10}{12} \\\\ \\hline & & & & 2 \\frac{5}{12} \\end{array}\\nonumber$ Now we can subtract. Hence, $$8 \\frac{1}{4} − 5 \\frac{5}{6} = 2 \\frac{5}{12}$$. Exercise Simplify: $$7 \\frac{1}{14} - 2", "# How do you solve 1/2+7x/10=13/20? Jul 28, 2015 answer 1: color(blue)(x=-137/2 answer 2 :color(blue)(x=3/14 #### Explanation: 1. Assuming that $7 \\left(\\frac{x}{10}\\right)$is a mixed fraction: $\\frac{1}{2} + 7 \\left(\\frac{x}{10}\\right) = \\frac{13}{20}$ Solving the mixed fraction: $\\frac{1}{2} + \\frac{10 \\cdot 7 + x}{10} = \\frac{13}{20}$ $\\frac{1}{2} + \\frac{70 + x}{10} = \\frac{13}{20}$ The L.C.M of the expression is $20$ $\\frac{1 \\cdot 10}{2 \\cdot 10} + \\frac{\\left(70 + x\\right) \\cdot 2}{10 \\cdot 2} = \\frac{13}{20}$ $\\frac{10}{\\cancel{20}} + \\frac{140 + 2 x}{\\cancel{20}} = \\frac{13}{\\cancel{20}}$ $10 + 140 + 2 x = 13$ $2 x = 13 - 150$ answer 1: color(blue)(x=-137/2 1. Taking $7 \\frac{x}{10}$ as a normal fraction: $\\frac{1}{2} + 7 \\frac{x}{10} = \\frac{13}{20}$ The L.C.M of the expression is $20$ $\\frac{1 \\cdot 10}{2 \\cdot 10} + \\frac{7 x \\cdot 2}{10 \\cdot 2} = \\frac{13}{20}$ $\\frac{10}{\\cancel{20}} + \\frac{14 x}{\\cancel{20}} = \\frac{13}{\\cancel{20}}$ $10 + 14 x = 13$ $14 x = 3$ answer 2 :color(blue)(x=3/14", "# How do you simplify 2/3 + 1/6 ? Jul 26, 2017 $\\frac{5}{6}$ #### Explanation: $\\text{before we can add the fractions we require them to have}$ $\\text{a \"color(blue)\"common denominator}$ $\\text{the lowest common multiple (LCM) of 6 and 3 is 6}$ $\\Rightarrow \\frac{2}{3} = \\frac{2}{3} \\times \\frac{2}{2} = \\frac{4}{6}$ $\\Rightarrow \\frac{2}{3} + \\frac{1}{6} = \\frac{4}{6} + \\frac{1}{6} = \\frac{5}{6}$ Jul 26, 2017 $\\frac{5}{6}$ #### Explanation: First, to really simplify things, we need a LCM of teh denominators. $\\therefore \\frac{2 \\cdot 2}{3 \\cdot 2} + \\frac{1}{6}$ $= \\frac{4 + 1}{6}$ $= \\frac{5}{6}$", "multiplied by the first term. Here, the first term is $-\\frac 1e$ Hence, $s(1)=-\\frac 1e \\cdot \\frac{e}{1+e}=-\\frac{1}{1+e}$", "# How do you simplify 2/(3sqrt63) ? Mar 18, 2018 Shown below... #### Explanation: We need to rationalise the denominator, in this case multiply both numerator and denominator by $\\sqrt{63}$ $\\implies \\frac{2}{3 \\sqrt{63}} \\cdot \\frac{\\sqrt{63}}{\\sqrt{63}}$ $\\implies \\frac{2 \\sqrt{63}}{3 \\cdot 63}$ $\\implies \\frac{2 \\cdot \\sqrt{3 \\cdot 3 \\cdot 7}}{189}$ $\\implies \\frac{6 \\sqrt{7}}{189}$ $\\implies \\frac{2 \\sqrt{7}}{63}$ Mar 18, 2018 $\\frac{2 \\sqrt{7}}{63}$ #### Explanation: $\\frac{2}{3 \\sqrt{63}}$ $\\therefore = \\frac{2}{3 \\sqrt{3 \\cdot 3 \\cdot 7}}$ $\\therefore \\sqrt{3} \\cdot \\sqrt{3} = 3$ $\\therefore = \\frac{2}{3 \\cdot 3 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$ $\\therefore \\sqrt{7} \\times \\sqrt{7} = 7$ $\\therefore = \\frac{2 \\sqrt{7}}{9 \\times 7}$ $\\therefore = \\frac{2 \\sqrt{7}}{63}$", "=5+\\frac1{1+\\frac1{6+\\frac1{\\frac53}}}=5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac23}}} =5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{\\frac32}}}}=5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{1+\\frac12}}}}$$ So, the last but one convergent is $$5+\\frac1{1+\\frac1{6+\\frac1{1+\\frac1{1}}}}=\\frac{88}{15}$$ So, using the convergent property of continued fraction, $223\\cdot15-38\\cdot88=1\\implies (38)^{-1}\\equiv -88\\pmod{223}\\equiv (223-88)=135$ So, $c\\equiv (38)^{-1}a\\pmod{223}\\equiv 135a$ $(-41)^{-1}$ can be calculated similarly. -", "1. Simplifying expressions I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 2. 1 - 1 x + h x 3. 2 x^2 10 x^5 4. 2x - 1 - 8 x^2-6x+9 x+1 x^2-2x-3 2. Originally Posted by journeygirl280 1. x^3-8 x-2 Hint: Difference two two cubes in numerator. 2. 1 - 1 x + h x With common denominator, $\\frac{x}{x(x+h)} - \\frac{x+h}{x(x+h)}$. Now continue. The last two do not make any sense. 3. Hello, journeygirl280! I think I deciphered the last two problems . . . $3)\\;\\;\\frac{\\frac{2}{x^2}}{\\frac{10}{x^5}}$ We have: . $\\frac{2}{x^2} \\div \\frac{10}{x^5} \\;= \\;\\frac{2}{x^2}\\cdot\\frac{x^5}{10} \\;=\\;\\frac{x^3}{5}$ $4)\\;\\;\\frac{2x}{x^2-6x+9} - \\frac{1}{x+1} - \\frac{8}{x^2-2x-3}$ We have: . $\\frac{2x}{(x-3)^2} - \\frac{1}{x+1} - \\frac{8}{(x+1)(x-3)}$ Get a common denominator: . . $\\frac{2x}{(x-3)^2}\\cdot{\\color{blue}\\frac{x+1}{x+1}} \\,-\\,\\frac{1}{x+1}\\cdot{\\color{blue}\\frac{(x-3)^2}{(x-3)^2}} \\,-\\,\\frac{8}{(x+1)(x-3)}\\cdot{\\color{blue}\\frac{x-3}{x-3}}$ . . $= \\;\\frac{2x(x+1)-(x-3)^2 - 8(x-3)}{(x-3)^2(x+1)} \\;= \\;\\frac{2x^2 + 2x - x^2 + 6x - 9 - 8x + 24}{(x-3)^2(x+1)} $ . . $= \\;\\frac{x^2+15}{(x-3)^2(x+1)}$ 4. Originally Posted by journeygirl280 I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 Different hint: $x^3-8=0$ when $x=2$ so $x-2$ is a factor of $x^3-8$. (well really the same hint in disguise) RonL", "$\\frac{1}{3} , \\frac{2}{9} , \\frac{3}{27} , \\ldots$ What does the $\\ldots$ signify? One possibility is that the standard term is $\\frac{n}{3} ^ n$, giving us: $\\frac{1}{3} , \\frac{2}{9} , \\frac{3}{27} , \\frac{4}{81} , \\frac{5}{243} , \\frac{6}{729} , \\ldots , \\frac{n}{3} ^ n , \\ldots$ Another is that this is simply an arithmetic sequence with initial term $\\frac{1}{3} = \\frac{3}{9}$ and common difference $- \\frac{1}{9}$, writable as: $\\frac{3}{9} , \\frac{2}{9} , \\frac{1}{9} , \\frac{0}{9} , - \\frac{1}{9} , - \\frac{2}{9} , \\ldots , \\frac{4 - n}{9} , \\ldots$", "\\frac{1}{4} = \\frac{12}{4} + \\frac{1}{4} = \\frac{13}{4}$. Therefore, the original expression, once simplified is as follows: $\\frac{3}{4} n + 3 m - \\frac{1}{3} n + \\frac{1}{4} m = \\left(\\frac{3}{4} - \\frac{1}{3}\\right) \\cdot n + \\left(3 + \\frac{1}{4}\\right) \\cdot m =$ $= \\frac{5}{12} n + \\frac{13}{4} m$.", "+0 Help 0 242 3 Simplify: $\\frac{1}{\\sqrt{2}+\\frac{1}{\\sqrt{8}+\\sqrt{200}+\\frac{1}{\\sqrt{18}}}}$. Jun 13, 2019 #1 +224 +1 $$\\frac{1}{\\sqrt{2}+\\frac{1}{\\sqrt{8}+\\sqrt{200}+\\frac{1}{\\sqrt{18}}}}$$ First, simplify the radicles in the expression $$\\sqrt{8}=\\sqrt{2 \\cdot 2\\cdot 2} =2\\sqrt{2}$$ $$\\sqrt{200}=\\sqrt{10\\cdot20\\cdot2}=10\\sqrt{2}$$ $$\\sqrt{18}=\\sqrt{3\\cdot3\\cdot2}=3\\sqrt{2}$$ $$\\frac{1}{\\sqrt{2}+\\frac{1}{2\\sqrt{2}+10\\sqrt{2}+\\frac{1}{3\\sqrt{2}}}}$$ The two radicles can be added together because they have the same base $$\\frac{1}{\\sqrt{2}+\\frac{1}{12\\sqrt{2}+\\frac{1}{3\\sqrt{2}}}}$$ Now start simplifying from the innermost fraction Find a common denominator to add the two on the bottom $$\\frac{1}{\\sqrt{2}+\\frac{1}{\\frac{73}{3\\sqrt{2}}}}$$ $$\\frac{1}{\\sqrt{2}+\\frac{3\\sqrt{2}}{73}}$$ Find a common denominator again $$\\frac{1}{\\frac{73\\sqrt{2}}{73}+\\frac{3\\sqrt{2}}{73}}$$ $$\\frac{1}{\\frac{76\\sqrt{2}}{73}}$$ $$\\frac{73}{76\\sqrt{2}}$$ Multiply by $$\\frac{\\sqrt{2}}{\\sqrt{2}}$$ to get the radicle out of the denominator $$\\boxed{\\frac{73\\sqrt{2}}{152}}$$ Jun 14, 2019 edited by power27 Jun 14, 2019 #2 +1 You got the answer right at this step: 73 / 76 * sqrt(2) . How did you \"factor out\" 73? Jun 14, 2019 #3 +224 +1 Oh oops thanks for catching that you can't do that. power27 Jun 14, 2019", "2 && (\\text{divide} \\ 30 \\div \\ 15 \\ \\text{to get} \\ 2) \\\\& 11, \\frac{11} {2}, \\frac{11} {4}, \\frac{11} {8}, \\frac{11} {16},\\ldots && r = \\frac{1} {2} && \\left (\\text{divide} \\ \\frac{1} {2} \\div 11 \\ \\text{to get} \\ \\frac{1} {2}\\right ) \\\\& 25, -5, 1, -\\frac{1} {5}, - \\frac{1} {25},\\ldots && r = - \\frac{1} {5} && \\left (\\text{divide} \\ 1 \\div (-5) \\ \\text{to get} - \\frac{1} {5}\\right )$ If we know the common ratio, we can find the next term in the sequence just by multiplying the last term by it. Also, if there are any terms missing in the sequence, we can find them by multiplying the terms before the gap by the common ratio. Example 1 Fill is the missing terms in the geometric sequences. a) $1,\\underline{\\;\\;\\;\\;\\;\\;\\;\\;}, 25, 125,\\underline{\\;\\;\\;\\;\\;\\;\\;\\;}$ b) $20,\\underline{\\;\\;\\;\\;\\;\\;\\;\\;}, 5,\\underline{\\;\\;\\;\\;\\;\\;\\;\\;}, 1.25$ Solution a) First we can find the common ratio by dividing 125 by 25 to obtain $r=5$. To find the $1^{st}$ missing term we multiply 1 by the common ratio $1 \\cdot 5 = 5$ To find the $2^{nd}$ missing term we multiply 125 by the common ratio $125 \\cdot 5 = 625$ Answer Sequence (a) becomes 1, 5, 25, 125, 625. b) We first need to find the common ratio, but we run into difficulty because we have", "1} + \\frac{5}{x - 2} = \\frac{2}{x - 1}\\cdot \\frac{x - 2}{x - 2} + \\frac{5}{x - 2} \\cdot \\frac{x - 1}{x - 1}$ $= \\frac{2(x - 2)}{(x - 1)(x - 2)} + \\frac{5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2(x - 2) + 5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2x - 4 + 5x - 5}{(x - 1)(x - 2)}$ $= \\frac{7x - 9}{(x - 1)(x - 2)}$. So $\\frac{\\frac{1}{x - 2} + \\frac{3}{x - 1}}{\\frac{2}{x - 1} + \\frac{5}{x - 2}} = \\frac{\\frac{4x - 7}{(x - 1)(x - 2)}}{\\frac{7x - 9}{(x - 1)(x - 2)}}$ $= \\frac{4x - 7}{(x - 1)(x - 2)}\\cdot \\frac{(x - 1)(x - 2)}{7x - 9}$ $= \\frac{4x - 7}{7x - 9}$. Solve: 3. $y - \\frac{14}{y} = 5$ 4. $\\frac{4}{a-7} = \\frac{-2a}{a+3}$ Again, for these I'd have to find a common denominator, correct? And than multiply everything by it? So number 3 the denominator would by Y and for number 4 the denominator would be... A? Maybe I'm no where close... 3. $y - \\frac{14}{y} = 5$. Multiply both sides of the equation by $y$. $y^2 - 14 = 5y$ $y^2 - 5y - 14 = 0$ $(y - 7)(y + 2) = 0$ $y - 7 = 0$ or $y + 2 = 0$ $y = 7$", "# How do you find the missing terms in the geometric sequence 32, __, 72, __, 162? Nov 12, 2015 Here is the sequence: $32 , 48 , 72 , 108 , 162$ #### Explanation: In a geometric series, we go from one term to another by multiplying by some number $r$. So if we call our first missing term $x$, we know the following: $32 \\cdot r = x$ $x \\cdot r = 72$ substitute to find the following... $32 \\cdot r \\cdot r = 72$ ${r}^{2} = \\frac{72}{32}$ ${r}^{2} = \\frac{9}{4}$ $r = \\frac{3}{2}$ now we can find the two missing terms... $32 \\cdot \\frac{3}{2} = 48$ $72 \\cdot \\frac{3}{2} = 108$" ]

[ "## Prealgebra (7th Edition) If the top third is removed, $\\frac{2}{3}$ will remain. $8.4\\times\\frac{2}{3}=5.6$", "x}(a, b)& \\frac{\\partial f}{\\partial y}(a,b)}$, and $Df(a, b)$ is just left multiplication by $f'(a, b)$ – Vectk Aug 10 '12 at 0:43", "that case, you leave it till the following step: $$\\frac{1}{4}\\left(1-\\frac{1}{n}\\right)^2$$", "# Question #ba7a0 Mar 15, 2017 $\\left(\\frac{6.3}{100}\\right)$ is written out (if you don’t use a calculator) as:", "Subbing $- \\frac{17}{2}$ we get $- 3.4 = 0.4 \\left(- \\frac{17}{2}\\right)$ which $- 3.4 = - 3.4$", "the subtraction of these fractions: $\\frac{4-3}{6}$ = $\\frac{1}{6}$", "What should be subtracted from Question: What should be subtracted from $\\left(\\frac{3}{4}-\\frac{2}{3}\\right)$ to get $\\frac{-1}{6} ?$ Solution: Let $x$ be subtracted. $\\therefore\\left(\\frac{3}{4}-\\frac{2}{3}\\right)-x=\\frac{-1}{6}$ $\\Rightarrow\\left(\\frac{9}{12}-\\frac{8}{12}\\right)-x=\\frac{-1}{6}$ $\\Rightarrow\\left(\\frac{9-8}{12}\\right)-x=\\frac{-1}{6}$ $\\Rightarrow x=\\frac{1}{12}-\\frac{-1}{6}$ $\\Rightarrow x=\\frac{1}{12}-\\frac{-2}{12}$ $\\Rightarrow x=\\frac{1-(-2)}{12}$ $\\Rightarrow x=\\frac{1+2}{12}$ $\\Rightarrow x=\\frac{3}{12}$ $\\Rightarrow x=\\frac{1}{4}$", "# What is (4^2-8)*2^3-8*5-:10? Jun 7, 2018 $\\left[\\left(16 - 8\\right) \\times 8\\right] - \\frac{40}{10} = 64 - 4 = 60$ #### Explanation: It's actually extremely simple if you know the order of operations. PEMDAS! • Parentheses • Exponent • Multiplication • Division • Subtraction It also goes left to right! So beginning with the parentheses, we focus on what is in the parentheses $\\left({4}^{2} - 8\\right)$ meaning $16$, which is the square of $4$, subtracted by $8$. which turns in to $8$. $\\left({4}^{2} - 8\\right) = 16 - 8 = 8$ After the paratheses, you multiply by $8$ the cube of $2$, which is then subtracted by $4$. $8 \\times {2}^{3} = 8 \\times 8 = 64$ Since, if you follow PEMDAS and multiply then divide from left to right, it's $\\frac{40}{10}$, which is $4$. So the answer, in the end, is $64 - \\frac{40}{10} = 64 - 4 = 60$", "then subtraction are carried out from left to right. $\\frac{1}{7}$ + [$\\frac{2}{9} - \\frac{2}{9}$] $\\frac{1}{7}$ + [ 0 ] Step 3: Simplify $\\frac{1}{7}$", "result of subtracting $3\\frac{1}{10}$ from $8\\frac{2}{5}$. Solution", "# A number subtracted from 16 is -3. What is the number? $16 - x = - 3$ $16 - x - 16 = - 3 - 16$ $- x = - 19$ $\\frac{- 1 x}{-} 1 = \\frac{- 19}{-} 1$ $x = 19$", "Home » » What fraction must be subtracted from the sum of $$2\\frac{1}{6}$$ and $$3\\frac{1}{4}$$?... # What fraction must be subtracted from the sum of $$2\\frac{1}{6}$$ and $$3\\frac{1}{4}$$?... ### Question What fraction must be subtracted from the sum of $$2\\frac{1}{6}$$ and $$3\\frac{1}{4}$$? ### Options A) $$\\frac{1}{3}$$ B) $$\\frac{1}{2}$$ C) $$1\\frac{1}{6}$$ D) $$1\\frac{1}{2}$$", "the $15$ total parts, and so has $1$ part out of $15$ (or $\\frac{1}{15}$) remaining to learn.", "\\left( 1 + y^2 \\right)$ now replace $y$ with $x^{ \\frac 12}$", "# Question #ba7a0 $\\left(\\frac{6.3}{100}\\right)$ is written out (if you don’t use a calculator) as:", "exercise towards showing that $\\;\\left(\\frac1{1+x}\\right)'=-\\frac1{(1+x)^2}\\;$ . At this level, I can't understand how the OP can't do or doesn't understand sum/substraction of fractions...+1 , of course. – DonAntonio Mar 1 '14 at 17:10", "the subproblem sizes left at depth $j$, $$A(n) \\le \\frac{1}{n^{k / 2}} \\sum_j\\sum_i \\frac{1}{a}.$$", "I do with terms like $f\\left( \\frac{x+1}{1-3x} \\right)$? What happens if you add them?", "# What is 1/8 divided by 1/4? $\\frac{1}{2}$ $\\frac{\\frac{1}{8}}{\\frac{1}{4}}$ becomes $\\frac{1}{8} \\times \\frac{4}{1}$ $\\implies$ $\\frac{4}{8}$=$\\frac{1}{2}$", "then, is $\\left(\\frac{3}{2} , 2\\right)$." ]

[ "ScottyClermont 5 # what is larger 3/8 or 5/16 $\\frac{3}{8} ... \\frac{5}{16}$ First, find the least common denominator between the two. Since 8 is a factor of 16, the LCD is 16. $\\frac{3*2}{8*2}$ $\\frac{3}{8}= \\frac{6}{16}$ Now, you can compare the numerators. $6>5$ Since 6 is greater than 5, you know that $\\frac{6}{16} > \\frac{5}{16}$. Therefore, we now know that $\\frac{3}{8}$ is larger. $\\frac{3}{8}=\\frac{3\\codt2}{8\\cdot2}=\\frac{6}{16} > \\frac{5}{16}\\\\\\\\Answer:\\frac{3}{8}\\ is\\ larger\\ than\\ \\frac{5}{16}.$", "## JazzyPoowh 3 years ago place the fraction in order from least to greatest , 1. JazzyPoowh 3/7 1/3 2/5 and 3/8 2. myininaya $\\frac{3}{7}, \\frac{1}{3}, \\frac{2}{5}, \\frac{3}{8}$ You could find a common denominator What do 7,3,5, and 8 divide? You could force a common denominator and just take all of their denominators and multiply them to find a common denominator like so: $\\frac{3}{7}=\\frac{3}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{1}{3}=\\frac{1}{3} \\cdot \\frac{7}{7} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{2}{5}=\\frac{2}{5} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{8}{8}$ $\\frac{3}{8}=\\frac{3}{8} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5}$ Since all the denominators are the same, all you have to do is compare the tops. The tops of: $\\frac{3(3)(5)(8)}{7(3)(5)(8)} , \\frac{1(7)(5)(8)}{3(7)(5)(8)}, \\frac{2(7)(3)(8)}{5(7)(3)(8)}, \\frac{3(7)(3)(5)}{8(7)(3)(5)}$ 3. myininaya So which is the least: 3(3)(5)(8) , 1(7)(5)(8), 2(7)(3)(8) , 3(7)(3)(5) 4. myininaya Multiply first. This may be more revealing to you. 5. myininaya You could also write each fraction as a decimal. These are the two ways I prefer.", "How do you simplify (2k)/8 + (3k)/6? Jun 4, 2016 Start simplifying the number in front of $k$ with the denominator $\\frac{2 k}{8} = \\frac{2 k}{2 \\cdot 4} = \\frac{k}{4}$ $\\frac{3 k}{6} = \\frac{3 k}{2 \\cdot 3} = \\frac{k}{2}$ Now we have $\\frac{2 k}{8} + \\frac{3 k}{6} = \\frac{k}{4} + \\frac{k}{2}$ We use as common denominator $4$. $\\frac{k}{4} + \\frac{k}{2} = \\frac{k}{4} + \\frac{2 k}{4}$ $= \\frac{k + 2 k}{4}$ $= \\frac{3 k}{4}$. You cannot simplify more because 3 and 4 do not share any factor.", "1. Simplifying expressions I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 2. 1 - 1 x + h x 3. 2 x^2 10 x^5 4. 2x - 1 - 8 x^2-6x+9 x+1 x^2-2x-3 2. Originally Posted by journeygirl280 1. x^3-8 x-2 Hint: Difference two two cubes in numerator. 2. 1 - 1 x + h x With common denominator, $\\frac{x}{x(x+h)} - \\frac{x+h}{x(x+h)}$. Now continue. The last two do not make any sense. 3. Hello, journeygirl280! I think I deciphered the last two problems . . . $3)\\;\\;\\frac{\\frac{2}{x^2}}{\\frac{10}{x^5}}$ We have: . $\\frac{2}{x^2} \\div \\frac{10}{x^5} \\;= \\;\\frac{2}{x^2}\\cdot\\frac{x^5}{10} \\;=\\;\\frac{x^3}{5}$ $4)\\;\\;\\frac{2x}{x^2-6x+9} - \\frac{1}{x+1} - \\frac{8}{x^2-2x-3}$ We have: . $\\frac{2x}{(x-3)^2} - \\frac{1}{x+1} - \\frac{8}{(x+1)(x-3)}$ Get a common denominator: . . $\\frac{2x}{(x-3)^2}\\cdot{\\color{blue}\\frac{x+1}{x+1}} \\,-\\,\\frac{1}{x+1}\\cdot{\\color{blue}\\frac{(x-3)^2}{(x-3)^2}} \\,-\\,\\frac{8}{(x+1)(x-3)}\\cdot{\\color{blue}\\frac{x-3}{x-3}}$ . . $= \\;\\frac{2x(x+1)-(x-3)^2 - 8(x-3)}{(x-3)^2(x+1)} \\;= \\;\\frac{2x^2 + 2x - x^2 + 6x - 9 - 8x + 24}{(x-3)^2(x+1)} $ . . $= \\;\\frac{x^2+15}{(x-3)^2(x+1)}$ 4. Originally Posted by journeygirl280 I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 Different hint: $x^3-8=0$ when $x=2$ so $x-2$ is a factor of $x^3-8$. (well really the same hint in disguise) RonL", "method will sometimes lead you to work with much bigger numbers than you need to: $\\frac{9}{16}-\\frac{5}{8}=\\frac{9\\cdot8}{16\\cdot8}-\\frac{5\\cdot16}{8\\cdot16}=\\frac{72}{128}-\\frac{80}{128}=\\frac{-8}{128}=\\frac{8\\cdot-1}{8\\cdot16}=-\\frac{1}{16}$ The following could have been solved more easily if we found a smaller common denominator: $\\frac{9}{16}-\\frac{5}{8}=\\frac{9}{16}-\\frac{5\\cdot2}{8\\cdot2}=\\frac{9}{16}-\\frac{10}{16}=-\\frac{1}{16}$ If you can see a simpler way to reach common denominators you can save time by avoiding the first approach, but when in doubt, just go that route. Practice Questions Solve and reduce to lowest terms. 1.) $\\frac{3}{4}\\cdot\\frac{1}{4}$ 2.) $\\frac{4}{3}\\cdot\\frac{3}{16}$ 3.) $-\\frac{2}{7}\\cdot-\\frac{4}{5}$ 4.) $\\frac{-3}{4}\\cdot\\frac{4}{5}$ 5.) $\\frac{3}{4}/\\frac{1}{4}$ 6.) $\\frac{8}{3}/\\frac{7}{6}$ 7.) $\\frac{-3}{8}/\\frac{-2}{3}$ 8.) $\\frac{-2}{5}/\\frac{1}{6}$ 9.) $\\frac{3}{4}+\\frac{1}{4}$ 10.) $\\frac{3}{4}+\\frac{2}{5}$ 11.) $\\frac{6}{7}+\\frac{5}{9}$ 12.) $\\frac{-3}{5}+\\frac{7}{10}$ 13.) $\\frac{7}{10}-\\frac{3}{5}$ 14.) $\\frac{3}{4}-\\frac{5}{4}$ 15.) $\\frac{11}{5}-\\frac{5}{3}$ 16.) $\\frac{-2}{7}-\\frac{-4}{3}$", "Math Help - subtract 1. subtract PROBLEM: 6/8 - 2/6 = 10 1/4 + 13 3/6= 2. Originally Posted by breanna2471 PROBLEM: 6/8 - 2/6 = 10 1/4 + 13 3/6= You need to get a common denominator. The most \"efficient\" denominator is the Least Common Multiple (LCM) of the two denominators, but if you can't figure out what that might be you can simply use the product of the two denominators. $\\frac{6}{8} - \\frac{2}{6}$ Frankly I'd express these in lowest terms before I subtracted. This isn't necessary, but makes the numbers a little nicer. So.... $\\frac{6}{8} - \\frac{2}{6} = \\frac{3}{4} - \\frac{1}{3}$ The LCM of 4 and 3 is 12. So we want the denominator in each fraction to become 12. For the $\\frac{3}{4}$ that means multiplying the denominator by 3. But what we do in the denominator we must do in the numerator as well: $\\frac{3}{4} = \\frac{3}{4} \\cdot \\frac{3}{3} = \\frac{9}{12}$ Similarly for $\\frac{1}{3}$ we need to mulitply the denominator by 4: $\\frac{1}{3} = \\frac{1}{3} \\cdot \\frac{4}{4} = \\frac{4}{12}$. Thus $\\frac{6}{8} - \\frac{2}{6} = \\frac{3}{4} - \\frac{1}{3} = \\frac{9}{12} - \\frac{4}{12}$ $= \\frac{9 - 4}{12} = \\frac{5}{12}$ ==================================== $10 ~ \\frac{1}{4} + 13 ~ \\frac{3}{6}$ I'm going to reorder this so that I'm adding whole numbers to whole numbers and adding fractions to fractions. (Or", "# How do you solve -\\frac { 5} { 3} = - \\frac { 5} { 6} + s? Apr 13, 2017 $- \\frac{5}{6} = s$ #### Explanation: Add $\\frac{5}{6}$ to both sides, $\\frac{5}{6} - \\frac{5}{3} = \\frac{5}{6} - \\frac{5}{6} + s$ $\\frac{5}{6} - \\frac{5}{3} = \\cancel{\\frac{5}{6}} - \\cancel{\\frac{5}{6}} + s$ Create a common denominator for $\\frac{5}{6}$ and $\\frac{5}{3}$ $\\frac{5}{6} - \\frac{5 \\cdot 2}{3 \\cdot 2} = s$ $\\frac{5}{6} - \\frac{10}{6} = s$ $- \\frac{5}{6} = s$", "## JazzyPoowh Group Title place the fraction in order from least to greatest , one year ago one year ago 1. JazzyPoowh Group Title 3/7 1/3 2/5 and 3/8 2. myininaya Group Title $\\frac{3}{7}, \\frac{1}{3}, \\frac{2}{5}, \\frac{3}{8}$ You could find a common denominator What do 7,3,5, and 8 divide? You could force a common denominator and just take all of their denominators and multiply them to find a common denominator like so: $\\frac{3}{7}=\\frac{3}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{1}{3}=\\frac{1}{3} \\cdot \\frac{7}{7} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{2}{5}=\\frac{2}{5} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{8}{8}$ $\\frac{3}{8}=\\frac{3}{8} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5}$ Since all the denominators are the same, all you have to do is compare the tops. The tops of: $\\frac{3(3)(5)(8)}{7(3)(5)(8)} , \\frac{1(7)(5)(8)}{3(7)(5)(8)}, \\frac{2(7)(3)(8)}{5(7)(3)(8)}, \\frac{3(7)(3)(5)}{8(7)(3)(5)}$ 3. myininaya Group Title So which is the least: 3(3)(5)(8) , 1(7)(5)(8), 2(7)(3)(8) , 3(7)(3)(5) 4. myininaya Group Title Multiply first. This may be more revealing to you. 5. myininaya Group Title You could also write each fraction as a decimal. These are the two ways I prefer.", "1} + \\frac{5}{x - 2} = \\frac{2}{x - 1}\\cdot \\frac{x - 2}{x - 2} + \\frac{5}{x - 2} \\cdot \\frac{x - 1}{x - 1}$ $= \\frac{2(x - 2)}{(x - 1)(x - 2)} + \\frac{5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2(x - 2) + 5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2x - 4 + 5x - 5}{(x - 1)(x - 2)}$ $= \\frac{7x - 9}{(x - 1)(x - 2)}$. So $\\frac{\\frac{1}{x - 2} + \\frac{3}{x - 1}}{\\frac{2}{x - 1} + \\frac{5}{x - 2}} = \\frac{\\frac{4x - 7}{(x - 1)(x - 2)}}{\\frac{7x - 9}{(x - 1)(x - 2)}}$ $= \\frac{4x - 7}{(x - 1)(x - 2)}\\cdot \\frac{(x - 1)(x - 2)}{7x - 9}$ $= \\frac{4x - 7}{7x - 9}$. Solve: 3. $y - \\frac{14}{y} = 5$ 4. $\\frac{4}{a-7} = \\frac{-2a}{a+3}$ Again, for these I'd have to find a common denominator, correct? And than multiply everything by it? So number 3 the denominator would by Y and for number 4 the denominator would be... A? Maybe I'm no where close... 3. $y - \\frac{14}{y} = 5$. Multiply both sides of the equation by $y$. $y^2 - 14 = 5y$ $y^2 - 5y - 14 = 0$ $(y - 7)(y + 2) = 0$ $y - 7 = 0$ or $y + 2 = 0$ $y = 7$", "# How do you find the least common denominator for rational expressions? Nov 2, 2014 The least common denominator is the least common multiple of denominators. Example Find the least common denominator to compute: $\\frac{1}{3} + \\frac{1}{4} - \\frac{1}{6}$ Let us observe: Multiples of $3$: $3 , 6 , 9$, 12,$15 , \\ldots$ Multiples of $4$: $4 , 8$,12, $16 , \\ldots$ Multiples of $6$: $6$,12,$18 , \\ldots$, So, the least common multiple of $3$,$4$, and $6$ is $12$. Hence, $\\frac{1}{3} + \\frac{1}{4} - \\frac{1}{6} = \\frac{4}{12} + \\frac{3}{12} - \\frac{2}{6} = \\frac{4 + 3 - 2}{12} = \\frac{5}{12}$ ... I hope that this was helpful.", "and work inside-out. $\\sin \\frac{\\pi}{14} \\sin \\frac{6\\pi}{14} = \\frac{1}{2}\\cos \\frac{5\\pi}{14}$ $\\sin \\frac{2\\pi}{14} \\sin\\frac{5\\pi}{14} = \\frac{1}{2}\\cos \\frac{3\\pi}{14}$ $\\sin \\frac{3\\pi}{14}\\sin \\frac{4\\pi}{13} = \\frac{1}{2}\\cos \\frac{\\pi}{14}$. Thus, $\\cos \\frac{\\pi}{14} \\cos \\frac{3\\pi}{14} \\cos \\frac{5\\pi}{14} = \\frac{\\sqrt{7}}{2^3}$. Generalize! If $n$ is odd then: $\\boxed{ \\cos \\frac{\\pi}{2n} \\cos \\frac{3\\pi}{2n} \\cdot ... \\cdot \\cos \\frac{(n-2)\\pi}{2n} = \\frac{\\sqrt{n}}{2^{(n-1)/2}} }$.", "while it's possible to try and work the addition of the whole numbers and fractions as two separate processes, the one-size-fits-all approach of following the usual steps and working with improper fractions will guarantee a smooth ride for any problem you ever tackle.$$5 \\, \\frac{7}{16} \\longrightarrow \\frac{87}{16}$$$$6 \\, \\frac{7}{12} \\longrightarrow \\frac{79}{12}$$Now we need a common denominator.Common denominator:$\\mathrm{LCM(}16, \\, 12\\mathrm{)} = 48$$\\frac{87}{16} \\cdot \\frac{3}{3} = \\frac{261}{48}$$$$\\frac{79}{12} \\cdot \\frac{4}{4} = \\frac{316}{48}$$Therefore,$$5 \\, \\frac{7}{16} + 6 \\, \\frac{7}{12} = \\frac{261}{48} + \\frac{316}{48}$$$$=\\frac{577}{48}$$Obtain the final answer via long division.$$577 \\div 48 = 12 \\,\\, \\mathrm{R} \\, 1$$$$\\longrightarrow 5 \\, \\frac{7}{16} + 6 \\, \\frac{7}{12} = 12 \\, \\frac{1}{48}$$ Example 13$$\\frac{11}{8} + \\frac{15}{7}$$ Show solution $\\blacktriangleright$ Once again, we look to convert these fractions to have the same denominator.Common denominator: $\\mathrm{LCM(}8, \\, 7\\mathrm{)} = 56$$\\frac{11}{8} \\cdot \\frac{7}{7} = \\frac{77}{56}$$$$\\frac{15}{7} \\cdot \\frac{8}{8} = \\frac{120}{56}$$$$\\frac{77}{56} + \\frac{120}{56}$$$$=\\frac{197}{56}$$ Example 14$$\\frac{41}{60} - \\frac{17}{24}$$ Show solution$\\blacktriangleright$Convert the fractions to a common denominator.Common denominator:$\\mathrm{LCM(}60, \\, 24\\mathrm{)} = 120$$\\frac{41}{60} \\cdot \\frac{2}{2} = \\frac{82}{120}$$$$\\frac{17}{24} \\cdot \\frac{5}{5} = \\frac{85}{120}$$Now we can subtract.$$\\frac{82}{120} - \\frac{85}{120}$$$$=-\\frac{3}{120}$$Finally, make sure you simplify! Don't risk losing points!$$=-\\frac{1}{40}$$ Example 15Subtract. Leave your answer as an improper fraction.$$3 \\, \\frac{3}{4} - 4 \\, \\frac{13}{18}$$ Show solution $\\blacktriangleright$ Let's work with improper fractions entirely.$$3 \\, \\frac{3}{4} \\longrightarrow \\frac{15}{4}$$$$4 \\, \\frac{13}{18} \\longrightarrow \\frac{85}{18}$$Now we'll need a common denominator.Common denominator: \\$\\mathrm{LCM(}4, \\,", "of $$t$$. This means that if we find $$\\frac{dy}{dt}$$ and $$\\frac{dx}{dt}$$ at $$t=2$$, we can divide $$\\frac{dy}{dt}$$ by $$\\frac{dx}{dt}$$ to solve for $$\\frac{dy}{dx}$$. Let’s start by finding $$\\frac{dx}{dt}$$ and $$\\frac{dy}{dt}$$ by differentiating $$x$$ and $$y$$ with respect to $$t$$: For $$\\frac{dx}{dt}$$, we’ll need to use the division rule: $$\\frac{dx}{dt}=\\frac{(t+1)6-6t(1)}{(t+1)^2}=\\frac{6t+6-6t}{(t+1)^2}=\\frac{6}{(t+1)^2}$$ So, at $$t=2$$, $$\\frac{dx}{dt}=\\frac{6}{(2+1)^2}=\\frac{6}{3^2}=\\frac{6}{9}=\\frac{2}{3}$$. Next, we can rewrite $$y(t)$$ as $$y(t)=-8(t^2 +4)^{-1}$$ so that we can use the power rule (and chain rule) to differentiate: $$\\frac{dy}{dt}=-8(-1)(t^2 +4)^{-2}\\cdot 2t=\\frac{16t}{(t^2 +4)^2}$$. Then, at $$t=2$$, $$\\frac{dy}{dt}=\\frac{16(2)}{(2^2 +4)^2}=\\frac{32}{8^2}=\\frac{32}{64}=\\frac{1}{2}$$. Now we can divide, and we get that: $$\\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}=\\frac{1/2}{2/3}=\\frac{3}{4}$$. ### Question 2 Recall that Euler’s method involves the following formula: $$y_n =y_{n-1}+\\Delta x \\cdot f'(x_{n-1},y_{n-1})$$ Essentially, what this formula means is that if we’re taking steps of $$\\Delta x=1$$, our next $$y$$-value ($$y_n$$) is equal to our previous $$y$$-value ($$y_{n-1}$$) plus the step multiplied by the derivative evaluated at $$y_{n-1}$$ and $$x_{n-1}$$. So, let’s start with $$y_0 =f(0)=1$$: $$y_0 =f(0)=1$$ $$y_1=f(0.5)=f(0)+(0.5)f'(0,1)=1+(0.5)(1+2(1))=1+(0.5)(3)=2.5$$ $$y_2=f(1)=f(0.5)+(0.5)f'(0.5,2.5)=2.5+(0.5)(1+2(2.5))=2.5+(0.5)(6)=5.5.$$ Since we’ve reached $$x=1$$, our Euler’s method approximation is $$f(1)=5.5$$. ### Question 3 To evaluate integrals with infinity, we typically use limits like as follows: $$\\lim_{b \\to \\infty}\\int_{0}^{b} kxe^{-2x}\\:dx$$ This way, we can integrate the function first, and then worry about the bounds. For this particular equation, we’ll need to do integration by parts. Also, remember to just treat $$k$$ as a constant.", "\\frac{6\\pi}7$ Which was shown in the first part to equal $-\\frac{1}2$, so $b = -\\frac{1}2$ Part 3: solving for c Notice that $\\cos \\frac{6\\pi}7 = \\cos \\frac{8\\pi}7$ $c$ is the negation of the product of roots by Vieta's formulas $c = -\\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ Multiply by $8 \\sin{\\frac{2\\pi}{7}}$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -8 \\sin{\\frac{2\\pi}{7}} \\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ Then use sine addition formula backwards: $2 \\sin \\frac{2\\pi}7 \\cos \\frac{2\\pi}7 = \\sin \\frac{4\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -4 \\sin \\frac{4\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -2 \\sin \\frac{8\\pi}7 \\cos \\frac{8\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -\\sin \\frac{16\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -\\sin \\frac{2\\pi}7$ $c = -\\frac{1}8$ Finally multiply $abc = \\frac{1}2 * - \\frac{1}2 * -\\frac{1}8 = \\frac{1}{32}$ or $\\boxed{D) \\frac{1}{32}}$. ~Tucker ## Solution 2 (Approximation) Letting the roots be $p$, $q$, and $r$, Vietas gives $$p + q + r = a$$ $$pq + qr + pq = -b$$ $$pqr = c$$ We use the Taylor series for $\\cos x$, $$\\cos x = \\sum_{k = 0}^{\\infty} (-1)^k \\frac{x^{2k}}{(2k)!}$$ to approximate the roots. Taking the sum up to $k = 3$ yields a close approximation, so we have $$\\cos\\left(\\frac{2\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{2\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{6}}{720} \\simeq 0.623$$ $$\\cos\\left(\\frac{4\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{4\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{4\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{4\\pi}{7}\\right)^{6}}{720} \\simeq -0.225$$ $$\\cos\\left(\\frac{6\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{6\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{6\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{6\\pi}{7}\\right)^{6}}{720} \\simeq -0.964.$$ Note that these approximations", "pile is changing at the instant $r = 4\\text{,}$ we use the value $r = 4$ along with $\\frac{dV}{dt} = 10$ in Equation (3.5.1), and hence find that \\begin{equation*} 10 = \\frac{1}{3} \\pi 4^2 \\cdot \\frac{1}{2} \\left. \\frac{dr}{dt} \\right|_{r=4} + \\frac{2}{3} \\pi 4 \\cdot \\frac{1}{2}4 \\cdot \\left. \\frac{dr}{dt} \\right|_{r=4} = \\frac{8}{3}\\pi \\left. \\frac{dr}{dt} \\right|_{r=4} + \\frac{16}{3} \\pi \\left. \\frac{dr}{dt} \\right|_{r=4}. \\end{equation*} With only the value of $\\left. \\frac{dr}{dt} \\right|_{r=4}$ remaining unknown, we solve for $\\left. \\frac{dr}{dt} \\right|_{r=4}$ and find that $10 = 8 \\pi \\left. \\frac{dr}{dt} \\right|_{r=4}\\text{,}$ so that \\begin{equation*} \\left. \\frac{dr}{dt} \\right|_{r=4} = \\frac{10}{8\\pi} \\approx 0.39789 \\end{equation*} feet per second. Because we were interested in how fast the height of the pile was changing at this instant, we want to know $\\frac{dh}{dt}$ when $r = 4\\text{.}$ Since $\\frac{dh}{dt} = \\frac{1}{2} \\frac{dr}{dt}$ for all values of $t\\text{,}$ it follows \\begin{equation*} \\left. \\frac{dh}{dt} \\right|_{r=4} = \\frac{5}{8\\pi} \\approx 0.19894 \\ \\text{ft/min} . \\end{equation*} Note particularly how we distinguish between the notations $\\frac{dr}{dt}$ and $\\left. \\frac{dr}{dt} \\right|_{r=4}\\text{.}$ The former represents the rate of change of $r$ with respect to $t$ at an arbitrary value of $t\\text{,}$ while the latter is the rate of change of $r$ with respect to $t$ at a particular moment, in fact the moment $r = 4\\text{.}$ While we don't know the exact value of $t\\text{,}$ because", "- 4 i ^ { 2 } } \\\\ & = \\frac { - 16 i + 6 ( - 1 ) } { - 4 ( - 1 ) } \\\\ & = \\frac { - 16 i - 6 } { 4 } \\\\ & = \\frac { - 6 - 16 i } { 4 }\\\\ & = \\frac{-6}{4} - \\frac{-16i}{4} \\\\ & = - \\frac { 3 } { 2 } - 4 i \\end{aligned} Because the denominator is a monomial, we could multiply numerator and denominator by $$1$$ in the form of $$\\frac{i}{i}$$ and save some steps reducing in the end. \\begin{aligned} \\frac { 8 - 3 i } { 2 i } & = \\frac { ( 8 - 3 i ) } { ( 2 i ) } \\cdot \\color{Cerulean}{\\frac { i } { i } }\\\\ & = \\frac { 8 i - 3 i ^ { 2 } } { 2 i ^ { 2 } } \\\\ & = \\frac { 8 i - 3 ( - 1 ) } { 2 ( - 1 ) } \\\\ & = \\frac { 8 i + 3 } { - 2 } \\\\ & = \\frac { 8 i } { - 2 } + \\frac { 3 }", "Then we calculate the $latex LCM$ of the denominators: $latex lcm(1; y-6)= y-6$ Therefore: $$5y^2+\\frac{9y}{y-6}=\\frac {5y^2}{1}+\\frac{9y}{y-6}$$ $$=\\frac {5y^2(y-6)+9y}{y-6}$$ $$=\\frac {5y^3-30y^2+9y}{y-6}$$ $$=\\frac {y(5y^2-30y+9)}{y-6}$$ ### EXAMPLE 10 $$\\frac{x}{x^2+2x-8}+\\frac{x+1}{x^2-3x+2}$$ First, if possible, the denominators are factored: $$x^2+2x-8 =(x+4)\\cdot(x-2)$$ $$x^2-3x+2 =(x-2)\\cdot(x-1)$$ And the expression is rewritten: $$\\frac{x}{x^2+2x-8}+\\frac{x+1}{x^2-3x+2}=\\frac{x}{(x+4)(x-2)}+\\frac{x+1}{(x-2)(x-1)}$$ Thus, the least common multiple of the denominators is: $$lcm[(x+4)(x-2);(x-2)(x-1)] =(x-2)(x+4)(x-1)$$ Therefore: $$\\frac{x}{(x+4)(x-2)}+\\frac{x+1}{(x-2)(x-1)}=\\frac{x(x-1)+(x+1)(x+4)}{(x-2)(x+4)(x-1)}$$ Now the products that appear in the numerator are expanded: $latex x(x-1)=x^2-x$ $latex (x+1)(x+4)=x^2+5x+4$ And they are added together, reducing like terms:", "$8$ and $2$ is number $8$ and thats is the denominator of the difference. $\\frac{8}{8} = 1; 1 \\cdot 5 = 5, \\frac{8}{2}= 4; 4 \\cdot 1 = 4$ $\\frac{5}{8} – \\frac{1}{2} = \\frac{(5-4)}{8} = \\frac{1}{8}$ Example 2: Subtract $\\frac{1}{2} – \\frac{5}{8}$ $\\frac{1}{2} – \\frac{5}{8} = \\frac{(4-5)}{8} = – \\frac{1}{8}$ Reminder: if you are unsure of which number is the least common multiple of your denominators, you can put any multiple, and shorten the fraction afterwards: $\\frac{1}{2} – \\frac{5}{8} = \\frac{(8-10)}{16} = \\frac{(8-10)}{16} = – \\frac{2}{16}$ ## Multiplying fractions Multiplying fractions is the easiest mathematical operation with fractions; you simply multiply the numerator of the first number with the numerator of the second number, and denominator of the first number with the denominator of the second number. $\\frac{a}{b} \\cdot \\frac{c}{d} = \\frac{ac}{bd}$ Example 1: Multiply. $\\frac{2}{7} \\cdot \\frac{3}{5} = ?$ $\\frac{2}{7} \\cdot \\frac{3}{5} = \\frac{(2 \\cdot 3)}{(7 \\cdot 5)} = \\frac{21}{35}$ Things can even go simpler. You can shorten them before multiplication, so you get smaller numbers. You can only shorten the numerators with denominators, never numerator with numerator or denominator with denominator. But you can take any denominator and any numerator in a multiplication. Example: Shorten the fraction before multiplication. $\\frac{2}{7} \\cdot \\frac{14}{8} = ?$ If it’s easier for you, solve this in a certain order. First", "# Prove that Question: Find $n$, If ${ }^{n+5} P_{n+1}=\\frac{11}{2}(n-1) \\cdot{ }^{n+3} P_{n}$, find $n .$ Solution: To find: the value of n Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\\frac{n !}{(n-r) !}$ ${ }^{n+5} P_{n+1}=\\frac{11}{2}(n-1) \\cdot{ }^{n+3} P_{n}$ $\\frac{(n+5) !}{4 !}=\\frac{11}{2}(n-1) \\frac{(n+3) !}{3 !}$ $\\frac{(n+5)(n+4)(n+3) !}{4 \\times 3 !}=\\frac{11}{2}(n-1) \\frac{(n+3) !}{3 !}$ $\\frac{(n+5)(n+4)}{2}=11(n-1)$ $n^{2}+9 n+20=22 n-22$ $n^{2}-13 n+42=0$ $(n-7)(n-6)=0$ $n=7,6$ Hence, values of n are 7 & 6", "\\frac{3\\cdot 7}{7\\cdot 8} = \\frac{16}{56} + \\frac{21}{56} = \\frac{37}{56}. \\end{align*} But again, there is nothing magic about using the LCM. In the first example, if you wanted to use a common denominator of $24$ and just cross-multiply to find the numerators, that would work fine: $$\\frac{1}{4} + \\frac{5}{6} = \\frac{1\\cdot 6}{4\\cdot 6} + \\frac{5\\cdot 4}{4\\cdot 6} = \\frac{6}{24} + \\frac{20}{24} = \\frac{26}{24} = \\frac{13}{12}.$$ • You can use any denominator that is a multiple of both of the denominators in question. The LCM is the least of these, but as you point out, it may be simpler just to cross-multiply to avoid having to find the LCM. As you can see above, they both give the same answer. And the reason is that, for example, $\\frac{6}{24} = \\frac{3}{12}$ in the above --- using a larger common denominator simply means you are multiplying numerator and denominator by a larger number, which of course does not change its value. – rogerl Oct 16 '14 at 21:51" ]

[ "ScottyClermont 5 # what is larger 3/8 or 5/16 $\\frac{3}{8} ... \\frac{5}{16}$ First, find the least common denominator between the two. Since 8 is a factor of 16, the LCD is 16. $\\frac{3*2}{8*2}$ $\\frac{3}{8}= \\frac{6}{16}$ Now, you can compare the numerators. $6>5$ Since 6 is greater than 5, you know that $\\frac{6}{16} > \\frac{5}{16}$. Therefore, we now know that $\\frac{3}{8}$ is larger. $\\frac{3}{8}=\\frac{3\\codt2}{8\\cdot2}=\\frac{6}{16} > \\frac{5}{16}\\\\\\\\Answer:\\frac{3}{8}\\ is\\ larger\\ than\\ \\frac{5}{16}.$", "\\frac{6\\pi}7$ Which was shown in the first part to equal $-\\frac{1}2$, so $b = -\\frac{1}2$ Part 3: solving for c Notice that $\\cos \\frac{6\\pi}7 = \\cos \\frac{8\\pi}7$ $c$ is the negation of the product of roots by Vieta's formulas $c = -\\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ Multiply by $8 \\sin{\\frac{2\\pi}{7}}$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -8 \\sin{\\frac{2\\pi}{7}} \\cos \\frac{2\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ Then use sine addition formula backwards: $2 \\sin \\frac{2\\pi}7 \\cos \\frac{2\\pi}7 = \\sin \\frac{4\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -4 \\sin \\frac{4\\pi}7 \\cos \\frac{4\\pi}7 \\cos \\frac{8\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -2 \\sin \\frac{8\\pi}7 \\cos \\frac{8\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -\\sin \\frac{16\\pi}7$ $c \\cdot 8 \\sin{\\frac{2\\pi}{7}} = -\\sin \\frac{2\\pi}7$ $c = -\\frac{1}8$ Finally multiply $abc = \\frac{1}2 * - \\frac{1}2 * -\\frac{1}8 = \\frac{1}{32}$ or $\\boxed{D) \\frac{1}{32}}$. ~Tucker ## Solution 2 (Approximation) Letting the roots be $p$, $q$, and $r$, Vietas gives $$p + q + r = a$$ $$pq + qr + pq = -b$$ $$pqr = c$$ We use the Taylor series for $\\cos x$, $$\\cos x = \\sum_{k = 0}^{\\infty} (-1)^k \\frac{x^{2k}}{(2k)!}$$ to approximate the roots. Taking the sum up to $k = 3$ yields a close approximation, so we have $$\\cos\\left(\\frac{2\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{2\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{2\\pi}{7}\\right)^{6}}{720} \\simeq 0.623$$ $$\\cos\\left(\\frac{4\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{4\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{4\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{4\\pi}{7}\\right)^{6}}{720} \\simeq -0.225$$ $$\\cos\\left(\\frac{6\\pi}{7}\\right) \\simeq 1-\\frac{\\left(\\frac{6\\pi}{7}\\right)^{2}}{2}+\\frac{\\left(\\frac{6\\pi}{7}\\right)^{4}}{24}-\\frac{\\left(\\frac{6\\pi}{7}\\right)^{6}}{720} \\simeq -0.964.$$ Note that these approximations", "method will sometimes lead you to work with much bigger numbers than you need to: $\\frac{9}{16}-\\frac{5}{8}=\\frac{9\\cdot8}{16\\cdot8}-\\frac{5\\cdot16}{8\\cdot16}=\\frac{72}{128}-\\frac{80}{128}=\\frac{-8}{128}=\\frac{8\\cdot-1}{8\\cdot16}=-\\frac{1}{16}$ The following could have been solved more easily if we found a smaller common denominator: $\\frac{9}{16}-\\frac{5}{8}=\\frac{9}{16}-\\frac{5\\cdot2}{8\\cdot2}=\\frac{9}{16}-\\frac{10}{16}=-\\frac{1}{16}$ If you can see a simpler way to reach common denominators you can save time by avoiding the first approach, but when in doubt, just go that route. Practice Questions Solve and reduce to lowest terms. 1.) $\\frac{3}{4}\\cdot\\frac{1}{4}$ 2.) $\\frac{4}{3}\\cdot\\frac{3}{16}$ 3.) $-\\frac{2}{7}\\cdot-\\frac{4}{5}$ 4.) $\\frac{-3}{4}\\cdot\\frac{4}{5}$ 5.) $\\frac{3}{4}/\\frac{1}{4}$ 6.) $\\frac{8}{3}/\\frac{7}{6}$ 7.) $\\frac{-3}{8}/\\frac{-2}{3}$ 8.) $\\frac{-2}{5}/\\frac{1}{6}$ 9.) $\\frac{3}{4}+\\frac{1}{4}$ 10.) $\\frac{3}{4}+\\frac{2}{5}$ 11.) $\\frac{6}{7}+\\frac{5}{9}$ 12.) $\\frac{-3}{5}+\\frac{7}{10}$ 13.) $\\frac{7}{10}-\\frac{3}{5}$ 14.) $\\frac{3}{4}-\\frac{5}{4}$ 15.) $\\frac{11}{5}-\\frac{5}{3}$ 16.) $\\frac{-2}{7}-\\frac{-4}{3}$", "Math Help - subtract 1. subtract PROBLEM: 6/8 - 2/6 = 10 1/4 + 13 3/6= 2. Originally Posted by breanna2471 PROBLEM: 6/8 - 2/6 = 10 1/4 + 13 3/6= You need to get a common denominator. The most \"efficient\" denominator is the Least Common Multiple (LCM) of the two denominators, but if you can't figure out what that might be you can simply use the product of the two denominators. $\\frac{6}{8} - \\frac{2}{6}$ Frankly I'd express these in lowest terms before I subtracted. This isn't necessary, but makes the numbers a little nicer. So.... $\\frac{6}{8} - \\frac{2}{6} = \\frac{3}{4} - \\frac{1}{3}$ The LCM of 4 and 3 is 12. So we want the denominator in each fraction to become 12. For the $\\frac{3}{4}$ that means multiplying the denominator by 3. But what we do in the denominator we must do in the numerator as well: $\\frac{3}{4} = \\frac{3}{4} \\cdot \\frac{3}{3} = \\frac{9}{12}$ Similarly for $\\frac{1}{3}$ we need to mulitply the denominator by 4: $\\frac{1}{3} = \\frac{1}{3} \\cdot \\frac{4}{4} = \\frac{4}{12}$. Thus $\\frac{6}{8} - \\frac{2}{6} = \\frac{3}{4} - \\frac{1}{3} = \\frac{9}{12} - \\frac{4}{12}$ $= \\frac{9 - 4}{12} = \\frac{5}{12}$ ==================================== $10 ~ \\frac{1}{4} + 13 ~ \\frac{3}{6}$ I'm going to reorder this so that I'm adding whole numbers to whole numbers and adding fractions to fractions. (Or", "jamessinatraaa 2021-12-27 How do you evaluate $8{C}_{2}$? autormtak0w Expert Explanation: We have, $n{C}_{r}=\\frac{n!}{\\left(n-r\\right)!r!}$ $\\therefore 8{C}_{2}=\\frac{8!}{\\left(8-2\\right)!\\cdot 2!}=\\frac{8!}{6!\\cdot 2!}=\\frac{6!\\cdot 7\\cdot 8}{6!\\cdot 2!}=\\frac{7\\cdot 8}{2!}=\\frac{56}{1\\cdot 2}=28$ Barbara Meeker Expert In the given problem, we have to evaluate $8{C}_{2}$ Here, we will use the formula of combination which is $n{C}_{r}=\\frac{n!}{\\left(n-r\\right)!r!}$ You can easily identify that n is 8 and r is 2 here. $⇒8{C}_{2}=\\frac{8!}{\\left(8-2\\right)!2!}$ $=\\frac{8!}{6!2!}$ $=\\frac{8×7×6!}{6!2!}$ After calculating the above e\\times pression, we get $=\\frac{8×7}{2!}$ $=\\frac{8×7}{2×1}$ $=\\frac{56}{2}$ $=28$ Therefore, $8{C}_{2}=28$ Vasquez Expert", "of $$t$$. This means that if we find $$\\frac{dy}{dt}$$ and $$\\frac{dx}{dt}$$ at $$t=2$$, we can divide $$\\frac{dy}{dt}$$ by $$\\frac{dx}{dt}$$ to solve for $$\\frac{dy}{dx}$$. Let’s start by finding $$\\frac{dx}{dt}$$ and $$\\frac{dy}{dt}$$ by differentiating $$x$$ and $$y$$ with respect to $$t$$: For $$\\frac{dx}{dt}$$, we’ll need to use the division rule: $$\\frac{dx}{dt}=\\frac{(t+1)6-6t(1)}{(t+1)^2}=\\frac{6t+6-6t}{(t+1)^2}=\\frac{6}{(t+1)^2}$$ So, at $$t=2$$, $$\\frac{dx}{dt}=\\frac{6}{(2+1)^2}=\\frac{6}{3^2}=\\frac{6}{9}=\\frac{2}{3}$$. Next, we can rewrite $$y(t)$$ as $$y(t)=-8(t^2 +4)^{-1}$$ so that we can use the power rule (and chain rule) to differentiate: $$\\frac{dy}{dt}=-8(-1)(t^2 +4)^{-2}\\cdot 2t=\\frac{16t}{(t^2 +4)^2}$$. Then, at $$t=2$$, $$\\frac{dy}{dt}=\\frac{16(2)}{(2^2 +4)^2}=\\frac{32}{8^2}=\\frac{32}{64}=\\frac{1}{2}$$. Now we can divide, and we get that: $$\\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}=\\frac{1/2}{2/3}=\\frac{3}{4}$$. ### Question 2 Recall that Euler’s method involves the following formula: $$y_n =y_{n-1}+\\Delta x \\cdot f'(x_{n-1},y_{n-1})$$ Essentially, what this formula means is that if we’re taking steps of $$\\Delta x=1$$, our next $$y$$-value ($$y_n$$) is equal to our previous $$y$$-value ($$y_{n-1}$$) plus the step multiplied by the derivative evaluated at $$y_{n-1}$$ and $$x_{n-1}$$. So, let’s start with $$y_0 =f(0)=1$$: $$y_0 =f(0)=1$$ $$y_1=f(0.5)=f(0)+(0.5)f'(0,1)=1+(0.5)(1+2(1))=1+(0.5)(3)=2.5$$ $$y_2=f(1)=f(0.5)+(0.5)f'(0.5,2.5)=2.5+(0.5)(1+2(2.5))=2.5+(0.5)(6)=5.5.$$ Since we’ve reached $$x=1$$, our Euler’s method approximation is $$f(1)=5.5$$. ### Question 3 To evaluate integrals with infinity, we typically use limits like as follows: $$\\lim_{b \\to \\infty}\\int_{0}^{b} kxe^{-2x}\\:dx$$ This way, we can integrate the function first, and then worry about the bounds. For this particular equation, we’ll need to do integration by parts. Also, remember to just treat $$k$$ as a constant.", "## JazzyPoowh 3 years ago place the fraction in order from least to greatest , 1. JazzyPoowh 3/7 1/3 2/5 and 3/8 2. myininaya $\\frac{3}{7}, \\frac{1}{3}, \\frac{2}{5}, \\frac{3}{8}$ You could find a common denominator What do 7,3,5, and 8 divide? You could force a common denominator and just take all of their denominators and multiply them to find a common denominator like so: $\\frac{3}{7}=\\frac{3}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{1}{3}=\\frac{1}{3} \\cdot \\frac{7}{7} \\cdot \\frac{5}{5} \\cdot \\frac{8}{8}$ $\\frac{2}{5}=\\frac{2}{5} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{8}{8}$ $\\frac{3}{8}=\\frac{3}{8} \\cdot \\frac{7}{7} \\cdot \\frac{3}{3} \\cdot \\frac{5}{5}$ Since all the denominators are the same, all you have to do is compare the tops. The tops of: $\\frac{3(3)(5)(8)}{7(3)(5)(8)} , \\frac{1(7)(5)(8)}{3(7)(5)(8)}, \\frac{2(7)(3)(8)}{5(7)(3)(8)}, \\frac{3(7)(3)(5)}{8(7)(3)(5)}$ 3. myininaya So which is the least: 3(3)(5)(8) , 1(7)(5)(8), 2(7)(3)(8) , 3(7)(3)(5) 4. myininaya Multiply first. This may be more revealing to you. 5. myininaya You could also write each fraction as a decimal. These are the two ways I prefer.", "pile is changing at the instant $r = 4\\text{,}$ we use the value $r = 4$ along with $\\frac{dV}{dt} = 10$ in Equation (3.5.1), and hence find that \\begin{equation*} 10 = \\frac{1}{3} \\pi 4^2 \\cdot \\frac{1}{2} \\left. \\frac{dr}{dt} \\right|_{r=4} + \\frac{2}{3} \\pi 4 \\cdot \\frac{1}{2}4 \\cdot \\left. \\frac{dr}{dt} \\right|_{r=4} = \\frac{8}{3}\\pi \\left. \\frac{dr}{dt} \\right|_{r=4} + \\frac{16}{3} \\pi \\left. \\frac{dr}{dt} \\right|_{r=4}. \\end{equation*} With only the value of $\\left. \\frac{dr}{dt} \\right|_{r=4}$ remaining unknown, we solve for $\\left. \\frac{dr}{dt} \\right|_{r=4}$ and find that $10 = 8 \\pi \\left. \\frac{dr}{dt} \\right|_{r=4}\\text{,}$ so that \\begin{equation*} \\left. \\frac{dr}{dt} \\right|_{r=4} = \\frac{10}{8\\pi} \\approx 0.39789 \\end{equation*} feet per second. Because we were interested in how fast the height of the pile was changing at this instant, we want to know $\\frac{dh}{dt}$ when $r = 4\\text{.}$ Since $\\frac{dh}{dt} = \\frac{1}{2} \\frac{dr}{dt}$ for all values of $t\\text{,}$ it follows \\begin{equation*} \\left. \\frac{dh}{dt} \\right|_{r=4} = \\frac{5}{8\\pi} \\approx 0.19894 \\ \\text{ft/min} . \\end{equation*} Note particularly how we distinguish between the notations $\\frac{dr}{dt}$ and $\\left. \\frac{dr}{dt} \\right|_{r=4}\\text{.}$ The former represents the rate of change of $r$ with respect to $t$ at an arbitrary value of $t\\text{,}$ while the latter is the rate of change of $r$ with respect to $t$ at a particular moment, in fact the moment $r = 4\\text{.}$ While we don't know the exact value of $t\\text{,}$ because", "1} + \\frac{5}{x - 2} = \\frac{2}{x - 1}\\cdot \\frac{x - 2}{x - 2} + \\frac{5}{x - 2} \\cdot \\frac{x - 1}{x - 1}$ $= \\frac{2(x - 2)}{(x - 1)(x - 2)} + \\frac{5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2(x - 2) + 5(x - 1)}{(x - 1)(x - 2)}$ $= \\frac{2x - 4 + 5x - 5}{(x - 1)(x - 2)}$ $= \\frac{7x - 9}{(x - 1)(x - 2)}$. So $\\frac{\\frac{1}{x - 2} + \\frac{3}{x - 1}}{\\frac{2}{x - 1} + \\frac{5}{x - 2}} = \\frac{\\frac{4x - 7}{(x - 1)(x - 2)}}{\\frac{7x - 9}{(x - 1)(x - 2)}}$ $= \\frac{4x - 7}{(x - 1)(x - 2)}\\cdot \\frac{(x - 1)(x - 2)}{7x - 9}$ $= \\frac{4x - 7}{7x - 9}$. Solve: 3. $y - \\frac{14}{y} = 5$ 4. $\\frac{4}{a-7} = \\frac{-2a}{a+3}$ Again, for these I'd have to find a common denominator, correct? And than multiply everything by it? So number 3 the denominator would by Y and for number 4 the denominator would be... A? Maybe I'm no where close... 3. $y - \\frac{14}{y} = 5$. Multiply both sides of the equation by $y$. $y^2 - 14 = 5y$ $y^2 - 5y - 14 = 0$ $(y - 7)(y + 2) = 0$ $y - 7 = 0$ or $y + 2 = 0$ $y = 7$", "How do you simplify (2k)/8 + (3k)/6? Jun 4, 2016 Start simplifying the number in front of $k$ with the denominator $\\frac{2 k}{8} = \\frac{2 k}{2 \\cdot 4} = \\frac{k}{4}$ $\\frac{3 k}{6} = \\frac{3 k}{2 \\cdot 3} = \\frac{k}{2}$ Now we have $\\frac{2 k}{8} + \\frac{3 k}{6} = \\frac{k}{4} + \\frac{k}{2}$ We use as common denominator $4$. $\\frac{k}{4} + \\frac{k}{2} = \\frac{k}{4} + \\frac{2 k}{4}$ $= \\frac{k + 2 k}{4}$ $= \\frac{3 k}{4}$. You cannot simplify more because 3 and 4 do not share any factor.", "$ The denominators are not the same... We can't simply subtract like we did before... So... lets make the denominators the same by multiplying and dividing by the number that's missing $ \\frac{6}{8} - \\frac{2}{6} = \\frac{6}{8}*\\frac{6}{6} - \\frac{2}{6}*\\frac{8}{8} $ (Note that in the above, I multiplied and divided by 6 for one number, and by 8 for the other number. By doing so, I did not change the value of the expression) Now, we have $ \\frac{6}{8}*\\frac{6}{6} - \\frac{2}{6}*\\frac{8}{8} = \\frac{6*6}{8*6} - \\frac{2*8}{6*8} = \\frac{36}{48} - \\frac{16}{48} $ Since the denominator is 48 in both the cases, we can proceed to simplify it as follows $ = \\frac{36-16}{48} = \\frac{20}{48} = \\frac{5}{12} $ Hope this helps Mahurshi Akilla 4. Originally Posted by breanna2471 $\\frac34-\\frac13=\\frac{12}{12}\\cdot\\left(\\frac34-\\frac13\\right)=\\frac1{12}\\cdot(9-4)=\\frac5{12}$", "# How do you solve -\\frac { 5} { 3} = - \\frac { 5} { 6} + s? Apr 13, 2017 $- \\frac{5}{6} = s$ #### Explanation: Add $\\frac{5}{6}$ to both sides, $\\frac{5}{6} - \\frac{5}{3} = \\frac{5}{6} - \\frac{5}{6} + s$ $\\frac{5}{6} - \\frac{5}{3} = \\cancel{\\frac{5}{6}} - \\cancel{\\frac{5}{6}} + s$ Create a common denominator for $\\frac{5}{6}$ and $\\frac{5}{3}$ $\\frac{5}{6} - \\frac{5 \\cdot 2}{3 \\cdot 2} = s$ $\\frac{5}{6} - \\frac{10}{6} = s$ $- \\frac{5}{6} = s$", "and work inside-out. $\\sin \\frac{\\pi}{14} \\sin \\frac{6\\pi}{14} = \\frac{1}{2}\\cos \\frac{5\\pi}{14}$ $\\sin \\frac{2\\pi}{14} \\sin\\frac{5\\pi}{14} = \\frac{1}{2}\\cos \\frac{3\\pi}{14}$ $\\sin \\frac{3\\pi}{14}\\sin \\frac{4\\pi}{13} = \\frac{1}{2}\\cos \\frac{\\pi}{14}$. Thus, $\\cos \\frac{\\pi}{14} \\cos \\frac{3\\pi}{14} \\cos \\frac{5\\pi}{14} = \\frac{\\sqrt{7}}{2^3}$. Generalize! If $n$ is odd then: $\\boxed{ \\cos \\frac{\\pi}{2n} \\cos \\frac{3\\pi}{2n} \\cdot ... \\cdot \\cos \\frac{(n-2)\\pi}{2n} = \\frac{\\sqrt{n}}{2^{(n-1)/2}} }$.", "$8$ and $2$ is number $8$ and thats is the denominator of the difference. $\\frac{8}{8} = 1; 1 \\cdot 5 = 5, \\frac{8}{2}= 4; 4 \\cdot 1 = 4$ $\\frac{5}{8} – \\frac{1}{2} = \\frac{(5-4)}{8} = \\frac{1}{8}$ Example 2: Subtract $\\frac{1}{2} – \\frac{5}{8}$ $\\frac{1}{2} – \\frac{5}{8} = \\frac{(4-5)}{8} = – \\frac{1}{8}$ Reminder: if you are unsure of which number is the least common multiple of your denominators, you can put any multiple, and shorten the fraction afterwards: $\\frac{1}{2} – \\frac{5}{8} = \\frac{(8-10)}{16} = \\frac{(8-10)}{16} = – \\frac{2}{16}$ ## Multiplying fractions Multiplying fractions is the easiest mathematical operation with fractions; you simply multiply the numerator of the first number with the numerator of the second number, and denominator of the first number with the denominator of the second number. $\\frac{a}{b} \\cdot \\frac{c}{d} = \\frac{ac}{bd}$ Example 1: Multiply. $\\frac{2}{7} \\cdot \\frac{3}{5} = ?$ $\\frac{2}{7} \\cdot \\frac{3}{5} = \\frac{(2 \\cdot 3)}{(7 \\cdot 5)} = \\frac{21}{35}$ Things can even go simpler. You can shorten them before multiplication, so you get smaller numbers. You can only shorten the numerators with denominators, never numerator with numerator or denominator with denominator. But you can take any denominator and any numerator in a multiplication. Example: Shorten the fraction before multiplication. $\\frac{2}{7} \\cdot \\frac{14}{8} = ?$ If it’s easier for you, solve this in a certain order. First", "# How do you write y=3/8x-3/4 in standard form? Jun 28, 2015 $y = \\frac{3}{8} x - \\frac{3}{4}$ in standard form is $3 x - 8 y = 6$. #### Explanation: Standard form for a linear equation is $A x + B y = C$. $y = \\frac{3}{8} x - \\frac{3}{4}$ Simplify $\\frac{3}{8} x$ to $\\frac{3 x}{8}$. Subtract $\\frac{3 x}{8}$ from both sides of the equation. $\\frac{- 3 x}{8} + y = - \\frac{3}{4}$ Multiply both sides by $- 1$. $\\frac{- 3 x}{8} \\left(- 1\\right) + y \\left(- 1\\right) = - \\frac{3}{4} \\left(- 1\\right)$ = $\\frac{3 x}{8} - y = \\frac{3}{4}$ Multiply both sides by $8$. $\\frac{\\left(3 x\\right) \\cdot 8}{8} - y \\left(8\\right) = \\frac{\\left(3\\right) \\cdot 8}{4}$ = $\\frac{\\left(3 x\\right) \\cdot {\\cancel{8}}^{1}}{{\\cancel{8}}^{1}} - y \\left(8\\right) = \\frac{\\left(3\\right) \\cdot {\\cancel{8}}^{2}}{{\\cancel{4}}^{1}}$ = $3 x - 8 y = 6$", ", - 8\\right)$ will satisfy the equation. $\\therefore - 8 = \\frac{5}{6} \\cdot \\left(- 2\\right) + c \\therefore c = \\frac{10}{6} - 8 = - \\frac{38}{6} = - \\frac{19}{3} = - 6 \\frac{1}{3} \\therefore y = \\frac{5}{6} x - 6 \\frac{1}{3}$ The equation of the line is $y = \\frac{5}{6} x - 6 \\frac{1}{3}$ [Ans]", "# What is 8 1/4 - 3 2/3? Dec 8, 2016 $\\frac{55}{12}$ #### Explanation: First we need to turn these mixed fractions and turn them into improper fractions: $\\left(\\left(\\frac{4}{4}\\right) \\cdot 8 + \\frac{1}{4}\\right) - \\left(\\left(\\frac{3}{3}\\right) \\cdot 3 + \\frac{2}{3}\\right)$ $\\left(\\frac{32}{4} + \\frac{1}{4}\\right) - \\left(\\frac{9}{3} + \\frac{2}{3}\\right)$ $\\frac{33}{4} - \\frac{11}{3}$ Now we need to get each of these fractions over a common denominator so we can subtract the numerators by multiplying each by the necessary form of $1$. The common denominator in this case can be $12 \\to 3 x 4$. $\\left(\\frac{3}{3} \\cdot \\frac{33}{4}\\right) - \\left(\\frac{4}{4} \\cdot \\frac{11}{3}\\right)$ $\\frac{99}{12} - \\frac{44}{12}$ $\\frac{99 - 44}{12}$ $\\frac{55}{12}$", "while it's possible to try and work the addition of the whole numbers and fractions as two separate processes, the one-size-fits-all approach of following the usual steps and working with improper fractions will guarantee a smooth ride for any problem you ever tackle.$$5 \\, \\frac{7}{16} \\longrightarrow \\frac{87}{16}$$$$6 \\, \\frac{7}{12} \\longrightarrow \\frac{79}{12}$$Now we need a common denominator.Common denominator:$\\mathrm{LCM(}16, \\, 12\\mathrm{)} = 48$$\\frac{87}{16} \\cdot \\frac{3}{3} = \\frac{261}{48}$$$$\\frac{79}{12} \\cdot \\frac{4}{4} = \\frac{316}{48}$$Therefore,$$5 \\, \\frac{7}{16} + 6 \\, \\frac{7}{12} = \\frac{261}{48} + \\frac{316}{48}$$$$=\\frac{577}{48}$$Obtain the final answer via long division.$$577 \\div 48 = 12 \\,\\, \\mathrm{R} \\, 1$$$$\\longrightarrow 5 \\, \\frac{7}{16} + 6 \\, \\frac{7}{12} = 12 \\, \\frac{1}{48}$$ Example 13$$\\frac{11}{8} + \\frac{15}{7}$$ Show solution $\\blacktriangleright$ Once again, we look to convert these fractions to have the same denominator.Common denominator: $\\mathrm{LCM(}8, \\, 7\\mathrm{)} = 56$$\\frac{11}{8} \\cdot \\frac{7}{7} = \\frac{77}{56}$$$$\\frac{15}{7} \\cdot \\frac{8}{8} = \\frac{120}{56}$$$$\\frac{77}{56} + \\frac{120}{56}$$$$=\\frac{197}{56}$$ Example 14$$\\frac{41}{60} - \\frac{17}{24}$$ Show solution$\\blacktriangleright$Convert the fractions to a common denominator.Common denominator:$\\mathrm{LCM(}60, \\, 24\\mathrm{)} = 120$$\\frac{41}{60} \\cdot \\frac{2}{2} = \\frac{82}{120}$$$$\\frac{17}{24} \\cdot \\frac{5}{5} = \\frac{85}{120}$$Now we can subtract.$$\\frac{82}{120} - \\frac{85}{120}$$$$=-\\frac{3}{120}$$Finally, make sure you simplify! Don't risk losing points!$$=-\\frac{1}{40}$$ Example 15Subtract. Leave your answer as an improper fraction.$$3 \\, \\frac{3}{4} - 4 \\, \\frac{13}{18}$$ Show solution $\\blacktriangleright$ Let's work with improper fractions entirely.$$3 \\, \\frac{3}{4} \\longrightarrow \\frac{15}{4}$$$$4 \\, \\frac{13}{18} \\longrightarrow \\frac{85}{18}$$Now we'll need a common denominator.Common denominator: \\$\\mathrm{LCM(}4, \\,", "# Prove that Question: Find $n$, If ${ }^{n+5} P_{n+1}=\\frac{11}{2}(n-1) \\cdot{ }^{n+3} P_{n}$, find $n .$ Solution: To find: the value of n Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\\frac{n !}{(n-r) !}$ ${ }^{n+5} P_{n+1}=\\frac{11}{2}(n-1) \\cdot{ }^{n+3} P_{n}$ $\\frac{(n+5) !}{4 !}=\\frac{11}{2}(n-1) \\frac{(n+3) !}{3 !}$ $\\frac{(n+5)(n+4)(n+3) !}{4 \\times 3 !}=\\frac{11}{2}(n-1) \\frac{(n+3) !}{3 !}$ $\\frac{(n+5)(n+4)}{2}=11(n-1)$ $n^{2}+9 n+20=22 n-22$ $n^{2}-13 n+42=0$ $(n-7)(n-6)=0$ $n=7,6$ Hence, values of n are 7 & 6", "1. Simplifying expressions I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 2. 1 - 1 x + h x 3. 2 x^2 10 x^5 4. 2x - 1 - 8 x^2-6x+9 x+1 x^2-2x-3 2. Originally Posted by journeygirl280 1. x^3-8 x-2 Hint: Difference two two cubes in numerator. 2. 1 - 1 x + h x With common denominator, $\\frac{x}{x(x+h)} - \\frac{x+h}{x(x+h)}$. Now continue. The last two do not make any sense. 3. Hello, journeygirl280! I think I deciphered the last two problems . . . $3)\\;\\;\\frac{\\frac{2}{x^2}}{\\frac{10}{x^5}}$ We have: . $\\frac{2}{x^2} \\div \\frac{10}{x^5} \\;= \\;\\frac{2}{x^2}\\cdot\\frac{x^5}{10} \\;=\\;\\frac{x^3}{5}$ $4)\\;\\;\\frac{2x}{x^2-6x+9} - \\frac{1}{x+1} - \\frac{8}{x^2-2x-3}$ We have: . $\\frac{2x}{(x-3)^2} - \\frac{1}{x+1} - \\frac{8}{(x+1)(x-3)}$ Get a common denominator: . . $\\frac{2x}{(x-3)^2}\\cdot{\\color{blue}\\frac{x+1}{x+1}} \\,-\\,\\frac{1}{x+1}\\cdot{\\color{blue}\\frac{(x-3)^2}{(x-3)^2}} \\,-\\,\\frac{8}{(x+1)(x-3)}\\cdot{\\color{blue}\\frac{x-3}{x-3}}$ . . $= \\;\\frac{2x(x+1)-(x-3)^2 - 8(x-3)}{(x-3)^2(x+1)} \\;= \\;\\frac{2x^2 + 2x - x^2 + 6x - 9 - 8x + 24}{(x-3)^2(x+1)} $ . . $= \\;\\frac{x^2+15}{(x-3)^2(x+1)}$ 4. Originally Posted by journeygirl280 I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions. 1. x^3-8 x-2 Different hint: $x^3-8=0$ when $x=2$ so $x-2$ is a factor of $x^3-8$. (well really the same hint in disguise) RonL" ]

[ "# There exist arbitrarily large prime gaps For every positive integer , there exists a sequence of consecutive integers all of which are positive. Thus, there exists a prime gap between consecutive primes that is greater than . Let . Consider the integers . For each , is divisible by and strictly larger than , hence is composite. Further, the sequence has length .", "is not divisible by 3. Therefore it is not the sum of three consecutive numbers.", "# Thread: Ten consecutive positive integers 1. ## Ten consecutive positive integers Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others.", "# The x is smaller! The positive number $$x$$ is the product of four different positive prime numbers. If the sum of these four prime numbers is a prime number greater than 20, what is the least positive value of $$x$$ ? ×", "4 Consecutive Integers cannot be Square-Free Theorem Let $n, n + 1, n + 2, n + 3$ be four consecutive positive integers. At least one of these is not square-free. Proof Exactly one of $n, n + 1, n + 2, n + 3$ is divisible by $4 = 2^2$. Thus, by definition, one of these is not square-free. $\\blacksquare$", "sum three numbers sum can be even or odd. But here sum of even power numbers is even. any sum of three numbers is divisible by 3. Take sum of 4^6 and 6^4. first take 6^4 is 36*36. both numbers are divisible by3. Now look 4^6. sum is 16*16*16. all three numbers not divisibly by 3. So option B is not a sum of consecutive numbers.. Re: Which of the following cannot be the sum of two or more cons [#permalink] 13 Sep 2015, 17:13 Similar topics Replies Last post Similar Topics: If a and b are prime numbers, which of the following CANNOT be the sum 0 05 Jul 2015, 14:32 4 Which of following cannot be true if sum of k consecutive 5 14 Jan 2014, 04:46 3 Which of the following is NOT the sum of the squares of two 5 09 Jan 2014, 06:54 16 Which of the following CANNOT be the sum of two prime numbers? 5 17 Feb 2011, 14:10 1 Which of the following CANNOT be the median of the four cons 10 27 Sep 2009, 17:35 Display posts from previous: Sort by", "positive integer can be written as the sum of 4 or fewer perfect squares. Three squares are not sufficient for numbers of the form 4k(8m + 7). A positive integer can be represented as a sum of two squares precisely if its prime factorization contains no odd powers of primes of the form 4k + 3. This is generalized by Waring's problem. A square number can only end with digits 00,1,4,6,9, or 25 in base 10, as follows: 1. If the last digit of a number is 0, its square ends in 00 and the preceding digits must also form a square. 2. If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four. 3. If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even. 4. If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four. 5. If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd. 6. If the last digit of a number", "# Which Four Integers? True or false: $\\quad$ The product of any 4 consecutive integers is always divisible by 4. ×", "respectively leave remainders of 1 and 2. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. the sum of three consecutive integers --. Prove that only one out of three consecutive positive integers is divisible. Thus it is divisible by both 3 and 2, which means it is divisible by 6. asked Feb 9, 2018 in Class X Maths by priya12 ( -12,636 points) real numbers. If p is a prime number, how many factors does p3 have? (A) One (B) Two (C) Three (D) Four (E) Five 5. [2] Name: Total Marks: Rebecca Simkins. What is the least positive integer n for which 165 × 513 + 10n is a multiple of. Verify this statement with the help of some examples. A set of integers such that each integer in the set differs from the integer immediately before by a difference of 2 and each integer is divisible by 2 Example 2, 4, 6, 8, 10, 12, 14,. As the product of three consecutive integers is divisible by 3, so is the product of four consecutive integers. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case –", "# Primes in consecutive numbers Is there a set of 2015 consecutive positive integers containing exactly 15 prime numbers? ×", "# Prove That The Product Of Three Consecutive Integers Is Divisible By 3 They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. Prove that the product of three consecutive positive integer is divisible by 6. gl/9WZjCW prove that the product of three consecutive positive integers is divisible by 6. Therefore, n = 3p or 3p + 1 or 3p + 2 , where p is some integer. 1 3 + 2 3 + 3 3 +. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. By the laws of divisibility, anything divisible by 2 and 3 is divisible by 6. Whenever a number is divided by 3 , the remainder obtained is either 0,1 or 2. Find the smallest number that, when. Essentially, it says that we can divide by a number that is relatively prime to. Let the three consecutive positive integers be n , n + 1 and n + 2. Let n be a positive integer. 1", "order. The example of non-consecutive odd integers, if someone went from 3 straight to 7, these are not consecutive. Try some examples: , ,. Question 1039608: Prove that one of every three consecutive positive numbers is divisible by 3 Answer by addingup(3677) (Show Source): You can put this solution on YOUR website! Let 3 consecutive positive integers be n, n+1 and n+2 Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2. 5, 7 The product of three consecutive numbers is always divisible by 6. Take the 3 consecutive integers, 2,3,4 their sum is 9 and you are done. Thus, either x, y, or z is a multiple of 3 and therefore has 3 as one of its prime factors. Suppose that for every sequence of n elements from H, some consecutive subsequence has the property that the product of its elements is the. In any set of 3 CONSECUTIVE numbers, there will always be one number that is divisible by 3, and at least one number that is divisible by 2. Hence the product is divisible y 2. Problem III. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. , 1000}, what is the probability", "? Free Version Moderate # Divisibility of Three Consecutive Integers GMAT-4ASG5V If $x$, $y$, and $z$ are a set of positive, consecutive integers where $x< y< z$, what number is $x+y+z$ always divisible by? A $2$ B $3$ C $4$ D All even numbers E None of the above", "# Is there a maximum number of consecutive sexy prime pair sums that are all divisible by $10$? I was finding the sums of pairs of sexy primes (prime numbers that differ by 6) and noticed that there are a lot of pairs who's sum is divisible by $$10$$. Ex. $$(7, 13)$$ as $$7+13=20$$ and $$20$$ is divisible by $$10$$. I then wondered if there are consecutive sexy prime pairs who's sum is divisible by $$10$$ and ran some PARI/GP code to find these kind of consecutive pairs. I found upto $$17$$ consecutive pairs divisible by $$10$$ on PARI and tested for a search limit of $$10^{12}$$. Here is the output I got after running my code, the first column displays the smallest prime of the lowest pair of consecutive sexy primes and the second column displays the number of consecutive pairs divisible by $$10$$. 7 1 167 2 2237 3 2267 4 108187 5 1004057 6 3281777 7 32895377 8 65947927 9 569959037 10 602817437 11 5476396897 12 16842019627 13 16842019637 14 17004549137 15 312318208577 16 382560132847 17 If there are an infinite number of sexy primes, then is there a maximum number of consecutive sexy prime pair sums that are all divisible by $$10$$? • Sum $10$ can only occur , if the smaller prime has", "### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### 14 Divisors What is the smallest number with exactly 14 divisors? ### Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? For how many primes $p$ is the number $p^3+3$ also prime?", "# POJ2739 Sum of Consecutive Prime Numbers（AC代码 + 详解） • Sum of Consecutive Prime Numbers Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. Your mission is to write a program that reports the number of representations for the given positive integer. • Input The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero. • Output The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.", ". One number less than a square (m − 1) is always the product of \\sqrt{m}-1 and \\sqrt{m}+1 (for example, 8 × 6 equals 48, while 7 equals 49). Thus, 3 is the only prime number one less than a square. A square number is also the sum of two consecutive triangular numbers. The sum of two consecutive square numbers is a centered square number. Every odd square is also a centered octagonal number. Another property of a square number is that (except 0) it has an odd number of positive divisors, while other natural numbers have an even number of positive divisors. An integer root is the only divisor that pairs up with itself to yield the square number, while other divisors come in pairs. Lagrange's four-square theorem states that any positive integer can be written as the sum of four or fewer perfect squares. Three squares are not sufficient for numbers of the form . A positive integer can be represented as a sum of two squares precisely if its prime factorization contains no odd powers of primes of the form . This is generalized by Waring's problem. In base 10, a square number can end only with digits 0, 1, 4, 5, 6 or 9, as follows: • if the last digit of a number", "# A Four Product The product of any 4 consecutive integers must be divisible by $\\text{\\_\\_\\_\\_\\_\\_\\_\\_\\_}.$ ×", "14 Divisors What is the smallest number with exactly 14 divisors? Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? Dozens Do you know a quick way to check if a number is a multiple of two? How about three, four or six? Weekly Problem 32 - 2015 Stage: 3 Challenge Level: In this Multiplication Magic Square, the product of the three numbers in each row, column and each of the diagonals is $1$. What is the value of $r+s$? This problem is taken from the UKMT Mathematical Challenges. View the previous week's solution View the current weekly problem", "## Sums of Positive Consecutive Integers: Proof In my previous post, I tackled this problem: Try to express positive integers in terms of the sum of two or more consecutive positive integers. For instance, 3 ... ## Sums of Consecutive Positive Integers Edit: The bit about the larger prime numbers was due to an error in my VBA programming, but it lead to a better understanding of the problem. Don’t take t..." ]

[ "pair:$$(445\\,953\\,248\\,528\\,881\\,275,659\\,008\\,669\\,204\\,392\\,325)$$such that the smallest prime factor of the first term is $3$, whereas the smallest prime factor of the second term is $5$.", "the smallest representable number is around $$10^{-300}$$.", "$499 \\triangleleft p$. Hence we may assume $z \\in \\{6\\}$, and thus $p \\in 94\\{6\\}9$, and the smallest prime $p \\in 94\\{6\\}9$ is $946669$. $\\blacksquare$", "by breaking out the calculator and got a multiple of 3. Chetan2u makes it look so easy! There is an error in your concept: Smallest prime factor of 2*5*11 is 2 - Correct Smallest prime factor of 2 + 5 is neither 2 nor 5 but actually 7 - a new prime number. When you add, you don't know the prime factors you will get. Here is how you will recombine: $$5^8 + 10^6 - 50^3$$ (You have the negative sign misplaced in your expression) $$5^8 + 2^6*5^6 - 2^3*5^6$$ $$5^6 * (25 + 64 - 8)$$ $$81 * 5^6$$ $$3^4 * 5^6$$ The smallest prime is 3. _________________ Karishma Veritas Prep GMAT Instructor Senior Manager Joined: 28 Feb 2014 Posts: 291 Location: United States Concentration: Strategy, General Management Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] ### Show Tags 23 Jan 2015, 21:18 Bunuel wrote: What is the smallest prime factor of 5^8+10^6–50^3? A. 2 B. 3 C. 5 D. 7 E. 11 Kudos for a correct solution. (5^8)+(10^6)–(50^3) Breaking each component down gives: (5^8) smallest prime is 5 (2*5)^6 smallest prime is 2 (2*5^2)^3 smallest prime is 2 Senior Manager Joined: 28 Feb 2014 Posts: 291 Location: United States Concentration: Strategy, General Management Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] ###", "to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\\boxed{260}$.", "# Think small What is the smallest (positive) prime number that divides $1 + 12 + 123 + 1234 + 12345 + 123456\\\\ + 1234567 + 12345678 + 123456789 + 1234567890?$ Details and assumptions There are 10 terms in the expression, and the last number is 1234567890. To ensure that you see the entire equation, scroll till you see the question mark (?). ×", "is the smallest prime factor of $$5^8+10^6–50^3$$? A. 2 B. 3 C. 5 D. 7 E. 11 Kudos for a correct solution. $$5^8 + 10^6 - 50^3$$ $$5^8 = (5^2)^4$$ $$10^6 = ((2*5)^2)^3$$ $$50^3 = (2*5^2)^3$$ $$(5^2)^3 + (5^2)^3 - (5^2)^3 [5^2 + 2^6 - 2^3] = 5^6 * 81 = 5^6 * 3^4.$$ So, $$3$$ is the smallest prime factor. Ans - B. Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] 15 Jan 2018, 23:06 Display posts from previous: Sort by", "is an article from Wikipedia for Deficient number. – user243301 Jun 5 '18 at 15:11 $1)$ The smallest counterexample is $12\\ 285$. There are many more. $2)$ Every prime of the form $105k-1$ does the job, and there are infinite many such primes (Dirichlet's theorem)", "same- what is the smallest prime number $k$ such that $k\\times 2^n+1$ is composite? The smallest proven ‘prime’ Sierpinski number is 271,129. In order to prove that it is the smallest prime Sierpinski number, it has to be shown that every single ‘prime’ $k$ less than 271,129 is not a Sierpinski number, and to do that, some $n$ must be found that makes $k\\times2^n+1$ prime.", "The list below shows the first ten numbers together with their divisors (factors): 1. $1$ 2. $1$, $2$ 3. $1$, $3$ 4. $1$, $2$, $4$ 5. $1$, $5$ 6. $1$, $2$, $3$, $6$ 7. $1$, $7$ 8. $1$, $2$, $4$, $8$ 9. $1$, $3$, $9$ 10. $1$, $2$, $5$, $10$ What is the smallest number with exactly twelve divisors? What is the smallest number with exactly fourteen divisors?", "# 1,122,659 Previous ... Next ## Number $1 \\, 122 \\, 659$ (one million, one hundred and twenty-two thousand, six hundred and fifty-nine) is: The $87 \\, 359$th prime number The $1$st term of the smallest Cunningham chain of the first kind of length $7$: $1 \\, 122 \\, 659$, $2 \\, 245 \\, 319$, $4 \\, 490 \\, 639$, $8 \\, 981 \\, 279$, $17 \\, 962 \\, 559$, $35 \\, 925 \\, 119$, $71 \\, 850 \\, 239$", "where the digits add to$19\\$ and that's 199. So the smallest where a (n)=4 will be the smallest where the digits add to 199. (It will have at least 22 digits so prime factors is definitely a useless idea). – fleablood Jan 30 at 2:50 Prime factorizations don't matter, digits do. So look at the smallest number such that $$a(n)>0$$. Note that $$T(d)=d$$ for all single digits $$d$$, so try $$T(10)=1$$, so $$a(10)=1$$. Going from here, we now know that if $$T(n)=10$$, then $$a(n)=2$$, since 10 is not a fixed point. To do that, simply make the digits sum to 10, and make the smallest number possible, which is $$19$$. Next, if $$T(n)=19$$ then we know $$a(n)=3$$. I'll leave you to figure out what the smallest number that has a digit sum of 19 is.", "Courses Courses for Kids Free study material Free LIVE classes More # What is the H.C.F. of the smallest prime number and the smallest composite number? Last updated date: 29th Mar 2023 Total views: 310.5k Views today: 4.87k Smallest prime number $= 2$ Smallest composite number $= 4$ Prime factors of $2$is itself $4$ Now consider the prime factors of $4$ i.e. $4 = 2 \\times 2$ In $2$, there are one $2$while in $4$, there are two $2s$. So, in this case the common factors are one $2$. Thus, the H.C.F. of the smallest prime number and smallest composite number is $2$. Note: Always remember that prime numbers start from $2$ and composite numbers starts from $4$. And prime numbers have only two factors i.e. one and itself. But in case of composite numbers the factors are more than two.", "of $4$. $5$ is the only odd number for which every number ending in it is composite. $6$ is the smallest perfect number $7$ is the smallest number that cannot be represented as the sum of three integer squares (e.g. $6=2^2+1^2+1^2$). Or $\\frac n7$ always features the same digits in the same order with the start varying. $8$ is the only Fibonacci Number besides $1$ that is a perfect cube. $9$ is roughly $2\\pi+e$ $10$ is the sum of the first three primes, the first four factorials, and the first four positive integers. The powers of $11$ can be found by Pascal's Triangle. $12$ is the smallest sublime number, of which there are only two known. $13$ is the smallest emirp. $14$ is the smallest satisfier of the Shapiro Inequality. $15$ is $1|5$ and $1+2+3+4+5$ $16$ is the smallest perfect fourth power besides $0$ and $1$ $17\\approx12\\sqrt2$ and so we can use it to approximate $\\sqrt2$ $18$ is the first non-square number expressible as $p\\cdot q^2$, with $p,q$ integers and $q>p$ Any natural number can be made by summing $19$ $4$th powers. The product of the divisors $(1,2,4,5,10)$ with the proper divisors $(2,4,5,10)$ of $20$ is $20$. $21$ is a triangular number. $22\\approx7\\pi$, thus $\\pi\\approx\\frac{22}{7}$ is used where a fractional approximation is needed. $23$ is the smallest number", "# At least you got this in i-land Find the smallest prime factor $$p> 3$$ of $$200!+200!!+2!!!+1$$. Review double factorials and multifactorials to understand what $$200!!$$ and $$2!!!$$ refers to. ×", "coz its 1 less than 1729 . The smallest number to represented as a sum of 2 cubes . $0! = 1$ —- Coz 0! = 1 …. $\\left ( 0010\\right )_{10}$ —- 0010 in base 10 is 2 . $FermatPrime\\left[ 0\\right ] = 3$ —- 3 is the first Fermat Prime . All numbers satisfying $F_{n} = 2^{2^{ \\overset{n} {}}} + 1$ $2 \\uparrow\\uparrow\\uparrow 2 = 4$ —- thats the way 4 represented in Knuth’s Up Arrow Notation . $\\left ( 2\\varphi - 1 \\right )^2 = 5$ —- where $\\varphi$ is the golden ratio $\\varphi = \\frac{1+\\sqrt{5}}{2}$ $\\mid S_{3} \\mid = 6$ —- 6 is the cardinality of the smallest non-abelian group . $M_{3} = 7$ —- 7 is the 3 Mersenne Prime . $F_{6} = 8$ —- 8 is the 6th Fibonacci Number . $3^{2^{1}} = 9$ —- 9 is an exponential factorial . $\\left ( 1010\\right )_2 = 10$ —- 1010 in binary is 10 in decimal (base 10) $\\frac{\\Phi_2(10)}{\\gcd(\\Phi_2(10),2)} = 11^{c} , {c} \\in \\mathbb{N}$ —- 11 is the second unique prime…. Finally, after the assembly it looked like this …… a better pic sometime sooner ….. # Beauty of Numbers I was stumblingupon sites when i came across something really beautiful . Maybe all of u must be aware of it anyways", "smallest prime number? Answer: 2 is the smallest prime number.", "\\times &&&&& 3 \\\\ \\hline 4&2&8&5& 7 & 1 \\end{array}$ The smallest such number is $142857.$ Really nice work, but I think you have first to check it for the 3,4 and 5 digits", "# Binary Primes Let $$f$$ be a function in the natural numbers, such that $$f(n)$$ gives the binary representation of $$n$$ in base 10. For example, $$f(5) = 101$$, because 5 is represented as 101 in binary. A number $$n$$ is called good if $$n$$ and $$f(n)$$ are both prime numbers. For example, 5 is good, because 5 and 101 are prime numbers. 7 is not good, since $$f(7) = 111 = 3 \\times 37$$ is not prime. The smallest good number is 3. The second smallest good number is 5. Find the $$19^\\text{th}$$ smallest good number. ×", "+0 # Math +1 70 1 +203 What is the smallest possible product of one one-digit prime and two distinct two-digit primes? Dec 9, 2021 #1 +132 0 The answer relies on minimizing all of the primes. The smallest one-digit prime is 2. The smallest two-digit primes are 11 and 13. $$2 \\cdot 11 \\cdot 13 = 286$$. Dec 9, 2021" ]

[ "# There exist arbitrarily large prime gaps For every positive integer , there exists a sequence of consecutive integers all of which are positive. Thus, there exists a prime gap between consecutive primes that is greater than . Let . Consider the integers . For each , is divisible by and strictly larger than , hence is composite. Further, the sequence has length .", "when written in that base. Output the sum of all the odd numbers between firstNum and secondNum inclusive. C / C++ Forums on Bytes. The sum of the digits is divisible by 4. Well, I'm tryin to make a for loop that computes the sum of the odd numbers in the range from 0 to 100. Now, the sum of the digits from 1 to 9 is odd, so the difference in any such number must be either 11 or 33. The sum of the integers would. The sum of the number: 234168. Similarly, the first digit cannot be 7 or 9, so this entire case does not work. Divisible by 5 if the last digit is 0 or 5 (9905). Question 2. (the smallest fortunate triangular number) 3 (A005235) (the smallest weird number)/(the only prime one less than a cube) 70 (A006037) /7 (a 3-1 is divisible by a-1, so it can be prime only for a = 2) = 10 (the third most probable product of the numbers showing when two standard six-sided dice are rolled). If there is no such maximum sum of the numbers, the program should print -1 as output. There are$3 \\times 3 = 9$total possibilities for the two dice, which makes the probability of getting a multiple of three$3/9 = 1/3$.", "numbers without leaving any remainder. So the factors of the smallest composite number $4 = 1 \\times 2 \\times 2$. And the factors of the smallest prime number 2 $= 1 \\times 2$. So, from above we clearly say that the largest positive integer number that divides both the numbers 4 and 2 without leaving any remainder is 2. So, the g.c.d of the smallest prime number and the smallest composite number is 2. Hence option (b) is correct. Note – Whenever we face such types of questions the key concept we have to remember is that always recall the property of composite number, prime number and GCD respectively which is stated above then use this property first find out the smallest prime and composite number respectively, then find the value of GCD using the property of GCD which is the required answer.", "Number Theory # Prime Factorization Warmup What is the smallest positive integer divisible by each of 10, 20, and 30? What is $a \\times b \\, ?$ A group of explorers discovers a chest containing 5040 gold coins. When they try to divide the coins so that each explorer receives the same number of coins, they find that they are unable to do so. Which of these could be the number of explorers in the group? Hint: $5040 = 7! = 1 \\times 2 \\times 3 \\times 4 \\times 5 \\times 6 \\times 7.$ What is the smallest number that is divisible by 1, 2, 3, 4, 5 and 6? Which of these statements are true? I. Even numbers are never divisible by odd numbers. II. Odd numbers are never divisible by even numbers. ×", "# The sum of $101$ consecutive positive integers is $p^3$, where $p$ is a prime. Find the smallest of the 101 integers. The sum of $101$ consecutive positive integers is $p^3$ where $p$ is a prime no. What is the smallest of the 101 integers? My Approach: $$p^3 = (n)+(n+1)+......+(n+100) = 101n + {(100)(101)/2} = 101(n + 50).$$ Now I can't find the characteristics of $p$. Can anybody help? - Well, $101$ is prime. So the smallest possible value for $n+50$ is...? – TonyK Jan 26 '14 at 10:40 @TonyK \"So the only value for ...\" – Henry Jan 26 '14 at 10:41 (n+50) = (101)^2 . Is it right? Then, is n = 10151 ? – Anish Bhattacharya Jan 26 '14 at 10:43 Yes, that's right. – TonyK Jan 26 '14 at 22:01", "# Is there a maximum number of consecutive sexy prime pair sums that are all divisible by $10$? I was finding the sums of pairs of sexy primes (prime numbers that differ by 6) and noticed that there are a lot of pairs who's sum is divisible by $$10$$. Ex. $$(7, 13)$$ as $$7+13=20$$ and $$20$$ is divisible by $$10$$. I then wondered if there are consecutive sexy prime pairs who's sum is divisible by $$10$$ and ran some PARI/GP code to find these kind of consecutive pairs. I found upto $$17$$ consecutive pairs divisible by $$10$$ on PARI and tested for a search limit of $$10^{12}$$. Here is the output I got after running my code, the first column displays the smallest prime of the lowest pair of consecutive sexy primes and the second column displays the number of consecutive pairs divisible by $$10$$. 7 1 167 2 2237 3 2267 4 108187 5 1004057 6 3281777 7 32895377 8 65947927 9 569959037 10 602817437 11 5476396897 12 16842019627 13 16842019637 14 17004549137 15 312318208577 16 382560132847 17 If there are an infinite number of sexy primes, then is there a maximum number of consecutive sexy prime pair sums that are all divisible by $$10$$? • Sum $10$ can only occur , if the smaller prime has", "50 million\". That's funny, I just tested all primes up to $3$ and found one whose base $7$ representation has a sum of digits divisible by $3$. Maybe you meant to exclude the \"trivial\" case? – Oscar Lanzi Jun 9 at 19:12 Let $a = \\sum_{i = 0}^n a_i 7^i$ be a number written in base $7$, that is, $0 \\leq a_i \\leq 6$. Note that $7^i = (1 + 3 \\cdot 2)^i = 1 + 3 b_i$ for some $b_i \\geq 0$. Hence $a = \\sum_{i = 0}^n a_i + 3 \\sum_{i = 0}^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$. You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three. So what is about the base of $7$: Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same.", "Divisible Probability... The sum of the digits of a $$9$$ digit number is $$77$$. The probability that this number is divisible by 11 can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are co-prime positive integers. Find the value of $$a+b$$. ×", "# Think small What is the smallest (positive) prime number that divides $1 + 12 + 123 + 1234 + 12345 + 123456\\\\ + 1234567 + 12345678 + 123456789 + 1234567890?$ Details and assumptions There are 10 terms in the expression, and the last number is 1234567890. To ensure that you see the entire equation, scroll till you see the question mark (?). ×", "$$N = \\frac{a! * b! * c! * d!}{e!}$$ where a, b, c, and d, are four distinct positive integers, which are greater than 1, and a, e are two consecutive numbers, which are prime. What is the least possible value of $$\\frac{(a + b + c + d)}{e}$$, if N is divisible by cube of product of the three smallest odd prime numbers? A. $$\\frac{26}{3}$$ B. 9 C. $$\\frac{21}{2}$$ D. 13 E. $$\\frac{27}{2}$$ _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. e-GMAT Representative Joined: 04 Jan 2015 Posts: 2942 Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................) [#permalink] ### Show Tags 02 Jan 2019, 02:07 Solution Given: • $$N = \\frac{a! * b! * c! * d!}{e!}$$ • a, b, c and d are four distinct positive integers, which are greater than 1 • a, e are two consecutive numbers, which are prime • N is divisible by cube of product of the three smallest odd primes To find: • The least possible value of $$\\frac{(a + b + c + d)}{e}$$ Approach and Working: • (a, e) = (2. 3). Since, (2, 3) is the only pair of numbers, which are prime And, we are given that N is divisible by", "the sum of its divisors is a perfect square. $$1 + 2 + 3 + 6 + 11 + 22 + 33 + 66 = 12^2 = 144$$ There are only three year dates (to my knowledge) for which the sum of the divisors is a square. They all three occur in this MathFacts page, 66, 70, and 81. 66 and 70 even have the same total. If you wrote out all the numbers on a 12 hour clock, (HrMin, so 6:03 would be 603, etc.), there would be 66 of them that are prime. The smallest number that can be expressed as a sum of two two-digit primes, each ending with the digit three, in two different ways (66 = 13 + 53 = 23 + 43) and the smallest number that can be expressed as the sum of two primes in six different ways. 66 is the largest day number of the year which does not have a letter \"e\" in its English spelling. Sixty-six is the 19th such numbers in the year, but the next number without an \"e\" is 2000. 66 however is not the sum of two squares. To check if an arbitrary number is the sum of two squares, factor it. If any factor p^a + 1 is divisible by four, then", "# What are the factors of 29? $29$ has positive factors $1$ and $29$ $29$ is only divisible by the positive numbers $1$ and $29$, so it is a prime number.", "Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. 14 Divisors What is the smallest number with exactly 14 divisors? Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? Stage: 3 Short Challenge Level: The number is less than $100$, but five times the number is greater than $100$, so it must be between $20$ and $100$. Reversing the digits makes a prime number, so the first digit must be $1$, $3$, $7$ or $9$, as all two-digit prime numbers are odd and not divisible by $5$. The number must be at least 20, so this rules out $1$. Since the digits add to a prime number, the possibilities are: First Digit Second Digit $3$ $2,4,8$ $7$ $4,6$ $9$ $2,4,8$ The number must be one more than a multiple of $3$, so the digits must sum to give one more than a multiple of $3$, as multiples of $3$ have digit sums that are multiples of $3$. This leaves $34$, $76$ and $94$. The number must have exactly one prime digit, which rules out $94$. The number must have exactly four factors. $34$ has", "# What is the smallest $n \\in \\mathbb{N}$ such that $n$ is divisible by $2,3,5$, is square and a fifth power Can anyone help me prove what is the smallest $n \\in \\mathbb{N}$ such that $n$ is divisible by $2,3,5$, is square and a fifth power I have so far, for $n,y,q,p,z\\in \\mathbb{N}$ $n=30q$ , $n=y^2$ $\\Rightarrow q=\\frac{p^2}{30}$ And obviously $n=z^5$ - The answer is $30^{10}.$ Think about prime factorizations. – Ragib Zaman Apr 22 '12 at 16:56 Clearly $n=(2\\cdot3\\cdot5)^{10}$. If $n$ divisible by a prime $p$, and it's a square, it must be divisible by $p^2$. Similarly, if it's a fifth power, it must be divisible by $p^5$ and hence by $p^{10}$.", "32 is the ninth happy number. - 5 years, 3 months ago 33 is the smallest sum of two different positive numbers, each of which raised to the fifth power. 33 is the largest positive integer that cannot be expressed as a sum of different triangular numbers. 33 is the smallest integer such that it and the next two integers all have the same number of divisors. - 5 years, 3 months ago 34 is the ninth Fibonacci number - 5 years, 3 months ago 35 is the sum of the first five triangular numbers, making it a tetrahedral number. 35 is the highest number one can count to on one's fingers using base 6. - 5 years, 3 months ago 36 is the square of the first perfect number. - 5 years, 3 months ago 37 - Every positive integer is the sum of at most 37 fifth powers. See Waring's Problem. - 5 years, 3 months ago 38 is the sum of the squares of the first three primes. Largest even number which cannot be written as sum of two odd composite numbers. - 5 years, 3 months ago 39 is the smallest natural number which has three partitions into three parts which all give the same product when multiplied: {25, 8, 6}, {24, 10, 5},", "least possible value of $$\\frac{(a + b + c + d)}{e}$$, if N is divisible by cube of product of the three smallest odd prime numbers? A. $$\\frac{26}{3}$$ B. 9 C. $$\\frac{21}{2}$$ D. 13 E. $$\\frac{27}{2}$$ _________________ If you liked my solution then please give Kudos. Kudos encourage active discussions. e-GMAT Representative Joined: 04 Jan 2015 Posts: 2432 Re: Question of the week - 29 (N = a! * b! * c! * d!/e! ................) [#permalink] Show Tags 02 Jan 2019, 01:07 Solution Given: • $$N = \\frac{a! * b! * c! * d!}{e!}$$ • a, b, c and d are four distinct positive integers, which are greater than 1 • a, e are two consecutive numbers, which are prime • N is divisible by cube of product of the three smallest odd primes To find: • The least possible value of $$\\frac{(a + b + c + d)}{e}$$ Approach and Working: • (a, e) = (2. 3). Since, (2, 3) is the only pair of numbers, which are prime And, we are given that N is divisible by $$3^3 * 5^3 * 7^3$$ • Thus, b, c and d must contain at least one 7, one 5 and one 3 each So, the minimum value that (b, c, d) can take is (7, 8 , 9), since,", "either of the form {0}4{0}1 or contain only 0 and 5, thus must be divisible by 5 2022-04-07, 03:25 #322 sweety439 \"99(4^34019)99 palind\" Nov 2016 (P^81993)SZ base 36 338010 Posts Since finding the set T requiring finding the set M(Lb) first, finding T is much harder than finding M(Lb) For b = 8 and b = 9 and b = 12, T contains infinitely many numbers: * For b = 8, T contains the smallest prime factor of 8^(2^n)+1 for all n * For b = 9, T contains the smallest prime factor of (9^n-1)/8 for all prime n, also contain the smallest prime factor of (25*9^n-1)/8, 4*9^n-49, 4*9^n-1, 9^n-4 for all n * For b = 12, T contains the smallest prime factor of 12^n-25 for all even n (for all odd n, the smallest prime factor is 13) Theorem: For all base b, T must contain all primes <= b Proof: For primes p < b, consider the combination of the digits divisible by p (including 0), and for prime p = b, consider the family 1{0} We find the T for b = 10: * T must contain {2,3,5,7} * The smallest prime of the form 2{0}21 is 20021 --> 221 (=13*17) and 2021 (=43*47) must be removed, thus either 13 or 17 is needed,", "Let $$S$$ be the set of integers between 1 and $$2^{40}$$ whose binary expansions have exactly two 1's. If a number is chosen at random from $$S,$$ the probability that it is divisible by 9 is $$p/q,$$ where $$p$$ and $$q$$ are relatively prime positive integers. Find $$p+q.$$ (第二十二届AIME2 2004 第10题)", "as much fun reading this article as I’ve had writing it! # A Solution to Google Treasure Hunt 2008 Problem 4 Find the smallest number that can be expressed as the sum of 3 consecutive prime numbers, the sum of 5 consecutive prime numbers, the sum of 517 consecutive prime numbers, the sum of 551 consecutive prime numbers, and is itself a prime number. For example, 41 is the smallest prime number that can be expressed as the sum of 3 consecutive primes (11 + 13 + 17 = 41) and the sum of 6 consecutive primes (2 + 3 + 5 + 7 + 11 + 13 = 41). I wrote my solution to this problem in ~60 lines of python. In a nutshell: 1. Get a precomputed list of (1,000,000) primes. 2. Find all possible summations of 551 consecutive prime numbers. If the sum is prime, store this in a list of ‘candidates’. 3. Find all possible summations of 517 consecutive prime numbers. If the sum is in the candidate list from step 2, store it in a new candidate list. 4. Repeat step 3 using this new candidate list for summations of 5 and 3 consecutive prime numbers. 5. The smallest number in the final candidates list is the answer. The script ran pretty quickly", "n. Excluding the possibility of very small numbers ($<5$), as you mentioned, 3 consecutive numbers have at least 1 of them even. So it must be the format of $2p$, where $p$ is a prime. Since $4|2p-2, 2p+2$, the only possibility is $2p-1, 2p, 2p+1$. During the search, you can also rule out the pair ($2p, 2p+1$) of safe primes. As you said one of the numbers must be divisible by $2$ and one must be divisible by $3$. You argued that they have to be different. And you argued that $2$ cannot divide more than one of the numbers. The third number has to be of the form $p^3$ or $pq$. The smallest values $p$ respectively $p,q$ can take are $5$ and $7$. So the smallest possible value the third number can take is $5 \\cdot 7$ as $5^3 >> 5 \\cdot 7$. Thus the largest of the numbers has to be at least 35, you can start your search here." ]

[ "at 22:46 $$32\\left(1.05 + \\dfrac{2.576(\\sqrt{0.4475})}{\\sqrt{32}}\\right) = 43.348... \\rightarrow 44$$", "\\left(156 - 72 \\ln \\left(3\\right)\\right)$", "99.5) \\\\ =&P\\left(Z \\ge \\dfrac{99.5-1316}{623} \\right) \\\\ =&P\\left(Z \\ge -1.9526 \\right) \\\\ =&P\\left(Z \\le 1.9526 \\right) \\\\ =&0.9744 \\end{array}$", "\\right).$", "\\right).$", "\\right]$", "\\right ]$", "+ 15\\right)$", "is $\\dfrac {1}{ 5 }$.", "1\\right)$", "1\\right)$", "1\\right)$", "\\right \\}}$.", "2\\right)$", "2\\right)$", "2\\right)$", "2\\right)$", "\\dfrac{1}{64974} \\approx 0.00001539\\ldots$.", "\\left(\\frac{J}{s}\\right)} = 0.0044 s$", "\\right\\} .$" ]

[ "L^nx$ is nonzero?", "Simplify: Explanation: We know that any non-zero number raised to the power of is equal to Now factor: ← Previous 1 3", "## Thinking Mathematically (6th Edition) $1$ $a^0=1, a \\ne 0$ Thus, any nonzero real number raised to the zero power is equal to 1.", "has no nonzero zero divisors.", "# Definition:Nonzero Integer ## Definition A nonzero integer is an element of the set: $\\Z \\setminus \\set 0$ where: $\\Z$ denotes the set of integers $\\setminus$ denotes set difference. Thus: $\\Z \\setminus \\set 0 = \\set {\\ldots, -3, -2, -1, 1, 2, 3, \\ldots}$ and can be denoted $\\Z_{\\ne 0}$. ## Also known as The term can be seen hyphenated: non-zero integer.", "an exponent $x0=1$; any nonzero base raised to the 0 power is defined to be 1. Zero factor property Given any real number a, $a⋅0=0⋅a=0.$ Zero factor property Given any real number a, $a⋅0=0⋅a=0.$ zero-product property Any product is equal to zero if and only if at least one of the factors is zero. zero-product property Any product is equal to zero if and only if at least one of the factors is zero.", "So, what condition on a set of numbers $\\{a_1,a_2,\\dots,a_n\\}$ will ensure that their product is nonzero?", "# How much is zero high worth at zero power? We know that every nonzero number, raised to zero, is equal to 1. But what if the number is zero? The mathematical expression 00 is considered as an indetermination in mathematics. In calculation, as it is a widely used expression, it is considered by convention to be equal to 1.", "# Which of the following answer choices lists numbers in decreasing order? $$4^{1}, 4^{0}, 0,-4$$ Explanation Any number to the zero power $$(4^{0})$$ is 1 and any number to the first power $$4^{1}$$ is that number. So, $$4^{1}, 4^{1}, 0,-4$$ and is in the correct order from largest to smallest number. Visit our website for other ASVAB topics now!", "## College Algebra (11th Edition) Any real number raised to the 0th power equals 1, so $6^{0}$ and $(-6)^{0}$ = 1. If there is a negative sign in front that is not in parentheses, such as -$6^{0}$ it is not raised to the 0th power, so it would be -1.", "exist some nonzero X such that $$X^tAX>=0$$.", "## College Algebra (6th Edition) Any non-zero base to the power of zero is equal to one. However, this base is only the three, the negative sign is not included in the bases unless indicated by parentheses. $-3^{0}=-(3^{0})=-1$", "## Intermediate Algebra (12th Edition) According to page 267, $a^{0}=1$ (where $a$ is a nonzero real number). In other words, we know that any nonzero real number raised to a power of 0 will equal 1. Therefore, $-8^{0}-k^{0}=-(8^{0})-(k^{0})=-(1)-(1)=-1-1=-2$. Recall that the bases of the exponents are 8 and k, respectively, and not -8 and –k.", "} } = x ^ { n - n } = x ^ { 0 }$$ This leads us to the definition of zero as an exponent107, $$x ^ { 0 } = 1\\: x \\neq 0$$ It is important to note that $$0^{0}$$ is indeterminate. If the base is negative, then the result is still positive one. In other words, any nonzero base raised to the zero power is defined to be equal to one. In the following examples assume all variables are nonzero. Example $$\\PageIndex{7}$$: Simplify: 1. $$(−2x)^{0}$$ 2. $$−2x^{0}$$ Solution: a. Any nonzero quantity raised to the zero power is equal to $$1$$. $$( - 2 x ) ^ { 0 } = 1$$ b. In the example, $$−2x^{0}$$ , the base is $$x$$, not $$−2x$$. \\begin{aligned} - 2 x ^ { 0 } & = - 2 \\cdot x ^ { 0 } \\\\ & = - 2 \\cdot 1 \\\\ & = - 2 \\end{aligned} Noting that $$2^{0} = 1$$ we can write, $$\\color{Cerulean}{\\frac { 1 } { 2 ^ { 3 } }}\\color{Black}{ = \\frac { 2 ^ { 0 } } { 2 ^ { 3 } } = 2 ^ { 0 - 3 } =}\\color{Cerulean}{ 2 ^ { - 3 }}$$ In general, given any nonzero real number $$x$$", "# How do you simplify 500^0? ##### 1 Answer Jan 2, 2017 ${500}^{0}$ = 1 #### Explanation: Any number raised to the zeroth power is one, which is illustrated by the image below:", "Intermediate Algebra (12th Edition) According to page 267, $a^{0}=1$ (where $a$ is a nonzero real number). In other words, we know that any nonzero real number raised to a power of 0 will equal 1. Therefore, $-4^{0}-m^{0}=-(4^{0})-(m^{0})=-(1)-(1)=-1-1=-2$. Recall that the bases of the exponents are 4 and m, respectively, and not -4 and –m.", "where $nnz$ is the number of non-zeros.", "$m+n=\\boxed{831}$.", "## Elementary Algebra The zero exponent rule states that any number raised to the $0$ power is $1$. For example, if we raise $4$ to the $0$ power, we obtain: $4^0=1$.", "non-zero, left annihilator of any element of the ultrapower." ]

[ "$\\dfrac{x+y}{2}$, the desired answer is $\\boxed{450}$.", "99.5) \\\\ =&P\\left(Z \\ge \\dfrac{99.5-1316}{623} \\right) \\\\ =&P\\left(Z \\ge -1.9526 \\right) \\\\ =&P\\left(Z \\le 1.9526 \\right) \\\\ =&0.9744 \\end{array}$", "{1 + \\dfrac 1 n}$", "# Which of the following answer choices lists numbers in decreasing order? $$4^{1}, 4^{0}, 0,-4$$ Explanation Any number to the zero power $$(4^{0})$$ is 1 and any number to the first power $$4^{1}$$ is that number. So, $$4^{1}, 4^{1}, 0,-4$$ and is in the correct order from largest to smallest number. Visit our website for other ASVAB topics now!", "at 22:46 $$32\\left(1.05 + \\dfrac{2.576(\\sqrt{0.4475})}{\\sqrt{32}}\\right) = 43.348... \\rightarrow 44$$", "or quotient rule, note that $\\dfrac{\\mathrm d}{\\mathrm dx}\\left(\\dfrac{1}{n+1}\\right)=0$, so it doesn't make any difference. – Prasun Biswas Apr 11 '15 at 17:09 • I would like to say that yes you can use the product or quotient rule! There's no restriction saying you can only use it if $u,v$ are non constant. – user223391 Apr 11 '15 at 17:24", "(1-1/(n\\ln n))^n <(\\ln (n-1)/\\ln n)^n < (1-1/[(n+1)\\ln n])^n.$$ Write the power $n$ as $[(n+1)\\ln n]\\cdot n/[(n+1)\\ln n]$ to see the right hand side $\\to (1/e)^0 = 1.$ Same idea for the left hand side. So the limit is $1.$", "the set of non-positive numbers $\\left(- \\infty , 0\\right]$ because that's the set of possible outputs. The graph is shown below. Make sure you think about all these answers in relation to the graph. graph{-x^2+6x-9 [-40, 40, -20, 20]}", "were not explained above. Let us look at them below: The one that we used in the end is - ${x^{ - a}} = \\dfrac{1}{{{x^a}}}$. For example: ${x^{ - 4}} = \\dfrac{1}{{{x^4}}}$. This rule is called the negative exponent rule. There are many other rules of powers. They are as follows: 1) Product rule: ${x^a} \\times {x^b} = {x^{a + b}}$ 2) Quotient rule: $\\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$ 3) Power rule: ${\\left( {{x^a}} \\right)^b} = {x^{ab}}$ 4) Zero rules: ${x^0} = 1$ There is another rule which says that any number raised to the power “one” equals itself and another rule related to one is that – the number “one” raised to power any number gives us “one” itself. These rules are very handy and help to solve questions easily.", "The power of 2 $\\left(1 - x^{32}\\right)\\left(\\frac{1}{1 - x} + \\frac{1}{x + 1} + \\frac{2}{x^2 + 1} + \\frac{4}{1 + x^4} + \\frac{8}{1 + x^8} + \\frac{16}{1 + x^{16}}\\right) = ?$ This is part of the series: It's easy, believe me! ×", "= \\dfrac{8\\sqrt3}{9}\\left(\\dfrac{27}{32}\\right) = \\boxed{\\dfrac{3\\sqrt 3}{4}}$$ . Apr 14, 2020", "is $\\dfrac {1}{ 5 }$.", "61 $\\dfrac{a^{-2}}{a^{-6}}=a^{4}$ ###### 62 $\\left(a^{-2}\\right)^{-3}=a^6$", "+0 # Any tricks for this? -1 125 1 +236 If $f(x) = 8x^3 - 6x^2 + 12x - 0$, what is the value of $f\\left(\\dfrac{3}{4}\\right)$? Aug 3, 2022", "# How do you simplify (-245x^31y^23)^0 ? ##### 1 Answer Apr 11, 2018 $= 1$ #### Explanation: We know ${a}^{0} = 1$ So anything to the power of 0 is 1 $\\implies {\\left(- 245 {x}^{31} {y}^{23}\\right)}^{0} = 1$", "Thus, $\\left(\\dfrac{13}{p}\\right) = -1$. Thus, we have $$\\left(\\dfrac{4719}{p}\\right) = (-1)(-1) = 1.$$ QED", "How do you simplify (5x^2)^0 ? Apr 28, 2018 $1$ Explanation: Anything to the power of $0$ is $1$ ${\\left(5 {x}^{2}\\right)}^{0}$ ${5}^{0} \\cdot {x}^{2 \\cdot 0}$ $1 \\cdot {x}^{0}$ $1 \\cdot 1$ $1$", "+ $\\frac{1}{2}$ a box left over. $\\frac{1}{2}$ a box is $\\left[\\frac{1}{2} \\times 18\\right] = 9$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Check: $\\left[50 \\times 18\\right] + 9 = 909$", "# Power = Weak? Algebra Level 4 $\\left(\\dfrac{a}{b}+\\dfrac{b}{a}+\\dfrac{b}{c}+\\dfrac{c}{b}+\\dfrac{a}{c}+\\dfrac{c}{a}+3\\right)\\left(3^{(a^2+b^2+c^2)}\\right)^{(a^3+b^3+c^3)}$ For the above expression, let $a,b,c$ be the roots of the equation $x^3-6x^2+18x-36=0$. What is the value of the said expression above? ×", "a $+254,$ we must add a 0 (the $+$ sign $)$ to the leftmost position of the number. $+254=001111110$" ]

[ "5x is equivalent to ` 5 x!", "3.3333 …. 10x – x = 9x = 3", "five numbers is 4, then their product is ${4^5}$.", "that $X_1 X_2 X_3 X_4$ ends in 1, 3, 7, or 9 is $P(X_1, X_2, X_3 \\text{ and } X_4 \\text{ are not divisible by 2 or 5}) = (4/10)^4$.", "# What is x if 5x - 2 = 2x + 7? Jun 5, 2018 $x = 3$ #### Explanation: $5 x - 2 = 2 x + 7$ $5 x - 2 x = 7 + 2$ $3 x = 9$ $x = \\frac{9}{3}$ $x = 3$", "+ 5$ or $x = - 5 + 5$ $x = 10$ or $x = 0$", "4 + 3 + 2 + 1 = 10}.", "3x + 4​...", "expression when x=4 and y=3 $5x + y^{2}- xy$", "2. $(f+h)(x) = f(x) + h(x)$. Therefore, given $f(x) = 4x$ and $h(x) = x + 5$, $(f+h)(x) = 4x + x + 5 = 5x + 5$.", "Answer: x = 3.5", "can comfortably note that $$x^2-4x-5=(x-2)^2 -9.$$ Finally, take the negative of this. We get $9-(x-2)^2$. The rest has been well done by others: let $x-2=3\\sin\\theta$, or, more slowly, let $x-2=u$ and then let $u=3\\sin\\theta$. -", "What is x if 3x+5=32? Oct 26, 2015 $x = 9$ Explanation: $3 x + 5 = 32$ $3 x = 32 - 5$ $3 x = 27$ $x = \\frac{27}{3}$ $x = 9$ Aug 10, 2018 $x = 9$ Explanation: Let's start by adding $5$ to both sides to get $3 x = 27$ To completely isolate $x$, we can divide both sides by $3$ to get $x = 9$ Hope this helps!", "+ 5 = 9. $($(foobar). and then we have 2x=10. some-var = You got me That seems pretty ok. I obviously have not tested it extensively, but it looks pretty promising.", "$3, 7, 11...$? 1. $f(n)=4n-1$ 2. $f(n)=3+2n$ 3. $f(x)=2n-3$ 4. $f(n)=4x-1$", "1$ and $3^{5x}9^y= 1$, click on the formulas.", "usually not a problem. We have $$f'(x) = 4x - 3$$ by simply using that $$(x^n)' = nx^{n-1}$$ for any $n \\in \\mathbb{N}$. Then you just plug $a$ in for $x$, so $$f'(a) = 4a - 3$$", "10^4$ $5x + 3y = 1305\\times 10^4$ 3. Thanks so much! It's perfectly clear now.", "$$x_1 = 6$$ and $$x_2 = 3$$, then $$x_3 + x_4 + x_5 = 11$$. 5. If $$x_1 = 8$$ and $$x_2 = 4$$, then $$x_3 + x_4 + x_5 = 8$$. 6. If $$x_1 = 10$$ and $$x_2 = 5$$, then $$x_3 + x_4 + x_5 = 5$$. 7. If $$x_1 = 12$$ and $$x_2 = 6$$, then $$x_3 + x_4 + x_5 = 2$$.", "x=4/5." ]

[ "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "$\\frac{13 - b}{7 - b} = 2$ Now we simplify the fraction : $13 - b = 2(7 - b)$ $13 - b = 14 - 2b$ $- b + 2b = 14 - 13$ $\\boxed{b = 1}$ 2. Find $a$ Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ : $13 = 4a + b$ Since $b = 1$, we get : $13 = 4a + 1$ $12 = 4a$ $3 = a$ $\\boxed{a = 3}$ 3. Conclusion You are done : you have found the linear relationship between $x$ and $f(x)$ : $\\boxed{f(x) = 3x + 1}$ Let's check those results : $f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ --> $f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ --> $f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ --> $f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ --> Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ : $f(1) = 1 \\times 3", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "equation,$x = 18/2 =9$, thus, the answer is $\\boxed{\\textbf{(C)}\\ 9}$. ---LarryFlora", "### Home > CC3 > Chapter 3 > Lesson 3.1.5 > Problem3-50 3-50. For the following equations, simplify and solve for the variable. Record your work. 1. $2 x - 7 = - 2 x + 1$ Flip subtraction to addition. Add $2x$ to both sides. Add $7$ to both sides. Divide by $4$ on both sides. $2x+(−7)=−2x+1$ $4x+(−7)=1$ $4x=8$ $x=2$ 1. $- 2 x - 5 = - 4 x + 2$ See the steps in part (a).", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$", "# How do you solve 9e+4=-5e+14+13e How do you solve $9e+4=-5e+14+13e$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it intacte87 First, group and combine like terms on the right side of the equation: $9e+4=-5e+13e+14$ $9e+4=\\left(-5+13\\right)e+14$ $9e+4=8e+14$ Now, subtract 4 and 8e from each side of the equation to solve for e while keeping the equation balanced: $9e-8e+4-4=8e-8e+14-4$ $\\left(9-8\\right)e+0=0+10$ $1e=10$ $e=10$ ###### Not exactly what you’re looking for? Hattie Schaeffer Move all the expressions to the left side of the equation. $9e+4+5e-14-13e=0$ Simplify $9e+4+5e-14-13e$ Add $9e$ and $5e$ $14e+4-14-13e=0$ Subtract $13e$ from $14e$ $e+4-14=0$ Subtract 14 from 4 $e-10=0$ Add $9e$ and $5e$ $14e+4-14-13e=0$ Subtract 14 from 4 $e-10=0$ Subtract 10 from 2.718281182 $-7.28171817=0$ Since $-7.2817187\\ne 0$ there are no solutions.", "# Question #f116a Nov 9, 2017 $x = 0$ #### Explanation: $5 + 4 x - 7 = 4 x - 2 - x$ Collect like terms $5 - 7 + 4 x = 4 x - x - 2$ Combine like terms $- 2 + 4 x = 3 x - 2$ Subtract $3 x$ from both sides to get all the $x$'s on the same side $- 2 + x = - 2$ Add $- 2$ to both sides to isolate $x$ $x = 0$ .................................. Check Sub in $0$ in the place of $x$ in the original equation $5 + 4 x - 7 = 4 x - 2 - x$ $5 + 4 \\cdot 0 - 7 = 4 \\cdot 0 - 2 - 0$ $5 + 0 - 7$ should equal $0 - 2 - 0$ $- 2$ does equal $- 2$ Check!", "# Can you help • April 22nd 2006, 05:32 PM maxmac97 Can you help translate into an equation then solve the sum of x and five subtracted from eight times x is the same as thirteen plus the product of two and the difference of x and two • April 22nd 2006, 05:46 PM ThePerfectHacker Quote: Originally Posted by maxmac97 translate into an equation then solve the sum of x and five subtracted from eight times x is the same as thirteen plus the product of two and the difference of x and two $x+(8x-5)=13+(2)\\cdot (x-2)$ Thus, watch those signs :eek: $9x-5=13+2x-4$ Subtract $2x$ from both sides, $7x-5=9$ $7x=14$ Thus, $x=14/7=2$. ----- I am confused by Quote: difference of x and two It can mean $2-x$ and even, $|x-2|$ That is the wording that holds me in confusion. • April 23rd 2006, 10:35 AM earboth Quote: Originally Posted by maxmac97 translate into an equation then solve the sum of x and five subtracted from eight times x is the same as thirteen plus the product of two and the difference of x and two Hello, maybe you mean $8x-(x+5)=13+2(x-2)$ (?) Expand both sides $8x-x-5=13+2x-4$ $7x-5=2x+9$ $5x=14\\ \\Longleftrightarrow \\ x=\\frac{14}{5}$ Greetings EB", "### Home > MC1 > Chapter 9 > Lesson 9.3.4 > Problem9-160 9-160. Find the value of x that would make each equation true. Show all of your work. 1. $19 = 3x + 7$ Refer to problem 9-9 if you need help with this problem. $x = 4$ 1. $37 - x = 16$ What do you subtract from $37$ to get $16$?", "# Question #bcc3f If you are trying to find $m$, then the answer is $m = 38$. So the first step is to add $25$ to both sides. The two $25$'s on the right automatically cancel each other out. When you add the $25$ and the $13$ together, you get $38$. If you are being really precise, you would flip the $38$ and the $m$ around so then the variable comes before the number. So $m = 38$", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "solutions. So write it as $x^2 = \\tfrac12(x^3-13x-10).$ Now take the square root to get $\\boxed{x = \\sqrt{\\tfrac12(x^3-13x-10)}}.$", "# Question #b318a Oct 26, 2017 The equation is $4 \\times \\left(x + 5\\right) = 54$ and by solving, we get $x = 8.5$ meaning each salad costs $8.50 #### Explanation: Let $x$be the cost of a salad. Then each day, she spends $x + 5$dollars. After doing this for four days, she has spent $4 \\times \\left(x + 5\\right)$dollars, which totals$54. Putting this together, we have $4 \\times \\left(x + 5\\right) = 54$ Now, to solve, multiply each term in the brackets by 4: $4 x + 20 = 54$ Collect like terms (by subtracting 20 from each side) $4 x = 34$ Divide each side by 4 $x = 8.5$", "can see that the result, 1463, has a 4 itself; therefore, since the true answer to the problem does not count $4$'s, we subtract $1$ from the result to get the true answer to the problem. $1463-1=\\boxed{1462}$" ]

[ "$9x - 5x = x(9-5) = 4x$ $4x = 52$ 3. $9x + (-5x) = 52$ (the + (-5x) would just be -5x.. positive times negative = negative) $9x - 5x = 52$ $4x = 52$ (subtracting like terms) $\\frac {4x}{4} = \\frac {52}{4}$ (isolating x by divinding by 4 on both sides) $x=13$", "# Question #f116a Nov 9, 2017 $x = 0$ #### Explanation: $5 + 4 x - 7 = 4 x - 2 - x$ Collect like terms $5 - 7 + 4 x = 4 x - x - 2$ Combine like terms $- 2 + 4 x = 3 x - 2$ Subtract $3 x$ from both sides to get all the $x$'s on the same side $- 2 + x = - 2$ Add $- 2$ to both sides to isolate $x$ $x = 0$ .................................. Check Sub in $0$ in the place of $x$ in the original equation $5 + 4 x - 7 = 4 x - 2 - x$ $5 + 4 \\cdot 0 - 7 = 4 \\cdot 0 - 2 - 0$ $5 + 0 - 7$ should equal $0 - 2 - 0$ $- 2$ does equal $- 2$ Check!", "you get $3 x + 21$. Multiply the $4$ to the $x$ and the $+ 3$ and you get $4 x + 12$. You can only combine $3 x$ and $4 x$ and $12$ and $21$. You cannot combine $4 x$ and $21$ because $21$ does not have a variable. So, add $3 x$ and $4 x$ together to get $7 x$ and then add $21$ and $12$ together to get $33$. Thus, $7 x + 33$. Sorry if I dumbed this down, I wanted to make sure you understood every part since you said you weren't good with math. Hope this helped and good luck!", "# How do you evaluate 3x+4 for x=3? Apr 22, 2016 $3 x + 4 = 3 \\times 3 + 4 = 13$ just substitute $x$ for $3$ in this linear equation, as $3 \\cdot x + 4 = 3 \\cdot 3 + 4 = 9 + 4 = 13$", "values of $x$ satisfy the original equation, $\\tfrac{a}{b} = \\tfrac54 + 1 = \\tfrac94,$ so $a+b = \\boxed{13}.$", "$$x_1 = 6$$ and $$x_2 = 3$$, then $$x_3 + x_4 + x_5 = 11$$. 5. If $$x_1 = 8$$ and $$x_2 = 4$$, then $$x_3 + x_4 + x_5 = 8$$. 6. If $$x_1 = 10$$ and $$x_2 = 5$$, then $$x_3 + x_4 + x_5 = 5$$. 7. If $$x_1 = 12$$ and $$x_2 = 6$$, then $$x_3 + x_4 + x_5 = 2$$.", "$3m + 4 = 3m - 9$ is equivalent to $4 = -9$ which which is not true for all $m$. Same goes for other equations/inequalities that you have mentioned. • Yeah, that might be right, right? You just subtracted $3m$ from both sides, which gets rid of both of those terms and leaves you with 4=-9, which isn't true. – Mathster Oct 28 '14 at 23:10", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "+ 1$ $= -9x^2(x^2 + 1) + 1(x^2 + 1)$ $= (x^2 + 1)(1 - 9x^2)$ $= (x^2 + 1)[1^2 - (3x)^2]$ $= (x^2 + 1)(1 - 3x)(1 + 3x)$. Double check the answer you have been given. 2) $(x + 2)^2 - 5(x + 2)$ Take out a common factor of $x + 2$ $= (x + 2)(x + 2 - 5)$ $= (x + 2)(x - 3)$. 3) $3(x^2+10x+25) - 4(x+5) = 3(x + 5)^2 - 4(x + 5)$ $= (x + 5)[3(x + 5) - 4]$ $= (x + 5)(3x + 15 - 4)$ $= (x + 5)(3x + 11)$.", "# How do you evaluate 3x+4 for x=3? $3 x + 4 = 3 \\times 3 + 4 = 13$ just substitute $x$ for $3$ in this linear equation, as $3 \\cdot x + 4 = 3 \\cdot 3 + 4 = 9 + 4 = 13$", "$\\frac{13 - b}{7 - b} = 2$ Now we simplify the fraction : $13 - b = 2(7 - b)$ $13 - b = 14 - 2b$ $- b + 2b = 14 - 13$ $\\boxed{b = 1}$ 2. Find $a$ Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ : $13 = 4a + b$ Since $b = 1$, we get : $13 = 4a + 1$ $12 = 4a$ $3 = a$ $\\boxed{a = 3}$ 3. Conclusion You are done : you have found the linear relationship between $x$ and $f(x)$ : $\\boxed{f(x) = 3x + 1}$ Let's check those results : $f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ --> $f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ --> $f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ --> $f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ --> Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ : $f(1) = 1 \\times 3", "use of the identity: $(x-a)(x+a) = x^2 - a^2$. Well done! Keone, from Sage Ridge School in Reno, started with four consecutive whole numbers: The product of the first and last numbers is always $2$ less than the product of the middle two numbers. Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$. So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$. Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$. So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$. When he considered five consecutive whole numbers he found that: The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers. Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$. So the product of the first and last numbers is $x(x+4) = x^2 + 4x$. Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$. So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$. And with $n$ consecutive whole numbers", "# How do you solve 9e+4=-5e+14+13e How do you solve $9e+4=-5e+14+13e$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it intacte87 First, group and combine like terms on the right side of the equation: $9e+4=-5e+13e+14$ $9e+4=\\left(-5+13\\right)e+14$ $9e+4=8e+14$ Now, subtract 4 and 8e from each side of the equation to solve for e while keeping the equation balanced: $9e-8e+4-4=8e-8e+14-4$ $\\left(9-8\\right)e+0=0+10$ $1e=10$ $e=10$ ###### Not exactly what you’re looking for? Hattie Schaeffer Move all the expressions to the left side of the equation. $9e+4+5e-14-13e=0$ Simplify $9e+4+5e-14-13e$ Add $9e$ and $5e$ $14e+4-14-13e=0$ Subtract $13e$ from $14e$ $e+4-14=0$ Subtract 14 from 4 $e-10=0$ Add $9e$ and $5e$ $14e+4-14-13e=0$ Subtract 14 from 4 $e-10=0$ Subtract 10 from 2.718281182 $-7.28171817=0$ Since $-7.2817187\\ne 0$ there are no solutions.", "decimal into “whole number” and “decimal” parts. $$8.9=9.3 - 2 - 0.9$$ Now just do the two subtractions one at a time. $$8.9=9.3 - 2 - 0.9$$ $$8.9=7.3 - 0.9$$ $$8.9=6.4$$ 8.9 does not equal 6.4, so $x=5.6$ does not satisfy the equation. We can use the same technique when adding two decimals. ## Example Questions #### Question 1: Does $x=3.7$ satisfy the equation $5.7=2+x$? $$5.7=2+x$$ $$5.7=2+3.7$$ $$5.7=5.7$$ 5.7 does equal 5.7, so $x=3.7$ satisfies the equation! #### Question 2: Does $x=6.4$ satisfy the equation $9.8=x+3.5$? $$9.8=x+3.5$$ $$9.8=6.4+3.5$$ $$9.8=9.9$$ Close but no cigar! 9.8 does not equal 9.9, so $x=6.4$ does not satisfy the equation!", "fourth numbers is $x(x+3) = x^2 + 3x$. Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$. So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$. When he considered five consecutive whole numbers he found that: The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers. Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$. So the product of the first and last numbers is $x(x+4) = x^2 + 4x$. Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$. So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$. And with $n$ consecutive whole numbers he found that: The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers. Let us consider the general case where there are n consecutive whole numbers. As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on. The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will", "gives you $$x$$. So, the result that you get is: $$4=x$$ The order in which you read an equation does not matter. It's the same thing either way you look at. So, the result above states that $$x=4$$. And that's your final answer. You can check your work by replacing $$x$$ with $$4$$: $$12=3\\cdot4\\\\ 12=12$$ As you can see both sides contain the same number, as it should be. • The 0.25 comes from moving the 3 to the other side I should think. Then the 12 drops down (and why not, if things move?) and you have 3/12. It could just as easily have been 9 or -12/3 too. – Paul Feb 12 at 13:57", "sub it in, which gives $m= -4-0 =-4$ So, at the point $(2,0)$, $m=-4$ You can also verify this by differentiating $f(x)$ as alexmahone did: $f' (x)=-2x$ $f'(2)=-2(2) = -4$ Which is the same as what we obtained before, although this doesn't directly use the first principles or limits.", "Then the larger number is $4 \\times x = 4x$. The product of $x$ and $4x$ is $x \\times 4x = 4x \\times x = 4x^2$, and this is equal to $36$. So $4x^2 = 36$ Divide both sides by $4$: $x^2 = 9$ Now x is a natural (counting) number, so $x = 3$. So the numbers are 3 and 12, and their sum is 15. But for someone who hasn't done this sort of algebra, the easiest alternative is probably to use trial and error. So, guess a number that seems about right for the smaller number; say 5. 4 x 5 = 20, and 20 x 5 = 100. That's much too big. So, now try, say, 2. 2 x 4 = 8, and 8 x 2 = 16. Too small. Guess 3. 3 x 4 = 12 and 12 x 3 = 36. Correct! I hope this helps. Four-fifths of 50 = 40 = number of cars containing only one person. Of these two-fifths are driven by men. So the number driven by men = $\\frac{2}{5}\\times 40 = 16$.", "6x^3 + 16x^2 - 25x + 10}{x^2 - 2x + k} = x^2 - 4x + 8 - k + \\frac{(2k - 9)x + k^2 - 8k + 10}{x^2 - 2x + k}$. So the remainder is: $x + a = (2k - 9)x + k^2 - 8k + 10$. Equating like co-efficients of $x$ gives $2k - 9 = 1$ and $k^2 - 8k + 10 = a$. Solving for $k$ in the first equation gives: $2k = 10$ $k = 5$. Substituting into the second equation gives: $5^2 - 8\\cdot 5 + 10 = a$ $25 - 40 + 10 = a$ $a = -5$. So $a = -5$ and $k = 5$.", "$4|x$, from the second however we have $3|x$ , from which we can conclude that $12|x$, but this contradicts the first." ]

[ "At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 &nbs [#permalink] 21 Aug 2018, 06:33 Display posts from previous: Sort by", "share his gum with two friends. He asks his friends to pay him for their share. Including Aarav, how much does each person spend on gum? Solution: The cost of each gum pack that Aarav buys is $\\$ 1.50$. He buys$5$packs costing him$5 \\times 1.50$=$ \\$7.50$ The number of gums that he buys is then distributed among his two friends including him. The price that he paid for the pack of gums is shared between his two friends and he himself. So, total number of heads to be counted who shared the cost is $2$ + $1$ = $3$. If $3$ persons paid $\\$7.50$for the$5$pack of gums, then the amount which each person paid for the pack of gums is$ \\ $\\frac{7.50 }{3}$ = $\\$2.50$. Thus, the amount each person spend on the gum is$ \\$2.50$ Example 5: At the end of the week Rima decided to open her piggy bank and count the amount she collected for a week. On counting the coins that was there in the piggy bank she found the amount to be $\\$6.60$and the number of coins to be$66$. The coins were of different denominations consisting of dimes, nickels and quarters. The relation between the different denominations is such that there are three times as many nickels as quarters and one – half", "÷ 0.01 is 600. d. 1.7 ÷ 0.1 1.7 ÷ 0.1 = 17. Explanation: The division of 1.7 ÷ 0.1 is 17. e. 31 ÷ 0.01 31 ÷ 0.01 = 3,100. Explanation: The division of 31 ÷ 0.01 is 3,100. f. 11 ÷ 0.01 11 ÷ 0.01 = 1,100. Explanation: The division of 11 ÷ 0.01 is 1,100. g. 125 ÷ 0.1 125 ÷ 0.1 = 1,250. Explanation: The division of 125 ÷ 0.1 is 1,250. h. 3.74 ÷ 0.01 3.74 ÷ 0.01 = 374. Explanation: The division of 3.74 ÷ 0.01 is 374. i. 12.5 ÷ 0.01 12.5 ÷ 0.01 = 1,250. Explanation: The division of 12.5 ÷ 0.01 is 1,250. Question 3. Yung bought $4.60 worth of bubble gum. Each piece of gum cost$0.10. How many pieces of bubble gum did Yung buy? Here, Yung bought $4.60 worth of bubble gum, and each piece of gum cost$0.10. So the number of pieces of bubble gum did Yung bought is $4.60 ÷$0.10 = $46. So Yung bought 46 pieces of gum. Question 4. Cheryl solved a problem: 84 ÷ 0.01 = 8,400. Jane said, “Your answer is wrong because when you divide, the quotient is always smaller than the whole amount you start with, for example, 6 ÷ 2 = 3 and 100 ÷ 4 = 25.”", "## Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to h Question Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to help you. This image shows 3 hundredths grid next to each other. The first grid is entirely shaded in blue, second grid has 0.19 nineteen hundredths shaded in blue and remaining 0.81 eighty one hundredths shaded in yellow and third grid has 0.14 fourteen hundredths shaded in yellow and remaining are empty.", "of a UCLA student. I am posting below recording of the webinar for those who could't attend this session. ### Show Tags 21 Aug 2018, 05:30 00:00 Difficulty: 15% (low) Question Stats: 89% (01:58) correct 11% (02:40) wrong based on 28 sessions ### HideShow timer Statistics At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 _________________ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3325 Location: India GPA: 3.12 Re: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 [#permalink] ### Show Tags 21 Aug 2018, 05:35 Bunuel wrote: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Let the cost of a gumball be G, the cost of a chocolate bar be C, and", "## At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat ##### This topic has expert replies Legendary Member Posts: 1622 Joined: 01 Mar 2018 Followed by:2 members ### At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat by Gmat_mission » Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$ lollipop is $$\\0.92.$$ The total cost of $$1$$ gumball and $$1$$ chocolate bar is $$\\1.24.$$ What is the cost in dollars of $$1$$ chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Source: Magoosh ### GMAT/MBA Expert GMAT Instructor Posts: 16072 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1268 members GMAT Score:770 ### Re: At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ choc by [email protected] » Wed Dec 01, 2021 7:09 am Gmat_mission wrote: Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$", "answers ### How do accessory pigments chlorophyll? How do accessory pigments chlorophyll?... 3 answers ### What was some political responses to the great depression in america and germany? What was some political responses to the great depression in america and germany?... 5 answers ### Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african americans the option to own land. it failed to give african americans the right to vote. it failed to end the bitterness between the north and the south.... 1 answer ### You go to the store with$20. you find that packages of gum are on sale for $0.75 per pack. set up an You go to the store with$20. you find that packages of gum are on sale for \\$0.75 per pack. set up an inequality to how how many packages of gum you can buy. $You go to the store with 20. you find that packages of gum are on sale for 0.75 per pack. set up a$... I NEED HELP WOTH THIS ASAP $I NEED HELP WOTH THIS ASAP$...", "of gelatin. How many packets of gelatin do you buy? Given, You buy 2 boxes of cherry gelatin, 4 boxes of strawberry gelatin, and 3 boxes of orange gelatin. Each box contains 2 packets of gelatin. 1 box – 2 packets of gelatin 2 box – 4 packets of gelatin 4 boxes – 4 × 2 = 8 packets of gelatin 3 boxes – 3 × 2 = 6 packets of gelatin 4 + 8 + 6 = 18 packets of gelatin. Therefore you can buy 18 packets of gelatin. Question 4. A pack of gum has 5 pieces. You have 3 packs of gum and give4 friends each 1 piece. How many pieces of gum do you have left? Given, A pack of gum has 5 pieces. You have 3 packs of gum. Total number of packs = 5 × 3 = 15 pieces and give4 friends each 1 piece. 15 – 4 = 11 pieces Thus 11 pieces of gum have left. Think and Grow: Modeling Real Life Descartes has four $10 bills, seven$5 bills, and one $2 bill. How much money does he have in all? Understand the problem: Make a plan: Solve: Descartes has$ _____. Given, Descartes has four $10 bills, seven$5 bills, and one $2 bill. 4 ×$10 = $40 7 ×$5 =", "are 100 - 42 = 58 green gumballs left. The second phase thus lasts for 58/2 = 29 rounds, and so costs 29 pennies. Finally, the machine enters its third phase, wherein it dispenses 1 pink gumball for each penny, and continues to do this until there are no pink gumballs left. After the first round, there were 98 pink gumballs left, and this remained unchanged during the second round, which dispensed no pink gumballs, so the third phase will last for 98 rounds, and so cost 98 pennies. So, Putting all the phases together, we have 42 + 29 + 98 pennies to release all the gumballs, which is 169 pennies. 2) Let $N_p$ be the initial number of pink gumballs, $N_g$ be the initial number of green gumballs, $P$ be the percentage setting, $D_{p1}$ be the number of pink dispensed per penny in the pink-and-green phase, $D_{g1}$ be the number of green dispensed per penny in the pink-and-green phase, $D_{p2}$ be the number of pink dispensed per penny in the pink-only phase, and $D_{g2}$ be the number of green dispensed per penny in the green-only phase. If $N_p$ is 0, presumably the machine will skip the pink-and-green phase, go straight into the green-only phase, continue till all green gumballs are dispensed, and then skip the pink-only", "were sold and the gumballs cost the candy store 0.10$ each, how much money will the store lose when cleaning the machine? in progress 0 6 months 2021-07-26T08:15:05+00:00 1 Answers 22 views 0", "# You want to buy some rice a 9-ounce package costs $3.33 a 12ounce costs$4.32 A 22 ounce package costs 8.58 which package is the best buy​ ###### Question: you want to buy some rice a 9-ounce package costs $3.33 a 12ounce costs$4.32 A 22 ounce package costs 8.58 which package is the best buy​ ### A container has peppermint gum and spearmint gum. There are 32 pieces of peppermint gum in the container and the ratio of peppermint gum to spearmint gum is 8 to 7. Find the number of pieces of spearmint gum in the container. A container has peppermint gum and spearmint gum. There are 32 pieces of peppermint gum in the container and the ratio of peppermint gum to spearmint gum is 8 to 7. Find the number of pieces of spearmint gum in the container.... ### Richie's average on his math tests was 64%. On the next test he had a mark of 76%, and this raised his overall average to 67%. When he wrote the next test after this his average improved again by 1%. What mark did he get on the last test? A) 68% B) 69% C) 70% D) 71% E) 72% Richie's average on his math tests was 64%. On the next test he had a mark of 76%,", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. 1. 12 pack of soda for $3.50 Reply 2. ngocdiep Answer: 12 pack of soda for$3.50", "the cost of different numbers of packs of baseball cards. Answer: Explanation: Given cost of favorite of baseball cards are$1.70 so, cost of different numbers are 2 X $1.70 =$3.40; 4 X $1.70 =$6.80; 8 X $1.70 =$13.60; 9 X $1.70 =$15.30 ; 10 X $1.70 =$17; 15 X $1.70 =$25.50 ; 25 X $1.70 =$42.50. Question 2. Dante’s favorite gum costs $0.60 a pack. Fill in the table below to show the cost of different numbers of packs of gum. Answer: Explanation: Given cost of favorite gum$0.60 then cost of different numbers are 2 X $0.60 =$1.20; 5 X $0.60 =$3.00; 9 X $0.60 =$5.40; 10 X $0.60 =$6.00 ; 19 X $0.60 = 11.40; 20 X$0.60 = $12.00; 25 X$0.60 = $15.00. Question 3. Bouncy balls come in packages that cost$3.15 each. Fill in the table below to show the cost of different numbers of packs of bouncy balls. Explanation: Given that cost of bouncy balls in packages is $3.15 then cost of different numbers are 2 X$3.1 = $6.30; 3 X$3.15 = $9.45; 6 X$3.15 = $18.90 ; 9 X$3.15 = $28.35; 10 X$3.15 = $31.50 ; 12 X$3.15 = $37.80; 20 X$3.15 = $63.00. Question 4. Dante decided to spend only$20.00 of his allowance and save the rest for later. a. Can he buy", "number of pieces of gum does Phil have now = 35 pieces. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 1. What is the hidden question you need to answer before you can solve the problem? Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 2. Solve the problem. Complete the bar diagrams. Show the equations you used. 7 x 5 = 35. 5 + 7 = 12. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Independent Practice Model with Math Jen bought 4 tickets. Amber bought 5 tickets. The tickets cost $2 each. How much did", "of Luke is mentioned in the preface, and given that the ... Please show correct answer The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let mu denote the true average reflectometer reading for a new type of paint under consideration. A test Listening & Reading_ANSWER SHEET new.pdf (p.5). How an ancient custom became big business. Chewing gum contains fewer than ten calories per stick, but it is C Research continues on new textures and flavors. Glycerin and other vegetable oil products are now used to blend the gum base. May 27, 2020 · U.S. gross domestic product was cut in half, from $103 billion to$55 billion, due partly to deflation.   The Consumer Price Index fell 27% between November 1929 to March 1933, according to the Bureau of Labor Statistics.   About half the cards in the stack should show a, and the other half should show letters that students have not yet learned, such as x or o. Now let's play a game. We're going to try to go through this stack of cards as fast as we can, saying either /aa/ or not-/aa/ for each card. 50 Years Later: Poverty and The Other America 50 Years Later: Poverty and The Other America Maurice Isserman", "per gram (Calcium Lactate Powder, 2017) 250 mg calcium (25% DV) = $.008 per piece (NSP Web Team, 2017) -Magnesium→$3.30=$.11 per gram (True Magnesium Berry flavored Powder Mix) 125 mg magnesium (30% DV) =$ .014 per piece (NSP Web Team, 2017) -Potassium → $1.15=$ .04 per gram (BulkSupplements, 2017) 90mg Potassium (3% DV)= $.0016 per piece (Nature Made, 2017) -Sodium→$1.90=$.067 per gram (Abrams, 2017) 40mg (2% DV)=$ .002 per piece So, the final price of the gum derived from the amylase, mineral additions and the estimate is about 2 cents/ piece. 3. What would be the anticipated average sale price (ASP)? Justify. -The anticipated average sale price for the gum would be $1.99. Each pack would have 20 pieces of gum so each pack would cost roughly 40 cents to make, so we then are selling it for 5 times the price to make it. 4. Using initial market size analysis in Lab 1, determine the market size in dollars per year. Assume 5% penetrance of the market (Sales price X 0.05 X total market size). -($1.99 x .05 x $2865.5 million) The projected market size in dollars per year is$285 million/year. 5. Using the fundability worksheet, determine if your prototype should be funded. Justify why or why not. -Since our product is projected to earn around 5", "# Gummy Worms Sale Calvin was at his local grocery store, and saw that they had gummy worms for sale. To satisfy his foreseeable sugar craving, he bought 72 bags of gummy worms. Upon exiting the store, he noticed that the ink on the receipt was smudged, and he could see the price as $$\\_ 3.4\\_$$. What is the price (in cents) of 1 bag of gummy worms? Details and assumptions A bag of gummy worms cost a whole number of cents. The tens digit of the dollars is missing, and the units digit of the cents is missing. The receipt was for less that \\$100. ×", "to dollars. One half of a dollar is $0.50. Thus, the discounted rate for the sodas is$.50/1 soda. Here is a helpful video on simplifying rates and ratios:", "100 cents per dollar.", "costing. ### Review Problem 4.2 Chewy Gum Corporation produces bubble gum in large batches and uses a process costing system. Three departments—Mixing, Rolling, and Packaging—are involved in the production process. Chewy Gum has the following transactions: 1. Direct materials totaling $20,000—$6,000 for the Mixing department, $5,000 for the Rolling department, and$9,000 for the Packaging department—are requisitioned and placed in production. 2. Each production department incurs the following direct labor costs (wages payable): Mixing $2,500 Rolling$4,600 Packaging $2,200 3. Manufacturing overhead costs are applied to each department as follows: Mixing$10,000 Rolling $7,000 Packaging$ 7,500 4. Products with a cost of $5,500 are transferred from the Mixing department to the Rolling department. 5. Products with a cost of$6,400 are transferred from the Rolling department to the Packaging department. 6. Products with a cost of $9,100 are completed and transferred from the Packaging department to the finished goods warehouse. 7. Products with a cost of$8,300 are sold to customers. Perform the following steps for each transaction: 1. Prepare a journal entry to record the transaction. 2. Summarize the flow of costs through T-accounts. Use the format presented in Figure 4.2 \"Flow of Product Costs in a Process Costing System\" (no need to include T-accounts for raw materials inventory, wages payable, or manufacturing overhead). Assume there are no beginning balances in the" ]

[ "share his gum with two friends. He asks his friends to pay him for their share. Including Aarav, how much does each person spend on gum? Solution: The cost of each gum pack that Aarav buys is $\\$ 1.50$. He buys$5$packs costing him$5 \\times 1.50$=$ \\$7.50$ The number of gums that he buys is then distributed among his two friends including him. The price that he paid for the pack of gums is shared between his two friends and he himself. So, total number of heads to be counted who shared the cost is $2$ + $1$ = $3$. If $3$ persons paid $\\$7.50$for the$5$pack of gums, then the amount which each person paid for the pack of gums is$ \\ $\\frac{7.50 }{3}$ = $\\$2.50$. Thus, the amount each person spend on the gum is$ \\$2.50$ Example 5: At the end of the week Rima decided to open her piggy bank and count the amount she collected for a week. On counting the coins that was there in the piggy bank she found the amount to be $\\$6.60$and the number of coins to be$66$. The coins were of different denominations consisting of dimes, nickels and quarters. The relation between the different denominations is such that there are three times as many nickels as quarters and one – half", "of gelatin. How many packets of gelatin do you buy? Given, You buy 2 boxes of cherry gelatin, 4 boxes of strawberry gelatin, and 3 boxes of orange gelatin. Each box contains 2 packets of gelatin. 1 box – 2 packets of gelatin 2 box – 4 packets of gelatin 4 boxes – 4 × 2 = 8 packets of gelatin 3 boxes – 3 × 2 = 6 packets of gelatin 4 + 8 + 6 = 18 packets of gelatin. Therefore you can buy 18 packets of gelatin. Question 4. A pack of gum has 5 pieces. You have 3 packs of gum and give4 friends each 1 piece. How many pieces of gum do you have left? Given, A pack of gum has 5 pieces. You have 3 packs of gum. Total number of packs = 5 × 3 = 15 pieces and give4 friends each 1 piece. 15 – 4 = 11 pieces Thus 11 pieces of gum have left. Think and Grow: Modeling Real Life Descartes has four $10 bills, seven$5 bills, and one $2 bill. How much money does he have in all? Understand the problem: Make a plan: Solve: Descartes has$ _____. Given, Descartes has four $10 bills, seven$5 bills, and one $2 bill. 4 ×$10 = $40 7 ×$5 =", "At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 &nbs [#permalink] 21 Aug 2018, 06:33 Display posts from previous: Sort by", "answers ### How do accessory pigments chlorophyll? How do accessory pigments chlorophyll?... 3 answers ### What was some political responses to the great depression in america and germany? What was some political responses to the great depression in america and germany?... 5 answers ### Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african americans the option to own land. it failed to give african americans the right to vote. it failed to end the bitterness between the north and the south.... 1 answer ### You go to the store with$20. you find that packages of gum are on sale for $0.75 per pack. set up an You go to the store with$20. you find that packages of gum are on sale for \\$0.75 per pack. set up an inequality to how how many packages of gum you can buy. $You go to the store with 20. you find that packages of gum are on sale for 0.75 per pack. set up a$... I NEED HELP WOTH THIS ASAP $I NEED HELP WOTH THIS ASAP$...", "## Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to h Question Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to help you. This image shows 3 hundredths grid next to each other. The first grid is entirely shaded in blue, second grid has 0.19 nineteen hundredths shaded in blue and remaining 0.81 eighty one hundredths shaded in yellow and third grid has 0.14 fourteen hundredths shaded in yellow and remaining are empty.", "÷ 0.01 is 600. d. 1.7 ÷ 0.1 1.7 ÷ 0.1 = 17. Explanation: The division of 1.7 ÷ 0.1 is 17. e. 31 ÷ 0.01 31 ÷ 0.01 = 3,100. Explanation: The division of 31 ÷ 0.01 is 3,100. f. 11 ÷ 0.01 11 ÷ 0.01 = 1,100. Explanation: The division of 11 ÷ 0.01 is 1,100. g. 125 ÷ 0.1 125 ÷ 0.1 = 1,250. Explanation: The division of 125 ÷ 0.1 is 1,250. h. 3.74 ÷ 0.01 3.74 ÷ 0.01 = 374. Explanation: The division of 3.74 ÷ 0.01 is 374. i. 12.5 ÷ 0.01 12.5 ÷ 0.01 = 1,250. Explanation: The division of 12.5 ÷ 0.01 is 1,250. Question 3. Yung bought $4.60 worth of bubble gum. Each piece of gum cost$0.10. How many pieces of bubble gum did Yung buy? Here, Yung bought $4.60 worth of bubble gum, and each piece of gum cost$0.10. So the number of pieces of bubble gum did Yung bought is $4.60 ÷$0.10 = $46. So Yung bought 46 pieces of gum. Question 4. Cheryl solved a problem: 84 ÷ 0.01 = 8,400. Jane said, “Your answer is wrong because when you divide, the quotient is always smaller than the whole amount you start with, for example, 6 ÷ 2 = 3 and 100 ÷ 4 = 25.”", "are 100 - 42 = 58 green gumballs left. The second phase thus lasts for 58/2 = 29 rounds, and so costs 29 pennies. Finally, the machine enters its third phase, wherein it dispenses 1 pink gumball for each penny, and continues to do this until there are no pink gumballs left. After the first round, there were 98 pink gumballs left, and this remained unchanged during the second round, which dispensed no pink gumballs, so the third phase will last for 98 rounds, and so cost 98 pennies. So, Putting all the phases together, we have 42 + 29 + 98 pennies to release all the gumballs, which is 169 pennies. 2) Let $N_p$ be the initial number of pink gumballs, $N_g$ be the initial number of green gumballs, $P$ be the percentage setting, $D_{p1}$ be the number of pink dispensed per penny in the pink-and-green phase, $D_{g1}$ be the number of green dispensed per penny in the pink-and-green phase, $D_{p2}$ be the number of pink dispensed per penny in the pink-only phase, and $D_{g2}$ be the number of green dispensed per penny in the green-only phase. If $N_p$ is 0, presumably the machine will skip the pink-and-green phase, go straight into the green-only phase, continue till all green gumballs are dispensed, and then skip the pink-only", "the cost of different numbers of packs of baseball cards. Answer: Explanation: Given cost of favorite of baseball cards are$1.70 so, cost of different numbers are 2 X $1.70 =$3.40; 4 X $1.70 =$6.80; 8 X $1.70 =$13.60; 9 X $1.70 =$15.30 ; 10 X $1.70 =$17; 15 X $1.70 =$25.50 ; 25 X $1.70 =$42.50. Question 2. Dante’s favorite gum costs $0.60 a pack. Fill in the table below to show the cost of different numbers of packs of gum. Answer: Explanation: Given cost of favorite gum$0.60 then cost of different numbers are 2 X $0.60 =$1.20; 5 X $0.60 =$3.00; 9 X $0.60 =$5.40; 10 X $0.60 =$6.00 ; 19 X $0.60 = 11.40; 20 X$0.60 = $12.00; 25 X$0.60 = $15.00. Question 3. Bouncy balls come in packages that cost$3.15 each. Fill in the table below to show the cost of different numbers of packs of bouncy balls. Explanation: Given that cost of bouncy balls in packages is $3.15 then cost of different numbers are 2 X$3.1 = $6.30; 3 X$3.15 = $9.45; 6 X$3.15 = $18.90 ; 9 X$3.15 = $28.35; 10 X$3.15 = $31.50 ; 12 X$3.15 = $37.80; 20 X$3.15 = $63.00. Question 4. Dante decided to spend only$20.00 of his allowance and save the rest for later. a. Can he buy", "the total cost of a water bottle and a bag of chips. After solving, write the complete number sentence in decimal form. 89 cents = 0.89. 5 dimes = 0.05. Explanation: In the above-given question, given that, 89 cents. 5 dimes. f. Brian and Sonya each have a container. They mark their containers to show tenths. Brian and Sonya both fill their containers with 0.7 units of juice. However, Brian has more juice in his container. Explain how this is possible.", "of Luke is mentioned in the preface, and given that the ... Please show correct answer The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let mu denote the true average reflectometer reading for a new type of paint under consideration. A test Listening & Reading_ANSWER SHEET new.pdf (p.5). How an ancient custom became big business. Chewing gum contains fewer than ten calories per stick, but it is C Research continues on new textures and flavors. Glycerin and other vegetable oil products are now used to blend the gum base. May 27, 2020 · U.S. gross domestic product was cut in half, from $103 billion to$55 billion, due partly to deflation.   The Consumer Price Index fell 27% between November 1929 to March 1933, according to the Bureau of Labor Statistics.   About half the cards in the stack should show a, and the other half should show letters that students have not yet learned, such as x or o. Now let's play a game. We're going to try to go through this stack of cards as fast as we can, saying either /aa/ or not-/aa/ for each card. 50 Years Later: Poverty and The Other America 50 Years Later: Poverty and The Other America Maurice Isserman", "## At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat ##### This topic has expert replies Legendary Member Posts: 1622 Joined: 01 Mar 2018 Followed by:2 members ### At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat by Gmat_mission » Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$ lollipop is $$\\0.92.$$ The total cost of $$1$$ gumball and $$1$$ chocolate bar is $$\\1.24.$$ What is the cost in dollars of $$1$$ chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Source: Magoosh ### GMAT/MBA Expert GMAT Instructor Posts: 16072 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1268 members GMAT Score:770 ### Re: At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ choc by [email protected] » Wed Dec 01, 2021 7:09 am Gmat_mission wrote: Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. 1. 12 pack of soda for $3.50 Reply 2. ngocdiep Answer: 12 pack of soda for$3.50", "number of pieces of gum does Phil have now = 35 pieces. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 1. What is the hidden question you need to answer before you can solve the problem? Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 2. Solve the problem. Complete the bar diagrams. Show the equations you used. 7 x 5 = 35. 5 + 7 = 12. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Independent Practice Model with Math Jen bought 4 tickets. Amber bought 5 tickets. The tickets cost $2 each. How much did", "were sold and the gumballs cost the candy store 0.10$ each, how much money will the store lose when cleaning the machine? in progress 0 6 months 2021-07-26T08:15:05+00:00 1 Answers 22 views 0", "# Math Help - math prroblem 1. ## math prroblem sally bought three cholate bars and a pack of gum and paid $1.75. Jake bought two cholate bars and four packs of gum and paid$2.00. Find the cost of a chocolate bar and the cost of a pack of gum 2. Hello, lizard4! This is a very basic problem with a system of equations. Assuming you're new at this, I'll baby-step through it. Sally bought three chocolate bars and a pack of gum and paid $1.75. Jake bought two chocolate bars and four packs of gum and paid$2.00. Find the cost of a chocolate bar and the cost of a pack of gum Let $C$ = cost of a chocolate bar. Let $G$ = cost of a pack of bum. Sally bought 3 chocolate bars at $C$ cents each. . . They cost her: . $3C$ cents. She bought 1 pack of gum at $G$ cents each. . . This cost her: . $G$ cents. So she spent: .3C + G[/tex] cents. . . But we are told that she spent 175 cents. There is one equation: . $3C + G \\:=\\:175$ .[1] Jake bought two chocolate bars at $C$ cents each. . . This cost him: .[ $2C$ cents. He bought 4 packs of gum at $G$", "of a UCLA student. I am posting below recording of the webinar for those who could't attend this session. ### Show Tags 21 Aug 2018, 05:30 00:00 Difficulty: 15% (low) Question Stats: 89% (01:58) correct 11% (02:40) wrong based on 28 sessions ### HideShow timer Statistics At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 _________________ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3325 Location: India GPA: 3.12 Re: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 [#permalink] ### Show Tags 21 Aug 2018, 05:35 Bunuel wrote: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Let the cost of a gumball be G, the cost of a chocolate bar be C, and", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. Step-by-step explanation: In this question, the unit rate refers to the cost of 1 can of soda. divide the cost by the quantity to get the unit rate. the 12 pack is better because if you times the six pack by 2 which is the same as the 12 pack you get $3.60 which is 10 cents more than the 12 pack. the price for each can is$29c", "100 cents per dollar.", "for a cost of \\$9.56, much more than the \\$7 it would take to get a gallon of cider.) 4. The unit rates we used in parts a and b are dollars per gallon and dollars per pack of eight 4.23-ounce boxes. 5. Additional unit rates could be dollars per ounce, cents per ounce, ounces per dollar, dollars per 4.23-ounce box, etc.", "package would contain $$24 \\times 210 = 5040$$ staples. So this means that either there are not exactly 210 staples in each strip or there are not exactly 5000 staples in the package." ]

[ "share his gum with two friends. He asks his friends to pay him for their share. Including Aarav, how much does each person spend on gum? Solution: The cost of each gum pack that Aarav buys is $\\$ 1.50$. He buys$5$packs costing him$5 \\times 1.50$=$ \\$7.50$ The number of gums that he buys is then distributed among his two friends including him. The price that he paid for the pack of gums is shared between his two friends and he himself. So, total number of heads to be counted who shared the cost is $2$ + $1$ = $3$. If $3$ persons paid $\\$7.50$for the$5$pack of gums, then the amount which each person paid for the pack of gums is$ \\ $\\frac{7.50 }{3}$ = $\\$2.50$. Thus, the amount each person spend on the gum is$ \\$2.50$ Example 5: At the end of the week Rima decided to open her piggy bank and count the amount she collected for a week. On counting the coins that was there in the piggy bank she found the amount to be $\\$6.60$and the number of coins to be$66$. The coins were of different denominations consisting of dimes, nickels and quarters. The relation between the different denominations is such that there are three times as many nickels as quarters and one – half", "of gelatin. How many packets of gelatin do you buy? Given, You buy 2 boxes of cherry gelatin, 4 boxes of strawberry gelatin, and 3 boxes of orange gelatin. Each box contains 2 packets of gelatin. 1 box – 2 packets of gelatin 2 box – 4 packets of gelatin 4 boxes – 4 × 2 = 8 packets of gelatin 3 boxes – 3 × 2 = 6 packets of gelatin 4 + 8 + 6 = 18 packets of gelatin. Therefore you can buy 18 packets of gelatin. Question 4. A pack of gum has 5 pieces. You have 3 packs of gum and give4 friends each 1 piece. How many pieces of gum do you have left? Given, A pack of gum has 5 pieces. You have 3 packs of gum. Total number of packs = 5 × 3 = 15 pieces and give4 friends each 1 piece. 15 – 4 = 11 pieces Thus 11 pieces of gum have left. Think and Grow: Modeling Real Life Descartes has four $10 bills, seven$5 bills, and one $2 bill. How much money does he have in all? Understand the problem: Make a plan: Solve: Descartes has$ _____. Given, Descartes has four $10 bills, seven$5 bills, and one $2 bill. 4 ×$10 = $40 7 ×$5 =", "At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 &nbs [#permalink] 21 Aug 2018, 06:33 Display posts from previous: Sort by", "÷ 0.01 is 600. d. 1.7 ÷ 0.1 1.7 ÷ 0.1 = 17. Explanation: The division of 1.7 ÷ 0.1 is 17. e. 31 ÷ 0.01 31 ÷ 0.01 = 3,100. Explanation: The division of 31 ÷ 0.01 is 3,100. f. 11 ÷ 0.01 11 ÷ 0.01 = 1,100. Explanation: The division of 11 ÷ 0.01 is 1,100. g. 125 ÷ 0.1 125 ÷ 0.1 = 1,250. Explanation: The division of 125 ÷ 0.1 is 1,250. h. 3.74 ÷ 0.01 3.74 ÷ 0.01 = 374. Explanation: The division of 3.74 ÷ 0.01 is 374. i. 12.5 ÷ 0.01 12.5 ÷ 0.01 = 1,250. Explanation: The division of 12.5 ÷ 0.01 is 1,250. Question 3. Yung bought $4.60 worth of bubble gum. Each piece of gum cost$0.10. How many pieces of bubble gum did Yung buy? Here, Yung bought $4.60 worth of bubble gum, and each piece of gum cost$0.10. So the number of pieces of bubble gum did Yung bought is $4.60 ÷$0.10 = $46. So Yung bought 46 pieces of gum. Question 4. Cheryl solved a problem: 84 ÷ 0.01 = 8,400. Jane said, “Your answer is wrong because when you divide, the quotient is always smaller than the whole amount you start with, for example, 6 ÷ 2 = 3 and 100 ÷ 4 = 25.”", "## Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to h Question Traci spent $1.19 on a pack of gum and$0.95 on lip balm. Part A What was the total amount Traci spent? Use the model to help you. This image shows 3 hundredths grid next to each other. The first grid is entirely shaded in blue, second grid has 0.19 nineteen hundredths shaded in blue and remaining 0.81 eighty one hundredths shaded in yellow and third grid has 0.14 fourteen hundredths shaded in yellow and remaining are empty.", "answers ### How do accessory pigments chlorophyll? How do accessory pigments chlorophyll?... 3 answers ### What was some political responses to the great depression in america and germany? What was some political responses to the great depression in america and germany?... 5 answers ### Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african Which of the following was a major failure of reconstruction? it failed to change the constitution. it failed to give african americans the option to own land. it failed to give african americans the right to vote. it failed to end the bitterness between the north and the south.... 1 answer ### You go to the store with$20. you find that packages of gum are on sale for $0.75 per pack. set up an You go to the store with$20. you find that packages of gum are on sale for \\$0.75 per pack. set up an inequality to how how many packages of gum you can buy. $You go to the store with 20. you find that packages of gum are on sale for 0.75 per pack. set up a$... I NEED HELP WOTH THIS ASAP $I NEED HELP WOTH THIS ASAP$...", "## At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat ##### This topic has expert replies Legendary Member Posts: 1622 Joined: 01 Mar 2018 Followed by:2 members ### At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolat by Gmat_mission » Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$ lollipop is $$\\0.92.$$ The total cost of $$1$$ gumball and $$1$$ chocolate bar is $$\\1.24.$$ What is the cost in dollars of $$1$$ chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Source: Magoosh ### GMAT/MBA Expert GMAT Instructor Posts: 16072 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1268 members GMAT Score:770 ### Re: At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ choc by [email protected] » Wed Dec 01, 2021 7:09 am Gmat_mission wrote: Sun Nov 28, 2021 7:20 am At Joe's candy store, the total cost of $$1$$ gumball and $$1$$ lollipop is $$\\0.74.$$ The total cost of $$1$$ chocolate bar and $$1$$", "of a UCLA student. I am posting below recording of the webinar for those who could't attend this session. ### Show Tags 21 Aug 2018, 05:30 00:00 Difficulty: 15% (low) Question Stats: 89% (01:58) correct 11% (02:40) wrong based on 28 sessions ### HideShow timer Statistics At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 _________________ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3325 Location: India GPA: 3.12 Re: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is$0 [#permalink] ### Show Tags 21 Aug 2018, 05:35 Bunuel wrote: At Joe's candy store, the total cost of 1 gumball and 1 lollipop is $0.74. The total cost of 1 chocolate bar and 1 lollipop is$0.92. The total cost of 1 gumball and 1 chocolate bar is $1.24. What is the cost in dollars of 1 chocolate bar? A. 0.53 B. 0.59 C. 0.63 D. 0.67 E. 0.71 Let the cost of a gumball be G, the cost of a chocolate bar be C, and", "costing. ### Review Problem 4.2 Chewy Gum Corporation produces bubble gum in large batches and uses a process costing system. Three departments—Mixing, Rolling, and Packaging—are involved in the production process. Chewy Gum has the following transactions: 1. Direct materials totaling $20,000—$6,000 for the Mixing department, $5,000 for the Rolling department, and$9,000 for the Packaging department—are requisitioned and placed in production. 2. Each production department incurs the following direct labor costs (wages payable): Mixing $2,500 Rolling$4,600 Packaging $2,200 3. Manufacturing overhead costs are applied to each department as follows: Mixing$10,000 Rolling $7,000 Packaging$ 7,500 4. Products with a cost of $5,500 are transferred from the Mixing department to the Rolling department. 5. Products with a cost of$6,400 are transferred from the Rolling department to the Packaging department. 6. Products with a cost of $9,100 are completed and transferred from the Packaging department to the finished goods warehouse. 7. Products with a cost of$8,300 are sold to customers. Perform the following steps for each transaction: 1. Prepare a journal entry to record the transaction. 2. Summarize the flow of costs through T-accounts. Use the format presented in Figure 4.2 \"Flow of Product Costs in a Process Costing System\" (no need to include T-accounts for raw materials inventory, wages payable, or manufacturing overhead). Assume there are no beginning balances in the", "the cost of different numbers of packs of baseball cards. Answer: Explanation: Given cost of favorite of baseball cards are$1.70 so, cost of different numbers are 2 X $1.70 =$3.40; 4 X $1.70 =$6.80; 8 X $1.70 =$13.60; 9 X $1.70 =$15.30 ; 10 X $1.70 =$17; 15 X $1.70 =$25.50 ; 25 X $1.70 =$42.50. Question 2. Dante’s favorite gum costs $0.60 a pack. Fill in the table below to show the cost of different numbers of packs of gum. Answer: Explanation: Given cost of favorite gum$0.60 then cost of different numbers are 2 X $0.60 =$1.20; 5 X $0.60 =$3.00; 9 X $0.60 =$5.40; 10 X $0.60 =$6.00 ; 19 X $0.60 = 11.40; 20 X$0.60 = $12.00; 25 X$0.60 = $15.00. Question 3. Bouncy balls come in packages that cost$3.15 each. Fill in the table below to show the cost of different numbers of packs of bouncy balls. Explanation: Given that cost of bouncy balls in packages is $3.15 then cost of different numbers are 2 X$3.1 = $6.30; 3 X$3.15 = $9.45; 6 X$3.15 = $18.90 ; 9 X$3.15 = $28.35; 10 X$3.15 = $31.50 ; 12 X$3.15 = $37.80; 20 X$3.15 = $63.00. Question 4. Dante decided to spend only$20.00 of his allowance and save the rest for later. a. Can he buy", "are 100 - 42 = 58 green gumballs left. The second phase thus lasts for 58/2 = 29 rounds, and so costs 29 pennies. Finally, the machine enters its third phase, wherein it dispenses 1 pink gumball for each penny, and continues to do this until there are no pink gumballs left. After the first round, there were 98 pink gumballs left, and this remained unchanged during the second round, which dispensed no pink gumballs, so the third phase will last for 98 rounds, and so cost 98 pennies. So, Putting all the phases together, we have 42 + 29 + 98 pennies to release all the gumballs, which is 169 pennies. 2) Let $N_p$ be the initial number of pink gumballs, $N_g$ be the initial number of green gumballs, $P$ be the percentage setting, $D_{p1}$ be the number of pink dispensed per penny in the pink-and-green phase, $D_{g1}$ be the number of green dispensed per penny in the pink-and-green phase, $D_{p2}$ be the number of pink dispensed per penny in the pink-only phase, and $D_{g2}$ be the number of green dispensed per penny in the green-only phase. If $N_p$ is 0, presumably the machine will skip the pink-and-green phase, go straight into the green-only phase, continue till all green gumballs are dispensed, and then skip the pink-only", "were sold and the gumballs cost the candy store 0.10$ each, how much money will the store lose when cleaning the machine? in progress 0 6 months 2021-07-26T08:15:05+00:00 1 Answers 22 views 0", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. 1. 12 pack of soda for $3.50 Reply 2. ngocdiep Answer: 12 pack of soda for$3.50", "of Luke is mentioned in the preface, and given that the ... Please show correct answer The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let mu denote the true average reflectometer reading for a new type of paint under consideration. A test Listening & Reading_ANSWER SHEET new.pdf (p.5). How an ancient custom became big business. Chewing gum contains fewer than ten calories per stick, but it is C Research continues on new textures and flavors. Glycerin and other vegetable oil products are now used to blend the gum base. May 27, 2020 · U.S. gross domestic product was cut in half, from $103 billion to$55 billion, due partly to deflation.   The Consumer Price Index fell 27% between November 1929 to March 1933, according to the Bureau of Labor Statistics.   About half the cards in the stack should show a, and the other half should show letters that students have not yet learned, such as x or o. Now let's play a game. We're going to try to go through this stack of cards as fast as we can, saying either /aa/ or not-/aa/ for each card. 50 Years Later: Poverty and The Other America 50 Years Later: Poverty and The Other America Maurice Isserman", "number of pieces of gum does Phil have now = 35 pieces. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 1. What is the hidden question you need to answer before you can solve the problem? Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Question 2. Solve the problem. Complete the bar diagrams. Show the equations you used. 7 x 5 = 35. 5 + 7 = 12. Explanation: In the above-given question, given that, A pack of gum contains 5 pieces. Phil had 7 packs of gum before he lost 2 pieces. 5 + 2 = 7. 7 x 5 = 35. so the number of pieces of gum does phil have now = 35. Independent Practice Model with Math Jen bought 4 tickets. Amber bought 5 tickets. The tickets cost $2 each. How much did", "# You want to buy some rice a 9-ounce package costs $3.33 a 12ounce costs$4.32 A 22 ounce package costs 8.58 which package is the best buy​ ###### Question: you want to buy some rice a 9-ounce package costs $3.33 a 12ounce costs$4.32 A 22 ounce package costs 8.58 which package is the best buy​ ### A container has peppermint gum and spearmint gum. There are 32 pieces of peppermint gum in the container and the ratio of peppermint gum to spearmint gum is 8 to 7. Find the number of pieces of spearmint gum in the container. A container has peppermint gum and spearmint gum. There are 32 pieces of peppermint gum in the container and the ratio of peppermint gum to spearmint gum is 8 to 7. Find the number of pieces of spearmint gum in the container.... ### Richie's average on his math tests was 64%. On the next test he had a mark of 76%, and this raised his overall average to 67%. When he wrote the next test after this his average improved again by 1%. What mark did he get on the last test? A) 68% B) 69% C) 70% D) 71% E) 72% Richie's average on his math tests was 64%. On the next test he had a mark of 76%,", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. Step-by-step explanation: In this question, the unit rate refers to the cost of 1 can of soda. divide the cost by the quantity to get the unit rate. the 12 pack is better because if you times the six pack by 2 which is the same as the 12 pack you get $3.60 which is 10 cents more than the 12 pack. the price for each can is$29c", "Gum Corporation produces bubble gum in large batches and uses a process costing system. Three departments—Mixing, Rolling, and Packaging—are involved in the production process. Chewy Gum has the following transactions: 1. Direct materials totaling $20,000—$6,000 for the Mixing department, $5,000 for the Rolling department, and$9,000 for the Packaging department—are requisitioned and placed in production. 2. Each production department incurs the following direct labor costs (wages payable): Mixing $2,500 Rolling$4,600 Packaging $2,200 3. Manufacturing overhead costs are applied to each department as follows: Mixing$10,000 Rolling $7,000 Packaging$ 7,500 4. Products with a cost of $5,500 are transferred from the Mixing department to the Rolling department. 5. Products with a cost of$6,400 are transferred from the Rolling department to the Packaging department. 6. Products with a cost of $9,100 are completed and transferred from the Packaging department to the finished goods warehouse. 7. Products with a cost of$8,300 are sold to customers. Perform the following steps for each transaction: 1. Prepare a journal entry to record the transaction. 2. Summarize the flow of costs through T-accounts. Use the format presented in Figure 3.2 “Flow of Product Costs in a Process Costing System” (no need to include T-accounts for raw materials inventory, wages payable, or manufacturing overhead). Assume there are no beginning balances in the work-in-process inventory, finished goods inventory, and", "the total cost of a water bottle and a bag of chips. After solving, write the complete number sentence in decimal form. 89 cents = 0.89. 5 dimes = 0.05. Explanation: In the above-given question, given that, 89 cents. 5 dimes. f. Brian and Sonya each have a container. They mark their containers to show tenths. Brian and Sonya both fill their containers with 0.7 units of juice. However, Brian has more juice in his container. Explain how this is possible.", "for a cost of \\$9.56, much more than the \\$7 it would take to get a gallon of cider.) 4. The unit rates we used in parts a and b are dollars per gallon and dollars per pack of eight 4.23-ounce boxes. 5. Additional unit rates could be dollars per ounce, cents per ounce, ounces per dollar, dollars per 4.23-ounce box, etc." ]

[ "# Problem #1356 1356 All sides of the convex pentagon $ABCDE$ are of equal length, and $\\angle A= \\angle B = 90^\\circ.$ What is the degree measure of $\\angle E?$ $\\textbf{(A) } 90 \\qquad\\textbf{(B) } 108 \\qquad\\textbf{(C) } 120 \\qquad\\textbf{(D) } 144 \\qquad\\textbf{(E) } 150$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Angles in an equilateral pentagon Geometry Level 4 Let $ABCDE$ be an equilateral convex pentagon such that $\\angle ABC=136^\\circ$ and $\\angle BCD=104^\\circ$. What is the measure (in degrees) of $\\angle AED$? Note: An equilateral polygon is a polygon whose sides are all of the same length. It does not imply that all the internal angles are equal, nor that the polygon is cyclic. ×", "### Abcde Is Regular Pentagon ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°. ∠BAM = $\\frac { 1 }{ 2 }$ x 108° = 54°", "× # Geometry is hard Let $$ABCDE$$ be a convex pentagon in which $$\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$$ and side lengths are five consecutive integers in some order. Find all the possible values of $$AB + BC + CD$$. Note by Raj Mantri 1 month ago Sort by: This is from INMO 2017 · 4 weeks ago", "when put together, the measures of angles: 1 three angles or an! 65 0 + 115 0 = 90 … 7 angles ABC are complementary angles put! The last statement of a regular pentagon ) 25° ( b ) supplement of an angle!", "# ACE Geometry Level 2 $$ABCDE$$ is a regular pentagon. What is the measure (in degrees) of $$\\angle ACE$$? ×", "A Pentagon and a Square Walk Into a Bar Geometry Level 2 $$ABCDE$$ is a regular pentagon. On the outside of $$AB$$, we construct a square $$ABFG$$. What is the measure (in degrees) of $$\\angle EAG$$? Details and assumptions The measure of an angle is given as a value from $$0^\\circ$$ to $$180^\\circ$$. × Problem Loading... Note Loading... Set Loading...", "# The Pentagon Geometry Level 5 In the pentagon $$ABCDE$$, $$\\angle ABC=\\angle AED = 90^\\circ$$, $$AB=CD=AE=BC+DE=1$$. FInd the area of $$ABCDE$$.", "# The Pentagon Geometry Level 5 In the pentagon $$ABCDE$$, $$\\angle ABC=\\angle AED = 90^\\circ$$, $$AB=CD=AE=BC+DE=1$$. FInd the area of $$ABCDE$$. Give your answer to 2 decimal places. ×", "Share # In a Regular Pentagon Abcde, Inscribed in a Circle; Find Ratio Between Angle Eda and Angle Adc. - Mathematics Course #### Question In a regular pentagon ABCDE, Inscribed in a circle; find ratio between angle EDA and angle ADC. #### Solution Arc AE subtends ∠AOE at the centre and ∠ADE at the remaining part of the circle. ∴ ∠ADE = 1/2 xx 72° = 36 [central angle is a regular pentagon at O] = 36° + 36° + 72° Is there an error in this question or solution?", "# Show that if a convex pentagon is equilateral, and two of its adjacent inner angles measure 108 degree, then the pentagon is equiangular. Show that if a convex pentagon is equilateral, and two of its adjacent inner angles measure $108$ degrees, then the pentagon is equiangular. My Attempt. Consider the figure below. Let $\\angle CDE = \\angle BCD = 180$ degrees. Currently, I don't really have a good idea on where to start, but my initial attempt was to show that I can draw a pentagram and show that $\\triangle ABC, \\triangle BCD, \\triangle CDE, \\triangle DEF,\\triangle EFA$ are congruent and hence, the figure is equiangular. It is rather easy to show that $\\triangle BCD$ and $\\triangle CDE$ are congruent by SAS congruence, however, I am unable to show congruences with other triangles. I feel like I am on the wrong tract. Any suggestions? Look at $BCDE$. One can complete this quadrilateral into a regular pentagon $A'BCDE$. All one has to do is prove that $A'=A$. But the triangles $ABE$ and $A'BE$ both have the same side-lengths, and so are congruent. As both $A$ and $A'$ lie above the line $BE$ then $A=A'$.", "× # Geometry is hard Let $$ABCDE$$ be a convex pentagon in which $$\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$$ and side lengths are five consecutive integers in some order. Find all the possible values of $$AB + BC + CD$$. Note by Raj Mantri 3 months ago Sort by: It's more #Discrete Mathematics than #Geometry. · 3 weeks, 2 days ago This is from INMO 2017 · 2 months, 3 weeks ago", "# Area of a Pentagon! Geometry Level 4 Let $$S$$ be the area of the convex pentagon $$ABCDE$$ having the following specifications: • $$AB = BC$$; $$CD = DE$$ • $$BD = 2$$ • $$\\angle ABC = 150^\\circ$$; $$\\angle CDE = 30^\\circ$$ Find the value of $$S^2$$. ×", "What is the degree of ∠CBF? Line l is parallel with line m Quantity B BC x is an angle If the complementary angle of x < x < half of the supplementary angle of x, then whats the range of x? What is the measure of the unknown angle? Quantity A The sum of interior angles of a pentagon plus 90° Quantity B The sum of interior angles of a hexagon Quantity A The average degree of all interior angles in a trapezoid Quantity B The average degree of all interior angles in a pentagon If the sum of the degree measures of the interior angles of regular polygon P is 360 degrees more than that of regular polygon Q, then P has how many more sides than Q? The perimeter of a regular pentagon is twice the perimeter of an equilateral triangle. Quantity A The side of the pentagon Quantity B The side of the equilateral triangle In the figure below is a regular octagon. Each side has a length of 1. What is the area of △ABC? 25000 +道题目 136本备考书籍", "$ABCDE$ is a regular pentagon of side length one unit. $BC$ produced meets $ED$ produced at $F$. Show that triangle $CDF$ is congruent to triangle $EDB$. Find the length of $BE$.", "Management, Entrepreneurship Schools: Sauder 'Dec 18 GPA: 3.79 WE: General Management (Energy and Utilities) Re: The pentagon ABCDE is inscribed in a circle with center O. How many de [#permalink] ### Show Tags 29 Oct 2017, 08:42 (1) The pentagon ABCDE is a regular pentagon, which means that all sides are the same length and all interior angles are equal and definite value. Sufficient (2) The radius of the circle is 5 inches. Not sufficient A Manager Joined: 15 Aug 2016 Posts: 57 The pentagon ABCDE is inscribed in a circle with center O. How many de [#permalink] ### Show Tags 29 Oct 2017, 10:31 IMO: A Statement 1- (n-2)*180 will give sum of all angles. Since all angles are equal we will get measure of each angle.No Need to calculate. Statement 2: We don't know anything about Polygon, whether it's regular or not.=>Insufficient. Intern Joined: 19 Oct 2015 Posts: 30 Re: The pentagon ABCDE is inscribed in a circle with center O. How many de [#permalink] ### Show Tags 02 Nov 2017, 09:44 Option 1: As we know that its regular pentagon, it has 5 sides. By applying formula to find interior angle (n-2)180/n - Where n = number of sides we can find the angle. Sufficient. Option 2: But just mere radius we cannot find the", "determine what each interior angle in this regular pentagon equals. You've probably already learned that a 5-sided convex polygon's interior angles add up to 3 x 180 degrees = 540 degrees. So, since the pentagon is regular: 540 / 5 = 108 degrees. That's how large each interior angles is. But, the angle we're interested in (the one between \"c\" and \"a\") is exactly half of 108 degrees or 54 degrees. Ok, so here's your picture: |\\ | \\ c |__\\ a In this picture \"b\" is the vertical line on the left. Remember, side \"b\" is the \"apothem\". You know the length of \"a\" and the angle between \"a\" and \"c\". Finally, using trigonometry, specifically tangent, you can find the length of \"b\". tan 54 = b / 2.5 I'm getting b = 3.44 (rounded to two decimal places) Hope this helps!", "shape below is a regular pentagon. Work out the size of the interior angle, x. [2 marks] Level 4-5 GCSE. This shape has 5 sides, so its interior angles add up to,. A regular pentagon has five equal angles, which all measure 108 degrees. The side length is all equal as well. The sum of the interior angles of a regular pentagon is 540 degrees. Many formulas. The angle at the pentagon's center is always 36º. (Starting with a full 360º center, you could divide it into 10 of these smaller triangles. 360 ÷ 10 = 36, so the angle at one triangle is 36º.) 5 Calculate the height of the triangle. The height of this triangle is the side at right angles to the pentagon's edge, leading to the center. Regular Pentagons Angles in Isosceles Triangles Given a triangle ABC with AB = AC, the angles opposite the equal sides are equal. Let a = angle BAC and let b = angle ABC = angle ACB. Using the angle sum theorem, if a is known, then b is determined, and if b is given, then a is determined. Write down the relations: a = b = Some important examples. polygon is not regular, it is called irregular. A polygon is concave if any part of a diagonal", "# Abcde is a Regular Pentagon. the Bisector of Angle a of the Pentagon Meets the Side Cd in Point M. Show that ∠Amc = 90°. - Mathematics Sum ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°. #### Solution Given: ABCDE is a regular pentagon. The bisector ∠A of the pentagon meets the side CD at point M. To prove : ∠AMC = 90° Proof: We know that the measure of each interior angle of a regular pentagon is 108°. ∠BAM = x 108° = 54° Since, we know that the sum of a quadrilateral is 360° ∠BAM + ∠ABC + ∠BCM + ∠AMC = 360° 54° + 108° + 108° + ∠AMC = 360° ∠AMC = 360° – 270° ∠AMC = 90° Is there an error in this question or solution? #### APPEARS IN Selina Concise Mathematics Class 8 ICSE Chapter 16 Understanding Shapes Exercise 16 (C) | Q 14 | Page 188", "Share Books Shortlist # In a Regular Pentagon Abcde, Inscribed in a Circle; Find Ratio Between Angle Eda and Angle Adc. - ICSE Class 10 - Mathematics ConceptArc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle #### Question In a regular pentagon ABCDE, Inscribed in a circle; find ratio between angle EDA and angle ADC. #### Solution Arc AE subtends ∠AOE at the centre and ∠ADE at the remaining part of the circle. ∴ ∠ADE = 1/2 xx 72° = 36 [central angle is a regular pentagon at O]" ]

[ "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "lie on the circle, and $\\angle ADC=\\angle CDQ$. $\\angle EAD=\\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$A$$\",A,W); label(\"$$B$$\",B,N); label(\"$$C$$\",C,N); label(\"$$D$$\",D,NE); label(\"$$E$$\",E,SE); label(\"$$F$$\",F,S); label(\"$$P$$\",(0.3,0.8),S); label(\"$$Q$$\",(-0.15,0.4),S); [/asy]$ Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$. Putting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$. Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html", "$AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry. We notice that $\\triangle ACD \\cong \\triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\\triangle AED \\sim \\triangle ADC,$ with $\\angle EAD \\cong \\angle DAC$ and $\\angle CDA \\cong \\angle DEA.$ This similarity means that we can create a proportion: $\\frac{AD}{AB}=\\frac{DE}{CD}.$ We plug in $AD=\\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\\frac{8}{17} \\cdot 15= \\boxed{\\textbf{(B) }\\frac{120}{17}}.$ (By the way, we could also use $\\triangle DEC \\sim \\triangle ADC.$) ## Solution 4: Inscribed Circle $[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(\"B\",B,SW); label(\"D\",D,SE); label(\"A\",A,N); label(\"M\",M,S); label(\"C\",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$ Call this triangle $ABD$ and let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of triangle $ABD$ is $\\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean", "lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$A$$\",A,W); label(\"$$B$$\",B,N); label(\"$$C$$\",C,N); label(\"$$D$$\",D,NE); label(\"$$E$$\",E,SE); label(\"$$F$$\",F,S); label(\"$$P$$\",(0.3,0.8),S); label(\"$$Q$$\",(-0.15,0.4),S); [/asy]$ Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$. Putting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$. Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)$$ $$PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "# Difference between revisions of \"Mock AIME 1 2006-2007 Problems/Problem 8\" ## Problem Let $ABCDE$ be a convex pentagon with $\\frac{AB}{\\sqrt{2}}=BC=CD=DE$, $\\angle ABC=150^\\circ$, $\\angle BCD=75^\\circ$, and $\\angle CDE=165^\\circ$. If $\\angle ABE=\\frac{m}{n}^\\circ$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. ## Solution $[asy]defaultpen(fontsize(8)); pair A=expi(pi*5/12)+expi(0)+expi(pi/2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0); draw(A--B--C--D--E--A);draw(B--P--E--B);draw(D--P--A); label(\"A\",A,(1,0));label(\"B\",B,(-1,0));label(\"C\",C,(-1,0));label(\"D\",D,(1,-1)); label(\"E\",E,(1,0));label(\"P\",P,(1,0)); dot(A^^B^^C^^D^^E^^P);[/asy]$ Let $P$ be a point in $ABCDE$ such that $\\angle ABP=\\angle BAP=45^\\circ$. We see that $\\angle CBP=115^\\circ$ and thus $BP||CD$. Since $BP=BC=CD$, we have that $BCDP$ is a rhombus. Therefore $\\angle CDP=115^\\circ$ so $\\angle PDE=60^\\circ$. Since $PD=CD=DE$ we have that $\\triangle PDE$ is equilateral. $\\angle BPE=75+60^\\circ=180^\\circ-2\\angle PBE\\implies \\angle PBE=\\frac{45}{2}$. $\\angle ABE=\\angle ABP+\\angle PBE=45+\\frac{45}{2}^\\circ=\\frac{135}{2}^\\circ\\implies m+n=\\boxed{137}$.", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "# 2005 AIME I Problems/Problem 7 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem In quadrilateral $ABCD,\\ BC=8,\\ CD=12,\\ AD=10,$ and $m\\angle A= m\\angle B = 60^\\circ.$ Given that $AB = p + \\sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$ ## Solution ### Solution 1 $[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label(\"A\",(0,0),SW); label(\"B\",(20.87,0),SE); label(\"E\",(15.87,8.66),NE); label(\"D\",(5,8.66),NW); label(\"P\",(5,0),S); label(\"Q\",(15.87,0),S); label(\"C\",(16.87,7),E); label(\"12\",(10.935,7.794),S); label(\"10\",(2.5,4.5),W); label(\"10\",(18.37,4.5),E); [/asy]$ Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\\angle CED = 120^{\\circ}$, using Law of Cosines on $\\bigtriangleup CED$ gives $$12^2=DE^2+4-2(2)(DE)(\\cos 120^{\\circ})$$ which gives $$144-4=DE^2+2DE$$. Adding $1$ to both sides gives $141=(DE+1)^2$, so $DE=\\sqrt{141}-1$. $\\bigtriangleup DAP$ and $\\bigtriangleup EBQ$ are both $30-60-90$, so $AP=5$ and $BQ=5$. $PQ=DE$, and therefore $AB=AP+PQ+BQ=5+\\sqrt{141}-1+5=9+\\sqrt{141} \\rightarrow (p,q)=(9,141) \\rightarrow \\boxed{150}$. ### Solution 2 Draw the perpendiculars from $C$ and $D$ to $AB$, labeling the intersection points as $E$ and $F$. This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$. Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$, we find another right triangle $\\triangle DGC$. $DG = DE - CF = 5\\sqrt{3} - 4\\sqrt{3} = \\sqrt{3}$. The", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B" ]

[ "# Problem #2229 2229 As shown below, convex pentagon $ABCDE$ has sides $AB=3$, $BC=4$, $CD=6$, $DE=3$, and $EA=7$. The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$-axis. The pentagon is then rolled clockwise to the right along the $x$-axis. Which side will touch the point $x=2009$ on the $x$-axis? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,0), Ep=7*dir(105), B=3*dir(0); pair D=Ep+B; pair C=intersectionpoints(Circle(D,6),Circle(B,4))[1]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(B--C--D--Ep--A); draw((6,6)..(8,4)..(8,3),EndArrow(3)); xaxis(\"x\",-8,14,EndArrow(3)); label(\"E\",Ep,NW); label(\"D\",D,NE); label(\"C\",C,E); label(\"B\",B,SE); label(\"(0,0)=A\",A,SW); label(\"3\",midpoint(A--B),N); label(\"4\",midpoint(B--C),NW); label(\"6\",midpoint(C--D),NE); label(\"3\",midpoint(D--Ep),S); label(\"7\",midpoint(Ep--A),W); [/asy]$ $\\text{(A) } \\overline{AB} \\qquad \\text{(B) } \\overline{BC} \\qquad \\text{(C) } \\overline{CD} \\qquad \\text{(D) } \\overline{DE} \\qquad \\text{(E) } \\overline{EA}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so:", "# Difference between revisions of \"Mock AIME 1 2006-2007 Problems/Problem 8\" ## Problem Let $ABCDE$ be a convex pentagon with $\\frac{AB}{\\sqrt{2}}=BC=CD=DE$, $\\angle ABC=150^\\circ$, $\\angle BCD=75^\\circ$, and $\\angle CDE=165^\\circ$. If $\\angle ABE=\\frac{m}{n}^\\circ$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. ## Solution $[asy]defaultpen(fontsize(8)); pair A=expi(pi*5/12)+expi(0)+expi(pi/2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0); draw(A--B--C--D--E--A);draw(B--P--E--B);draw(D--P--A); label(\"A\",A,(1,0));label(\"B\",B,(-1,0));label(\"C\",C,(-1,0));label(\"D\",D,(1,-1)); label(\"E\",E,(1,0));label(\"P\",P,(1,0)); dot(A^^B^^C^^D^^E^^P);[/asy]$ Let $P$ be a point in $ABCDE$ such that $\\angle ABP=\\angle BAP=45^\\circ$. We see that $\\angle CBP=115^\\circ$ and thus $BP||CD$. Since $BP=BC=CD$, we have that $BCDP$ is a rhombus. Therefore $\\angle CDP=115^\\circ$ so $\\angle PDE=60^\\circ$. Since $PD=CD=DE$ we have that $\\triangle PDE$ is equilateral. $\\angle BPE=75+60^\\circ=180^\\circ-2\\angle PBE\\implies \\angle PBE=\\frac{45}{2}$. $\\angle ABE=\\angle ABP+\\angle PBE=45+\\frac{45}{2}^\\circ=\\frac{135}{2}^\\circ\\implies m+n=\\boxed{137}$.", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "# 1972 USAMO Problems/Problem 5 ## Problem A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property. $[asy] size(80); defaultpen(fontsize(7)); pair A=(0,7), B=(5,4), C=(3,0), D=(-3,0), E=(-5,4), P; P=extension(B,D,C,E); draw(D--E--A--B--C--D--B--E--C); label(\"A\",A,(0,1));label(\"B\",B,(1,0));label(\"C\",C,(1,-1));label(\"D\",D,(-1,-1));label(\"E\",E,(-1,0));label(\"P\",P,(0,1)); [/asy]$ ## Solution Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\\overline{A_{n - 1}A_{n + 1}}\\parallel\\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5). Proof: For the \"only if\" direction, since $[A_0A_1A_2] = [A_1A_2A_3]$, $A_0$ and $A_3$ are equidistant from $\\overline{A_1A_2}$, and since the pentagon is convex, $\\overline{A_0A_3}\\parallel\\overline{A_1A_2}$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed. An image is supposed to go here. You can help us out by creating one and editing it in. Thanks. Let $A'B'C'D'E'$ be the inner pentagon, labeled so that $A$ and $A'$ are opposite each other, and let $a, b, c, d, e$ be the side lengths of the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "#### Triangles and their classification ##### Angle Classifacation Acute Triangle All angles are less than 90° Right Triangle Has a right angle (90°) Obtuse Triangle Has an angle more than 90° ##### Side Classifacation Equilateral Triangle Three equal sides Isosceles Triangle Two equal sides Two equal angles Scalene Triangle No equal sides No equal angles This part shows what Proves ussually look like ## Proof that any nonperfect square positive integer is irrational Let us assume that $\\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\\frac{p}{q}=\\sqrt{n} \\Rightarrow \\frac{p^2}{q^2}=n \\Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\\sqrt{n}$ is irrational. ## Proof of the Pythagorean Theorem $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Proof", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "# Difference between revisions of \"Circumradius\" The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ## Formula for a Triangle Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply $R=\\frac{abc}{4A}$. Also, $A=\\frac{abc}{4R}$ ## Proof $[asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label(\"A\", A, W); label(\"B\", B, N); label(\"C\", C, E); label(\"D\", D, S); label(\"O\", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label(\"E\", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]$ We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\\angle ADB = \\angle BCA$ because they both subtend arc $AB$. Therefore, $\\triangle BAD \\sim \\triangle BEC$ by AA similarity, so we have $$\\frac{BD}{BA} = \\frac{BC}{BE},$$ or $$\\frac {2R} c = \\frac ah.$$ However, remember that area $\\triangle ABC = \\frac {bh} 2$, so $h=\\frac{2 \\times \\text{Area}}b$. Substituting this in gives", "lie on a second line, parallel to the first, with $EF=1$. A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle? $\\text{(A) } 3 \\qquad \\text{(B) } 4 \\qquad \\text{(C) } 5 \\qquad \\text{(D) } 6 \\qquad \\text{(E) } 7$ ## Problem 13 As shown below, convex pentagon $ABCDE$ has sides $AB=3$, $BC=4$, $CD=6$, $DE=3$, and $EA=7$. The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$-axis. The pentagon is then rolled clockwise to the right along the $x$-axis. Which side will touch the point $x=2009$ on the $x$-axis? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,0), Ep=7*dir(105), B=3*dir(0); pair D=Ep+B; pair C=intersectionpoints(Circle(D,6),Circle(B,4))[1]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(B--C--D--Ep--A); draw((6,6)..(8,4)..(8,3),EndArrow(3)); xaxis(\"x\",-8,14,EndArrow(3)); label(\"E\",Ep,NW); label(\"D\",D,NE); label(\"C\",C,E); label(\"B\",B,SE); label(\"(0,0)=A\",A,SW); label(\"3\",midpoint(A--B),N); label(\"4\",midpoint(B--C),NW); label(\"6\",midpoint(C--D),NE); label(\"3\",midpoint(D--Ep),S); label(\"7\",midpoint(Ep--A),W); [/asy]$ $\\text{(A) } \\overline{AB} \\qquad \\text{(B) } \\overline{BC} \\qquad \\text{(C) } \\overline{CD} \\qquad \\text{(D) } \\overline{DE} \\qquad \\text{(E) } \\overline{EA}$ ## Problem 14 On Monday, Millie puts a quart of seeds, $25\\%$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $25\\%$ of the millet in the feeder,", "# Difference between revisions of \"Circumradius\" The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ## Formula for a Triangle Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply $R=\\frac{abc}{4A}$. Also, $A=\\frac{abc}{4R}$ ## Proof Proof: [asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label(\"$A$\", A, W); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$O$\", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label(\"$E$\", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy] We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\\angle ADB = \\angle BCA$ because they both subtend arc $AB$. Therefore, $\\triangle BAD \\sim \\triangle BEC$ by AA similarity, so we have $$\\frac{BD}{BA} = \\frac{BC}{BE},$$ or $$\\frac {2R} c = \\frac ah.$$ However, remember that area $\\triangle ABC = \\frac {bh} 2$, so $h=\\frac{2 \\times \\text{Area}}b$. Substituting", "+ .75*dir(360-65-115-55-30)); label(\"65^\\circ\",B + .75*dir(180-32.5)); label(\"x^\\circ\",A + .5*dir(42.5)); label(\"5^\\circ\",D + 2.5*dir(360-60-2.5)); label(\"60^\\circ\",D + .75*dir(360-30)); label(\"60^\\circ\",E + .5*dir(360-150)); label(\"5^\\circ\",B + 2.5*dir(180-65-2.5)); [/asy]$ Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles. ~Someonenumber011", "lie on the circle, and $\\angle ADC=\\angle CDQ$. $\\angle EAD=\\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$A$$\",A,W); label(\"$$B$$\",B,N); label(\"$$C$$\",C,N); label(\"$$D$$\",D,NE); label(\"$$E$$\",E,SE); label(\"$$F$$\",F,S); label(\"$$P$$\",(0.3,0.8),S); label(\"$$Q$$\",(-0.15,0.4),S); [/asy]$ Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$. Putting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$. Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)$$ $$PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "# 2004 AIME II Problems/Problem 13 ## Problem Let $ABCDE$ be a convex pentagon with $AB \\parallel CE, BC \\parallel AD, AC \\parallel DE, \\angle ABC=120^\\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ ## Solution Let the intersection of $\\overline{AD}$ and $\\overline{CE}$ be $F$. Since $AB \\parallel CE, BC \\parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\\triangle ABC \\cong \\triangle CFA$. Also, as $AC \\parallel DE$, it follows that $\\triangle ABC \\sim \\triangle EFD$. $[asy] pointpen = black; pathpen = black+linewidth(0.7); pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); D(MP(\"A\",A,(1,0))--MP(\"B\",B,N)--MP(\"C\",C,NW)--MP(\"D\",D)--MP(\"E\",E)--cycle); D(D--A--C--E); D(MP(\"F\",F)); MP(\"5\",(B+C)/2,NW); MP(\"3\",(A+B)/2,NE); MP(\"15\",(D+E)/2); [/asy]$ By the Law of Cosines, $AC^2 = 3^2 + 5^2 - 2 \\cdot 3 \\cdot 5 \\cos 120^{\\circ} = 49 \\Longrightarrow AC = 7$. Thus the length similarity ratio between $\\triangle ABC$ and $\\triangle EFD$ is $\\frac{AC}{ED} = \\frac{7}{15}$. Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $\\triangle ABC, \\triangle BDE$ to $AC, DE$ respectively. Then, the ratio of the areas $\\frac{[ABC]}{[BDE]} = \\frac{\\frac 12 \\cdot h_{ABC} \\cdot AC}{\\frac 12 \\cdot h_{BDE} \\cdot DE} = \\frac{7}{15}", "the radius, $120=17r$, we can just solve for the radius. We get $r=\\boxed{\\textbf{(B) }\\frac{120}{17}}$. This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself! ## Solution 7 Let us draw altitude $\\overline{CD}.$ This cuts our base into line segments with length $\\dfrac{16}{2}=8.$ Finding the area of the resulting triangles gives $[\\triangle ADC] = [\\triangle BDC] = \\dfrac{8 \\cdot 15}{2} = 60.$ Since $m\\angle CDB = m\\angle CDA = 90^\\circ,$ we use the Pythagorean Theorem to find length $\\overline{AC}$: \\begin{align*} (\\overline{AD})^2+(\\overline{CD})^2 &= (\\overline{AC})^2 \\\\ 8^2+15^2 &= (\\overline{AC})^2 \\\\ 64+225 &= (\\overline{AC})^2 \\\\ 289 &= (\\overline{AC})^2 \\\\ \\sqrt{289} &= \\sqrt{(\\overline{AC})^2} \\\\ 17 &= AC. \\end{align*} Thus we have $\\overline{AC}=\\overline{BC}=17.$ $[asy] pair A, B, C, D; A=(0,0); B=(16,0); C=(8,15); D=B/2; draw(A--B--C--cycle); draw(C--D); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); [/asy]$ Letting $17$ be the base of the triangle makes our height the radius of the semicircle: $[asy] pair A, B, C, D, EE; A=(0,0); B=(16,0); C=(8,15); D=B/2; EE=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--EE); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,EE,D,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); label(\"r\",D--EE, N); [/asy]$ Let $r$ be the radius of the semicircle. Then we have $\\dfrac{17 \\cdot r}{2}=60 \\implies 17 \\cdot r = 120 \\implies \\dfrac{17r}{17} = \\dfrac{120}{17}", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "$AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry. We notice that $\\triangle ACD \\cong \\triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\\triangle AED \\sim \\triangle ADC,$ with $\\angle EAD \\cong \\angle DAC$ and $\\angle CDA \\cong \\angle DEA.$ This similarity means that we can create a proportion: $\\frac{AD}{AB}=\\frac{DE}{CD}.$ We plug in $AD=\\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\\frac{8}{17} \\cdot 15= \\boxed{\\textbf{(B) }\\frac{120}{17}}.$ (By the way, we could also use $\\triangle DEC \\sim \\triangle ADC.$) ## Solution 4: Inscribed Circle $[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(\"B\",B,SW); label(\"D\",D,SE); label(\"A\",A,N); label(\"M\",M,S); label(\"C\",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$ Call this triangle $ABD$ and let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of triangle $ABD$ is $\\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean", "# Difference between revisions of \"2010 USAMO Problems/Problem 1\" ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from", "lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$A$$\",A,W); label(\"$$B$$\",B,N); label(\"$$C$$\",C,N); label(\"$$D$$\",D,NE); label(\"$$E$$\",E,SE); label(\"$$F$$\",F,S); label(\"$$P$$\",(0.3,0.8),S); label(\"$$Q$$\",(-0.15,0.4),S); [/asy]$ Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$. Putting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$. Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html", "# 1998 APMO Problems/Problem 4 ## Problem (Răzvan Gelca) Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ be on a line through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$, respectively. Prove that $AN$ is perpendicular to $NM$. ## Solution We use directed angles mod $\\pi$. $[asy] size(200); defaultpen(1); pair B=(0,0), A=(1,3), C=(4,0), X=(4,4); pair D=foot(A,B,C); path O1=circumcircle(A,B,D), O2=circumcircle(A,C,D); pair E=IntersectionPoint(O1,D--X,1), F=IntersectionPoint(O2,D--X,0); pair M=(B+C)/2; pair N=(E+F)/2; draw(C--F--A--B--C--A); draw(B--E--A--D--E--F); draw(A--N--M--A,dotted); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,SSE); label(\"F\",F,E); label(\"M\",M,S); label(\"N\",N,ESE); [/asy]$ Since $\\angle ADB$ and $\\angle AEB$ are both right angles, points $ABDE$ are concyclic. It follows that $$\\angle CBA \\equiv \\angle DBA \\equiv \\angle DEA \\equiv \\angle FEA .$$ Similarly, the quadrilateral $ACDF$ is cyclic, so $$\\angle ACB \\equiv \\angle ACD \\equiv \\angle AFD \\equiv \\angle AFE.$$ Thus $ACB$ and $AFE$ are similar triangles, so $AFEN$ and $ACBM$ are similar figures. It follows that $$\\angle AND \\equiv \\angle ANE \\equiv \\angle AMB \\equiv \\angle AMD,$$ so points $AMND$ are concyclic. Since $ADM$ is a right angle, it then follows that $ANM$ is a right angle, as desired. $\\blacksquare$" ]

[ "radius larger than 25 degrees; we considered smaller circles.", "+0 # circles -3 63 1 +-291 How much greater, in square inches, is the area of a circle of radius 20 inches than a circle of diameter 20 inches? Express your answer in terms of $\\pi$. Apr 5, 2020 #1 +4569 +2 First circle: pi(20)^2=400pi inches squared: Second cirle\"pi(20/2)^2=100 pi inches squared. Thus, 400pi-100pi=300 pi inches squared(square inches). Apr 5, 2020", "area of the region between the circles is . What is the radius of the smaller circle?", "EXACT AREA OF A CIRCLE WITH A DIAMETER OF 23 INCHES", "0 # How many square inches in 16 inch round pan? Wiki User 2013-04-25 08:18:05 The area of a circle is pi x radius2. You know the radius already since it's diameter/2 which is 8 inches. Your answer then is pi times 8 inches squared which is 64 times pi inches squared. pi is about 3.1415 so multiply that by 64 if you need to resolve the pi. Wiki User 2013-04-25 08:18:05 🙏 0 🤨 0 😮 0 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.72 373 Reviews Anonymous Lvl 1 2020-04-03 17:12:21 256", "+0 HELP CIRCLES 0 48 1 The radius of a circle is 2 inches. When the radius is doubled, by how many square inches is the area increased? Express your answer in terms of $\\pi$. Apr 6, 2020 #1 +252 0 If the radius is 2 inches long, the area is 4pi square inches. Doubling the radius to 4, the area of that would be 16pi square inches. 16pi-4pi= 12pi square inches.' Apr 6, 2020", "$60 \\pi$ square feet $69 \\pi$ square feet $90 \\pi$ square feet 0 votes 1 answer 5 95 views What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches", "the smaller circle is, the greater it slides.", "square feet What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches Which of the following figures has the largest perimeter $(1 \\text{ foot} = 12 \\text{ inches})$ a square with a diagonal of $5$ feet a rectangle with sides of $3$ feet and $4$ feet an equilateral triangle with a side equal to $48$ inches a regular hexagon whose longest diagonal is $6$ feet", "radius squared!", "## 2yoututu11 4 years ago What is the exact area of a circle with a radius of 12 feet? A. 24π square feet B. 36π square feet C. 121π square feet D. 144π square feet $$\\ \\huge A= \\pi r^2$$. So, $$\\ \\huge A= \\pi \\times 12^2$$. That's $$\\ \\huge 144 \\pi feet^2$$.", "# Find the surface area of a sphere in terms of pi given the radius of 25 inches Only authorized users can leave an answer!", "## 2yoututu11 Group Title What is the exact area of a circle with a radius of 12 feet? A. 24π square feet B. 36π square feet C. 121π square feet D. 144π square feet 2 years ago 2 years ago $$\\ \\huge A= \\pi r^2$$. So, $$\\ \\huge A= \\pi \\times 12^2$$. That's $$\\ \\huge 144 \\pi feet^2$$.", "## 2yoututu11 What is the exact area of a circle with a radius of 12 feet? A. 24π square feet B. 36π square feet C. 121π square feet D. 144π square feet one year ago one year ago $$\\ \\huge A= \\pi r^2$$. So, $$\\ \\huge A= \\pi \\times 12^2$$. That's $$\\ \\huge 144 \\pi feet^2$$.", "can be taken to be a shape that is made up of infinite number of equal sides. A circle with an area of 250 square foot has a radius of sqrt(250/pi). To minimize costs the enclosed area should be a circle with a radius sqrt(250/pi) foot. Need a fast expert's response? Submit order and get a quick answer at the best price for any assignment or question with DETAILED EXPLANATIONS! #### Comments No comments. Be first! Tweet", "25 feet intersect so that the distance between their centers is 30 feet. What is the length of the side of the largest square inscribable within their intersecting arcs?", "radius \\ ( \\quantity 3!", "## Algebra and Trigonometry 10th Edition $x \\gt 0\\\\y \\gt x \\\\ \\pi y^2 \\gt \\pi x^2+10$. The radius of the smaller circle is positive; this means that $x \\gt 0$ The radius $y$ is for the larger circle; this means that $y \\gt x$ The area of the larger circle is at least $10 \\ sq \\ unit$ greater; this means that $\\pi y^2 \\gt \\pi x^2+10$. Thus, the system is: $x \\gt 0\\\\y \\gt x \\\\ \\pi y^2 \\gt \\pi x^2+10$.", "# GEOM 1 | Lesson 1 | Practice (Area of Circles) How many circles of radius $$=2$$ ft. would fit in your patio in the shape of a rectangle, parallelogram, triangle, or trapezoid? Draw a diagram. See a possible solution", "Length x Width 2. Base x Height 4. Pi x Diameter 5. 1/2 x Base x Height Line AB is 14 inches long. What is the approximate area of this circle? 1. 42 square inches 2. 615 square inches 3. 160 square inches 4. 154 square inches" ]

[ "cm results http://www.ehow.com/how_8592343_convert-diameter-s... If the diameter of a circle is 36 inches, the http://www.chacha.com/question/what-is-the-square-... A pizza with diameter 20 inches has radius 10 inches, and therefore area 100*pi square inches. That's approximately 5.09 cents per square inch. http://answers.yahoo.com/question/index?qid=201205... Top Related Searches", "just using the formula for the area of a circle: PI * Radius Squared If the diameter is 5\", the radius is 2.5\" 2.5\"^2 * PI 6.25 * 3.1415927 = 19.6349540849 (what Sandra said) To find out the area of a circle, we need to know its diameter which is the length of its widest part. area of a circle is equal to r equals diameter / 2 Circle 1 has a radius of 8/2 = 4 inches. This on the web one-way conversion tool converts surface area units from ∅ 1-foot circles ( ∅ 1 ft ) into square inches ( in 2 , sq in ) instantly online. So, the cost per square inch of pizza is about .13, or 13 cents. 1 circ mil ≈ 0.000 000 000 506 707 479 m 2. Area , A= πd²/4 =π×12²/4=36π =113.097 square inches For example, assume the diameter of the circular area to be 12 feet. The first one, a, the radius squared is 8*8=64. How many square inches ( in 2 , sq in ) are in 1 circle 1-foot diameter ( 1 ∅ 1 ft )? This will give you the cost per square inch of the pizza. Divide the price of the pizza by the number of square inches. So, we divide our diameter", "$32 m^2$ 4. $48 m^2$ Which of the following has the largest area? A circle with a diameter of 10 inches A rectangle that measures 12 inches by 11 inches A circle with a radius of 6 inches A square that is 11 inches on each side 1. Circle with a diameter of 10 inches 2. Rectangle that measures 12 inches by 11 inches 3. Circle with a radius of 6 inches 4. Square that is 11 inches on each side Why would you use the formula A = bh? 1. To get the area of a triangle 2. To get the area of a circle 3. To get the area of a parallelogram 4. To get the perimeter of a square A square has a perimeter of 24 inches. What is the area of the square? 1. 36 square inches 2. 144 square inches 3. 12 square inches 4. 28 square inches", "+0 # circles -3 63 1 +-291 How much greater, in square inches, is the area of a circle of radius 20 inches than a circle of diameter 20 inches? Express your answer in terms of $\\pi$. Apr 5, 2020 #1 +4569 +2 First circle: pi(20)^2=400pi inches squared: Second cirle\"pi(20/2)^2=100 pi inches squared. Thus, 400pi-100pi=300 pi inches squared(square inches). Apr 5, 2020", "0 # How many square inches in 16 inch round pan? Wiki User 2013-04-25 08:18:05 The area of a circle is pi x radius2. You know the radius already since it's diameter/2 which is 8 inches. Your answer then is pi times 8 inches squared which is 64 times pi inches squared. pi is about 3.1415 so multiply that by 64 if you need to resolve the pi. Wiki User 2013-04-25 08:18:05 🙏 0 🤨 0 😮 0 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.72 373 Reviews Anonymous Lvl 1 2020-04-03 17:12:21 256", "$60 \\pi$ square feet $69 \\pi$ square feet $90 \\pi$ square feet 0 votes 1 answer 5 95 views What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches", "increases as $n=3,4,5,...$ with the limit $1/(4\\pi)=0.079577...,$ which is the area of a circle of perimeter 1. Note that for the square, $A_4=0.0625$, while for the hexagon, $A_6=0.0721687...$. This is pretty. But it only shows that a circle is better than any regular $n$-gon. There are still a lot of curves to be accounted for. – AndrewG Dec 25 '12 at 8:54", "# Difference between revisions of \"2012 AMC 12B Problems/Problem 2\" ## Problem A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle? $\\textbf{(A)}\\ 50\\qquad\\textbf{(B)}\\ 100\\qquad\\textbf{(C)}\\ 125\\qquad\\textbf{(D)}\\ 150\\qquad\\textbf{(E)}\\ 200$ ## Solution If the radius is $5$, then the width is $10$, hence the length is $20$. $10\\times20=200$, $\\boxed{\\text{E}}$", "square feet What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches Which of the following figures has the largest perimeter $(1 \\text{ foot} = 12 \\text{ inches})$ a square with a diagonal of $5$ feet a rectangle with sides of $3$ feet and $4$ feet an equilateral triangle with a side equal to $48$ inches a regular hexagon whose longest diagonal is $6$ feet", "## anonymous one year ago ***WILL MEDAL AND FAN*** A 10\" diameter pumpkin pie is cut into six equal servings. What is the area of the top of each piece of pie? a) 3.33pi inches b) 4.17pi inches c) 16.66pi inches 1. anonymous im thinking its A cause im trying to work it out but it just comes close to ten 2. phi 10 inch diameter means 5 inch radius Do you know how to find the area of a circle with r=5 ? 3. anonymous Yea. The area would be 78.5 right? 4. anonymous @phi 5. phi I would not make it decimal (because the answer choices have \"pi\" in them can you find the area, but do not change pi to a decimal? 6. anonymous Hmmm` 7. phi first, what is the formula for Area of a circle? 8. anonymous pi x radius squared 9. phi and r= 5 r squared ( $$r^2$$ ) = 5*5= 25 so Area= 25 * 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058... pi goes on forever... rather than truncating to approximately 3.14 (which is close but not exactly correct, as you can see), and rather than typing all those numbers (which you can't because they go on forever), people just say... call that number pi Area= $$25 \\pi$$ and let's not multiply it out unless you", "+0 HELP CIRCLES 0 48 1 The radius of a circle is 2 inches. When the radius is doubled, by how many square inches is the area increased? Express your answer in terms of $\\pi$. Apr 6, 2020 #1 +252 0 If the radius is 2 inches long, the area is 4pi square inches. Doubling the radius to 4, the area of that would be 16pi square inches. 16pi-4pi= 12pi square inches.' Apr 6, 2020", "Length x Width 2. Base x Height 4. Pi x Diameter 5. 1/2 x Base x Height Line AB is 14 inches long. What is the approximate area of this circle? 1. 42 square inches 2. 615 square inches 3. 160 square inches 4. 154 square inches", "why scientists don’t need math, and Jeremy Fox on why they do. Tonight the God Plays Dice art department made blondies! These are supposed to be made, according to the recipe, in a pan which is an eight-inch square. But we have no such thing. We do have a nine-inch circular pan, though. Will that do? Well, what matters is that the two pans have the same area – and therefore that the same volume of batter will have the same thickness and cook roughly the same. (If you thought I was going to solve some PDEs and work out how the heat transfers, you haven’t been paying attention.) A nine-inch circle has area $\\pi (9/2)^2 = 81\\pi/4$ square inches, which is about 63.62. An eight-inch square, of course, has area 64 square inches. Not bad! What would it take for this approximation to be exactly correct? This would require that $81\\pi/4 = 64$ exactly; solving for $\\pi$ gives \\$\\pi = 256/81″, which is often credited as an Egyptian approximation to $\\pi$ as it implicitly appears in the Rhind papyrus, an ancient Egyptian document of,problems in mathematics. In fact the setting in which this is established there is almost exactly this one – a circle of diameter 9 and a square of side 8 are said to have", "can be taken to be a shape that is made up of infinite number of equal sides. A circle with an area of 250 square foot has a radius of sqrt(250/pi). To minimize costs the enclosed area should be a circle with a radius sqrt(250/pi) foot. Need a fast expert's response? Submit order and get a quick answer at the best price for any assignment or question with DETAILED EXPLANATIONS! #### Comments No comments. Be first! Tweet", "# Thread: One more quick question 1. ## One more quick question The circumference of a circle varies directly as the diameter. A circular pizza pie with a diameter of 1 foot has a circumference of 3.14 feet. What will be the circumference of a pizza pie with a diameter of 1.5 feet? Thank you all for the help i really appreciate it. 2. Originally Posted by Ravangerz The circumference of a circle varies directly as the diameter. A circular pizza pie with a diameter of 1 foot has a circumference of 3.14 feet. What will be the circumference of a pizza pie with a diameter of 1.5 feet? Thank you all for the help i really appreciate it. As you stated the proportion of diameter and radius of circles is a constant: $\\pi$. (Even though the question refers to pizze this constant number is definitely not pie but pi ) Now use the proportion: $\\dfrac{3.14'}{1'} = \\dfrac c{1.5'}$ Solve for c (= circumference)", "## Elementary Technical Mathematics The difference between the diameters of the circular ends of the taper is $0.532 in$. The difference between the diameter is the bigger diameter minus the smaller diameter. So, difference between the diameters $= 1.625 in - 1.093 in$ difference between the diameters $= 0.532 in$", "its circumference: Q's circumference: $$24\\pi = \\pi*d$$ Q's diameter, $$d = 24$$ P's diameter, $$d = 6$$ The difference between the two diameters is (24 - 6) = 18 Kudos [?]: 264 [1], given: 552 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 1648 Kudos [?]: 840 [1], given: 2 Location: United States (CA) Re: Circle P is inside Circle Q, and the two circles share the same center [#permalink] ### Show Tags 22 Sep 2017, 11:45 1 KUDOS Expert's post Circle P is inside Circle Q, and the two circles share the same center X. If the circumference of Q is four times the circumference of P, and the radius of Circle P is three, what is the difference between Circle Q’s diameter and Circle P’s diameter? A 6 B 9 C 12 D 18 E 24 Since the radius of Circle P (the smaller circle) is 3, the diameter and circumference of Circle P are 6 and 6π, respectively. Since the circumference of Circle Q is four times the circumference of Circle P, the circumference and diameter of Circle Q are 24π and 24, respectively. Thus, the difference between the diameters of Circles Q and P is 24 - 6 = 18. _________________ Scott Woodbury-Stewart Founder and", "inches Circumference $=$ πd $= 22/7 \\times 7$ $= 22$ inches So, the distance covered by the wheel in one rotation $= 22$ inches Total distance to be covered $= 110$ feet $= (110 \\times 12)$ inches $= 1320$ inches Hence, no. of rotations required$= 1320/22 = 60$ 3 ### A circular flowerbed has a diameter of 20 feet. Find the cost of fencing the flowerbed at the rate of $10 per feet. Take π$= 3.14$.$\\$$228 \\$$528 $\\$$628 \\$$728 CorrectIncorrect Correct answer is:$\\$$628 Diameter of the flowerbed (d) = 20 feet Fencing the circular flowerbed refers to the boundary of the circle, i.e., the circumference of the circle. Circumference of the flowerbed = πd = 3.14 \\times 20 m = 62.8 feet Now, the cost of fencing = \\$$10 per ft So, the cost of fencing $62.8$ feet $= 62.8 \\times$ $\\$$10 = \\$$628 4 ### The difference between a circle's circumference and diameter is 10 feet. Find its radius. Take π$= 3.14$. 1.33 feet 2.33 feet 3.33 feet 4.33 feet CorrectIncorrect Correct answer is: 2.33 feet Given: Circumference – Diameter$=$10 feet We know that: Circumference$= 2$πr Diameter$= 2$r So,$2$πr$-2$r$= 10$feet$2$r$(\\text{π}-1) = 10$feet$2$r$(3.14-1) = 10$feet$2$r$\\times 2.14 = 10$feet r = 10 / 4.28 r = 2.33 feet 5 ### The ratio of the circumference of two circles is", "A circular mat with diameter 20 inches is placed on a square tabletop, each of whose sides is 24 inches long. Which of the following is closest to the fraction of the tabletop covered by the mat? ~$\\frac{5}{12}$~ ~$\\frac{2}{5}$~ ~$\\frac{1}{2}$~ ~$\\frac{3}{4}$~ ~$\\frac{5}{6}$~", "the center of the circle is $(-9,-7)$ and the radius is $5$. Want to try more problems like this? Check out this exercise and this exercise." ]

[ "+0 # circles -3 63 1 +-291 How much greater, in square inches, is the area of a circle of radius 20 inches than a circle of diameter 20 inches? Express your answer in terms of $\\pi$. Apr 5, 2020 #1 +4569 +2 First circle: pi(20)^2=400pi inches squared: Second cirle\"pi(20/2)^2=100 pi inches squared. Thus, 400pi-100pi=300 pi inches squared(square inches). Apr 5, 2020", "$60 \\pi$ square feet $69 \\pi$ square feet $90 \\pi$ square feet 0 votes 1 answer 5 95 views What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches", "0 # How many square inches in 16 inch round pan? Wiki User 2013-04-25 08:18:05 The area of a circle is pi x radius2. You know the radius already since it's diameter/2 which is 8 inches. Your answer then is pi times 8 inches squared which is 64 times pi inches squared. pi is about 3.1415 so multiply that by 64 if you need to resolve the pi. Wiki User 2013-04-25 08:18:05 🙏 0 🤨 0 😮 0 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.72 373 Reviews Anonymous Lvl 1 2020-04-03 17:12:21 256", "+0 HELP CIRCLES 0 48 1 The radius of a circle is 2 inches. When the radius is doubled, by how many square inches is the area increased? Express your answer in terms of $\\pi$. Apr 6, 2020 #1 +252 0 If the radius is 2 inches long, the area is 4pi square inches. Doubling the radius to 4, the area of that would be 16pi square inches. 16pi-4pi= 12pi square inches.' Apr 6, 2020", "area of the region between the circles is . What is the radius of the smaller circle?", "square feet What is the area of a semicircle with a diameter of $16$ inches? $32 \\pi$ square inches $64 \\pi$ square inches $128 \\pi$ square inches $256 \\pi$ square inches Which of the following figures has the largest perimeter $(1 \\text{ foot} = 12 \\text{ inches})$ a square with a diagonal of $5$ feet a rectangle with sides of $3$ feet and $4$ feet an equilateral triangle with a side equal to $48$ inches a regular hexagon whose longest diagonal is $6$ feet", "$32 m^2$ 4. $48 m^2$ Which of the following has the largest area? A circle with a diameter of 10 inches A rectangle that measures 12 inches by 11 inches A circle with a radius of 6 inches A square that is 11 inches on each side 1. Circle with a diameter of 10 inches 2. Rectangle that measures 12 inches by 11 inches 3. Circle with a radius of 6 inches 4. Square that is 11 inches on each side Why would you use the formula A = bh? 1. To get the area of a triangle 2. To get the area of a circle 3. To get the area of a parallelogram 4. To get the perimeter of a square A square has a perimeter of 24 inches. What is the area of the square? 1. 36 square inches 2. 144 square inches 3. 12 square inches 4. 28 square inches", "## Algebra and Trigonometry 10th Edition $x \\gt 0\\\\y \\gt x \\\\ \\pi y^2 \\gt \\pi x^2+10$. The radius of the smaller circle is positive; this means that $x \\gt 0$ The radius $y$ is for the larger circle; this means that $y \\gt x$ The area of the larger circle is at least $10 \\ sq \\ unit$ greater; this means that $\\pi y^2 \\gt \\pi x^2+10$. Thus, the system is: $x \\gt 0\\\\y \\gt x \\\\ \\pi y^2 \\gt \\pi x^2+10$.", "radius larger than 25 degrees; we considered smaller circles.", "Question Kaitlin is choosing between two circular wall clocks. One has a radius of 5 inches while the other has a radius of 6 inches. How much more wall space will the clock with a radius of 6 inches cover? 34.5 sq in Step-by-step explanation: A = πr² find the difference between the area of the larger clock versus the smaller 36π – 25π = 11π 11π ≈ 34.5", "why scientists don’t need math, and Jeremy Fox on why they do. Tonight the God Plays Dice art department made blondies! These are supposed to be made, according to the recipe, in a pan which is an eight-inch square. But we have no such thing. We do have a nine-inch circular pan, though. Will that do? Well, what matters is that the two pans have the same area – and therefore that the same volume of batter will have the same thickness and cook roughly the same. (If you thought I was going to solve some PDEs and work out how the heat transfers, you haven’t been paying attention.) A nine-inch circle has area $\\pi (9/2)^2 = 81\\pi/4$ square inches, which is about 63.62. An eight-inch square, of course, has area 64 square inches. Not bad! What would it take for this approximation to be exactly correct? This would require that $81\\pi/4 = 64$ exactly; solving for $\\pi$ gives \\$\\pi = 256/81″, which is often credited as an Egyptian approximation to $\\pi$ as it implicitly appears in the Rhind papyrus, an ancient Egyptian document of,problems in mathematics. In fact the setting in which this is established there is almost exactly this one – a circle of diameter 9 and a square of side 8 are said to have", "area is $$\\pi 3^2$$ square inches, or $$9\\pi$$ square inches, which is approximately 28.3 square inches. • squared We use the word squared to mean “to the second power.” This is because a square with side length $$s$$ has an area of $$s \\boldcdot s$$, or $$s^2$$.", "increases as $n=3,4,5,...$ with the limit $1/(4\\pi)=0.079577...,$ which is the area of a circle of perimeter 1. Note that for the square, $A_4=0.0625$, while for the hexagon, $A_6=0.0721687...$. This is pretty. But it only shows that a circle is better than any regular $n$-gon. There are still a lot of curves to be accounted for. – AndrewG Dec 25 '12 at 8:54", "## anonymous one year ago ***WILL MEDAL AND FAN*** A 10\" diameter pumpkin pie is cut into six equal servings. What is the area of the top of each piece of pie? a) 3.33pi inches b) 4.17pi inches c) 16.66pi inches 1. anonymous im thinking its A cause im trying to work it out but it just comes close to ten 2. phi 10 inch diameter means 5 inch radius Do you know how to find the area of a circle with r=5 ? 3. anonymous Yea. The area would be 78.5 right? 4. anonymous @phi 5. phi I would not make it decimal (because the answer choices have \"pi\" in them can you find the area, but do not change pi to a decimal? 6. anonymous Hmmm` 7. phi first, what is the formula for Area of a circle? 8. anonymous pi x radius squared 9. phi and r= 5 r squared ( $$r^2$$ ) = 5*5= 25 so Area= 25 * 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058... pi goes on forever... rather than truncating to approximately 3.14 (which is close but not exactly correct, as you can see), and rather than typing all those numbers (which you can't because they go on forever), people just say... call that number pi Area= $$25 \\pi$$ and let's not multiply it out unless you", "can be taken to be a shape that is made up of infinite number of equal sides. A circle with an area of 250 square foot has a radius of sqrt(250/pi). To minimize costs the enclosed area should be a circle with a radius sqrt(250/pi) foot. Need a fast expert's response? Submit order and get a quick answer at the best price for any assignment or question with DETAILED EXPLANATIONS! #### Comments No comments. Be first! Tweet", "Length x Width 2. Base x Height 4. Pi x Diameter 5. 1/2 x Base x Height Line AB is 14 inches long. What is the approximate area of this circle? 1. 42 square inches 2. 615 square inches 3. 160 square inches 4. 154 square inches", "EXACT AREA OF A CIRCLE WITH A DIAMETER OF 23 INCHES", "## Area of Inscribed Circle The perimeter and area of a triangle are $20$ inches and $12$ square inches, respectively. What is the area in square inches of a circle inscribed in the triangle? Source: NCTM Mathematics Teacher, February 2006 SOLUTION area of $AIB=(1/2)cr\\qquad (1)$ area of $BIC=(1/2)ar\\qquad (2)$ area of $AIC=(1/2)br\\qquad (3)$ area of $ABC=(1/2)r(a+b+c)$ $12=(1/2)r(20)$ $r=6/5$ Area of inscribed circle $\\pi(6/5)^2=1.44\\pi$ Answer: $1.44\\pi$ square inches", "notice with a full demonstration of my device. A 20 ft diameter circle is all the space needed. Thats seems to be a contradiction if you are not in northern NJ. Comment Source:>I operate the invention only in Northern Nj and >I can be in your area with 3 days notice with a full demonstration of my device. A 20 ft diameter circle is all the space needed. Thats seems to be a contradiction if you are not in northern NJ. • Options 90. edited June 2014 Congratulations, Nathan! It's a sign of success. Comment Source:Congratulations, Nathan! It's a sign of success.", "+0 +1 198 1 +26 What is the area, in square units, of a square whose sides are the same length as the radius of a circle with a circumference of $12\\pi$ units? Jul 1, 2020 #1 +28025 +2 Circumference = pi X d 12 pi = pi X d d = 12 then r = 6 units <----- this is the square side length and width Area of square = l x w = 6 x 6 = 36 units2 Jul 1, 2020" ]

[ "(1 followed by 100 zeros)", "100 can't find large zeros, only with $\\Re(s)<2$.", "is: $$e^{-1000a}+0$$", "# Just ones and zeroes! How many ones are there in the expansion of $(101) \\times (10001) \\times (100000001) \\ .....\\times (100.... ( 2^7-1) zeroes ...0001) ?$ ×", "x * 1000, \"'\") } )", "# zeroes Find the number of trailing zeroes in the number of trailing zeroes of the number $$1000!$$. ×", "1}) - 1 = 0.$ Substituting $n = 1000$ we get your case.", "< X \\le 1000$$, $$X$$ = 1000.", "or $$n=1000$$, testing shows that this expression is accurate to within a factor of $$2$$.", "left out the one in $$10000$$. Therefore, the right answer is $$4+1=5$$. In general: $$10^k$$ has $$\\left\\lfloor k \\right\\rfloor+1$$ digits. - 3 years, 3 months ago", "one has $$N'={5+10-1\\choose 5-1}=1001\\ .$$ Now the solution $x_0=5$, all other $x_k=0$ is not acceptable, as it cannot be realized by a single five digit number; but all other solutions can be realized. It follows that $N=1000$.", "Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) Let the following hold: $|a-e| > |b-e|$. Then, it can be shown that $a^b < b^a$. In fact, for every $x \\ge 0$, we have $e^x > x^e$ (assuming $x \\ne e$).", "Let $$N$$ be the number of consecutive 0's at the right end of the decimal representation of the product $$1!2!3!4!\\cdots99!100!.$$ Find the remainder when $$N$$ is divided by 1000. (第二十四届AIME1 2006 第4题)", "$= 100 = 1.0 \\times {10}^{2}$, which is what we started with.", "## [1] 1000", "# Just ones and zeroes! Number Theory Level 4 How many ones are there in the expansion of $(101) \\times (10001) \\times (100000001) \\ .....\\times (100.... ( 2^7-1) zeroes ...0001) ?$ ×", "# Is this a coincidence too? If we write out the decimal expansion of the smaller root of $f(x)=1000000x^2-1000000x+1$, we get the following decimal: $0.000001 \\quad 000001\\quad 000002\\quad 000005 \\quad 000014 \\quad \\ldots$ which we notice as the first 5 Catalan numbers. How many distinct Catalan numbers occur in a row before this pattern stops? ×", "to $N=1000$, we get Pretty good.", "/ 109 + {y}^{2} / 100 = 1$", "subsequences that will appear? Examples: $11010$ occurs since it can be decomposed into $1|10|10$, and the sum of each 1-word is 1: $1+0=1+0$. $01100$ also occurs since it can be decomposed into $01|10|0$. $11011$ cannot occur, because neither $1|10|11$ nor $11|01|1$ satisfies the second condition. The other two examples also follow." ]

[ "where $0< m<10^{100-d}$ and $$100^{100}-n=100^{100}-m10^d=10^d(100^{100-d}-m)$$ which has $d$ terminating because $100^{100-d}-m>0$. • Oh, so since 100! has 10$^{24}$ as its max power of 10, but 100$^{100}$ can have more, such as 10$^{25}$, then 100! < 100$^{100}$, and therefore the terminating zeros are those of 100!? In this case being 24. How do you know that n and 100$^{100}$ - n will have the same terminating zeros, though? – Ivan Ortiz Jul 1 '17 at 8:50", "N3 - Multiplicative reasoning N3.1 In N3.1 we learn how to multiply by powers of 10. First of all, powers of ten are shown as $10^{x}$ here are the powers of ten (these are infinite) $10^{10 }$ = 10000000000 $10^{9 }$ = 1000000000 $10^{8 }$ = 100000000", "will use the $10^n$ for this number, but one Important Thing to notice is that the first three numbers after the decimal are 042 which are unique so, we will take an n = 3 for this case: n = 3, $10^n$ = $10^3$ = 1000 1000 x = 1042.42424242 Then we follow that up with the $10^{n-1}$ but given the nature of this problem, to Eliminate the decimal values we have to use $10^{n-2}$: n -2 = 3 – 2 = 1, $10^{n-1}$ = $10^1$ = 10 Subtracting 10x on both sides looks like: 1000x – 10x = 1042.42424242 – 10.42424242 = 1032 Hence, 990x = 1032, x = $\\frac{1032}{990}$ Finally, we have our solution.", "Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) Let the following hold: $|a-e| > |b-e|$. Then, it can be shown that $a^b < b^a$. In fact, for every $x \\ge 0$, we have $e^x > x^e$ (assuming $x \\ne e$).", "tens • starting at 10 • within 100 • within 1000 • within 10000 • starting at$n$, for$n = 20,30,40,...$• within 100 • within 1000 • by hundreds • starting at 100 • within 1000 • within 10000 • starting at$n$, for$n = 200,300,400,...$• within 1000 • by thousands • starting at 1000 • within 10000 • starting at$n$, for$n = 2000,3000,4000,...$• within 10000 • counting backward • by ones • starting at$n$, for$n = 2,3,4,...$• within 100 • within 1000 • by twos • starting at$n$, for$n = 4,6,8,...$• within 100 • within 1000 • by fives • starting at$n$, for$n = 10,15,20,...$• within 100 • within 1000 • by tens • starting at$n$, for$n = 20,30,40,...$• within 100 • within 1000 • by hundreds • starting at$n$, for$n = 200,300,400,...$• within 1000 • ordering • given a set of counting numbers, put them in increasing order • set presented as pictures • set presented as numerals • set presented as written words • given a set of counting numbers, put them in decreasing order • set presented as pictures • set presented as numerals • set presented as written words • identify positions of objects in sequences • first, second, third, fourth, fifth • sixth, seventh, eight, ninth, tenth • comparing sizes of groups • matching", "# How do you write 10^5 in decimal form? $100000$ A power is a reiterated multiplication. So, ${10}^{5} = 10 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10$. And as you know, multiplying by $10$ means to add a zero at the end of the number. This also means that ${10}^{k}$ is simply a one followed by $k$ zeroes.", "$3$, $1000_\\text{ten}=1101001_\\text{three}$; thus, the number of factors of $3$ in $1000!$ is $\\frac{1000-4}{3-1}=498$. Thus, the number of factors of $24$ in $1000!$ is $331$. - Yes - it is the same concept; essentially, you want to write $1000! = 3^{x}\\cdot 8^{y} \\cdot \\textrm{stuff}.$ Then take the minimum of $x$ and $y$ to get the number of zeros at the end. - You need to figure out how many times 24 divides 1000!. Since $24 = 2^3\\cdot 3$, figure out how many times $2$ and $3$ divide 1000!. This can be done in exactly the same way you did it above: What you found was the number of times $5$ divided $1000!$. -", "Virginia SOL 6 - 2020 Edition 1.11 Place value and powers of 10 Lesson ## Powers of ten Recall that an exponent (or power) tells us the number of times to multiply a certain number by itself. Let's review the patterns in the powers of ten with the following exploration. #### Exploration Move the slider in the applet below and think about the answers to the following questions. 1. What patterns do you see between the power of ten and its expanded form? 2. What patterns do you see between the power of ten and the number of zeros? 3. Why do you think $10^0=1$100=1? Notice that the power of ten is the same as the number of zeros before or after the decimal point. That's because of the place value system. The following table demonstrates another way to think of some of the powers of ten that we see in the applet. Power of Ten Meaning Value (basic numeral) In Words $10^5$105 $10\\times10\\times10\\times10\\times10$10×10×10×10×10 $100000$100000 One hundred thousand $10^4$104 $10\\times10\\times10\\times10$10×10×10×10 $10000$10000 Ten thousand $10^3$103 $10\\times10\\times10$10×10×10 $1000$1000 One thousand $10^2$102 $10\\times10$10×10 $100$100 One hundred $10^1$101 $10$10 $10$10 Ten $10^0$100 $1$1 $1$1 One #### Worked example ##### Question 1 What value should go in the space? $300=\\editable{}\\times10^2$300=×102 Think: What basic numeral is equivalent to $10^2$102? How will this move our", "or $$n=1000$$, testing shows that this expression is accurate to within a factor of $$2$$.", "# How do you express 1times10^12 in standard notation? $1 , 000 , 000 , 000 , 000$ $1 \\cdot {10}^{12}$ is just a one followed by twelve zeros. $\\therefore = 1 , 000 , 000 , 000 , 000$", "this problem, the numbers are given in exactly that form! We can work out that n/2= 10 so n= 20, and that $(n/2)^2- (m/3)^3= 100- (m/3)^3= 108$ so that (m/3)^3= -8, m/3= -2, and m= -6.", "# How do you express 1times10^12 in standard notation? Jan 15, 2018 $1 , 000 , 000 , 000 , 000$ $1 \\cdot {10}^{12}$ is just a one followed by twelve zeros. $\\therefore = 1 , 000 , 000 , 000 , 000$", "ten where power is equal to the position of digit counted from left to right starting from 0. For digits after decimals point, multiply each digit with the negative powers of ten where power is equal to the position of digit counted from right to left starting from 1. 1) $$10^{0}=1$$ 2) $$10^{1}=10$$ 3) $$10^{2}=100$$ 4) $$10^{3}=1000$$ 5) $$10^{4}=10000$$ And so on... 6) $$10^{-1}=0.1$$ 7) $$10^{-2}=0.01$$ 8) $$10^{-3}=0.001$$ 9) $$10^{-4}=0.0001$$ The given question is $$93.6-0.518$$", "# Indices law - law 2 ## a^-n=1/a^n To remember this put simple numbers in. ie Try simple numbers you know first. 10^-2=1/10^2=1/(10times10)=1/100=0.001 8^-1=1/8=0.125 8^-2=1/8^2=1/(8times8)=1/64=0.0156 Also try other examples you know like starting with 10^2 and remove 1 from the power and understand the answer you get: Examples 1. 5^-3 is the same as 1/5^3 2. x^-3 is the same as 1/x^3", "ten to base one million.) Let $n=dq+r$, where $0\\le r<d$. Then $n=999999(77520q)+r$, so $r\\equiv n\\equiv m\\pmod{999999}$, and it follows immediately that $r\\ge 100000$.", "# What this N could be? $$N = 1 + 11 + 101 + 1001 +10001 + \\cdots +\\overbrace{1000\\ldots00001}^{50\\text{ zeros}}$$, When $$N$$ is calculated and written as a single integer, the sum of its digits is $$\\text{___} .$$ ×", "# Powers Express the expression ${1000}^{21}\\cdot {10}^{101}\\cdot {10}^{145}$ as the n-th power of the base 10. Did you find an error or inaccuracy? Feel free to write us. Thank you! Showing 1 comment: Math student Help I need help I don’t get it", "$1 \\le q \\le 100$. • $1 \\le n \\le 1000$ • $1 \\le n \\le 10^9$ • $x_1=x_2$ and $y_1=y_2$ • $1 \\le n \\le 10^5$ • $1 \\le n \\le 10^9$ • $x_1=y_1=1$ • $1 \\le n \\le 10^9$", "# How do you write 10^-5 in decimal form? Oct 9, 2015 ${10}^{- 5} = 0.00001$ #### Explanation: By definition of negative power, ${A}^{- n} = \\frac{1}{{A}^{n}}$ Therefore, by definition, ${10}^{- 5} = \\frac{1}{{10}^{5}}$ Obviously, $\\frac{1}{{10}^{5}} = \\frac{1}{100000} = 0.00001$ Hence, ${10}^{- 5} = 0.00001$", "find a power number of two that starts with 999 when writing it in decimal: $$2 ^ n = 10 ^ a \\cdot b$$ $$b \\in [0.999, 1); a, n \\in N$$ Taking base 10 logarithms: $$n * 0.3010299956639811952... = a + c$$ $$c \\in [-0.00043451..., 0)$$ Which is easy to solve taking n as 100,000: $$2 ^ {100,000} = 0.999002093 ... \\cdot 10 ^ {30103} = 999.002093 ... \\cdot 10 ^ {30100}$$" ]

[ "or $$n=1000$$, testing shows that this expression is accurate to within a factor of $$2$$.", "where $0< m<10^{100-d}$ and $$100^{100}-n=100^{100}-m10^d=10^d(100^{100-d}-m)$$ which has $d$ terminating because $100^{100-d}-m>0$. • Oh, so since 100! has 10$^{24}$ as its max power of 10, but 100$^{100}$ can have more, such as 10$^{25}$, then 100! < 100$^{100}$, and therefore the terminating zeros are those of 100!? In this case being 24. How do you know that n and 100$^{100}$ - n will have the same terminating zeros, though? – Ivan Ortiz Jul 1 '17 at 8:50", "Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.) Let the following hold: $|a-e| > |b-e|$. Then, it can be shown that $a^b < b^a$. In fact, for every $x \\ge 0$, we have $e^x > x^e$ (assuming $x \\ne e$).", "< X \\le 1000$$, $$X$$ = 1000.", "N3 - Multiplicative reasoning N3.1 In N3.1 we learn how to multiply by powers of 10. First of all, powers of ten are shown as $10^{x}$ here are the powers of ten (these are infinite) $10^{10 }$ = 10000000000 $10^{9 }$ = 1000000000 $10^{8 }$ = 100000000", "$3$, $1000_\\text{ten}=1101001_\\text{three}$; thus, the number of factors of $3$ in $1000!$ is $\\frac{1000-4}{3-1}=498$. Thus, the number of factors of $24$ in $1000!$ is $331$. - Yes - it is the same concept; essentially, you want to write $1000! = 3^{x}\\cdot 8^{y} \\cdot \\textrm{stuff}.$ Then take the minimum of $x$ and $y$ to get the number of zeros at the end. - You need to figure out how many times 24 divides 1000!. Since $24 = 2^3\\cdot 3$, figure out how many times $2$ and $3$ divide 1000!. This can be done in exactly the same way you did it above: What you found was the number of times $5$ divided $1000!$. -", "# How do you express 1times10^12 in standard notation? $1 , 000 , 000 , 000 , 000$ $1 \\cdot {10}^{12}$ is just a one followed by twelve zeros. $\\therefore = 1 , 000 , 000 , 000 , 000$", "# How do you express 1times10^12 in standard notation? Jan 15, 2018 $1 , 000 , 000 , 000 , 000$ $1 \\cdot {10}^{12}$ is just a one followed by twelve zeros. $\\therefore = 1 , 000 , 000 , 000 , 000$", "In binary, we have $(11001_2, 1101_2, 100111_2)$. • Next, we XOR all our numbers: $11001_2 \\oplus 1101_2 \\oplus 100111_2 = 110011_2$ • To generate our winning strategy, we want to take out enough coins from the largest pile until the xor of the piles is 0. In other words, we want the largest pile to become $010100_2$, or to take out 19 from that pile. (To verify, $25 \\oplus 13 \\oplus 20 = 0$) ### Another Example Is $(27, 23, 22, 15)$ in $P$ or $N$? $(11011_2, 10111_2, 10110_2, 01111_2)$ gives a Nim Sum of $10101$. We can move $27$ to $14$. The remainder of the winning moves is left as an exercise for homework. ## Footnotes 1. If and only if", "$= 100 = 1.0 \\times {10}^{2}$, which is what we started with.", "will use the $10^n$ for this number, but one Important Thing to notice is that the first three numbers after the decimal are 042 which are unique so, we will take an n = 3 for this case: n = 3, $10^n$ = $10^3$ = 1000 1000 x = 1042.42424242 Then we follow that up with the $10^{n-1}$ but given the nature of this problem, to Eliminate the decimal values we have to use $10^{n-2}$: n -2 = 3 – 2 = 1, $10^{n-1}$ = $10^1$ = 10 Subtracting 10x on both sides looks like: 1000x – 10x = 1042.42424242 – 10.42424242 = 1032 Hence, 990x = 1032, x = $\\frac{1032}{990}$ Finally, we have our solution.", "tens • starting at 10 • within 100 • within 1000 • within 10000 • starting at$n$, for$n = 20,30,40,...$• within 100 • within 1000 • by hundreds • starting at 100 • within 1000 • within 10000 • starting at$n$, for$n = 200,300,400,...$• within 1000 • by thousands • starting at 1000 • within 10000 • starting at$n$, for$n = 2000,3000,4000,...$• within 10000 • counting backward • by ones • starting at$n$, for$n = 2,3,4,...$• within 100 • within 1000 • by twos • starting at$n$, for$n = 4,6,8,...$• within 100 • within 1000 • by fives • starting at$n$, for$n = 10,15,20,...$• within 100 • within 1000 • by tens • starting at$n$, for$n = 20,30,40,...$• within 100 • within 1000 • by hundreds • starting at$n$, for$n = 200,300,400,...$• within 1000 • ordering • given a set of counting numbers, put them in increasing order • set presented as pictures • set presented as numerals • set presented as written words • given a set of counting numbers, put them in decreasing order • set presented as pictures • set presented as numerals • set presented as written words • identify positions of objects in sequences • first, second, third, fourth, fifth • sixth, seventh, eight, ninth, tenth • comparing sizes of groups • matching", "$1 \\le q \\le 100$. • $1 \\le n \\le 1000$ • $1 \\le n \\le 10^9$ • $x_1=x_2$ and $y_1=y_2$ • $1 \\le n \\le 10^5$ • $1 \\le n \\le 10^9$ • $x_1=y_1=1$ • $1 \\le n \\le 10^9$", "# How do you write 10^5 in decimal form? $100000$ A power is a reiterated multiplication. So, ${10}^{5} = 10 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10$. And as you know, multiplying by $10$ means to add a zero at the end of the number. This also means that ${10}^{k}$ is simply a one followed by $k$ zeroes.", "is: $$e^{-1000a}+0$$", "and $n_1+\\ldots+n_m=100$, so $100-m=90$, and $m=10$. – Brian M. Scott Dec 9 '12 at 1:03", "Virginia SOL 6 - 2020 Edition 1.11 Place value and powers of 10 Lesson ## Powers of ten Recall that an exponent (or power) tells us the number of times to multiply a certain number by itself. Let's review the patterns in the powers of ten with the following exploration. #### Exploration Move the slider in the applet below and think about the answers to the following questions. 1. What patterns do you see between the power of ten and its expanded form? 2. What patterns do you see between the power of ten and the number of zeros? 3. Why do you think $10^0=1$100=1? Notice that the power of ten is the same as the number of zeros before or after the decimal point. That's because of the place value system. The following table demonstrates another way to think of some of the powers of ten that we see in the applet. Power of Ten Meaning Value (basic numeral) In Words $10^5$105 $10\\times10\\times10\\times10\\times10$10×10×10×10×10 $100000$100000 One hundred thousand $10^4$104 $10\\times10\\times10\\times10$10×10×10×10 $10000$10000 Ten thousand $10^3$103 $10\\times10\\times10$10×10×10 $1000$1000 One thousand $10^2$102 $10\\times10$10×10 $100$100 One hundred $10^1$101 $10$10 $10$10 Ten $10^0$100 $1$1 $1$1 One #### Worked example ##### Question 1 What value should go in the space? $300=\\editable{}\\times10^2$300=×102 Think: What basic numeral is equivalent to $10^2$102? How will this move our", "ten to base one million.) Let $n=dq+r$, where $0\\le r<d$. Then $n=999999(77520q)+r$, so $r\\equiv n\\equiv m\\pmod{999999}$, and it follows immediately that $r\\ge 100000$.", "Thus, because we can express $12$ integers in the range $(1, 20)$, we can cover $12*50 = \\boxed{600}$ from 1 to 1000. $-\\text{thinker123}$ ## Solution 7 After observing, we can see that there are $6$ values of can be evaluated through the expression every $10$ numbers, so our answer is $6*100=600$ ~bluesoul", "find a power number of two that starts with 999 when writing it in decimal: $$2 ^ n = 10 ^ a \\cdot b$$ $$b \\in [0.999, 1); a, n \\in N$$ Taking base 10 logarithms: $$n * 0.3010299956639811952... = a + c$$ $$c \\in [-0.00043451..., 0)$$ Which is easy to solve taking n as 100,000: $$2 ^ {100,000} = 0.999002093 ... \\cdot 10 ^ {30103} = 999.002093 ... \\cdot 10 ^ {30100}$$" ]

[ "be factors of $n$? Oct 6: If the four-digit number $72d2$ is divisible by 6, what is the largest possible value of the digit $d$?", "Divisible Probability... The sum of the digits of a $$9$$ digit number is $$77$$. The probability that this number is divisible by 11 can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are co-prime positive integers. Find the value of $$a+b$$. ×", "\\textcolor{g r e e n}{23} - 5 \\cdot \\textcolor{g r e e n}{8} = - 17$. Thus, $2584$ is divisible by $17$ • A number is divisible by $25$ if it ends with $00 , 25 , 50$ or $75$.", "number is divisible by$$16$$. The factors of $$16$$ are $$1, 2, 4, 8, 16$$. Therefore, the number is also divisible by $$1, 2, 4, 8$$.", "# Find my A and B If the 7-digit integer $$\\overline{3A54B10}$$ is divisible by 330, which of the options are the possible values of $$A$$ and $$B$$? ×", "than $10^{62}$ if less than $10^{9118}$ then it is divisible by the $6$th power of some prime.", "divisible by $4$; so $1/2$ times $1/2$, or $1/4$ of all numbers have this property. So: The calculation of the chance that $XYZ$ is divisible by $25$ is similar. In this case, the chance that a number has exactly one factor of $5$ is the chance that it has at least one (which is $1/5$) minus the chance that it has at least two (which is $1/25$). So: These two probabilities are independent, because every twenty-fifth number divisible by $4$ is also divisible by $25$ (and vice versa). Therefore the probability that $XYZ$ is divisible by both $4$ and $25$, that is, that it is divisible by $100$, is their product, which is exactly $.1243$.", "# All divisible? True or False? All integers which are in the form $12008,\\ 120308,\\ 1203308,\\ 12033308,\\ \\ldots,\\ 120\\underbrace{3333333\\ldots3}_{n \\text{ number of 3's}}08,\\ \\ldots$ are divisible by 19. ×", "Question 144 # Let N = $$55^3 + 17^3 - 72^3$$. N is divisible by: Solution $$55^3 + 17^3 - 72^3$$ = $$(55-72)k + 17^3$$. This is divisible by 17 Remainder when $$55^3$$ is divided by 3 = 1 Remainder when $$17^3$$ is divided by 3 = -1 Remainder when $$72^3$$ is divided by 3 = 0 So, $$55^3 + 17^3 - 72^3$$ is divisible by 3 So, the answer is d) 3 and 17", "of $S_1$ and $S_2$ is correct The number of integers between $1$ and $500$ (both inclusive) that are divisible by $3$ or $5$ or $7$ is ____________ .", "$2342$ are not (because $38$ and $42$ are not divisible by $4$). (Why does this work?) It could be a good idea to make a table to keep track of where the digits $1$ to $6$ could go. Where will the even numbers have to go? So what about the odd numbers? Where will the $5$ have to go?", "two digits, and they are all equally likely, so the probability that the number is divisible by 4 is $6/20 = \\boxed{3/10}$. ### Answer This Question First Name: School Subject: Answer: ### Related Questions More Related Questions Post a New Question", "# Divisible By 41 Consider all values of $n$ such that the number $111\\ldots 111$ in which the number 1 occurs $n$ times is divisible by 41. What is the largest integer which must be a factor of $n$? ×", "\\}$ and $4$ of them are divisible by $4$. I'll continue to update my answer if I can clean up my analysis, someone spots a mistake, or if I can generalize my proof to cover the condition where $x$ and $y$ are not relatively prime.", "# Elusive scenario If the six digit integer $$\\overline{739ABC}$$ is divisible by 7, 8, and 9. What is the sum of all possible values $$A$$ , $$B$$ and $$C$$? ×", "# Easy #2 Number Theory Level 3 How many four-digit numbers of the form $$6A4B$$ ($$A$$ and $$B$$ are digits) are divisible by $$36$$? ×", "# x, y, 24? Integers $x$ and $y$ are such that $xy+1$ is divisible by 24. Is $x+y$ also divisible by 24? ×", "4 \\times 5 = 120$$. >I. 120 is divisible by $$x - 1 = 4 - 1 = 3$$. >II. 120 is divisible by $$2x = 2 (4) = 8$$. >III. 120 is divisible by $$x! = 4! = 4 \\times 3 \\times 2 \\times 1 = 24$$. I only II only II and III only I and III only I, II, and III Continue", "($2 \\times 2$). So for example in the box $6 \\times 6$, each factor of $6$ is divisble by $2$ and so the product $6 \\times 6$ is divisible by $4$ (even though neither factor of $6$ is divisible by $4$). Since $2$ and $3$ are prime numbers they cannot be written as a product of smaller whole numbers and so these isolated shaded boxes do not occur for multiples of $2$ and $3$.", "had a slightly different method: I found the factors of the numbers, and spotted that all are divisible by $6$. When I had divided by $6$, the series was $36, 28, 21, 15, 10, 6$ and the differences went down by one each time. Very well done!" ]

[ "leaves the digits $1,2,6$, to be arranged so that the resulting number is divisible by $8$. The only option for that is $216$, hence we look at $9873216$. This isn't divisible by $7$ either. Next try, start with $9872$. To arrange the digits $1,3,6$ so that the resulting number is divisible by $8$, we have only the option $136$, but $9872136$ isn't divisible by $7$ either. Thus we try to start with $9871$. To get a multiple of $8$, the only candidate is $9871632$, but that is again not divisible by $7$. Thus we move towards numbers starting with $986$, and the largest even such number is $9867312$. Check: $$9867312 = 9\\cdot 1096368 = 8 \\cdot 1233414 = 7\\cdot 1409616.$$", "The number must be divisible by $$8000$$, because $$15!$$ contains $$8$$ and $$1000$$. $$d_2$$ is non-zero number, because $$15!$$ contains only three $$5$$s. It implies $$1d_10767436d_2$$ must be divisible by $$8$$. It implies $$36d_2$$ is divisible by $$8$$. Hence, $$d_2=8$$. Now you can use the divisibility by $$9$$ ($$d_1+d_2=11$$) and find $$d_1=3$$. $$15!=2^{11}\\cdot 5^3\\cdot 7^2\\cdot 11\\cdot 13=(1000)X$$ where $$X=2^8\\cdot 3^6\\cdot 7^2\\cdot 11\\cdot 13.$$ The last digits of $$2^8(=16^2), 3^6 (=9^3),7^2, 11,13$$ are, respectively $$6,9,9,1,3 .$$ Modulo $$10$$ we have $$6\\cdot 9\\cdot 9 \\cdot 1\\cdot 3\\equiv 6\\cdot(-1)^2\\cdot 3\\equiv 18\\equiv 8$$. So the last digit of $$X$$ is an $$8$$. Therefore the 2nd digit of $$15!$$ must be a $$3$$ in order for the sum of all its digits to be divisible by $$9$$.", "$17\\cdot20=\\boxed{\\textbf{(C) } 340},$ from which $$n=\\operatorname{lcm}(323,d)=17\\cdot19\\cdot20=6460.$$ ~MRENTHUSIASM ~bjhhar Since prime factorizing $323$ gives you $17 \\cdot 19$, the desired answer needs to be a multiple of $17$ or $19$, this is because if it is not a multiple of $17$ or $19$, $n$ will be more than a $4$ digit number. For example, if the answer were to instead be $324$, $n$ would have to be a multiple of $2^2\\cdot3^4\\cdot17\\cdot19$ for both $323$ and $324$ to be a valid factor, meaning $n$ would have to be at least $104652$, which is too big. Looking at the answer choices, $\\textbf{(A)}$ and $\\textbf{(B)}$ are both not a multiple of neither $17$ nor $19$, $\\textbf{(C)}$ is divisible by $17$. $\\textbf{(D)}$ is divisible by $19$, and $\\textbf{(E)}$ is divisible by both $17$ and $19$. Since $\\boxed{\\textbf{(C) } 340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$, a $4$-digit number. Note that $n$ is also divisible by $2$, one of the listed divisors of $n$. (If $n$ was not divisible by $2$, we would need to look for a different divisor.) -Edited by Mathandski Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$-digit $n$. So, we need a choice that shares", "in the range from $$2$$ to $$31$$: $$2,3,5,7,11,13,17,19,23,29,31;$$ so these are the only primes that \"some number\" might need to be divisible by. But some of these primes (the first three) also appear to higher powers in this range: we have $$2,4,8,16$$; $$3,9,27$$; and $$5,25$$. In each case, the largest one will take care of the rest, so the list of prime-power factors that need to be considered are these: $$7,11,13,16,17,19,23,25,27,29,31.$$ A number is divisible by all the numbers from $$2$$ to $$31$$ if it is divisible by all of these. Conversely, if it is not divisible by some number in the list, it is not divisible by one of these. So we want to find two of these to remove, but only one pair are consecutive numbers: $$16$$ and $$17$$. So the number we are looking for is divisible by all the other nine numbers listed, but not by $$16$$ or $$17$$. The smallest such number is the product of those nine, along with the next smaller power of $$2$$ (namely, $$8$$): $$7\\cdot8\\cdot11\\cdot13\\cdot19\\cdot23\\cdot25\\cdot27\\cdot29\\cdot31 = 2123581660200.$$", "= 2^{10} \\cdot 2^3 = 1024 \\cdot 8 = 8192$, and $3^8 = (3^4)^2 = 81^2 = 6561$. We have the 2's and 3's out of the way, and then we have $7 \\cdot 17 \\cdot 19 = 2261$. Now if we multiply all the values calculated, we get $1000 \\cdot 1001 \\cdot 8192 \\cdot 6561 \\cdot 2261 = 121,645,100,408,832,000$. Thus $T = 4, M = 8, H = 0$, and the answer $T + M + H = 12$, thus (C). ## Solution 3 We know that $9$ and $11$ are both factors of $19!$. Furthermore, we know that H is 0 because $19!$ ends in three zeroes. We can simply use the divisibility rules for $9$ and $11$ for this problem to find T and M. For $19!$ to be divisible by $9$, the sum of digits must be divisible by $9$. Summing the digits, we get that T + M + $33$ must be divisible by $9$. This leaves either A or C as our answer choice. Now we test for divisibility by $11$. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. $2728$, $2$-$7$+$2$-$8$ = -11 so $2728$ is divisible by $11$). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by", "($2 \\times 2$). So for example in the box $6 \\times 6$, each factor of $6$ is divisble by $2$ and so the product $6 \\times 6$ is divisible by $4$ (even though neither factor of $6$ is divisible by $4$). Since $2$ and $3$ are prime numbers they cannot be written as a product of smaller whole numbers and so these isolated shaded boxes do not occur for multiples of $2$ and $3$.", "be divisible by $4$. These digits are ($1$ or $3$) and ($4$ or $6$) Therefore the possible numbers are $14$, $16$, $34$, and $36$ Since only $16$ and $36$ are divisible by $4$, the fourth digit of the number must be $6$. By process of elimination the last number must therefore be $4$. So we are left with ($1$ or $3$),$2$,($1$ or $3$),$6$,$5$,$4$ This gives the two numbers $123654$ and $321654$, both of which are correct, as shown above. Hannah, Megan, Jess, Toby, Callum, Callum, Lydia, Nadia, Sam and Chris from Peterchurch Primary School also used very similar reasoning to start with, but used a different method again towards the end of their solution: $123654$ and $321654$ We worked this out by the following process: 1. We knew that the fifth digit must be a $5$ because that's the only number divisible by $5$ (because we had no zero) 2. We knew that the second, fourth and sixth digits had to be even to be divisible by $2$, $4$ and $6$. 3. The first and third digits had to odd - $1$ or $3$ 4. We worked out that the second digit must be $2$ because 143 or 163 would not be divisible by $3$ 5. This meant that we only had four combinations left: $123456$, $123654$, $321456$,", "= 55k$, or $9c - 9a = 5k$. Thus, k is divisible by $9$. Because $55 * 18 = 990$, $k$ must be $9$, and therefore $c - a = 5$. Because $a + b + c \\leq{7}$ and $a \\geq{1}$, $a = 1$, $c = 6$ and $b = 0$, and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$, or $\\boxed{D}$.", "have to count check as many numbers to see if they're divisible by $3$3 as well. When we have multiple numbers it's always easiest to think about the biggest number first. So, starting from $11$11, let's see which multiples of $11$11 are also divisible by $3$3 Multiples of $11$11 Is it divisible by $3$3? Numbers that have both $3$3 and $11$11 as factors $11$11 no $22$22 no $33$33 yes $33$33 $44$44 no $55$55 no $66$66 yes $66$66 $77$77 no $88$88 no $99$99 yes $99$99 So now we have found three numbers that have factors of both $11$11 and $3$3 They are $33$33, $66$66 and $99$99 #### Example ##### Question 1 What are the factors of $42$42? Separate the factors with commas. ##### Question 2 What factor of $30$30, other than $1$1, is also a factor of $27$27 and $21$21? ##### Question 3 What numbers from $1$1 to $100$100 have factors of $3$3, $5$5 and $10$10? 1. Write each number on the same line, separated by commas.", "Does such a value for n even exist (besides the n for our example $$F=527$$)? In Step 3 and 4 we want to examine for which values of n the expression $$(9 \\cdot 10^n - 41)$$ is divisible by 17. ------------------------------------------------------ STEP 3 We already know that $$F=527$$ fulfills the swap condition. We obtain $$F=527$$ for $$n=3$$: $$F = \\frac{1}{17} (9 \\cdot 10^3 -41) = \\frac{1}{17} (9000 - 41) = \\frac{1}{17}(8959) = 527$$ There is a rule for testing whether a number is divisible by 17, see here: http://www.egge.net/~savory/maths1.htm and http://primes.utm.edu/curios/page.php/17.html In order to illustrate the rule we test if 8959 is divisible by 17: Step 1: Split 8959 in 895 and 9 (last digit), multiply 9 (the last digit) by 5: $$9 \\cdot 5 = 40$$. Step 2: Substract this from the ramaining digits 895: $$895-9\\cdot5=850$$. Step 3: If the last two or three remaining digits are divisible by 17 then the original number 8959 is divisible by 17, too. In this case, 850 is divisible by 17, thus 8959 is divisible by 17, too. So, for $$n=3$$, F is an integer. What about $$n=4$$: $$F = \\frac{1}{17}(9 \\cdot 10^4 -41) = \\frac{1}{17}(90000-41) = \\frac{1}{17}(89959)$$ Let's test if 89959 is divisible by 17: Step 1: $$8995-9\\cdot 5 = 8955$$ Step 2: $$895 -5 \\cdot 5 = 870$$", "for $a_5$. There are then $9\\cdot 8\\cdot 6\\cdot 4\\cdot 2=3456$ ten digit numbers satisfying all of the desired properties. The largest number of which is formed with the largest selections available for $a_1,a_2,\\dots$ respectively and is then $9876501234$, the $10000$'s place being the $5$. Lemma: Any ten digit number of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is divisible by $11111$ if and only if $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ is divisible by $11111$. $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\equiv (9\\cdot 11111)a_1a_2a_3a_4a_5+a_1a_2a_3a_4a_5 + b_1b_2b_3b_4b_5$ $\\equiv 10^5a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\equiv a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5\\pmod{11111}$ Claim: In the set of ten digits numbers using all digits $0$-$9$ the number is divisible by $11111$ if and only if it is of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ where $a_i+b_i=9$ for all $i$. $\\Leftarrow)~~$ Suppose that $a_i+b_i=9$ for all $i$. Then $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=99999$ is divisible by $11111$. Then by the lemma, so too is $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$. $\\Rightarrow)~~$ Suppose that $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is a ten-digit number using all digits exactly once which is divisible by $11111$. By the lemma, we know then that $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=n\\cdot 11111$ for some $n$. As all digits are used exactly once, the sum of the digits is $1+2+\\dots+9=\\frac{9\\cdot 10}{2}=45$. As this is divisible by $9$ both $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ and $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ must also be divisible by $9$. (see sum of digits of a number is equivalent modulo 9 to the number itself proof elsewhere). As the digits are distinct $16047= 13579+02468\\leq a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\leq 97531+86420=183951$ as those selected", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "$49$, which is true because $6+8$ is divisible by $7$ and $$6^6-6^5\\cdot8+...+8^6\\equiv0(\\mod7).$$ Observe that $8^7\\equiv1\\pmod{49}$. Since $7|273$, so $8^{273}\\equiv1\\pmod{49}$. Observe also that $6^7\\equiv-1\\pmod{49}$. Since $7|273$, so $6^{273}\\equiv-1\\pmod{49}$. Adding,$8^{273}+ 6^{273}\\equiv0\\pmod{49}$.", "\\textcolor{g r e e n}{23} - 5 \\cdot \\textcolor{g r e e n}{8} = - 17$. Thus, $2584$ is divisible by $17$ • A number is divisible by $25$ if it ends with $00 , 25 , 50$ or $75$.", "# How many of the integers between $100$ and $200$ are divisible by $3$ or divisible by $2$ but not by $5$? How many of the integers between $$100$$ and $$200$$ are divisible by $$3$$ or divisible by $$2$$ but not by $$5$$? is the range of integer 200-100+1=101 or 100? $$A_5$$ number that is divisible by 5 \\begin{align*} A_1 & = \\left\\lfloor{\\frac{101}{3}} \\right\\rfloor = 33 && \\text{(divisible by 3)}\\\\ A_2 & = \\left\\lfloor{\\frac{101}{2}} \\right\\rfloor = 50 && \\text{(divisible by 2)}\\\\ \\\\ | A_1 \\cap A_5 | & = \\left\\lfloor{\\frac{101}{3 \\cdot 5}}\\right\\rfloor = 6\\\\ | A_2 \\cap A_5 | & = \\left\\lfloor{\\frac{101}{2 \\cdot 5}}\\right\\rfloor = 10\\\\ \\\\ | A_1 \\cap A_2 \\cap A_3 | & = \\left\\lfloor{\\frac{101}{2 \\cdot 3 \\cdot 5}}\\right\\rfloor = 3 \\end{align*} Therefore, by the principle exclusion inclusion theorem $$= 50 + 33 - (6 + 10) + 3 =70$$ Is this right? Im trying to count one by one such as, and enumerate for $$100-130$$ , number that divisible by 2,3 but not 5 $$\\{102,104,106,111,108,112,114,116,118,122,123,124,126,128\\}=14$$ number $$14*3=42$$ number $$(100-190)$$ for $$191-200=$${ $$192,194,196,198\\}=4$$ number $$42+4=46$$ i used program to check: 102 104 106 108 111 112 114 116 117 118 122 123 124 126 128 129 132 134 136 138 141 142 144 146 147 148 152 153 154 156 158 159 162 164 166 168", "\\cdot 3 = 28$$ to our total. $28$ is congruent to $1 \\pmod 9$. Thus our sum is congruent to $3 \\pmod 9$, and $k = 1 \\implies \\boxed{B}$. ## Solution 2 As our first trial with 3, We can say that $N\\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\\pmod{3}$. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that $N\\equiv 19+20+21+22+...+91+92\\pmod{3}$, and adding that up using the rainbow strategy, we get $N\\equiv (111\\times37)\\pmod{3}$. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is $\\boxed{B}$ ~mathpro12345", "is 6, so you use blocks of 3. Note that $10^3+1=1001 \\equiv 0 \\pmod {13}$ - Your professor is using the fact that $100000001=10^8+1$ is divisible by $17$. Given for example your $80$-digit number, you can subtract $98765432\\cdot 100000001=9876543298765432$, which will leave zeros in the last $16$ places. Slash the zeros, and repeat. After $5$ times you are left with the number $0$, which is divisible by $17$, and hence your $80$-digit number must also be divisible by $17$. When checking for divisibility by $17$, you can also subtract multiples of $102=6\\cdot 17$ in the same way. For divisibility by $7$, $11$, or $13$, you can subtract any multiple of the number $1001=7\\cdot 11\\cdot 13$ without affecting divisibility by these three numbers. For example, $6017-6\\cdot 1001=11$, so $6017$ is divisible by $11$, but not by $7$ or $13$. For divisibility by $19$, you can use the number $1000000001=10^9+1=7\\cdot 11\\cdot 13\\cdot 19\\cdot 52579$. By subtracting multiples of this number, you will be left with a number of at most $9$ digits, which you can test for divisibility by $19$ by performing the division. - The whole process can be explained by writing A^x/B = C where C can be +1 or -1 or 0. Here C is the remainder got on division of A^x by B. Now we operate", "it ends in $2$ it must be divisible by $2$, if it ends in $3$ it must be divisible by $3$ and if it ends in $4$ it must be divisible by $2$. So the numbers that can contend as possible primes, in base $6$, look like this (bracketed): $$(2),(3),4,(5),10,(11),12,13,14,(15),20,(21),22,23,24,(25),30,(31),32,33,34,(35),40,(41),42,43,44,(45),50,(51),52,53,54,(55),\\dots$$ Or in base 10: $$(2),(3),4,(5),6,(7),8,9,10,(11),12,(13),14,15,16,(17),18,(19),20,21,22,(23),24,(25),26,27,28,(29),30,(31),32,33,34,(35),\\dots$$ Notice the repeated skip a number skip three numbers pattern that we can observe. This is the reason why we cannot have strings longer than $3$. Again I stress the above bold words. I am not claiming numbers in brackets are prime, just that numbers not in bracket must, necessarily, not be primes. I hope this provides some intuition.", "So $1516$ is divisible by $4$, too. • $1035$ caramel cubes: Because the last digit is odd, this number can't be divided by $2$ and thus can't be divided by $4$. • $1600$ strawberry strips: $00$ is divisible by $4$ because $0/4 = 0$ and thus $1600$ is divisible by $4$. • #### Calculate in which cases numbers are divisible by $2$. ##### Tipps If a number is divided by $2$ and doesn't have a remainder it also means it's evenly divisible or divisible by $2$. Write down the multiples of $2$. The last digit of any number divisible by $2$ is $0$, $2$, $4$, $6$ or $8$. ##### Lösung The multiples of $2$ are: $2$, $4$, $6$, $8$, ..., $12$, $14$, $16$, ..., $102$, $104$, ... • Each of those numbers is even. • Or, in other words the last digit is divisible by $2$. • Or, each number ends with $0$, $2$, $4$, $6$ or $8$. • #### Find the correct statements for divisibility rules for 4, 5, 8, and 10. ##### Tipps For example: $35$ is divisible by $5$. On the other hand $40$ is divisible by $5$, as well as by $10$. $12$ is divisible by $4$, but you can't divide it by $8$. • $20\\div 2=6$ • $10\\div 2=3$ The second time we", "of Prime numbers: $2$$2$, $3$$3$, $5$$5$, $7$$7$, $13$$13$, $17$$17$, $\\cdots$$\\cdots$ Some numbers are divisible by at-least one number between $2$$2$ and one less than the number. eg: $6$$6$ is divisible by $2$$2$ and $3$$3$. And $63$$63$ is divisible by $3$$3$, $7$$7$, $9$$9$, and $21$$21$. Such numbers are called \"composite numbers\". They can be equivalently given as a product of other numbers. eg: $6=2×3$$6 = 2 \\times 3$ and $63=3×21$$63 = 3 \\times 21$ or $63=7×9$$63 = 7 \\times 9$. The word \"composite\" means \"made up of several parts or elements\". The word \"prime\" means \"having distinct or important properties\". Composite Numbers : Numbers that are divisible by at-least a number other than $1$$1$ and the number itself. Prime Numbers : Numbers that are divisible by only $1$$1$ and the number itself. examples We need to find if $59$$59$ is a prime or composite number. To do that, • Check for divisibility by numbers from $2$$2$ to $58$$58$ OR • Check for divisibility by numbers from $2$$2$ to $7$$7$ By the definition of prime numbers, the number can be checked for divisibility by numbers from $2$$2$ to $58$$58$. But, it is sufficient to check for divisibility by numbers from $2$$2$ to $7$$7$. The number $7$$7$ is chosen because the given number $59$$59$ lies between the perfect squares $7×7=49$$7 \\times" ]

[ "The number must be divisible by $$8000$$, because $$15!$$ contains $$8$$ and $$1000$$. $$d_2$$ is non-zero number, because $$15!$$ contains only three $$5$$s. It implies $$1d_10767436d_2$$ must be divisible by $$8$$. It implies $$36d_2$$ is divisible by $$8$$. Hence, $$d_2=8$$. Now you can use the divisibility by $$9$$ ($$d_1+d_2=11$$) and find $$d_1=3$$. $$15!=2^{11}\\cdot 5^3\\cdot 7^2\\cdot 11\\cdot 13=(1000)X$$ where $$X=2^8\\cdot 3^6\\cdot 7^2\\cdot 11\\cdot 13.$$ The last digits of $$2^8(=16^2), 3^6 (=9^3),7^2, 11,13$$ are, respectively $$6,9,9,1,3 .$$ Modulo $$10$$ we have $$6\\cdot 9\\cdot 9 \\cdot 1\\cdot 3\\equiv 6\\cdot(-1)^2\\cdot 3\\equiv 18\\equiv 8$$. So the last digit of $$X$$ is an $$8$$. Therefore the 2nd digit of $$15!$$ must be a $$3$$ in order for the sum of all its digits to be divisible by $$9$$.", "= 2^{10} \\cdot 2^3 = 1024 \\cdot 8 = 8192$, and $3^8 = (3^4)^2 = 81^2 = 6561$. We have the 2's and 3's out of the way, and then we have $7 \\cdot 17 \\cdot 19 = 2261$. Now if we multiply all the values calculated, we get $1000 \\cdot 1001 \\cdot 8192 \\cdot 6561 \\cdot 2261 = 121,645,100,408,832,000$. Thus $T = 4, M = 8, H = 0$, and the answer $T + M + H = 12$, thus (C). ## Solution 3 We know that $9$ and $11$ are both factors of $19!$. Furthermore, we know that H is 0 because $19!$ ends in three zeroes. We can simply use the divisibility rules for $9$ and $11$ for this problem to find T and M. For $19!$ to be divisible by $9$, the sum of digits must be divisible by $9$. Summing the digits, we get that T + M + $33$ must be divisible by $9$. This leaves either A or C as our answer choice. Now we test for divisibility by $11$. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. $2728$, $2$-$7$+$2$-$8$ = -11 so $2728$ is divisible by $11$). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by", "in the range from $$2$$ to $$31$$: $$2,3,5,7,11,13,17,19,23,29,31;$$ so these are the only primes that \"some number\" might need to be divisible by. But some of these primes (the first three) also appear to higher powers in this range: we have $$2,4,8,16$$; $$3,9,27$$; and $$5,25$$. In each case, the largest one will take care of the rest, so the list of prime-power factors that need to be considered are these: $$7,11,13,16,17,19,23,25,27,29,31.$$ A number is divisible by all the numbers from $$2$$ to $$31$$ if it is divisible by all of these. Conversely, if it is not divisible by some number in the list, it is not divisible by one of these. So we want to find two of these to remove, but only one pair are consecutive numbers: $$16$$ and $$17$$. So the number we are looking for is divisible by all the other nine numbers listed, but not by $$16$$ or $$17$$. The smallest such number is the product of those nine, along with the next smaller power of $$2$$ (namely, $$8$$): $$7\\cdot8\\cdot11\\cdot13\\cdot19\\cdot23\\cdot25\\cdot27\\cdot29\\cdot31 = 2123581660200.$$", "$17\\cdot20=\\boxed{\\textbf{(C) } 340},$ from which $$n=\\operatorname{lcm}(323,d)=17\\cdot19\\cdot20=6460.$$ ~MRENTHUSIASM ~bjhhar Since prime factorizing $323$ gives you $17 \\cdot 19$, the desired answer needs to be a multiple of $17$ or $19$, this is because if it is not a multiple of $17$ or $19$, $n$ will be more than a $4$ digit number. For example, if the answer were to instead be $324$, $n$ would have to be a multiple of $2^2\\cdot3^4\\cdot17\\cdot19$ for both $323$ and $324$ to be a valid factor, meaning $n$ would have to be at least $104652$, which is too big. Looking at the answer choices, $\\textbf{(A)}$ and $\\textbf{(B)}$ are both not a multiple of neither $17$ nor $19$, $\\textbf{(C)}$ is divisible by $17$. $\\textbf{(D)}$ is divisible by $19$, and $\\textbf{(E)}$ is divisible by both $17$ and $19$. Since $\\boxed{\\textbf{(C) } 340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$, a $4$-digit number. Note that $n$ is also divisible by $2$, one of the listed divisors of $n$. (If $n$ was not divisible by $2$, we would need to look for a different divisor.) -Edited by Mathandski Note that $323$ multiplied by any of the answer choices results in a $5$ or $6$-digit $n$. So, we need a choice that shares", "leaves the digits $1,2,6$, to be arranged so that the resulting number is divisible by $8$. The only option for that is $216$, hence we look at $9873216$. This isn't divisible by $7$ either. Next try, start with $9872$. To arrange the digits $1,3,6$ so that the resulting number is divisible by $8$, we have only the option $136$, but $9872136$ isn't divisible by $7$ either. Thus we try to start with $9871$. To get a multiple of $8$, the only candidate is $9871632$, but that is again not divisible by $7$. Thus we move towards numbers starting with $986$, and the largest even such number is $9867312$. Check: $$9867312 = 9\\cdot 1096368 = 8 \\cdot 1233414 = 7\\cdot 1409616.$$", "divisible by $4$; so $1/2$ times $1/2$, or $1/4$ of all numbers have this property. So: The calculation of the chance that $XYZ$ is divisible by $25$ is similar. In this case, the chance that a number has exactly one factor of $5$ is the chance that it has at least one (which is $1/5$) minus the chance that it has at least two (which is $1/25$). So: These two probabilities are independent, because every twenty-fifth number divisible by $4$ is also divisible by $25$ (and vice versa). Therefore the probability that $XYZ$ is divisible by both $4$ and $25$, that is, that it is divisible by $100$, is their product, which is exactly $.1243$.", "($2 \\times 2$). So for example in the box $6 \\times 6$, each factor of $6$ is divisble by $2$ and so the product $6 \\times 6$ is divisible by $4$ (even though neither factor of $6$ is divisible by $4$). Since $2$ and $3$ are prime numbers they cannot be written as a product of smaller whole numbers and so these isolated shaded boxes do not occur for multiples of $2$ and $3$.", "for $a_5$. There are then $9\\cdot 8\\cdot 6\\cdot 4\\cdot 2=3456$ ten digit numbers satisfying all of the desired properties. The largest number of which is formed with the largest selections available for $a_1,a_2,\\dots$ respectively and is then $9876501234$, the $10000$'s place being the $5$. Lemma: Any ten digit number of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is divisible by $11111$ if and only if $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ is divisible by $11111$. $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\equiv (9\\cdot 11111)a_1a_2a_3a_4a_5+a_1a_2a_3a_4a_5 + b_1b_2b_3b_4b_5$ $\\equiv 10^5a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\equiv a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5\\pmod{11111}$ Claim: In the set of ten digits numbers using all digits $0$-$9$ the number is divisible by $11111$ if and only if it is of the form $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ where $a_i+b_i=9$ for all $i$. $\\Leftarrow)~~$ Suppose that $a_i+b_i=9$ for all $i$. Then $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=99999$ is divisible by $11111$. Then by the lemma, so too is $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$. $\\Rightarrow)~~$ Suppose that $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ is a ten-digit number using all digits exactly once which is divisible by $11111$. By the lemma, we know then that $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5=n\\cdot 11111$ for some $n$. As all digits are used exactly once, the sum of the digits is $1+2+\\dots+9=\\frac{9\\cdot 10}{2}=45$. As this is divisible by $9$ both $a_1a_2a_3a_4a_5b_1b_2b_3b_4b_5$ and $a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5$ must also be divisible by $9$. (see sum of digits of a number is equivalent modulo 9 to the number itself proof elsewhere). As the digits are distinct $16047= 13579+02468\\leq a_1a_2a_3a_4a_5+b_1b_2b_3b_4b_5\\leq 97531+86420=183951$ as those selected", "$49$, which is true because $6+8$ is divisible by $7$ and $$6^6-6^5\\cdot8+...+8^6\\equiv0(\\mod7).$$ Observe that $8^7\\equiv1\\pmod{49}$. Since $7|273$, so $8^{273}\\equiv1\\pmod{49}$. Observe also that $6^7\\equiv-1\\pmod{49}$. Since $7|273$, so $6^{273}\\equiv-1\\pmod{49}$. Adding,$8^{273}+ 6^{273}\\equiv0\\pmod{49}$.", "\\textcolor{g r e e n}{23} - 5 \\cdot \\textcolor{g r e e n}{8} = - 17$. Thus, $2584$ is divisible by $17$ • A number is divisible by $25$ if it ends with $00 , 25 , 50$ or $75$.", "have to count check as many numbers to see if they're divisible by $3$3 as well. When we have multiple numbers it's always easiest to think about the biggest number first. So, starting from $11$11, let's see which multiples of $11$11 are also divisible by $3$3 Multiples of $11$11 Is it divisible by $3$3? Numbers that have both $3$3 and $11$11 as factors $11$11 no $22$22 no $33$33 yes $33$33 $44$44 no $55$55 no $66$66 yes $66$66 $77$77 no $88$88 no $99$99 yes $99$99 So now we have found three numbers that have factors of both $11$11 and $3$3 They are $33$33, $66$66 and $99$99 #### Example ##### Question 1 What are the factors of $42$42? Separate the factors with commas. ##### Question 2 What factor of $30$30, other than $1$1, is also a factor of $27$27 and $21$21? ##### Question 3 What numbers from $1$1 to $100$100 have factors of $3$3, $5$5 and $10$10? 1. Write each number on the same line, separated by commas.", "= 55k$, or $9c - 9a = 5k$. Thus, k is divisible by $9$. Because $55 * 18 = 990$, $k$ must be $9$, and therefore $c - a = 5$. Because $a + b + c \\leq{7}$ and $a \\geq{1}$, $a = 1$, $c = 6$ and $b = 0$, and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$, or $\\boxed{D}$.", "of Prime numbers: $2$$2$, $3$$3$, $5$$5$, $7$$7$, $13$$13$, $17$$17$, $\\cdots$$\\cdots$ Some numbers are divisible by at-least one number between $2$$2$ and one less than the number. eg: $6$$6$ is divisible by $2$$2$ and $3$$3$. And $63$$63$ is divisible by $3$$3$, $7$$7$, $9$$9$, and $21$$21$. Such numbers are called \"composite numbers\". They can be equivalently given as a product of other numbers. eg: $6=2×3$$6 = 2 \\times 3$ and $63=3×21$$63 = 3 \\times 21$ or $63=7×9$$63 = 7 \\times 9$. The word \"composite\" means \"made up of several parts or elements\". The word \"prime\" means \"having distinct or important properties\". Composite Numbers : Numbers that are divisible by at-least a number other than $1$$1$ and the number itself. Prime Numbers : Numbers that are divisible by only $1$$1$ and the number itself. examples We need to find if $59$$59$ is a prime or composite number. To do that, • Check for divisibility by numbers from $2$$2$ to $58$$58$ OR • Check for divisibility by numbers from $2$$2$ to $7$$7$ By the definition of prime numbers, the number can be checked for divisibility by numbers from $2$$2$ to $58$$58$. But, it is sufficient to check for divisibility by numbers from $2$$2$ to $7$$7$. The number $7$$7$ is chosen because the given number $59$$59$ lies between the perfect squares $7×7=49$$7 \\times", "this range, since the even numbers are obviously divisible by 2 and therefore not prime. 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 Numbers divisible by 3: 21, 27, 33, 39 Numbers divisible by 5: 25, 35 The remaining numbers are 23, 29, 31, and 37. A quick check will tell us that none of these numbers are divisible by anything other than themselves and 1. Therefore, 4 of the numbers between 20 and 40 are prime (Choice B). Question 5 ### If $\\, a = −1$ and $\\, b = 4$, what is the value of $\\, a^3b +3b$? A $7$ B $16$ C $9$ D $8$ E $6$ Question 5 Explanation: The correct answer is (D). Substitute the values for $a$ and $b$ into the expression: $a^3b + 3b$ $(-1)^3(4) + 3(4)$ $(-1)(4) + 12$ $-4 + 12$ $= 8$ Question 6 ### What is the least common multiple (the smallest integer that is divisible by all three numbers) of 3, 4, and 8? A $16$ B $12$ C $32$ D $96$ E $24$ Question 6 Explanation: The correct answer is (E). The simplest approach is to make a list of the multiples of each of the three numbers, and then find the least multiple that appears in all three lists: Multiples of 3:", "\\cdot 3 = 28$$ to our total. $28$ is congruent to $1 \\pmod 9$. Thus our sum is congruent to $3 \\pmod 9$, and $k = 1 \\implies \\boxed{B}$. ## Solution 2 As our first trial with 3, We can say that $N\\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\\pmod{3}$. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that $N\\equiv 19+20+21+22+...+91+92\\pmod{3}$, and adding that up using the rainbow strategy, we get $N\\equiv (111\\times37)\\pmod{3}$. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is $\\boxed{B}$ ~mathpro12345", "is 6, so you use blocks of 3. Note that $10^3+1=1001 \\equiv 0 \\pmod {13}$ - Your professor is using the fact that $100000001=10^8+1$ is divisible by $17$. Given for example your $80$-digit number, you can subtract $98765432\\cdot 100000001=9876543298765432$, which will leave zeros in the last $16$ places. Slash the zeros, and repeat. After $5$ times you are left with the number $0$, which is divisible by $17$, and hence your $80$-digit number must also be divisible by $17$. When checking for divisibility by $17$, you can also subtract multiples of $102=6\\cdot 17$ in the same way. For divisibility by $7$, $11$, or $13$, you can subtract any multiple of the number $1001=7\\cdot 11\\cdot 13$ without affecting divisibility by these three numbers. For example, $6017-6\\cdot 1001=11$, so $6017$ is divisible by $11$, but not by $7$ or $13$. For divisibility by $19$, you can use the number $1000000001=10^9+1=7\\cdot 11\\cdot 13\\cdot 19\\cdot 52579$. By subtracting multiples of this number, you will be left with a number of at most $9$ digits, which you can test for divisibility by $19$ by performing the division. - The whole process can be explained by writing A^x/B = C where C can be +1 or -1 or 0. Here C is the remainder got on division of A^x by B. Now we operate", "be divisible by $4$. These digits are ($1$ or $3$) and ($4$ or $6$) Therefore the possible numbers are $14$, $16$, $34$, and $36$ Since only $16$ and $36$ are divisible by $4$, the fourth digit of the number must be $6$. By process of elimination the last number must therefore be $4$. So we are left with ($1$ or $3$),$2$,($1$ or $3$),$6$,$5$,$4$ This gives the two numbers $123654$ and $321654$, both of which are correct, as shown above. Hannah, Megan, Jess, Toby, Callum, Callum, Lydia, Nadia, Sam and Chris from Peterchurch Primary School also used very similar reasoning to start with, but used a different method again towards the end of their solution: $123654$ and $321654$ We worked this out by the following process: 1. We knew that the fifth digit must be a $5$ because that's the only number divisible by $5$ (because we had no zero) 2. We knew that the second, fourth and sixth digits had to be even to be divisible by $2$, $4$ and $6$. 3. The first and third digits had to odd - $1$ or $3$ 4. We worked out that the second digit must be $2$ because 143 or 163 would not be divisible by $3$ 5. This meant that we only had four combinations left: $123456$, $123654$, $321456$,", "Divisible Probability... The sum of the digits of a $$9$$ digit number is $$77$$. The probability that this number is divisible by 11 can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are co-prime positive integers. Find the value of $$a+b$$. ×", "$2+0+1+9$ is divisible by $3$, so is $2019$. I got that $2019=3\\cdot 673$. However, on an exam, I need to motivate how I know that $673$ is a prime, no calculators allowed. How Do I now find the coprimes? – Parseval Feb 5 '19 at 15:45 • Well it suffices to check that it is not divisible by the primes less than $\\sqrt{673}<26$, which is a bit of work but not hard. Then a number is coprime to $2019$ if it is not divisible by $3$ and not divisible by $673$. I leave it to you to count how many such numbers there are. – Servaes Feb 5 '19 at 15:54 For the second question, your reasoning is correct. Indeed the inverse of $$32$$ in $$\\Bbb{Z}/2019\\Bbb{Z}$$ is $$-694$$, though perhaps a representative in the interval $$[0,2018]$$ is preferred. For the first question, it helps to first know the prime factors of $$2019$$. It is not hard to check that $$2019=3\\times673$$ and that $$3$$ and $$673$$ are prime. Then an integer in the interval $$[0,2018]$$ is coprime to $$2019$$ if and only if it is not divisible by $$3$$ and not divisible by $$673$$. I leave it to you (for now) to count how many such integers there are. • Great! So to go from $-694$ to somewhere positive,", "$4181$ is not divisible by $2, 3, 5, 7$, or $19$. • $4181 = 4400 - 219$, where $4400$ has prime factors $2, 5, 11$ and $219$ has prime factors $3, 73$. It follows that $4181$ is not divisible by $11$ or $73$. • $4181 = 3900 + 260 + 21 = 13(320) + 21$, where $13(320)$ has prime factors $2, 5, 13$ and $21$ has prime factors $3, 7$. It follows that $4181$ is not divisible by $13$. • $4181 = 3400 + 680 + 101 = 17(240) + 101$, where $17(240)$ has prime factors $2, 3, 5, 17$ and $101$ is prime. It follows that $4181$ is not divisible by $17$ or $101$. • $4181 = 3800 + 380 + 1 = 19(220) + 1$, where $19(220)$ has prime factors $2, 5, 11, 19$. It follows that $4181$ is not divisible by $19$. • $4181 = 4600 - 460 + 41 = 23(180) + 41$, where $23(180)$ has prime factors $2, 3, 5, 23$ and $41$ is prime. It follows that $4181$ is not divisible by $23$ or $41$. • $4181 = 2900 + 1160 + 121 = 29(140) + 121$, where $29(140)$ has prime factors $2, 5, 7, 29$ and $121 = 11^2$. It follows that $4181$ is not divisible by $29$. At this point" ]

[ "# Questions Archives ## Past Questions PSLE P4 & P5 SMOPS NMOS IQ 21/6/12 5/6/12 8/5/12 4/5/12 30/4/12 29/4/12 25/4/12 20/4/12 15/4/12 8/4/12 25/3/12 22/4/12 15/3/12 20/4/12 16/2/12 16/4/12 15/2/12 8/4/12 14/2/12 3/4/12 30/4/12 30/4/12 13/2/12 30/3/12 12/2/12 28/3/12 9/2/12 21/2/12 2/2/12 20/2/12 27/1/12 19/2/12 26/1/12 18/2/12 20/1/12 17/2/12 15/1/12 9/2/12 12/1/12 8/2/12 5/1/12 7/2/12 16/8/11 16/8/11 1/8/11 1/8/11 11/7/11 11/7/11 11/7/11 13/6/11 26/5/11 26/5/11 26/5/11 16/5/11 16/5/11 16/5/11 3/5/11 3/5/11 3/5/11 3/5/11 20/4/11 20/4/11 20/4/11 20/4/11 13/4/11 13/4/11 13/4/11 5/4/11 5/4/11 5/4/11 30/3/11 24/3/11 24/3/11 24/3/11 ### 10 Responses to “Questions Archives” 1. susanti April 8, 2011 at 4:18 pm # I WANT TO LEARN!!!!!! • Anne April 9, 2011 at 10:36 am # some marbeles are share among 3boys.Aaron’s share 3/4 of Brian’s share and Brian’s share is 4/5 of Charle’s share. If Aaron and Brian have a toatl 56 marbles altogether.How many marbles do the 3 boys have in all? 2. matharena April 9, 2011 at 11:54 am # A : B : C 3 : 4 : 5 A+B —> 7 u —-> 56 1 u —-> 8 Altogether: 12 u —> 12 x 8 = 96 marbles 3. claire tan September 19, 2011 at 8:00 pm # heyy auntie persis, im claire, i used to b in yr math class for P6? im", "26 The distance between two posts is 30 meters then total distance between posts is 26 x 30 = 780 meters The width of each post is 10 centimeters and then total posts width is 27 x 10 =270 centimeters = 2 meters and 70 centimeters Question 4. Kirsten has 64 coins in her piggy bank. She has$9.25 in dimes and quarters. How many dimes and how many quarters does she have? Answer: 45 dimes and 19 quarters Explanation: Let there be x dimes and 64-x quarters. 0.1x+0.25(64-x) = 9.25 0.1x+16-0.25x=9.25 -0.15x=-6.75 x= 45 64-x=64-45=19 so 45 dimes and 19 quarters Problem Solving Solve. Use any strategy. Question 1. Darcy, Jason, and Maria share $268. Jason has$20 more than Darcy and Maria has twice as much money as Jason. How much money do Darcy and Jason have altogether? Answer: Darcy and Jason have altogether=$124 Explanation: Let Darcy has money=x Jason has$20 more than Darcy Jason =20+x Maria has twice as much money as Jason Maria = 2(20+x)=40+2x Darcy,Jason and Maria share $268 x+20+x+40+2x=268 4x+60=268 4x=208 x=52 So Darcy has money=$52 Jason has money=20+x=20+52=$72 Darcy and Jason have altogether=$(52+72)=\\$124 Question 2. Juan and Rachel have the same number of marbles. Rachel gives away 10 marbles and Juan gives away 22 marbles. Rachel then has 3 times as many marbles", "had then to Diana. Finally, Diana gave 27of whatever she had to Isabelle. As a result, they each had the same number of marbles. How many marbles did Isabelle have at first? 5 m PSLE Wesley had 11 jars of balls. At first, each of the jars contained the same number of balls. He took out 24 marbles from each jar. After that, the total number of balls left in the 11 jars was equal to the total number of balls in 3 of the jars at first. What was the number of balls in each jar at first? 3 m Violet had 6 times as many erasers as Jane at first. After Violet gave 40 erasers to Jane, they each had the same number of erasers in the end. How many erasers did Violet have at first? 3 m A jar contains $2 and$5 notes with a total value of $518. 4 pieces of$5 were exchanged for $2 notes of similar value. In the end, the number of$2 notes was the same as the $5 notes. How many$5 notes were there at first? 3 m Sean and Cathy had some salary each. If Cathy gives Sean $49, Sean would have 3 times as much salary as Cathy. If Sean gives Cathy$20, he will have the same amount", "for Vinny. After Round 1, Johnny gave half his marbles to Vinny. So Johnny has V marbles, and Vinny has 2V marbles. After Round 2, Vinny gives 3/4s of his marbles to Johnny, so Vinny has 0.5V, and Johnny has V + 1.5V = 2.5V marbles. We are told Johnny now has 30 marbles, so 2.5V = 30, and V = 12. And J = 24. So, Johnny started and ended with 24 marbles, and Vinny with 12. • No, you answered faster than me! – Culver Kwan Apr 9 at 1:52 • @CulverKwan - Sorry. I know how that feels. I've been on your end of things many times! – Lanny Strack Apr 9 at 1:56 Let $$x$$ be the number of marbles Vinny started with. Then Johnny has $$2x$$ marbles in the beginning. After the first round, Johnny lost $$\\frac{2x}2$$ marbles and he has $$2x-x=x$$ marbles. So Vinny got $$x$$ marbles, which is $$2x$$ marbles. After the second round, Vinny lost $$2x\\cdot\\frac34=\\frac32x$$ marbles and he has $$2x-\\frac32x=\\frac12x$$ marbles left. Then Johnny has $$x+\\frac32x=\\frac52x$$ marbles. But the problem said that he has $$30$$ marbles. So $$\\frac52x=30, x=12$$. So Johnny has $$2(12)=24$$ marbles and Vinny has $$12$$ marbles. • Could someone help me mark spoiler properly? I can't do it. – Culver Kwan Apr 9 at 2:05 •", "The ratio of Liam's to Mark's money is 7 : 9. The ratio of Mark's to Ian's money is 8 : 1. If Liam has $112, how much less does Ian have than Mark? 2 m Level 3 Wendy and Anna have$294. Violet and Wendy have $493. The ratio of Anna to Violet is 1 : 2. How much does Wendy have? 3 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 There were some grapefruits and durians in Container A and Container B. In Container A, the ratio of the grapefruits to the number of durians was 3 : 1. In Container B, the ratio of the number of grapefruits to the number of durians was 3 : 2. There were 2 times as many fruits in Container A as in Container B. After 56 durians were put into Container B, the", "# Question 1 of 8 Three boys, Alex, Ben and Calvin have some marbles. The number of marbles Alex has is $\\frac{1}{4}$ the total number of marbles Ben and Calvin have. The number of marbles Ben has is $\\frac{1}{5}$ the total number of marbles Alex and Calvin have. If Calvin has 152 marbles, find the number of marbles Alex and Ben have respectively. A Alex: 200 Ben: 180 B Alex: 220 Ben: 190 C Alex: 240 Ben: 200 D Alex: 260 Ben: 210 E None of the above", "# 20 coins on the table Two boys , Brian and Gabe, decide to play a game. They lay 20 coins on the table. Each player took turns taking 1, 2, or 3 coins. Whoever took the last coin won. Brian thought if he went first, he'd win for sure since he had the advantage of making the first move. So Brian did go first. But Gabe won every time! Obviously Gabe had a winning strategy, but what was it? Could Brian have outsmarted it? • Great answer! Player 1 can win in any other case situation because he can just make the coins a multiple of 4. Remember that any amount of coins have the form $4n,4n+1,4n+2,4n+3$ – durum Oct 6 '14 at 10:27", "is a square number. We also know that he has the most marbles, so his number must be $>94$. The first square number that fits is $100$, so let's try that out. If Peter Piper has 100 marbles, and the amount of silver marbles is the square root of the total then we get the nice easy sum of $\\sqrt100$ which is 10. So Peter has the least amount of silver marbles which is 10. Okay, but what about the others? Our Mary Poppins has the most silver marbles. So the amount of silver marbles, must be $>10$, which is the least. Because there are three people, and she has the most, not the joint most, then she has to have an amount of silver marbles $≥12$, so there is at least one number in between her number and Peter's number for Paul the Dickie Bird to have which isn't equal to the most or least. Mary and her little lamb can therefore have a minimum of $12*7$ marbles which is $84$, as a seventh of her marbles are silver. I see, but what about Paul? We already know that Paul has 11 silver marbles if the most is 12 and the least is 10, but how many does he have total? Well Peter the Pumpkin Eater and", "How Cheenta works to ensure student success? Explore the Back-Story # Math Kangaroo (Benjamin) 2014 Problem No 24 Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180 marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180", "than John. Which is the best estimate of the amount of money that John′s sister had saved? a.$110 b. $50 c.$80 d. \\$90 Correct answer : (1) 7. Mathew′s collection has 220 red marbles, 175 green marbles, and 115 yellow marbles. Which is the best estimate of the total number of marbles in his collection? a. 700 b. 600 c. 500 d. 400 Correct answer : (3) 8. Ashley has 156 candies. She gave 79 to her brother. About how many candies does she have? a. 100 b. 70 c. 80 d. 90 Correct answer : (3) 9. John scored 423 points and his brother scored 326 points. About how many more points did John score? a. 150 b. 80 c. 100 d. 90 Correct answer : (4) 10. There are 477 quarts of water in a large water tank. Matt used 343 quarts of water. Estimate the quantity of water left in the tank. Round to the nearest ten. a. 820 quarts b. 900 quarts c. 140 quarts d. 100 quarts", "possible that another player has all the rest, in which case you have not won: that player has at least as many as you have. If you have $11$, however, there are only $9$ other marbles, so even if one player has all of them, you win. If you have $11$ marbles, you are certain to win, so your probability of winning is $1$. - Hi thanks for your input. I just edited the question – black sensei Jan 7 '13 at 22:57 (1) You are correct: The minimum number of marbles to be ASSURED of the victory = $11$ marbles. • Then the total number of marbles that the other three players can have is $20 - 11 = 9$. No matter how to split up $9$ among three players, even if one player has all the other 9 marbles, $11 > 9$, ($11 > 20/10 = 10$). So a player with 11 marbles is guaranteed to win... (2) So the probability that the player with $11$ marbles who is ASSURED of victory will win = $1$ • After all, the rules of the game are \"he who manage to get the absolute maximum of marbles is the winner\" (that means WIll certainly win, not might win.) - Hi thanks for your comments , i just edited", "Pizza Portions My friends and I love pizza. Can you help us share these pizzas equally? Forgot the Numbers On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they? Pies Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie? Andy's Marbles Stage: 2 Challenge Level: Andy and his friend Sam were walking along the road together. Andy had a big bag of marbles. Unfortunately the bottom of the bag split and all the marbles spilled out. Poor Andy! One third ($\\frac{1}{3}$) of the marbles rolled down the slope too quickly for Andy to pick them up. One sixth ($\\frac{1}{6}$) of all the marbles disappeared into the rain-water drain. Andy and Sam picked up all they could but half ($\\frac{1}{2}$) of the marbles that remained nearby were picked up by other children who ran off with them. Andy counted all the marbles he and Sam had rescued. He gave one third ($\\frac{1}{3}$) of these to Sam for helping him pick them up. Andy put his remaining marbles into his pocket. There were $14$ of them. How many marbles were there in Andy's bag before the bottom split? What", "Level 2 Violet had 6 times as many erasers as Jane at first. After Violet gave 40 erasers to Jane, they each had the same number of erasers in the end. How many erasers did Violet have at first? 3 m Level 3 A jar contained $2 and$5 notes with a total value of $518. 4 pieces of$5 were exchanged for $2 notes of similar value. In the end, the number of$2 notes was the same as the $5 notes. How many$5 notes were there at first? 3 m Level 3 Sean and Cathy had some salary each. If Cathy gives Sean $49, Sean would have 3 times as much salary as Cathy. If Sean gives Cathy$20, he will have the same amount of salary as her. How much salary was each man's salary respectively? 1. Sean? 2. Cathy? 3 m Level 2 The ratio of Tim's money to Carl's money is 3 : 5. If Carl has $845, how much must he give to Tim so both will have the equal amount of money? 2 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles", "marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "# Two naughty boys... Level pending Brenda has an infinite number of coins. Unfortunately, her 2 mischievous younger brothers, Brian and Billy, stole an increasing number of coins from her. The first day, Brian stole 1 coin from her, the second day, 2 , the third day, 3 and so on. Billy stole 1 coin on the second day, 2 coins on the fourth day, 3 coins on the sixth day and so on. Up to the nth day, Brenda lost $$\\frac{an^2 + bn}{c}$$ coins (this expression is in its simplest form). a+b+c= ? Assumptions: n is an even number. ×", "could earn on one of the other two tests? $$48$$ $$52$$ $$66$$ $$70$$ $$74$$ ###### Solution(s): To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a $$100$$ on her fourth test. The sum of the $$4$$ tests is then $76 + 94 + 87 + 100 = 357.$ For an average of $$81,$$ Shauna's test scores must add to $$5 \\cdot 81 = 405.$$ This means that she needs to get a $$405 - 357 = 48$$ on her last test. Thus, the correct answer is A. 8. Gilda has a bag of marbles. She gives $$20\\%$$ of them to her friend Pedro. Then Gilda gives $$10\\%$$ of what is left to another friend, Ebony. Finally, Gilda gives $$25\\%$$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $$20$$ $$33\\dfrac{1}{3}$$ $$38$$ $$45$$ $$54$$ ###### Solution(s): Assume that Gilda starts off with $$100$$ marbles. After giving $$20$$ marbles to Pedro, she has $$100 - 20 = 80$$ marbles left. She then gives $$80 \\times .1 = 8$$ marbles to Ebony, with her having $$80 - 8 = 72$$ marbles left. Finally, Gilda gives $$72 \\times .25 = 18$$ marbles to Jimmy. She has a", "# Marble Pickeroo! Peter has $$10$$ distinct marbles. He chooses $$5$$ marbles, then chooses $$3$$ of the $$5$$ marbles to give to Paul. How many combination of marbles can Paul receive? × Problem Loading... Note Loading... Set Loading...", "dollars? [size=150]Hi Reinout-g. Thanks for the puzzle[/size] Once again I will state from the outset that I am bound to be wrong BUT I would put just 1 white marble in one urn and the rest of the marbles in the other urn. So my chance of winning the money will be 1/2 * 1 + 1/2 * 49/99 = 1/2 + 49/198 = 148/198 = 74/99 That sounds like pretty good odds to me. Now you can tell me why i am wrong Haha, no actually you are right. Thanks for the detailed answer. I'm trying to find the level where you still need to work your brain a little but won't find it impossible to solve the puzzle. I'll post another one in a couple of minutes Mar 2, 2014 #12 +2353 +11 And here is the new puzzle! If you have trouble solving it, feel free to give the easier one a try. Easier one: There are 31,542 people attending a concert. One person is picked from the audience and if his birthday is a monday he will win $1.000.000 What are the odds that that persons birthday is a monday? Hard one (I'm curious to see whether anyone can find a different strategy than I found.) : Steve again wants to play a game", "so Hap has $3 \\times 20$, or 60 more marbles than Gus. Feb 23, 2012 Nov 05, 2014", "Three children, Alice, Brian, and Chris have a total of [#permalink] ### Show Tags 27 Mar 2012, 22:11 I have to disagree. In the case where each have 40c ie you have a set (40, 40, 40) they each individually have the most (ie the highest value = 40) and it so happens they each individually have the least, again 40. Since this provides two cases, Brian having the most when they all share the most (similar to tied for 1st place - they are equally best) and Brian not having the most when any other values are chosen, one requires both (1) and (2) to determine if Chris does/doesn't have the most. Clearly this is a definition debate around \"most\" and ties for most, and the question would likely (hopefully) be thrown out by the gmac folks! Math Expert Joined: 02 Sep 2009 Posts: 34475 Followers: 6288 Kudos [?]: 79775 [1] , given: 10022 Re: Three children, Alice, Brian, and Chris have a total of [#permalink] ### Show Tags 28 Mar 2012, 00:59 1 KUDOS Expert's post bschool83 wrote: Three children, Alice, Brian, and Chris have a total of \\$1.20 between them. Does Chris have the most money? (1) Alice has 35 cents. (2) Chris has 40 cents. This is a poor quality question because of its" ]

[ "have together? Let the number of marbles per friend be $$x_1, ~x_2, ~x_3$$ and $$x_4$$. \\begin{align*} \\frac{x_1 + x_2 + x_3 + x_4}{4} & = 10\\\\ x_1 + x_2 + x_3 + x_4 & = 40 \\end{align*} One friend leaves: \\begin{align*} x_1 + x_2 + x_3 & = 40 - 4 \\\\ x_1 + x_2 + x_3 & = 36 \\end{align*} Therefore the remaining friends have $$\\text{36}$$ marbles.", "could earn on one of the other two tests? $$48$$ $$52$$ $$66$$ $$70$$ $$74$$ ###### Solution(s): To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a $$100$$ on her fourth test. The sum of the $$4$$ tests is then $76 + 94 + 87 + 100 = 357.$ For an average of $$81,$$ Shauna's test scores must add to $$5 \\cdot 81 = 405.$$ This means that she needs to get a $$405 - 357 = 48$$ on her last test. Thus, the correct answer is A. 8. Gilda has a bag of marbles. She gives $$20\\%$$ of them to her friend Pedro. Then Gilda gives $$10\\%$$ of what is left to another friend, Ebony. Finally, Gilda gives $$25\\%$$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $$20$$ $$33\\dfrac{1}{3}$$ $$38$$ $$45$$ $$54$$ ###### Solution(s): Assume that Gilda starts off with $$100$$ marbles. After giving $$20$$ marbles to Pedro, she has $$100 - 20 = 80$$ marbles left. She then gives $$80 \\times .1 = 8$$ marbles to Ebony, with her having $$80 - 8 = 72$$ marbles left. Finally, Gilda gives $$72 \\times .25 = 18$$ marbles to Jimmy. She has a", "+ 2B + 3C = 92 8 + 2A + 2B + 3C = 92 Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her? Thanks Current Student Joined: 27 Jun 2012 Posts: 418 Concentration: Strategy, Finance Followers: 67 Kudos [?]: 591 [0], given: 183 Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] 15 Jan 2013, 00:54 Consider Bill has x marbles. Bill has 8 more than Allen, i.e. Allen has (x-8) marbles. Charles has 2 times more than Bill, i.e. Charles has 2x marbles. Daryl has 3 times more than Charles, i.e. Daryl has 3*2x = 6x marbles. All 4 kids have total 92 marbles. Hence $$x + (x-8) + 2x + 6x = 92$$ i.e. $$10x = 100$$ i.e. $$x = 10$$ Therefore, Charles has 2x=20 marbles. Calculating the rest, Allen=2, Bill=10, Charles=20, Daryl=60, Total=92 _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Moderator Joined: 02 Jul 2012 Posts: 1230 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 100", "solution. Example 1 Find the number of marbles in each bag. Write the equation represented by the model. Verify your answer using substitution. Find the answer... Write the equation and check the answer... Evaluate the equation for \\$b = 5\\$. \\$\\matrix{b+12&=&3b+2\\\\5+12&=&(3)(5)+2\\\\17&=&17\\\\&\\large\\surd~&}\\$ Example 2 Find the number of marbles in each bag. Write the equation represented by the model. Verify your answer using substitution. Find the answer... Write the equation and check the answer... Evaluate the equation for \\$b = 5\\$. \\$\\matrix{3b+2&=&2b+7\\\\(3)(5)+2&=&(2)(5)+7\\\\15+2&=&10+7\\\\17&=&17\\\\ &\\large\\surd~&}\\$ # Self-Check Question 1 Maria claims that the solution to this puzzle is each bag contains 3 marbles. Is she correct? [show answer] \\$b=3\\$ \\$2b+5=3b+2\\$ \\$(2)(3)+5=(3)(3)+2\\$ \\$6+5=9+2\\$ \\$11=11\\$ Maria is correct! Question 2 Martin says that this puzzle can be represented algebraically as \\$\\$2x+9=4x+1\\$\\$ Is he correct? [show answer] Yes he is correct. (By the way, the solution is \\$b = 4\\$) Question 3 Find the number of marbles in each bag. Write the algebraic equation that this puzzle models. [show answer] \\$b=3\\$ \\$3b+8=5b+2\\$", "solution. Example 1 Find the number of marbles in each bag. Write the equation represented by the model. Verify your answer using substitution. Find the answer... Write the equation and check the answer... [show answer] Evaluate the equation for $b = 5$. $\\matrix{b+12&=&3b+2\\\\5+12&=&(3)(5)+2\\\\17&=&17\\\\&\\large\\surd~&}$ Example 2 Find the number of marbles in each bag. Write the equation represented by the model. Verify your answer using substitution. Find the answer... Write the equation and check the answer... [show answer] Evaluate the equation for $b = 5$. $\\matrix{3b+2&=&2b+7\\$$3)(5)+2&=&(2)(5)+7\\\\15+2&=&10+7\\\\17&=&17\\\\ &\\large\\surd~&}$ # Self-Check Question 1 Maria claims that the solution to this puzzle is each bag contains 3 marbles. Is she correct? [show answer] $b=3$ $2b+5=3b+2$ $(2)(3)+5=(3)(3)+2$ $6+5=9+2$ $11=11$ Maria is correct! Question 2 Martin says that this puzzle can be represented algebraically as $$$2x+9=4x+1\\($ Is he correct? [show answer] Yes he is correct. (By the way, the solution is $b = 4$) Question 3 Find the number of marbles in each bag. Write the algebraic equation that this puzzle models. [show answer] $b=3$ $3b+8=5b+2$", "marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "How Cheenta works to ensure student success? Explore the Back-Story # Math Kangaroo (Benjamin) 2014 Problem No 24 Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180 marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180", "ls$45 - x$. After they gave away 5 marbles each to their sisters, now, Ram has$x - 5$marbles and Radha has$45 - x - 5 = 40 - x$marbles. According to the given information,$(x - 5)(40 - x) = 124$. We can solve this equation after we first write it in standard form: \\begin{eqnarray} (x - 5)(40 - x) &=& 124\\\\ 40x - x^2 - 200 + 5x &=& 124\\\\ - x^2 + 45x - 200 &=& 124\\\\ -x^2 + 45x - 324 &=& 0\\\\ x^2 - 45x + 320 &=& 0\\\\ (x - 36)(x - 9) &=& 0\\\\ x - 36 &=&= 0 \\implies x = 36\\\\ x - 9 &=& 0 \\implies x = 9\\\\ \\end{eqnarray} Answer: One of them had 36 marbles and the other had 9 marbles originally. Example 2: Product of two consecutive even integers is 840. Find the two integers. Solution: We note that the difference between two even or two odd consecutive integers is 2. So, if one even integer is$x$, then the next consecutive even integer is$x + 2$. Given that the product of these two is 840. So, we have the equation$x(x + 2) = 840$that models this situation. Rearranging the terms in standard form and solving: \\begin{eqnarray} x(x + 2)&=& 840\\\\ x^2 + 2x &=& 840\\\\ x^2 +", "the amount. Here is our proportion to solve. $\\frac{6}{40}=\\frac{p}{100}$ Now we can use cross products and solve. $40p & = 600\\\\p & = 15$ Jeremy has 25 marbles and 12 of them are cat’s eye marbles. What percent of his marbles are cat’s eye marbles? We can think of this problem as “What percent of 25 is 12?” 12 is the amount $(a)$ and 25 is the base $(b)$ . We need to find the percent $(p)$ . $\\frac{a}{b}& =\\frac{p}{100}\\\\\\frac{12}{25}& =\\frac{p}{100}\\\\25p& =1,200\\\\\\frac{25p}{25}& =\\frac{1,200}{25}\\\\p& =48$ Since $p = 48, \\frac{48}{100}$ , which is 48%. The answer is that 48% of Jeremy’s marbles are cat’s eye marbles. Use a proportion to find each percent. Example A What percent of 20 is 2? Solution: $10\\%$ Example B What percent of 30 is 6? Solution: $20\\%$ Example C What percent of 45 is 15? Solution: $33.3\\%$ Here is the original problem once again. Reread it and use what you have learned about percents to figure out the questions at the end of the problem. Taylor’s younger brother Max decided to visit her at the candy store. Max is only seven and can be a handful sometimes, so while Taylor loves to see him, she was a little hesitant to have him in the shop. Plus, what seven year old doesn’t love candy.", "\\text{ units} &= 16-4\\\\[2ex] &= 12\\\\[2ex] 1 \\text{ unit} &= 12 \\div 2\\\\[2ex] & = 6 \\end{align} Number of marbles Ryan had at first $$=$$ Number of marbles Wilson had at first Number of marbles Wilson had at first\\begin{align}\\\\[6ex] &= 1 \\text{ unit} +16\\\\[2ex] &= 6+16\\\\[2ex] &= 22 \\end{align} Ryan had 22 marbles at first. 22 marbles ## 2. Equal Stage In The End Both subjects have a different number of items at first. After some items are added/removed from each subject, they have the same number of items in the end. Question 1: Mrs Lim had four times as many biscuits as Mrs Goh. After Mrs Lim gave away 17 biscuits and Mrs Goh received another 4 biscuits, they had an equal number of biscuits. How many biscuits did Mrs Goh have in the end? Solution: Difference in units\\begin{align}\\\\[2ex] &= 4 \\text{ units} - 1 \\text{ unit}\\\\[2ex] &= 3 \\text{ units} \\end{align} \\begin{align}​​ 3 \\text{ units} &= 4+17\\\\[2ex] &= 21\\\\[2ex] 1 \\text{ unit} &= 21 \\div 3\\\\[2ex] & = 7 \\end{align} Number of biscuits Mrs Goh had in the end\\begin{align}\\\\[6ex] &= 1 \\text{ unit} +4\\\\[2ex] &= 7+4\\\\[2ex] &= 11 \\end{align} Mrs Goh had 11 biscuits in the end. 11 biscuits Question 2: Alvin had four times as many marbles as Wei Ming. After Alvin lost 24 marbles", "cups and saucers\\begin{align}\\\\[2ex] &= 3 + 7\\\\[2ex] &= 10 \\end{align} \\begin{align} \\text{Cups} &: \\text{Total number of cups and saucers}\\\\[2ex] 3 &: 10 \\end{align} $$3 : 10$$ Question 2: Solve the following, $$24 : 18 = 4 :$$__________ Solution: 3 Question 3: Solve the following, __________ $$: 15 = 35 : 21$$ Solution: Step 1: Find the simplest form of $$35 : 21$$. $$35 : 21 = 5 : 3$$ Step 2: Find the equivalent ratio of $$5 : 3$$. We get $$25 : 15 = 5 : 3$$ $$25$$ Question 4: Chris has $$\\displaystyle{\\frac{2}{5}}$$ as many marbles as John. 1. The ratio of the number of marbles Chris had to the number of marbles John had is __________ : __________. 2. The ratio of the number of marbles John had to the number of marbles Chris had is __________ : __________. 3. The ratio of the number of marbles Chris had to the total number of marbles the two boys had is __________ : __________. 4. The number of marbles John had is __________ of the number of marbles Chris had. 5. The number of marbles John had is __________ of the total number of marbles the two boys had. Solution: Number of marbles Chris had $$\\displaystyle{=\\frac{2}{5}}$$ number of marbles John had 1. The ratio of the number", "Kudos [?]: 1069 [0], given: 116 Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] 15 Jan 2013, 03:19 tara15225 wrote: 4 kids have a total of 92 marbles Bill has 8 more than Allen Charles has 2 times more than Bill Daryl has 3 times more than Charles How many marbles does Charles have? My daughter has gotten this far: Bill = 8 + A Allen = A Charles = 2B Daryl = 3C 8+A + A + 2B + 3C = 92 8 + 2A + 2B + 3C = 92 Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her? Thanks 2B = 2*(8 + A) = 16 + 2A 3C = 3*2B = 6B = 6*(8 + A) = 48 + 6A Picking up from where you left 8 + 2A + 2B + 3C = 92 8 + 2A + 16 + 2A + 48 + 6A = 92 10A + 72 = 92 10A = 92 - 72 = 20 A = 2 B = 2 + 8 = 10 C = 2*10 = 20 D = 3*20 = 60 _________________ Did you find this post helpful?... Please let me know through", "Difference between revisions of \"2019 AMC 8 Problems/Problem 8\" Problem 8 Gilda has a bag of marbles. She gives $20\\%$ of them to her friend Pedro. Then Gilda gives $10\\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $\\textbf{(A) }20\\qquad\\textbf{(B) }33\\frac{1}{3}\\qquad\\textbf{(C) }38\\qquad\\textbf{(D) }45\\qquad\\textbf{(E) }54$ Solution 1 After Gilda gives $20$% of the marbles to Pedro, she has $80$% of the marbles left. If she then gives $10$% of what's left to Ebony, she has $(0.8*0.9)$ = $72$% of what she had at the beginning. Finally, she gives $25$% of what's left to her brother, so she has $(0.75*0.72)$ $\\boxed{\\textbf{(E)}\\ 54}$. of what she had in the beginning left. Solution 2 Suppose Gilda has 100 marbles. Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles. Out of 80 marbles she gives 10% of 80 = 8 to Ebony. Thus she remains with 72 marbles. Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54. And $\\frac{54}{100}$=54%=$\\boxed{\\textbf{(E)}\\ 54}$ ~phoenixfire Solution 3 (Only if you have lots of time do it this way) Since she gave away 20% and 10% of", "for Vinny. After Round 1, Johnny gave half his marbles to Vinny. So Johnny has V marbles, and Vinny has 2V marbles. After Round 2, Vinny gives 3/4s of his marbles to Johnny, so Vinny has 0.5V, and Johnny has V + 1.5V = 2.5V marbles. We are told Johnny now has 30 marbles, so 2.5V = 30, and V = 12. And J = 24. So, Johnny started and ended with 24 marbles, and Vinny with 12. • No, you answered faster than me! – Culver Kwan Apr 9 at 1:52 • @CulverKwan - Sorry. I know how that feels. I've been on your end of things many times! – Lanny Strack Apr 9 at 1:56 Let $$x$$ be the number of marbles Vinny started with. Then Johnny has $$2x$$ marbles in the beginning. After the first round, Johnny lost $$\\frac{2x}2$$ marbles and he has $$2x-x=x$$ marbles. So Vinny got $$x$$ marbles, which is $$2x$$ marbles. After the second round, Vinny lost $$2x\\cdot\\frac34=\\frac32x$$ marbles and he has $$2x-\\frac32x=\\frac12x$$ marbles left. Then Johnny has $$x+\\frac32x=\\frac52x$$ marbles. But the problem said that he has $$30$$ marbles. So $$\\frac52x=30, x=12$$. So Johnny has $$2(12)=24$$ marbles and Vinny has $$12$$ marbles. • Could someone help me mark spoiler properly? I can't do it. – Culver Kwan Apr 9 at 2:05 •", "us work out the remaining number of coins. \\begin{align*} \\text{mean} & = \\frac {\\text{number of coins (after)}}{\\text{group size}} \\\\ \\text{number of coins (after)} & = \\text{mean} \\times \\text{group size} \\\\ \\text{number of coins (after)} & = (\\text{2,5}) \\times (4) \\\\ \\text{number of coins (after)} & = 10 \\end{align*} Now we can calculate how many coins were taken by the 5 friends who left the group. \\begin{align*} \\text{number of coins removed } (x) & = \\text{items before} - \\text{items after} \\\\ \\text{number of coins removed } (x) & = (36) - (10) \\\\ \\text{number of coins removed } (x) & = 26 \\end{align*} A group of 9 friends each have some marbles. They work out that the mean number of marbles they have is 3. Then 3 friends leave with an unknown number ($$x$$) of marbles. The remaining 6 friends work out that the mean number of marbles they have left is $$\\text{1,17}$$. When the 3 friends left, how many marbles did they take with them? If the mean number of marbles the group originally had was 3 then the total number of marbles must have been: \\begin{align*} \\text{mean} & = \\frac {\\text{number of marbles (before)}}{\\text{group size}} \\\\ \\text{number of marbles (before)} & = \\text{mean} \\times \\text{group size} \\\\ \\text{number of marbles (before)} & = (3) \\times (9) \\\\", "have 29 marbles. The balancing of the scales is similar to the balancing of the following equation. \\begin{align*}3x + 2 = 29 \\end{align*} “Three bags plus two marbles equals 29 marbles” To solve for \\begin{align*}x\\end{align*}, we need to first get all the variables (terms containing an \\begin{align*}x\\end{align*}) alone on one side of the equation. We’ve already got all the \\begin{align*}x\\end{align*}’s on one side; now we just need to isolate them. \\begin{align*}3x + 2 &= 29\\\\ 3x + 2 - 2 &= 29 - 2 \\qquad \\text{Get rid of the 2 on the left by subtracting it from both sides.}\\\\ 3x &= 27\\\\ \\frac{3x}{3} &= \\frac{27}{3} \\qquad \\quad \\ \\text{Divide both sides by 3.}\\\\ x &= 9\\end{align*} There are nine marbles in each bag. We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance. Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each. Three", "# Algebra Word Problems - Page 3 Algebra Word Problems • Page 3 21. Paula has 21 marbles in her marble collection. Her friend gave her 3 marbles. How many marbles does Paula have now? a. 21 b. 26 c. 24 d. 20 #### Solution: Let b be the number of marbles in Paula ’s collection now. Number of marbles in Paula ’s collection now = number of marbles she has + number of marbles her friend gave her. b = 21 + 3 [Substitute the number of marbles.] b = 24 So, Paula has to 24 marbles in her collection now. 22. There were 70 people in a movie theatre. How many children were there if 30 of them were adults? a. 45 b. 42 c. 40 d. 38 #### Solution: Let n be the number of children in the theatre. Number of children = total number of people - number of adults. n = 70 - 30 [Substitute the values.] n = 40 [Subtract.] So, 40 children were there in the theatre. 23. Sheila collected 14 seashells and Juan collected 8 seashells on an island. How many more seashells did Sheila collect than Juan? a. 9 b. 8 c. 6 d. 5 #### Solution: Let m be the number of seashells that Sheila collected more than", "there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each. Three bags of marbles balances three piles of nine marbles. So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles. #### Solving for Unknown Values 1. Solve 6(x+4)=12\\begin{align*}6(x + 4) = 12 \\end{align*}. This equation has the x\\begin{align*}x\\end{align*} buried in parentheses. To dig it out, we can proceed in one of two ways: we can either distribute the six on the left, or divide both sides by six to remove it from the left. Since the right-hand side of the equation is a multiple of six, it makes sense to divide. That gives us x+4=2\\begin{align*} x + 4 = 2\\end{align*}. Then we can subtract 4 from both sides to get x=2\\begin{align*}x = -2\\end{align*}. 2. Solve x35=7\\begin{align*}\\frac{x - 3}{5} = 7\\end{align*}. It’s always a good idea to get rid of fractions first. Multiplying both sides by 5 gives us x3=35\\begin{align*}x - 3 = 35\\end{align*}, and then we can add 3 to both sides to get x=38\\begin{align*}x = 38\\end{align*}. ### Example #### Example 1 Solve 59(x+1)=27\\begin{align*}\\frac{5}{9}(x + 1) =\\frac{2}{7}\\end{align*}. First, we’ll cancel the fraction on the left by multiplying", "SEARCH HOME Math Central Quandaries & Queries Question from Bryson: Yolanda, Brian, and Charlie have a total of in their wallets. Brian has less than Yolanda. Charlie has times what Brian has. How much do they have in their wallets? Hi Bryson, The numbers didn't come through in your question so I am going to supply my own. Yolanda, Brian, and Charlie have a total of $\\$27$in their wallets. Brian has$\\$2$ less than Yolanda. Charlie has $3$ times what Brian has. How much do they have in their wallets? Suppose Yolanda has $Y$ dollars, Brian has $B$ dollars, and Charlie has $C$ dollars. Now put the fact you know into equation form. Yolanda, Brian, and Charlie have a total of $\\$27$in their wallets. $Y + B + C = 27. \\mbox{ Equation 1}$ Brian has$\\$2$ less than Yolanda. $B = Y - 2. \\mbox{ Equation 2}$ Charlie has $3$ times what Brian has. $C = 3 \\times B. \\mbox{ Equation 3}$ Equation 1 has three variables, $Y, B$ and $C$ and I would like to transform it into an equation with one variable. Equation 2 says $B = Y - 2$ or $Y = B + 2$ and hence if I substitute $Y = B + 2$ into equation 1 I will have converted equation 1 to an", "Algebra Puzzle – Challenge 53 Try to solve this interesting puzzle and critical thinking problem to learn more tips & tricks in mathematics. Challenge: Nicole bought some marbles. If she had bought four times as many as she did, she would have 33 more. How many marbles did she buy? $11.99 Satisfied 123 Students The correct answer is 11. Let x be the number of marbles Nicole bought. Then: 4x = x + 33 → x = 11 Nicole bought 11 marbles. If she had bought four times as many as she did (4 × 11 = 44), she would have 33 more marbles. (11 + 33 = 44) More math articles What people say about \"Algebra Puzzle - Challenge 53\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE$5 It was $16.99 now it is$11.99" ]

[ "+ 2B + 3C = 92 8 + 2A + 2B + 3C = 92 Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her? Thanks Current Student Joined: 27 Jun 2012 Posts: 418 Concentration: Strategy, Finance Followers: 67 Kudos [?]: 591 [0], given: 183 Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] 15 Jan 2013, 00:54 Consider Bill has x marbles. Bill has 8 more than Allen, i.e. Allen has (x-8) marbles. Charles has 2 times more than Bill, i.e. Charles has 2x marbles. Daryl has 3 times more than Charles, i.e. Daryl has 3*2x = 6x marbles. All 4 kids have total 92 marbles. Hence $$x + (x-8) + 2x + 6x = 92$$ i.e. $$10x = 100$$ i.e. $$x = 10$$ Therefore, Charles has 2x=20 marbles. Calculating the rest, Allen=2, Bill=10, Charles=20, Daryl=60, Total=92 _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 VOTE: vote-best-gmat-practice-tests-excluding-gmatprep-144859.html PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Moderator Joined: 02 Jul 2012 Posts: 1230 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 100", "How Cheenta works to ensure student success? Explore the Back-Story # Math Kangaroo (Benjamin) 2014 Problem No 24 Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180 marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24. ## Math Kangaroo (Benjamin) 2014 | Problem No 24 Grandma gives 180", "could earn on one of the other two tests? $$48$$ $$52$$ $$66$$ $$70$$ $$74$$ ###### Solution(s): To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a $$100$$ on her fourth test. The sum of the $$4$$ tests is then $76 + 94 + 87 + 100 = 357.$ For an average of $$81,$$ Shauna's test scores must add to $$5 \\cdot 81 = 405.$$ This means that she needs to get a $$405 - 357 = 48$$ on her last test. Thus, the correct answer is A. 8. Gilda has a bag of marbles. She gives $$20\\%$$ of them to her friend Pedro. Then Gilda gives $$10\\%$$ of what is left to another friend, Ebony. Finally, Gilda gives $$25\\%$$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $$20$$ $$33\\dfrac{1}{3}$$ $$38$$ $$45$$ $$54$$ ###### Solution(s): Assume that Gilda starts off with $$100$$ marbles. After giving $$20$$ marbles to Pedro, she has $$100 - 20 = 80$$ marbles left. She then gives $$80 \\times .1 = 8$$ marbles to Ebony, with her having $$80 - 8 = 72$$ marbles left. Finally, Gilda gives $$72 \\times .25 = 18$$ marbles to Jimmy. She has a", "for Vinny. After Round 1, Johnny gave half his marbles to Vinny. So Johnny has V marbles, and Vinny has 2V marbles. After Round 2, Vinny gives 3/4s of his marbles to Johnny, so Vinny has 0.5V, and Johnny has V + 1.5V = 2.5V marbles. We are told Johnny now has 30 marbles, so 2.5V = 30, and V = 12. And J = 24. So, Johnny started and ended with 24 marbles, and Vinny with 12. • No, you answered faster than me! – Culver Kwan Apr 9 at 1:52 • @CulverKwan - Sorry. I know how that feels. I've been on your end of things many times! – Lanny Strack Apr 9 at 1:56 Let $$x$$ be the number of marbles Vinny started with. Then Johnny has $$2x$$ marbles in the beginning. After the first round, Johnny lost $$\\frac{2x}2$$ marbles and he has $$2x-x=x$$ marbles. So Vinny got $$x$$ marbles, which is $$2x$$ marbles. After the second round, Vinny lost $$2x\\cdot\\frac34=\\frac32x$$ marbles and he has $$2x-\\frac32x=\\frac12x$$ marbles left. Then Johnny has $$x+\\frac32x=\\frac52x$$ marbles. But the problem said that he has $$30$$ marbles. So $$\\frac52x=30, x=12$$. So Johnny has $$2(12)=24$$ marbles and Vinny has $$12$$ marbles. • Could someone help me mark spoiler properly? I can't do it. – Culver Kwan Apr 9 at 2:05 •", "SEARCH HOME Math Central Quandaries & Queries Question from Bryson: Yolanda, Brian, and Charlie have a total of in their wallets. Brian has less than Yolanda. Charlie has times what Brian has. How much do they have in their wallets? Hi Bryson, The numbers didn't come through in your question so I am going to supply my own. Yolanda, Brian, and Charlie have a total of $\\$27$in their wallets. Brian has$\\$2$ less than Yolanda. Charlie has $3$ times what Brian has. How much do they have in their wallets? Suppose Yolanda has $Y$ dollars, Brian has $B$ dollars, and Charlie has $C$ dollars. Now put the fact you know into equation form. Yolanda, Brian, and Charlie have a total of $\\$27$in their wallets. $Y + B + C = 27. \\mbox{ Equation 1}$ Brian has$\\$2$ less than Yolanda. $B = Y - 2. \\mbox{ Equation 2}$ Charlie has $3$ times what Brian has. $C = 3 \\times B. \\mbox{ Equation 3}$ Equation 1 has three variables, $Y, B$ and $C$ and I would like to transform it into an equation with one variable. Equation 2 says $B = Y - 2$ or $Y = B + 2$ and hence if I substitute $Y = B + 2$ into equation 1 I will have converted equation 1 to an", "marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get? • 19 • 20 • 21 • 22 • 23 Arithmetic Equation solving Algebra ## Suggested Book | Source | Answer Algebra by Gelfand Math Kangaroo (Benjamin), 2014 $23$ ## Try with Hints Hint 1 Here Number of children is $10$. Number of marbles is $180$. And Anna gets the most and no 2 children gets the same number of marbles. Hint 2 Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation Hint 3 The equation will be- $x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$ Now solve for $x$. Hint 4 So, Anna can have $23$ marbles. ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "\\text{ units} &= 16-4\\\\[2ex] &= 12\\\\[2ex] 1 \\text{ unit} &= 12 \\div 2\\\\[2ex] & = 6 \\end{align} Number of marbles Ryan had at first $$=$$ Number of marbles Wilson had at first Number of marbles Wilson had at first\\begin{align}\\\\[6ex] &= 1 \\text{ unit} +16\\\\[2ex] &= 6+16\\\\[2ex] &= 22 \\end{align} Ryan had 22 marbles at first. 22 marbles ## 2. Equal Stage In The End Both subjects have a different number of items at first. After some items are added/removed from each subject, they have the same number of items in the end. Question 1: Mrs Lim had four times as many biscuits as Mrs Goh. After Mrs Lim gave away 17 biscuits and Mrs Goh received another 4 biscuits, they had an equal number of biscuits. How many biscuits did Mrs Goh have in the end? Solution: Difference in units\\begin{align}\\\\[2ex] &= 4 \\text{ units} - 1 \\text{ unit}\\\\[2ex] &= 3 \\text{ units} \\end{align} \\begin{align}​​ 3 \\text{ units} &= 4+17\\\\[2ex] &= 21\\\\[2ex] 1 \\text{ unit} &= 21 \\div 3\\\\[2ex] & = 7 \\end{align} Number of biscuits Mrs Goh had in the end\\begin{align}\\\\[6ex] &= 1 \\text{ unit} +4\\\\[2ex] &= 7+4\\\\[2ex] &= 11 \\end{align} Mrs Goh had 11 biscuits in the end. 11 biscuits Question 2: Alvin had four times as many marbles as Wei Ming. After Alvin lost 24 marbles", "Kudos [?]: 1069 [0], given: 116 Re: Solving an Algebra I (8th Grade) Word Problem [#permalink] 15 Jan 2013, 03:19 tara15225 wrote: 4 kids have a total of 92 marbles Bill has 8 more than Allen Charles has 2 times more than Bill Daryl has 3 times more than Charles How many marbles does Charles have? My daughter has gotten this far: Bill = 8 + A Allen = A Charles = 2B Daryl = 3C 8+A + A + 2B + 3C = 92 8 + 2A + 2B + 3C = 92 Then she is stuck and so am I. Can you please help us complete the steps and solve showing work so I can try to explain it to her? Thanks 2B = 2*(8 + A) = 16 + 2A 3C = 3*2B = 6B = 6*(8 + A) = 48 + 6A Picking up from where you left 8 + 2A + 2B + 3C = 92 8 + 2A + 16 + 2A + 48 + 6A = 92 10A + 72 = 92 10A = 92 - 72 = 20 A = 2 B = 2 + 8 = 10 C = 2*10 = 20 D = 3*20 = 60 _________________ Did you find this post helpful?... Please let me know through", "# Questions Archives ## Past Questions PSLE P4 & P5 SMOPS NMOS IQ 21/6/12 5/6/12 8/5/12 4/5/12 30/4/12 29/4/12 25/4/12 20/4/12 15/4/12 8/4/12 25/3/12 22/4/12 15/3/12 20/4/12 16/2/12 16/4/12 15/2/12 8/4/12 14/2/12 3/4/12 30/4/12 30/4/12 13/2/12 30/3/12 12/2/12 28/3/12 9/2/12 21/2/12 2/2/12 20/2/12 27/1/12 19/2/12 26/1/12 18/2/12 20/1/12 17/2/12 15/1/12 9/2/12 12/1/12 8/2/12 5/1/12 7/2/12 16/8/11 16/8/11 1/8/11 1/8/11 11/7/11 11/7/11 11/7/11 13/6/11 26/5/11 26/5/11 26/5/11 16/5/11 16/5/11 16/5/11 3/5/11 3/5/11 3/5/11 3/5/11 20/4/11 20/4/11 20/4/11 20/4/11 13/4/11 13/4/11 13/4/11 5/4/11 5/4/11 5/4/11 30/3/11 24/3/11 24/3/11 24/3/11 ### 10 Responses to “Questions Archives” 1. susanti April 8, 2011 at 4:18 pm # I WANT TO LEARN!!!!!! • Anne April 9, 2011 at 10:36 am # some marbeles are share among 3boys.Aaron’s share 3/4 of Brian’s share and Brian’s share is 4/5 of Charle’s share. If Aaron and Brian have a toatl 56 marbles altogether.How many marbles do the 3 boys have in all? 2. matharena April 9, 2011 at 11:54 am # A : B : C 3 : 4 : 5 A+B —> 7 u —-> 56 1 u —-> 8 Altogether: 12 u —> 12 x 8 = 96 marbles 3. claire tan September 19, 2011 at 8:00 pm # heyy auntie persis, im claire, i used to b in yr math class for P6? im", "ls$45 - x$. After they gave away 5 marbles each to their sisters, now, Ram has$x - 5$marbles and Radha has$45 - x - 5 = 40 - x$marbles. According to the given information,$(x - 5)(40 - x) = 124$. We can solve this equation after we first write it in standard form: \\begin{eqnarray} (x - 5)(40 - x) &=& 124\\\\ 40x - x^2 - 200 + 5x &=& 124\\\\ - x^2 + 45x - 200 &=& 124\\\\ -x^2 + 45x - 324 &=& 0\\\\ x^2 - 45x + 320 &=& 0\\\\ (x - 36)(x - 9) &=& 0\\\\ x - 36 &=&= 0 \\implies x = 36\\\\ x - 9 &=& 0 \\implies x = 9\\\\ \\end{eqnarray} Answer: One of them had 36 marbles and the other had 9 marbles originally. Example 2: Product of two consecutive even integers is 840. Find the two integers. Solution: We note that the difference between two even or two odd consecutive integers is 2. So, if one even integer is$x$, then the next consecutive even integer is$x + 2$. Given that the product of these two is 840. So, we have the equation$x(x + 2) = 840$that models this situation. Rearranging the terms in standard form and solving: \\begin{eqnarray} x(x + 2)&=& 840\\\\ x^2 + 2x &=& 840\\\\ x^2 +", "# Algebra Word Problems - Page 3 Algebra Word Problems • Page 3 21. Paula has 21 marbles in her marble collection. Her friend gave her 3 marbles. How many marbles does Paula have now? a. 21 b. 26 c. 24 d. 20 #### Solution: Let b be the number of marbles in Paula ’s collection now. Number of marbles in Paula ’s collection now = number of marbles she has + number of marbles her friend gave her. b = 21 + 3 [Substitute the number of marbles.] b = 24 So, Paula has to 24 marbles in her collection now. 22. There were 70 people in a movie theatre. How many children were there if 30 of them were adults? a. 45 b. 42 c. 40 d. 38 #### Solution: Let n be the number of children in the theatre. Number of children = total number of people - number of adults. n = 70 - 30 [Substitute the values.] n = 40 [Subtract.] So, 40 children were there in the theatre. 23. Sheila collected 14 seashells and Juan collected 8 seashells on an island. How many more seashells did Sheila collect than Juan? a. 9 b. 8 c. 6 d. 5 #### Solution: Let m be the number of seashells that Sheila collected more than", "Level 2 Violet had 6 times as many erasers as Jane at first. After Violet gave 40 erasers to Jane, they each had the same number of erasers in the end. How many erasers did Violet have at first? 3 m Level 3 A jar contained $2 and$5 notes with a total value of $518. 4 pieces of$5 were exchanged for $2 notes of similar value. In the end, the number of$2 notes was the same as the $5 notes. How many$5 notes were there at first? 3 m Level 3 Sean and Cathy had some salary each. If Cathy gives Sean $49, Sean would have 3 times as much salary as Cathy. If Sean gives Cathy$20, he will have the same amount of salary as her. How much salary was each man's salary respectively? 1. Sean? 2. Cathy? 3 m Level 2 The ratio of Tim's money to Carl's money is 3 : 5. If Carl has $845, how much must he give to Tim so both will have the equal amount of money? 2 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles", "South African regulations? Justify your answer. 2. The heights of $\\scriptsize 11$ girls in a netball team are measured in centimetres. The data set is as follows: $\\scriptsize \\{151,\\ 171,\\ 153,\\ \\text{147},\\ 142,\\ 167,\\ 146,\\ 157,\\ 156\\text{, }157,\\ 158\\}$ 1. Calculate the mean, median and mode of the data. 2. Is the data skewed? If so, how? 3. A twelfth player is added to the squad who is exceptionally tall at $\\scriptsize 183\\ \\text{cm}$. Recalculate the mean, median and mode for the new dataset and explain any changes (or not) to these measures. 4. Is the new data set skewed? If so, how? The full solutions are at the end of the unit. ### Example 2.2 A group of $\\scriptsize 13$ friends each have some marbles. They work out that the mean number of marbles they have is $\\scriptsize 12$. Then seven friends leave with an unknown number ($\\scriptsize x$) of marbles. The remaining six friends work out that the mean number of marbles they have left is $\\scriptsize 15.5$. How many marbles did the seven friends take with them? Solution We can calculate the total number of marbles the group of $\\scriptsize 13$ had. \\scriptsize \\begin{align*}\\bar{x}=\\displaystyle \\frac{{{{x}_{1}}+{{x}_{2}}+...+{{x}_{{13}}}}}{{13}} & =12\\\\\\therefore {{x}_{1}}+{{x}_{2}}+...+{{x}_{{13}}} & =12\\times 13\\\\ & =156\\end{align*} After the seven leave, the mean number of marbles is $\\scriptsize 15.5$.", "Difference between revisions of \"2019 AMC 8 Problems/Problem 8\" Problem 8 Gilda has a bag of marbles. She gives $20\\%$ of them to her friend Pedro. Then Gilda gives $10\\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $\\textbf{(A) }20\\qquad\\textbf{(B) }33\\frac{1}{3}\\qquad\\textbf{(C) }38\\qquad\\textbf{(D) }45\\qquad\\textbf{(E) }54$ Solution 1 After Gilda gives $20$% of the marbles to Pedro, she has $80$% of the marbles left. If she then gives $10$% of what's left to Ebony, she has $(0.8*0.9)$ = $72$% of what she had at the beginning. Finally, she gives $25$% of what's left to her brother, so she has $(0.75*0.72)$ $\\boxed{\\textbf{(E)}\\ 54}$. of what she had in the beginning left. Solution 2 Suppose Gilda has 100 marbles. Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles. Out of 80 marbles she gives 10% of 80 = 8 to Ebony. Thus she remains with 72 marbles. Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54. And $\\frac{54}{100}$=54%=$\\boxed{\\textbf{(E)}\\ 54}$ ~phoenixfire Solution 3 (Only if you have lots of time do it this way) Since she gave away 20% and 10% of", "had then to Diana. Finally, Diana gave 27of whatever she had to Isabelle. As a result, they each had the same number of marbles. How many marbles did Isabelle have at first? 5 m PSLE Wesley had 11 jars of balls. At first, each of the jars contained the same number of balls. He took out 24 marbles from each jar. After that, the total number of balls left in the 11 jars was equal to the total number of balls in 3 of the jars at first. What was the number of balls in each jar at first? 3 m Violet had 6 times as many erasers as Jane at first. After Violet gave 40 erasers to Jane, they each had the same number of erasers in the end. How many erasers did Violet have at first? 3 m A jar contains $2 and$5 notes with a total value of $518. 4 pieces of$5 were exchanged for $2 notes of similar value. In the end, the number of$2 notes was the same as the $5 notes. How many$5 notes were there at first? 3 m Sean and Cathy had some salary each. If Cathy gives Sean $49, Sean would have 3 times as much salary as Cathy. If Sean gives Cathy$20, he will have the same amount", "the amount. Here is our proportion to solve. $\\frac{6}{40}=\\frac{p}{100}$ Now we can use cross products and solve. $40p & = 600\\\\p & = 15$ Jeremy has 25 marbles and 12 of them are cat’s eye marbles. What percent of his marbles are cat’s eye marbles? We can think of this problem as “What percent of 25 is 12?” 12 is the amount $(a)$ and 25 is the base $(b)$ . We need to find the percent $(p)$ . $\\frac{a}{b}& =\\frac{p}{100}\\\\\\frac{12}{25}& =\\frac{p}{100}\\\\25p& =1,200\\\\\\frac{25p}{25}& =\\frac{1,200}{25}\\\\p& =48$ Since $p = 48, \\frac{48}{100}$ , which is 48%. The answer is that 48% of Jeremy’s marbles are cat’s eye marbles. Use a proportion to find each percent. Example A What percent of 20 is 2? Solution: $10\\%$ Example B What percent of 30 is 6? Solution: $20\\%$ Example C What percent of 45 is 15? Solution: $33.3\\%$ Here is the original problem once again. Reread it and use what you have learned about percents to figure out the questions at the end of the problem. Taylor’s younger brother Max decided to visit her at the candy store. Max is only seven and can be a handful sometimes, so while Taylor loves to see him, she was a little hesitant to have him in the shop. Plus, what seven year old doesn’t love candy.", "The ratio of Liam's to Mark's money is 7 : 9. The ratio of Mark's to Ian's money is 8 : 1. If Liam has $112, how much less does Ian have than Mark? 2 m Level 3 Wendy and Anna have$294. Violet and Wendy have $493. The ratio of Anna to Violet is 1 : 2. How much does Wendy have? 3 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 There were some grapefruits and durians in Container A and Container B. In Container A, the ratio of the grapefruits to the number of durians was 3 : 1. In Container B, the ratio of the number of grapefruits to the number of durians was 3 : 2. There were 2 times as many fruits in Container A as in Container B. After 56 durians were put into Container B, the", "have together? Let the number of marbles per friend be $$x_1, ~x_2, ~x_3$$ and $$x_4$$. \\begin{align*} \\frac{x_1 + x_2 + x_3 + x_4}{4} & = 10\\\\ x_1 + x_2 + x_3 + x_4 & = 40 \\end{align*} One friend leaves: \\begin{align*} x_1 + x_2 + x_3 & = 40 - 4 \\\\ x_1 + x_2 + x_3 & = 36 \\end{align*} Therefore the remaining friends have $$\\text{36}$$ marbles.", "more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 Howard and Sam had some money at first. Howard received 4 times as much money as Sam every fortnight. At the end of 2 fortnights, both of them had the same amount of money. If Sam had$1500 at first and Howard had $1900 at the end of those 2 fortnights, what was the ratio of Howard's money to the total amount of money at first? 4 m Level 2 Olivia and Fiona shared a box of chocolates. Olivia had 72% of the total chocolates. if Olivia gave 66 of her chocolates to Fiona, both would have the same number of chocolates. How many chocolates did Olivia have? 2 m Level 2 Fill in the blanks in units (Eg 3 u) or unit equations (Eg 1 u + 2). Adam and Bryan have some stickers. When Bryan uses 13 of his stickers and Adam buys another 20 stickers, they have the same number of stickers each. Let the number of stickers that Bryan has in the end be 2 u. 1. Number of stickers that Adam has", "2). Adam, Bryan and Chris have some stickers. Chris has 23 as many as Adam. Bryan has 10 more than Chris. Let the number of stickers that Adam has = 3 u 1. Number of stickers that Chris has = _____ 2. Number of stickers that Bryan has = _____ 3. Total number of stickers = _____ 4. Number of more stickers that Adam has than Chris = _____ 4 m Level 3 PSLE Tina has 130 marbles more than Olivia. If Tina transfers 384 marbles to Olivia, Olivia will have 3 times as many as Tina. How many marbles does Tina have? 3 m Level 1 Look at the picture. Give the answer in number. (Eg 1) String _____ is longer. 2 m Level 2 Adam and Bryan have 200 stickers. 1. Who has less stickers? (2) Bryan 2. How many less? 3 m Level 2 Fill in the blanks in units (Eg 1 u) or unit equations (Eg 1 u + 2). Adam, Bryan and Chris have some stickers. Chris has 23 as many as Adam. Bryan has 10 less than Chris. Let the number of stickers that Adam has = 3 u 1. Number of stickers that Chris has = _____ 2. Number of stickers that Bryan has = _____ 3. Total number of stickers" ]

[ "## WC '15 Contest 1 J2 - Grouping Recruits View as PDF Points: 5 Time limit: 1.0s Memory limit: 16M Authors: Problem types ##### Woburn Challenge 2015-16 Round 1 - Junior Division Now that you know a little more about Woburn, let's learn a bit more about the amazing computer science enrichment opportunity that it holds. You might not know that every couple of years, Woburn's very own Programming Enrichment Group recruits elementary and middle school students who aspire to be talented little computer scientists. Indeed, training programmers from such a young age has been a key principle behind the club's success. Since the number of recruits is usually very high, the PEG leader(s) will often have trouble handling them alone and all at once. Thus, an age-old solution has been devised – splitting them up into groups, each of which will be taught by a senior member of the club! However, this is a more challenging task than it seems, since senior members tend to complain amongst themselves about how unfair it is that some of them have to deal with teaching significantly more recruits than others. Let's say that there are recruits and senior members within the club . We would like to \"evenly\" divide the recruits up into groups such that the numbers of recruits", "demographic of Mathcampers? Are there minorities (by gender, race, age, etc.) among the campers or is it more \"balanced\"? It's a big mix. Usually, about 1/3 girls, 2/3 boys; about 75% US, 10-15% Canadian, and 10-15% from other countries. The mean, median, and mode of the age distrituion are almost always 16. On the first day of Mathcamp 2015: 4 campers age 13 8 campers age 14 23 campers age 15 35 campers age 16 33 campers age 17 17 campers age 18 0 campers age 19 (That was slightly more young students than we usually see, and we do once in a while have 19-year-olds.) Math1331Math 2016-03-15 20:13:44 How are the classes run? Are they more proof-oriented or problem solving based? roed 2016-03-15 20:13:46 Classes vary based on the instructor, but more of them are about cool topics that you wouldn't normally see in high school than about problem solving. We usually have 1 or 2 problem solving classes each week. (One or two per day, that is, that continue throughout the week.) AllenWang314 2016-03-15 20:14:25 Are the courses aimed toward college or usamo/imo training? More towards college, but it's a mix. himaximushi 2016-03-15 20:14:41 What are the pepper things in this link? http://www.mathcamp.org/2015/academics/4WeekSchedule.pdf khanh93 2016-03-15 20:14:49 @himaximushi: Chili peppers! Some classes are \"spicier\" than others, which", "team for hosting us. We're here to talk about Mathcamp 2018, both the program and the application process. (You’ve got a captive audience with some admissions officers, so this is a great chance to get ready for application season.) First, I’ll say in a few short sentences what Mathcamp is. Then we’ll spend about 30 minutes discussing the program, and 30 minutes discussing the application process, and then we’ll wrap up around 9pm ET. So, Mathcamp: it’s a 5-week residential math summer program. This year, it runs Sunday, June 24th through Sunday, July 29th, and takes place at Colorado School of Mines in Golden, CO (right outside of Denver). gocooldude 2018-02-15 19:32:42 Wow, gotta say, you type very fast. I did copy/paste that intro. :-D Really really frequently asked questions, to get them out of the way: Who comes to Mathcamp? We have 120 students per year, from all over the world. It’s about 65-75 new campers and 45-55 returning alumni per year. The last two years have been 40% girls, 60% boys. What kind of math? All kinds. A cool thing about Mathcamp is that our course catalog has about 150 classes per summer, and you have complete freedom to design your own curriculum. What else happens? There are a zillion (ish) activities happening during the evenings", "different mathematical backgrounds. Many classes will have no prerequisites (other than high-school algebra or precalculus); others will require some pretty intense knowledge of advanced mathematics. (Though almost always, the appropriate prerequisites will be offered as Mathcamp classes earlier in camp!) The result is that you can fill a schedule that works for you and your math background, whatever it may be. To get a sense of the variety of class offerings, take a look at last year's class schedule and descriptions: https://www.mathcamp.org/2020/classes/ mutyala2006 2021-02-23 19:42:41 how long(weeks/days/hrs) will be the camp Camp runs for Saturday, July 3 to Sunday, August 8, 2021 this year. ingmom 2021-02-23 19:43:27 Joy/parent/Michigan: Daughter participated in Mathpath for the past 2 summers, how does Mathcamp compare? Mathcamp is similar in some ways to MathPath, but with a lot more flexibility in both academic and social time, and a much higher ceiling of challenging math classes. CrystalFlower 2021-02-23 19:44:10 what is the max age level? We always invite applications up to age 18. We have made exceptions for admitting 19-year-old before. SparklyFlowers 2021-02-23 19:44:27 What's the min age level? The minimum age is 13 (and we have only once ever made an exception to that; we're firm on the minimum). eagle702 2021-02-23 19:44:55 Is there a curfew? Definitely not for VMC, since participants are", "is optional, and we’ll address you any way you like.) Okay, ready? Let’s spend about half an hour on questions about the program itself: mathematical and residential life at Mathcamp; the people; the campus; the \"during\" and \"after\" of Mathcamp. (And then we’ll switch over to discussing the \"before\": the application process and financial aid.) mastermind.hk16 2018-02-15 19:36:11 Is it for overseas students too? About 75-80% of our students are from the U.S.; 5-10% from Canada; and 15% from everywhere else. (There's a big list of our students' home countries here: http://www.mathcamp.org/prospectiveapplicants/eligibility.php .) (In terms of U.S. states, California sends us the most students, but we've had campers from every state except South Dakota.) josephwidjaja 2018-02-15 19:37:00 Any age restriction? Students ages 13 - 18 are eligible to apply. stay_clean 2018-02-15 19:37:22 Mohith/13/PA: Typically, how old are the students that are attending Mathcamp? The average age (in fact, often the mean, median, *and* mode) at camp is usually 16. On the first day of Mathcamp 2017, there were: 3 campers age 13 7 campers age 14 21 campers age 15 36 campers age 16 34 campers age 17 15 campers age 18 3 campers age 19 That's pretty typical for us. Kreps 2018-02-15 19:37:55 How long is the camp for?(Sorry if this question has already been answered I", "things they already like and enjoy, and customizing the delivery so it carried personal significance. No matter the skill level of the student, it’s important to make it fun and tailored to them as an individual. Tougher concepts shouldn’t equal less fun! The summer math programs ensures to keep the interest of the students intact, teaching them difficult concepts in a fun manner. Let’s discuss the math programs you should consider participating this summer: ### Awesome Math Summer Program Awesome Math Summer Program (AMSP) is a 3-week rigorous summer camp that help students, passionate for Mathematics, strengthen their mathematical skills. Eligibility: Usually students of age 12-18 participate in the camp. But, in special cases, students less than 12 years old may participate in the camp, if they are have good mathematical and problem-solving skills. One has to clear an admission test to participate in the camp. Important dates: The dates for the Summer Camp will be available on the website at the end of December 2020. Canada/USA Math Camp is a 5-week long math program for the high school students, where they can explore undergraduate and even graduate-level topics while building their problem-solving skills. Eligibility: 13-18 years old students can participate in the math camp. Important dates: The dates for the Math Camp 2021 will be announced on", "this understanding. ### Ages 14-16 There are minimal statutory requirements for computing in the national curriculum beyond the age of 14, but computing does remain a compulsory subject for pupils at this age. The requirements can be satisfied through embedding computer science, information technology and digital literacy across the rest of the curriculum, or through providing a number of focus days or independent project work over the course of these two school years. In addition to this, there’s a requirement that pupils “must have the opportunity to study” IT and computer science in greater depth - this is interpreted as meaning that schools should offer courses leading to public exams in IT or computing at 16+, but not require pupils to enter such exams. Detailed specifications have been developed by English exam boards for General Certificate of Secondary Education (GCSE) qualifications in computer science21. These include aspects of computational thinking including algorithm design, knowledge of Boolean logic and binary representation and an extended practical programming project, most typically undertaken in Python. ### Ages 16-18 Between the ages of 16 and 18 computer science becomes an elective subject, with relative small, but growing, numbers of pupils continuing to study the subject at this age. Again, English exam boards have developed detailed specifications for qualifications at this level22. These include", "or five or six. So that Sandrov mess or the Central Office whole system will be at one over 12 and five over six, okay? #### Topics Applications of Integration Lectures Join Bootcamp", "do: improve access to computing, increase students’ confidence in their ability to succeed in computing, increase students’ knowledge of computing concepts, and change students’ attitudes about computing. The programs themselves sound a lot like all the other summer camps: fun for average or above average kids, but offering nothing for the passionate gifted kids who want something more than playing with Scratch or App Inventor. Where are the computer equivalents of Awesome Math Camp (which I blogged about a couple of year ago) or RSI (Research Science Institute)? I’ve not found them. Next Page »", "wrap up around 9pm ET. mathw0nder 2020-02-11 19:32:57 Hi charlizard714 2020-02-11 19:32:57 hi there yoyopianow 2020-02-11 19:32:57 Hi! Williamgolly 2020-02-11 19:32:57 Hello!! GrizzlyTurtle 2020-02-11 19:32:57 Hi' This is going to be fun I hope so! So, Mathcamp: it's a 5-week residential math summer program. This year, it runs Sunday, July 5 through Sunday, August 9, and takes place at Champlain College in Burlington, Vermont. Really really frequently asked questions (RRFAQ), to get them out of the way: Who comes to Mathcamp? We have 120 students per year, from all over the world. It's about 65 new campers and 55 returning alumni per year. The last two years have been 45–50% girls, 50–55% boys, with a few non-binary students. trainlover2027 2020-02-11 19:33:49 How old do you have to be to join Mathcamp? trainlover2027 2020-02-11 19:33:49 What are the age limits for Mathcamp? It's for students ages 13 - 18. mathleticguyyy 2020-02-11 19:34:05 And middle school as well intelligence_20 2020-02-11 19:34:05 cant 8 graders also do it Yep, middle schoolers are welcome to apply, starting at age 13. beaver531 2020-02-11 19:34:47 Is it ok if I'm younger? Awesome_360 2020-02-11 19:34:47 Can you be younger but still join? infinity_theory 2020-02-11 19:34:47 can I still apply if I am 12 mathw0nder 2020-02-11 19:34:47 I am going to turn 13 in July, is", "students. [57] In 1981, the BBC produced a micro-computer and classroom network and Computer Studies became common for GCE O level students (11–16-year-old), and Computer Science to A level students. Its importance was recognised, and it became a compulsory part of the National Curriculum, for Key Stage 3 & 4. In September 2014 it became an entitlement for all 7,000,000 pupils over the age of 4. [58] In the US, with 14,000 school districts deciding the curriculum, provision was fractured. [59] According to a 2010 report by the Association for Computing Machinery (ACM) and Computer Science Teachers Association (CSTA), only 14 out of 50 states have adopted significant education standards for high school computer science. [60] Israel, New Zealand, and South Korea have included computer science in their national secondary education curricula, [61] [62] and several others are following. [63] ### Challenges In many countries, there is a significant gender gap in computer science education. In 2012, only 20 percent of computer science degrees in the United States were awarded to women. [64] The gender gap is also a problem in other western countries. [65] The gap is smaller, or nonexistent, in some parts of the world. In 2011, women earned half of the computer science degrees in Malaysia. [66] In 2001, 55 percent of computer science graduates", "average kids, but offering nothing for the passionate gifted kids who want something more than playing with Scratch or App Inventor. Where are the computer equivalents of Awesome Math Camp (which I blogged about a couple of year ago) or RSI (Research Science Institute)? I’ve not found them. Next Page »", "InfoQ Homepage Articles How Did You Start Coding? # How Did You Start Coding? This item in chinese A study conducted in July 2015 and recently published surveyed more than 2,200 coders and developers, asking them to “recount personal traits, tendencies and preferences from their younger years” in an attempt to determine what makes kids to end up in a computer science profession. The study, commissioned by Code School, an online learn-to-code website, concluded that on average developers show an interest in computers before the age of 16. Other findings are: • Usually men start playing with computers at the age of 15 or early, while women do it at the age of 16 or later. • The top hobby of 83% of men surveyed was computers, with sports next for 61% and then music for 59%. Women still preferred music (63%) over computers (52%). • Women procrastinate less while some men (41%) wait until the last minute to do their school assignments. • Fewer women (7%) than men (14%) drop out of school. Also, more women end up with a bachelor’s degree (51%) or a graduate degree (30%) compared to men with 42% and respectively 27%. • When it comes to income, women tend to have a more steady income with 32% making $50K-$100K/year but only 17%", "the National Science Foundation Why are those numbers so low? This Planet Money podcast points the finger at personal computers, which were introduced in the 1980s. They were marketed towards boys, so lots of boys grew up playing with computers. As a result, young men had more experience and interest in computers and were more likely to study computer science in college. Most girls didn’t grow up playing with computers, so very few had any experience by the time they entered college. These young women felt that the men were far ahead of them in computer science classes and consequently dropped out. I started the CS program at UT Dallas at age 16. I got my bachelor’s degree in CS at age 19 in the spring of 2012. In my graduating class, less than 10% of the students were women. It didn’t start this way. About a third of the students in my first year classes were female. In my second year that number dropped significantly, and in my third and final year, I was either the only girl or one of two girls in the class. This didn’t come as a shock. I was constantly put down by my male peers (even though I consistently made the top test scores). No one took me seriously. Everyone just", "for hours, was not enough. Everyone 16-21 years old can apply (if you don’t fit into this gap, contact us and we’ll see what can be done). Yes, everyone can apply, as there won’t be any quiz to get in. The Winter Workshops won’t be free, but you can apply for financial aid if you would like to attend but financial circumstances would prevent you from doing so. If that doesn’t convince you, just come and check for yourself... Be fast though, because the places are selling like hotcakes! Book your place at our : website Make sure you like our Facebook fanpage so that you can hear the freshest news • +54 By kpw29, 3 years ago, Hey! Together with Meet IT Foundation, we're delighted to present Camp IT! So, what is it actually about? The main objective is to provide a place for development and networking for young programmers. This relates not only to algorithmic olympiads, but also to computer science activities of any kind! Yet another IOI camp? Of course not! Although you will spend some time during the day doing IOI-styled problems, there are also Practice Activities and Evening Activities which aim to show you cool parts of computer science and give entertainment after a day of hard work. Who can apply? You can", "cap for the program? Eligibility is by age, not grade in school. mastermind.hk16 2018-02-15 19:41:40 how will i take 5 weeks off from school? If your school is in session over the summer, then yes. Emberflurry 2018-02-15 19:41:51 John/14/WA: Can parents visit? Of course! (Though we cannot provide accommodation for visitors.) person314159 2018-02-15 19:42:22 So if there is an age cap, do university students who are within the age limit allowed to apply as well? Yes, we welcome students who have started college early if they're still under 18. Torterror1536 2018-02-15 19:42:40 So could you be in college and apply for Mathcamp? Indeed. mastermind.hk16 2018-02-15 19:42:58 When does the camp start? It's June 24 - July 29 this year. Geek_Prince 2018-02-15 19:43:27 Can someone from middle school attend it if they are going to ninth grade? Yes! Geek_Prince 2018-02-15 19:43:53 Can someone who is going to be 13 on July 30 be able to participate in the camp? can you still enroll if you are 12 You should be 13 by the first day of camp in order to apply. dangan 2018-02-15 19:44:16 Will all students undergo same lessons? This is an interesting question, and the answer is definitely no: we teach 5 classes per hour, 4-5 hours per day, so you have to choose which classes you'll", "# Introduction Formally known as the American Computer Science League, ACSL is a computer science competition consisting of 4 contests and a final All-Star contest (if our cumulative scores are high enough). There are 5 different levels of difficulty, although this team will focus on the Intermediate and Senior levels. ## Basics ACSL is a contest that focuses on around a dozen different concepts of computer science. While many of these concepts are more abstract in nature, ACSL also provides concrete programming problems. ## Contest Format There will always be 2 parts to any contest: short answer and programming. For the short answer section, you will have 30 minutes to answer 5 questions; we do this during some of our meetings. As for the programming section, you will have 3 days to complete one program (in any programming language you want) at home. The short answer and programming sections both contribute 5 points each, thus making the highest score you can get for any given contest a 10. For the All-Star contest, there are still 2 parts to it. However, for the programming section, you work with a team for 4 hours on a multitude of programs. As for the short answer portion, you will have 1 hour to do 12 questions instead; you will have to be", "it recommended to apply? What of you are in middle school and are 13? awesomeguy856 2020-02-11 19:34:53 so can 6th grade do it kylie.huang 2020-02-11 19:34:53 Lots of younger inquirers this time! We are pretty firm about the lower age limit: we have only once made an exception for a 12-year-old. Younger students should consider MathPath, which is a great program for students ages 11 – 14: http://www.mathpath.org . What kind of math? All kinds. A cool thing about Mathcamp is that we offer about 150 classes and projects over the course of a single summer, and you have complete freedom to design your own curriculum. What else happens? During the evenings and on the weekends, there are a zillion (ish) activities happening, and you choose how to spend your time. That flexibility and creativity – academic and nonacademic – are the defining characteristics of Mathcamp. How do I get in? Admission is competitive. (We expect to admit 10 - 15% of the applicants.) The two main ingredients are the Qualifying Quiz and your personal essay. You'll also submit recommendation letters and some short answers about your math background. For more info, see https://www.mathcamp.org/quiz/ and https://www.mathcamp.org/essay/ . When do I apply? The deadline to apply is a few weeks away: March 12th, 2020. (All applications received by the", "it counts.(maybe I can assume that after five years i will have 10 core at home, damn I will be 42 then, whlole life devoted to computers :\\ )", "Who comes to Mathcamp? We have 120 students per year, from all over the world. It's about 65 new campers and 55 returning alumni per year. (This is the same at our traditional residential program and our COVID-era online program.) The student body is about 50% girls and nonbinary students, and 50% boys. What kind of math? All kinds. A cool thing about Mathcamp is that we offer 100+ classes and projects over the course of a single summer, and you have complete freedom to design your own curriculum. What else happens? During the evenings and on the weekends, there are a zillion (ish) activities happening, and you choose how to spend your time. That flexibility and creativity – academic and nonacademic – are the defining characteristics of Mathcamp. How do I get in? Admission is competitive. (We typically admit 10 - 15% of the applicants.) The two main ingredients are the Qualifying Quiz and your personal essay. You'll also submit recommendation letters and some short answers about your math background. For more info, see https://www.mathcamp.org/quiz/ and https://www.mathcamp.org/essay/. When do I apply? The deadline to apply is a few weeks away: March 18th, 2021. (All applications received by the deadline get equal consideration, so there's no advantage to applying early.) Scholarships? Yes! Every student (US, Canadian, & International;" ]

[ "matches will be = $\\frac{276}{8} = 34.5$ The average weight of 21 boys is 64 Kgs. The average weight of 21 boys and the teacher is 65 Kgs. So, total weight of 21 boys is 21*64 = 1344 The weight of 21 boys and the teacher = 22*65 = 1430 Hence, the weight of the teacher is 1430 – 1344 = 86 Kgs There are 9 numbers in the series: Average =$\\frac{253+124+255+534+836+375+101+443+760}{9}$ = 3681/9 = 409 Since the ages of A, B, C, D are consecutive Let the ages of A, B, C, D be n, n+1,n+2,n+3 $\\frac{n+n+1+n+2+n+3}{4} = 49.5$ 4n+6 = 49.5*4 = 198 4n = 192 n = 48 Ages of A, B, C, D = 48, 49, 50 ,51 Product of ages of B and D = 49*51 = (50-1)(50+1) =$50^2-1$ = 2500-1 = 2449. We can calculate the average as follows = $\\frac{62+76+42+84+21+47+28}{7} = 51.42$ Let there be three groups of 25 girls each having averages a, b and c. 25a+25b+25c=1050 a+b+c=42 Now, a = 12 and b = 16 hence, c = 14 Therefore average age of remaining 25 girls = c = 14 years. Hence, option C is correct option. Average of the given scores can be calculated as $\\frac{48+47+64+91+72+68+93}{7}$ = $\\frac{483}{7}$ = 69 The average weight of 15 girls is 54", "= 465 => Required average = $\\frac{465}{6}=77.5$ => Ans – (A) Average height of 6 friends = 167 cm => Sum of height of 6 friends = $167\\times6=1002$ cm When the boy with height 162 cm left, sum of height of remaining 5 friends = $1002-162=840$ cm => Required average = $\\frac{840}{5}=168$ cm => Ans – (A) Sum of ages of husband and wife at the time of their marriage = $23\\times2=46$ years Sum of the family after 5 years = 5 years of husband + 5 years of wife + 1 year of child => Total age = $46+5+5+1=57$ years => Required average = $\\frac{57}{3}=19$ years => Ans – (B) Let the 8 numbers be $x_1,x_2,…..,x_8$ Sum of these 8 numbers = $x_1+x_2+…..+x_8=27\\times8=216$ ————–(i) If each number is multiplied by 8, => Numbers = $8x_1,8x_2,…..,8x_8$ Sum = $8(x_1+x_2+…..+x_8)$ = $8\\times216$ $\\therefore$ New average = $\\frac{8\\times216}{8}=216$ => Ans – (D) Sum of scores of all students=50*42=2100 Sum of 22 girls scores=35*22=770 Sum of scores of boys=2100-770 =1330 Average of boys=1330/28 =47.5", "= 384000Non - adults = 384000 $\\times$ $28 \\over 100$ = 107520 8 . Answer : Option D Explanation : 9 . Answer : Option D Explanation : $Total_B$ = 2400000 x $18\\over 100$ = 432000 $Male_B$ = $432000 \\over 9$ x 5 = 240000 $Female_B$ = 432000 - 240000 = 192000 Difference = 240000 - 192000 = 48000 10 . Answer : Option D Explanation : $Adult_E$ = $75\\over 100$[2400000 x $10\\over 100$] = 180000$Male_D$ = $2 \\over 5$[2400000 x $20\\over 100$] = 192000Reqd percentage = $180000\\over 192000$ x 100 = 93.75 %", "girl will be $7y-3x$. We are told that $x$ is half of this; so $2x=7y-3x$, which means that $x=\\frac{7}{5}y$. - Let the boy's age (in years) be $b$ and the girl's age be $g$. When the boy was the girl's age, the girl was $g+(g-b)=2g-b$, so the sum of their ages then was $3g-b$, and twice that sum is $6g-2b$. The boy will be that age in $6g-3b$ years, at which point the girl will be $g+(6g-3b)=7g-3b$. The boy is half that old now, so $b=\\frac12(7g-3b),$ from which we see (multiplying by $2$ and rearranging) that $5b=7g$. Thus, $b=\\frac75g$, so the boy is $\\frac75$ times the girl's age at present. -", "the last post, we discussed $$\\frac{w_1}{w_2} = \\frac{(A_2 – A_{avg})}{(A_{avg} – A_1)}$$ The ratio of their salaries $$\\frac{w_1}{w_2} = \\frac{56000}{72000} = \\frac{7}{9}$$ $$\\frac{7}{9} = \\frac{(40 – T_{avg})}{(T_{avg} – 30)}$$ $$T_{avg} = 35.6%$$ Imagine that! No long calculations! In the last post, when we wanted to find the average age of boys and girls – 10 boys with an average age of 17 yrs and 20 girls with an average age of 20 yrs, all we needed was the relative weights (relative number of people) in the two groups i.e. 1:2. It didn’t matter whether there were 10 boys and 20 girls or 100 boys and 200 girls. It’s exactly the same concept here. It doesn’t matter what the actual salaries are. We just need to find the ratio of the salaries. Also notice that the two tax rates are 30% and 40%. The average tax rate is 35.6% i.e. closer to 40% than to 30%. Doesn’t it make sense? Since the salary of Ingrid is \\$72,000, that is, more than salary of John, her tax rate of 40% ‘pulls’ the average toward itself. In other words, Ingrid’s tax rate has more ‘weight’ than John’s. Hence the average shifts from 35% to 35.6% i.e. toward Ingrid’s tax rate of 40%. This question is discussed HERE. Let’s now look at", "$$\\frac{w_1}{w_2} = \\frac{56000}{72000} = \\frac{7}{9}$$ $$\\frac{7}{9} = \\frac{(40 – T_{avg})}{(T_{avg} – 30)}$$ $$T_{avg} = 35.6%$$ Imagine that! No long calculations! In the last post, when we wanted to find the average age of boys and girls – 10 boys with an average age of 17 yrs and 20 girls with an average age of 20 yrs, all we needed was the relative weights (relative number of people) in the two groups i.e. 1:2. It didn’t matter whether there were 10 boys and 20 girls or 100 boys and 200 girls. It’s exactly the same concept here. It doesn’t matter what the actual salaries are. We just need to find the ratio of the salaries. Also notice that the two tax rates are 30% and 40%. The average tax rate is 35.6% i.e. closer to 40% than to 30%. Doesn’t it make sense? Since the salary of Ingrid is $72,000, that is, more than salary of John, her tax rate of 40% ‘pulls’ the average toward itself. In other words, Ingrid’s tax rate has more ‘weight’ than John’s. Hence the average shifts from 35% to 35.6% i.e. toward Ingrid’s tax rate of 40%. This question is discussed HERE. Let’s now look at PS question no. 148 from the Official Guide which is a beautiful example of the use of", "27 = 3 Question 6. Find the 8th term of an A.P -2, -4, -6 ……….. T = a + (n – 1)d T8 = -2 + (8 – 1)(-2) T8 = -2 – 14 = -16 Question 7. Solve: $$\\frac{x+2}{x-2}=\\frac{5}{2}$$ 2(x + 2) = 5(x – 1) 2x + 4 = 5x – 5 4 + 5 = 5x – 2x ⇒ 3x = 9 ⇒x = 3. Question 8. Define perpetuity. An annuity that is payable forever is called a perpetuity. Question 9. Convert 42% into a decimal. 42% = $$\\frac{42}{100}$$ = 0.42. Question 10. The average score of 35 girls is 80 and the average score of 25 boys is 68. Find the average score of both boys and girls together. 12 = $$\\frac{35 \\times 80+25 \\times 68}{35+25}=\\frac{2800+1700}{60}$$ = 75 Question 11. Question 12. If the distance between the points (3, -2) and (-1, a) is 5 units. Find the values of $$\\sqrt{(3+1)^{2}+(-2-a)^{2}}$$ = 5 42 + 4 + a2 + 4a = 5 a2 + 4a + 15 = 0 Using formula method we get the values of a. Part – B II. Answer any TEN questions. (10 × 2 = 20) Question 13. Find the L.C.M of 12, 21 and 24. ∴ L.C.M = 23 × 3 × 7 = 24 × 7", "of the old numbers of the remaining students is $54$. Divide these students into three arbitrary (non overlapping) groups. The average sum of three numbers in this group is $54/3 = 18$. Hence, one of the new numbers must be 18 or greater. -", "x $12 \\over 100$ x $11\\over 24$$Female_C$ = 30000 x $17 \\over 100$ x $5\\over 17$Ratio = $11\\over 10$ = 11 : 10 3 . Answer : Option C Explanation : $Male_B$ = 30000 x $18\\over 100$ x $11\\over 18$ = 3300$Female_E$ = 30000 x $22\\over 100$ x $4\\over 11$ = 2400Reqd % = $3300 - 2400 \\over 2400$ x 100 = $900 \\over 24$ = 37.5 % 4 . Answer : Option B Explanation : Total Females = $30000\\over 100$ x [21 x $4\\over 15$ + 18 x $7\\over 18$ + 17 x $5\\over 17$ + 12 x $13\\over 24$ + 22 x $4\\over 11$ + 10 x $11\\over 30$]= 300 [5.6 + 7 + 5 + 6.5 + 8 + $11\\over 3$]= 300 (32.1 + $11\\over 3$ )= 100 x 107.3 = 10730Reqd % = $10730 \\over 30000$ x 100 = 35.76 $\\cong$ 36 % 5 . Answer : Option D Explanation : $D_{Total}$ = 30000 x $12\\over 100$ = 3600$A_{Female}$ = 30000 x $21\\over 100$ x $4\\over 15$ = 1680 Required fraction = $1680 \\over 3600$ = $7\\over 15$ 6 . Answer : Option D Explanation : Total Males = $9600000\\over 10000$ [16 × 52 + 15 × 57 + 24 × 51 + 9 × 48 + 7 × 47 + 17 × 53", "$$\\{2,4,6\\}$$. For each of these there are six permutations of the girls, and for each of these there are also six permutations of the boys, so the answer should be $$4\\cdot 3!\\cdot 3! = 144$$. • The logic is fine, but $4*6*6 = 144$ – Doug M Oct 11 '18 at 19:40 • @DougM What's wrong with $6-5! \\times 2! + 4! \\times 3!$? I counted the ways I can have all girls together, which equals $4! \\times 3!$ and the ways I can have two girls together, $5! \\times 2!$ – Mark Oct 11 '18 at 19:41", "one book is 60% more than the price of the other, what is the price of each of these two books ? 16,12 10,16 23,9 8,7 None of these 29 . The average age of a group of persons going for picnic is 16 years. Twenty new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years. The number of persons initially going for picnic is : 15 20 23 24 None of these 30 . 16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of group B? $233 \\over 3$ $243 \\over 4$ $254 \\over 5$ $345 \\over 7$ None of these", "18 girls = 18 × 15 = 270Total age of remaining 12 girls 390 – 270 = 120Average age of remaining 12 girls =$\\frac{120}{12}=10$Using Trick: Average age of remaining 12 girls = $\\frac{30\\times 13-18\\times 15}{30-18}=10$Trick: If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y, then the average of remaining numbers will beAverage (A) = $\\frac{mx-ny}{m-n}$ Q13. The average height of the basketball team A is 5 ft 11 inches and that of B is 6 ft 2 inches. There are 20 players in team A and 18 players in team B. The overall average height is (a) 72.42 inches (b) 72 inches (c) 70.22 inches (d) 70 inches Answer: (a) 72.42 inchesSolution: overall average height = $\\frac{20\\times 71+18\\times 74}{20+18}=\\frac{1420+1332}{38}=72.4211$Rules: If the given observations (x) are occurring with certain frequency (A) then,Average = $\\frac{A_1x_1+A_2x_2+…….+A_nx_n}{x_1+x_2+…..+x_n}$ , where, A1, A2, A3. ………. An are frequencies Q14. The average of 13 results is 60. If the average of first 7 results is 59 and that of last 7 results is 61, what will be the seventh result? (a) 90 (b) 50 (c) 75 (d) 60 (e) None of the above Answer: (d) 60Solution: Total of 13 results = 13 × 60 = 780Total of first 7 results = 7 ×", "48, 50, 52, 54 The average of the two greatest of these integers 52 and 54 is $\\frac{(52 + 54)}{2}=$ = 53 Q 10 - The average height of students in a school is 5 foot 6 inches. If the 200 girls at the school average 5 foot 4 inches, what is the average height of the 180 boys at the school? ### Explanation Step 1: Let the average height of the boys = x Then total height of 180 boys = 180x Total height of 200 girls = 200 × $5\\frac{4}{12} = \\frac{3200}{3}$ Step 2: Total height of boys and girls = 180x + $\\frac{3200}{3}$ = Average x number of boys and girls = 5$\\frac{6}{12}$ × 380 180x + $\\frac{3200}{3}$ = 2090; 180x = $\\frac{(6270 − 3200)}{3} = \\frac{3070}{3}$; x = $\\frac{3070}{540}$ = 5.69 feet. computations_involving_mean_sample_size_and_sum_of_data_set.htm", "boys = (5.6 x 4) in = 22.4 in Total height of 3 boys = (5.2 + 5.4 + 5.5) in = 16.1 in Height of the 4th boy = (total height of 4 boys) – (total height of 3 boys) = (22.4 – 16.1) in = 6.3 in Hence, the height of the fourth boy is 6.3 in Example 6: The average of 5 numbers is 30. If each number is increased by 2, what will the new average be? Solution: Given that the average of 5 numbers = 30 Sum of the 5 numbers = 30 x 5 = 150 If each number is increased by 2, the total increase = 2 x 5 = 10 Then, the new sum = 150+10 = 160 and new Average = 160/10 = 16 Hence, the new average is 16. Example 7: The average age of four boys is 25 years and their ages are in proportion 2:4:6:8. What is the age in years of the youngest boy? Solution: Given, Average age of four boys = 25 years Sum of four boys = 25 x 4 = 100 Ages = 2x + 4x + 6x + 8x 2x + 4x + 6x + 8x = 100 20x = 100 x = 100/20 x = 5 years Age of the", "ages = 2A Total of the eight persons = ET When replaced (ET – 80 + 2A)/8 = ET/8 + 2 2A – 80 = 16 A = 96/2 = 48 The average of the three numbers is 60 The first number is one fourths of the sum of the others. or 4p = q + r $\\frac {p + q + r}{3}$ = 60 So, 5p/3 = 60 p = 180/5 = 36 Let his monthly salary be x rupees. So, 5 (x – 1200) + 7 (x-1300) = 2900 12x – 6000 – 9100 = 2900 So, x = 18000/12 = 1500 There are 5 Sundays in the month that starts with sunday and 25 other days. So, the average =$\\frac{ 5 * 510 + 25 * 240}{30}$ = 285 Let the employee be absent for x days. He is present for 25 – x days. His salary is (25 – x)*2 – .5x = 50 – 2.5x But 50 – 2.5x = 37.5 So, x = 5 Given, (f+m)/2=45 =>f+m=90……(1) Similarly, f+m+d=105……(2) Subracting (2) from (1) we get, d=15 Sum of scores of the 3 boys=avg of scores of the 3 boys×no of boys=30×3=90 Sum of scores including the 4th boy=25×4=100 Thus, the score of the 4th boy=100-90=10 Let the average age of the women", "and adults, the total cost will be too high. If there are only adults and children, we can find a solution but the answer is not a whole number. Therefore, there are adults, children and pensioners in the cinema hall. For every adult, the price is £2.50 more than the average price. For every child, the price is 15p less than the average price. The LCM of $250$p and $15$p is $750$p, i.e. £7.50. £2.50 $\\times3$ is £7.50 and $15$p $\\times 50$ is £7.50. Therefore, for every $50$ children, there are $3$ adults. We can conclude that there are $3$ adults, $47$ pensioners and $50$ children. 9MAT1A from Diocesan Girls School in New Zealand sent us this solution, which takes a similar approach using simultaneous equations: Initially we have two pairs of simultaneous equations and $3$ variables. If $A$ is the number of adults, $C$ the number of children and $P$ the number of pensioners then we find: $$3.5A + 0.85C + P = 100$$ $$A + C + P = 100$$ We realised that the number of pensioners would be equal to the amount the pensioners paid so really there are only two variables: $$3.5A +0.85C = 100-p = A + C$$ so $$2.5A = 0.15C$$ We decided to make the coefficients whole numbers, since $A, C$", "problem (average family size is $2$). Given $1000$ children split between $500$ families, and iterating through this procedure $10000$ times, we get that: $$\\text{average brothers per male} = 0.9963$$ $$\\text{average brothers per female} = 0.9984$$ This doesn't exactly answer the OP's question (in his/her view, the distribution of family sizes is entirely unknown), so take these results with a grain of salt. I think this model is realistic, however, and because of the high number of iterations and relative difference between the values, I'm heuristically convinced that girls have more brothers than boys. I'm going to play around with the $n$ and $f$ parameters a bit, and if I find anything interesting, I'll try to whip up some graphs. The difference between the boy/girl values seems to increase as $f$ decreases, with deterministic behavior (obviously) at $f=1$. Assuming that there is an amount of men and women that does not equal $0$ in a family (not including the parents), let's say $3$ men and $3$ women, obviously a woman has $3$ brothers whereas one of her brothers only has $2$ brothers. Conversely, every man has more sisters than a woman does. To generalize it, if there are $n$ men and $m$ women for an arbitrary $n,m \\in \\Bbb N$, a woman has $n$ brothers while a man has", "i.e; sum =6 x 15=90 (Write 30 pairs with sum 3) 12, 18, 13, 17, 14, 16 iii. Mean is 25 sum = 25 x 6 = 150 (Write 50 pairs with sum 3) 22, 28, 23, 27, 24, 26 HSSLive Statistics Question 2. The table below shows the children in a class, sorted according to their heights. Height(cm) Number of children 148 – 152 8 152 – 156 10 156 – 160 15 160 – 164 10 164 – 168 7 What is the mean height of a child in this class? Height (cm) No. of children Mid Value Total Height 148 -152 8 150 150 × 8 = 1200 152 -156 10 154 154 × 10 = 1540 156 -160 15 158 158 × 15 = 2370 160 -164 10 162 162 × 10 = 1620 164 -168 7 166 160 × 7 = 1162 Total 50 7892 Mean height = $$\\frac { Total height }{ No of children }$$ = $$\\frac { 7892 }{ 50 }$$ = 157.84 cm Natural Rubber Sheets Question 3. The teachers in a university are sorted according to their ages, as shown below. Age Number of Persons 25 – 30 6 30 – 35 14 35 – 40 16 40 – 45 22 45 – 50 5 50 – 55", "average expenditure of all the 19. What was the total money spent by them? A) Rs. 1398.96 B) Rs. 1457.09 C) Rs. 1662.35 D) Rs. 1536.07 Explanation: Let average expenditure was Rs. R 13 x 79 + 6x(R + 4) = 19R => R = Rs. 80.84 Total money = 19 x 80.84 = Rs. 1536.07. 21 1531 Q: The average age of a class is 19 years. While the average age of girls is 18 and that of boys is 21. If the number of girls in the class is 16, Find the number of Boys in the class? A) 12 B) 10 C) 8 D) 6 Explanation: Let the number of boys = x From the given data, => [21x + 16(18)]/(x+16) = 19 => 21x - 19x = 19(16) - 16(18) => 2x = 16 => x = 8 Therefore, the number of boys in the class = 8.", "six consecutive odd numbers. Given that their average is 52 Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6 6x + 30 = 312 x = 47 So Product = 47 × 57 = 2679 12 1145 Q: 19 friends went to a restaurant for a combined weekend dinner party. 13 of them spent Rs. 79 each on their dinner and the rest spent Rs. 4 more than the average expenditure of all the 19. What was the total money spent by them? A) Rs. 1398.96 B) Rs. 1457.09 C) Rs. 1662.35 D) Rs. 1536.07 Explanation: Let average expenditure was Rs. R 13 x 79 + 6x(R + 4) = 19R => R = Rs. 80.84 Total money = 19 x 80.84 = Rs. 1536.07. 19 1075 Q: The average age of a class is 19 years. While the average age of girls is 18 and that of boys is 21. If the number of girls in the class is 16, Find the number of Boys in the class? A) 12 B) 10 C) 8 D) 6 Explanation: Let the number of boys = x From the given data, => [21x + 16(18)]/(x+16) = 19 => 21x - 19x = 19(16) - 16(18) => 2x = 16 => x = 8 Therefore, the" ]

[ "make more than $100K. 25% of men make more than$100K but, on the other extreme, 20% of them make less than $25K. InfoQ has also conducted a survey (How did you start coding?) among our readers and promoted it through social media channels. The survey asked three questions -“What age were you when you started coding?”, “What was your first computer?” and “What do you love about coding?” – and was taken by ~120 readers. After a bit of analysis, the results of the survey are: • InfoQ readers started coding between age 5 and 30 • The majority of respondents (45%) started coding at ages between 10 and 14 years • The average age for starting coding is ~14 years (13.89) • Younger readers started working on Windows (36%), and older ones on Commodore (19%), Spectrum (14%) with a few (5%) on mainframes • When it comes to the reason they love to program, the respondents provided answers that can be reduced to: Creativity (35%), Problem Solving (16%), Magic (15%), etc. One of them said he likes programming for the fun of it, while the one who started coding at the age of 30 is doing it for the money. InfoQ editors are not journalists by trade, but software architects, team leaders or freelancers in the real", "the month begins with a Sunday, so there will be five Sundays in the month. Required average = ${(\\frac{510\\times5+240\\times25}{30})}$ = ${\\frac{8550}{30}}$ = 285 5. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is: • 35 years • 40 years • 50 years • None of these The Correct answer is :40 years Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years Therefore, Husband's present age = (90-50)years = 40 years 6. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team? • 23 years • 24 years • 25 years • None of these The Correct answer is :23 years Let the average age of the whole team by x years. Therefore,", "18 girls = 18 × 15 = 270Total age of remaining 12 girls 390 – 270 = 120Average age of remaining 12 girls =$\\frac{120}{12}=10$Using Trick: Average age of remaining 12 girls = $\\frac{30\\times 13-18\\times 15}{30-18}=10$Trick: If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y, then the average of remaining numbers will beAverage (A) = $\\frac{mx-ny}{m-n}$ Q13. The average height of the basketball team A is 5 ft 11 inches and that of B is 6 ft 2 inches. There are 20 players in team A and 18 players in team B. The overall average height is (a) 72.42 inches (b) 72 inches (c) 70.22 inches (d) 70 inches Answer: (a) 72.42 inchesSolution: overall average height = $\\frac{20\\times 71+18\\times 74}{20+18}=\\frac{1420+1332}{38}=72.4211$Rules: If the given observations (x) are occurring with certain frequency (A) then,Average = $\\frac{A_1x_1+A_2x_2+…….+A_nx_n}{x_1+x_2+…..+x_n}$ , where, A1, A2, A3. ………. An are frequencies Q14. The average of 13 results is 60. If the average of first 7 results is 59 and that of last 7 results is 61, what will be the seventh result? (a) 90 (b) 50 (c) 75 (d) 60 (e) None of the above Answer: (d) 60Solution: Total of 13 results = 13 × 60 = 780Total of first 7 results = 7 ×", "Vivian O. A summer camp wants to hire counselors and aides to fill it staffing needs at minimum cost. the average monthly salary of a counselor A summer camp wants to hire counselors and aides to fill it staffing needs at minimum cost. the average monthly salary of a counselor is $2400 and the average salary of an aids is$1100.The camp can accommodate up to 35 staff members and needs at least 20 to run property. They must have at least 10 aids, and may have up to 3 aids for every 2 counselor. How many counselors and how many aids should the camp have to minimize cost. By: Tutor 4.8 (24) Patient and effective UH grad for High school Math tutoring Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "If a, b, c, d, e are five consecutive odd numbers, their average is : A. $5(a+4)$ B. $\\Large\\frac{\\text { abcde }}{5}$ C. $5(\\mathbf{a}+\\mathbf{b}+\\mathbf{c}+\\mathbf{d}+\\mathbf{e})$ D. $\\mathbf{a}+\\mathbf{4}$ Answer: Option D Show… ## The average age of 40 students of a class is 15 years… The average age of 40 students of a class is 15 years. When 10 new students are admitted, the average is increased by 0.2 year. The average age of the… ## In an examination average marks obtained by the girls of a class is 85.. In an examination average marks obtained by the girls of a class is 85 and the average marks obtained by the boys of the same class is 87. If the… ## If the difference between the average of x, y and y, z If the difference between the average of x, y and y, z is 12, then the difference between x and z is : A. 6 B. 48 C. 24 D.… ## 4 boys and 3 girls spent Rs. 120 on an average 4 boys and 3 girls spent Rs. 120 on an average, of which boys spent Rs. 150 on the average. The the average amount spent by the girls is- A.…", "age of 20 boys is 14 years. A boy who is 16 years old left them and one new boy replaces him. Then the average becomes 13.8 years. Find the age of the new boy? Solution: 20*14 – 16+a= 13.8*20 280-16+a=276 a = 276-264 = 12 Shortcut: 16-0.2*20=12 Problem 3: The average temperature of Monday, Tuesday and Wednesday was $30^{\\circ}C$ and that of Tuesday, Wednesday and Thursday was $33^{\\circ}C$. If the temperature on Monday was $32^{\\circ}C$, then the temperature on Thursday was? Solution: T+W+TH = 33*3 M+T+W=30*3 TH-32=9 TH = 32+9 =41 ## Averages Part 6 Problem 1: The average weight of A, B, and C is 50kg. If the average weight of A and B be 48 kg and that of B and C be 45 kg, then find the weight of B? Solution: A+B+C = 50*3 =150 A+B = 48*2 =96 B+C = 45*2 = 90 A+B+B+C = 186 A+B+C=150 B=36 Problem 2: The average age of eleven players in a cricket team decreases by 2 months when two new players are included in the team replacing two players of ages 17 years and 20 years. The average age of new players is? Solution: $\\frac{17+20}{2}=\\frac{37}{2}$ $\\frac{2\\times&space;11}{2}=11$ 18years 6 months – 11 17 years 7 months Problem 3: The average wages of a worker for 15 consecutive", "My Math Forum Simplex Word Problem (Linear Equation Issue) Applied Math Applied Math Forum December 23rd, 2011, 07:57 AM #1 Member Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Simplex Word Problem (Linear Equation Issue) Okay, we're having an issue with this problem, and even our teacher is confused on how the logic is on 1 of the equations. So basically, the part in bold is the linear equation we're having trouble setting up. Quote: A summer camp wants to hire counselors and aids to fill its staffing needs at minimum cost. The average monthly salary of a counselor is $2400 and the average monthly salary of an aide is$1100. The camp accommodate up to 45 staff members and needs at least 30 to run properly. They must have at least 10 aides, and may have up to 3 aides for every 2 counselors. How many counselors and how many aides should the camp hire to minimize cost? The equation we're using for that in our simplex model is 3y <= 2x, but that causes the answer to be 18 counselors with 12 aides, but we know it's supposed to be 12 counselors with 18 aids. What do you guys think? ---- EDIT: Well we plugged in 2y <= 3x and got the correct answer. Anyone", "rest. What average interest rate does he pay on the total$35,000? (Round your answer to the nearest tenth of a percent.) ## Everyday Math Exercise $$\\PageIndex{45}$$ As the treasurer of her daughter’s Girl Scout troop, Laney collected money for some girls and adults to go to a 3-day camp. Each girl paid $75 and each adult paid$30. The total amount of money collected for camp was $765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp? Answer 9 girls, 3 adults Exercise $$\\PageIndex{46}$$ Laurie was completing the treasurer’s report for her son’s Boy Scout troop at the end of the school year. She didn’t remember how many boys had paid the$15 full-year registration fee and how many had paid the $10 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If$250 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee? ## Writing Exercises Exercise $$\\PageIndex{47}$$ Suppose you have six quarters, nine dimes, and four pennies. Explain how you find the total value of all the coins. Exercise $$\\PageIndex{48}$$ Do you find it helpful to use a", "matches will be = $\\frac{276}{8} = 34.5$ The average weight of 21 boys is 64 Kgs. The average weight of 21 boys and the teacher is 65 Kgs. So, total weight of 21 boys is 21*64 = 1344 The weight of 21 boys and the teacher = 22*65 = 1430 Hence, the weight of the teacher is 1430 – 1344 = 86 Kgs There are 9 numbers in the series: Average =$\\frac{253+124+255+534+836+375+101+443+760}{9}$ = 3681/9 = 409 Since the ages of A, B, C, D are consecutive Let the ages of A, B, C, D be n, n+1,n+2,n+3 $\\frac{n+n+1+n+2+n+3}{4} = 49.5$ 4n+6 = 49.5*4 = 198 4n = 192 n = 48 Ages of A, B, C, D = 48, 49, 50 ,51 Product of ages of B and D = 49*51 = (50-1)(50+1) =$50^2-1$ = 2500-1 = 2449. We can calculate the average as follows = $\\frac{62+76+42+84+21+47+28}{7} = 51.42$ Let there be three groups of 25 girls each having averages a, b and c. 25a+25b+25c=1050 a+b+c=42 Now, a = 12 and b = 16 hence, c = 14 Therefore average age of remaining 25 girls = c = 14 years. Hence, option C is correct option. Average of the given scores can be calculated as $\\frac{48+47+64+91+72+68+93}{7}$ = $\\frac{483}{7}$ = 69 The average weight of 15 girls is 54", "44 Q: The average monthly salary of the employees of a firm is Rs.60 out of the employees 12 officers get an average of Rs.400 per month and the rest an average of Rs. 56 per month , Find the number of emp Q: If the average height of boys in a hall is 180 cm and that of girls is 150 cm. If the average height of the gathering is 165 cm, then find the number of girls present in the hall if the number of boys present in the hall are 78 ? A) 56 B) 64 C) 87 D) 78 Explanation: Using Trial and error method, we get the number of girls in the hall = 78 11 239 Q: The average age of 7 boys is increased by 1 year when one of tem whose age is replaced by a girl. What is the age of the girl ? A) 33 yrs B) 27 yrs C) 21 yrs D) 29 yrs Explanation: Let the avg age of 7 boys be 'p' years Let the age of the girl be 'q' years From given data, The age of 7 boys = 7p years Now the new average = (p + 1) when 22 yrs is replaced by q Now the age of all 7 will", "7 years is Rs 77 lakhs and that of last 7 years is Rs 89 lakhs, find the revenue for the 7th year. A. Rs 98 lakhs B. Rs 96 lakhs C. Rs 94 lakhs D. Rs 92 lakhs Question 100 In an examination, a pupil’s average mark was 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination? A. 8 B. 9 C. 10 D. 11 Question 101 The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is- A. 20 B. 21 C. 22 D. 23 Question 102 The average of 7 consecutive numbers is 20. The largest of these numbers is- A. 20 B. 22 C. 23 D. 24 Question 103 The average ages of parents and two children are 30 years and 8 years respectively. The average age of the family is A. 16 years B. 19 years C. 18 years D. 17 years Question 104 The average height of 35 girls in a class was", "average age of the boys in the class is 14 yr and the average age of the girls in the class is 13 yr. What is the average age of the whole class ? (Rounded off to two digits after decimal) 13.50 13.53 12.51 13.42 None of these 5 . The average marks of 65 students in a class was calculated as 150. It was later realised that the marks of one of the students was calculated as 142, whereas his actual marks were 152. What is the actual average marks of the group of 65 students ? (Rounded off to two digits after decimal) 151.25 150.15 151.10 150.19 None of these 6 . The total ages of a class of 75 girls is 1050, the average age of 25 of them is 12 yre and that of another 25 is 16 yr. Find the average age of the remaining girls. 12 yr 13yr 14yr 15yr None of these 7 . Average score of Rahul, Manish and Suresh is 63. Rahul's score is 15 less than Ajay and 10 more than Manish. If Ajay scored 30 marks more than the average scores of Rahul, Manish and Suresh, what is the sum of Manish's and Suresh's scores ? 120 111 117 Cannot be determined None of these 8 .", "+ 5 * 2) = 50 years. Husband’s present age = (90 – 50) = 40 years. 2. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is: A. 20 B. 21 C. 22 D. 23 Explanation: Let the total number of workers be x. Then, 8000x = (12000 * 7) + 6000(x – 7) = 2000x = 42000 = x = 21. 3. In an examination a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination? A. 8 B. 9 C. 10 D. 11 Explanation: Let the number of papers be x. Then, 63x + 20 + 2 = 65x = 2x = 22 = x = 11. 4. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ration of the number of boys to the number of girls in the", "of the two digits a and b? A. 8 B. 6 C. 2 D. 4 Question 154 The average monthly salary of 19 members of a group is Rs. 16000. If one more members whose monthly salary is Rs. 20000 has joined the group, then the average salary of the group is A. Rs.18250 B. Rs.16200 C. Rs.18000 D. Rs.16250 Question 155 Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is- A. 9 B. 25 C. 30 D. None of these Question 156 The average of 80 boys in a class is 15. The average age of a group of 15 boys in the class is 16 and the average age of another 25 boys in the class is 14. What is the average age of the remaining boys in the class? A. 14 B. 14.75 C. 15.25 D. Cannot be determined Question 157 A shop of electronic goods is closed on Monday. The average daily sales for remaining six days of a week is Rs. 15,640 and the average sale of Tuesday to Saturday is Rs. 14,124. The sales on Sunday is - A. Rs. 20,188 C. Rs.", "start earlier in making math and science attractive to girls. But unlike many in the industry who give up at that point, he’s willing to put some money where his mouth is — to the tune of that \\$50,000 stipend. Hacker School is a three-month long retreat to help make people better programmers, and Etsy is hosting it this year. He and the founders have set a goal to accept 20 women into the overall class of 40 with Hacker School retaining full control over the admissions process. That’s a big goal, as he notes, and would be 20 times the number of women in the current batch. It’s going to require folks spreading the word to female programmers, making sure Hacker School is an environment that doesn’t alienate women (or maybe even welcomes them), and the financial support that Etsy is offering. Will people grumble about this? Hell yes. Will that stop Etsy and Hacker School? I hope not. The lack of women in engineering is a problem, as things like the rise of the brogrammer culture make clear. As Hedlund writes: Twenty is a small number, but it would roughly match the total number of female engineers I’ve hired in the past 17 years. Even a small change can have a large impact, given the severity", "demographic of Mathcampers? Are there minorities (by gender, race, age, etc.) among the campers or is it more \"balanced\"? It's a big mix. Usually, about 1/3 girls, 2/3 boys; about 75% US, 10-15% Canadian, and 10-15% from other countries. The mean, median, and mode of the age distrituion are almost always 16. On the first day of Mathcamp 2015: 4 campers age 13 8 campers age 14 23 campers age 15 35 campers age 16 33 campers age 17 17 campers age 18 0 campers age 19 (That was slightly more young students than we usually see, and we do once in a while have 19-year-olds.) Math1331Math 2016-03-15 20:13:44 How are the classes run? Are they more proof-oriented or problem solving based? roed 2016-03-15 20:13:46 Classes vary based on the instructor, but more of them are about cool topics that you wouldn't normally see in high school than about problem solving. We usually have 1 or 2 problem solving classes each week. (One or two per day, that is, that continue throughout the week.) AllenWang314 2016-03-15 20:14:25 Are the courses aimed toward college or usamo/imo training? More towards college, but it's a mix. himaximushi 2016-03-15 20:14:41 What are the pepper things in this link? http://www.mathcamp.org/2015/academics/4WeekSchedule.pdf khanh93 2016-03-15 20:14:49 @himaximushi: Chili peppers! Some classes are \"spicier\" than others, which", "summer camp? (a) 97 (b) 107 (c) 117 (d) 127 Explanation: At summer camp there are 52 boys, 47 girls, and 18 adults. 57 47 +18 52 + 47 + 18 = 117 Therefore 117 people are at summer camp. The correct answer is option C. Question 2 At camp, 32 children are swimming, 25 are fishing, and 28 are canoeing. How many children are swimming, fishing, or canoeing? (a) 75 (b) 85 (c) 95 (d) 105 Explanation: At camp, 32 children are swimming, 25 are fishing, and 28 are canoeing. 32 25 +28 Make a group of 10. 32 25 +28 8 + 2 = 10 1 will be carried to the tens place. 5 will be in the ones place. 30 + 20 + 20 + 10 = 80 80 + 5 = 85 The correct answer is option B. Spiral Review Question 3 Four students estimated the width of the door to their classroom. Who made the best estimate? (a) Ted: 1 foot (b) Hank: 3 feet (c) Ann: 10 feet (d) Maria: 15 feet Question 4 Four students estimated the height of the door to their classroom. Who made the best estimate? (a) Larry: 1 meter (b) Garth: 2 meters (c) Ida: 14 meters (d) Jill: 20 meters Question 5 Jeff’s dog weighs 76", "30 D. Rs. 40 Question 396 The average of seven numbers is 18. If one of the numbers is 17 and if it is replaced by 31, then the average becomes : A. 21.5 B. 19.5 C. 20 D. 21 Question 397 Out of 20 boys, 6 are each of 1 m 15 cm height, 8 are of 1 m 10 cm and rest of 1 m 12 cm. The average height of all of them is : A. 1 m 12.1 cm B. 1 m 21.1 cm C. 1 m 21 cm D. 1 m 12 cm Question 398 The average age of 11 players of a cricket team decreases by 2 months when two new players are included in the team replacing two players of age 17 years and 20 years. The average age of new players is : A. 17 years 1 month B. 17 years 7 month C. 17 years 11 month D. 18 years 3 month Question 399 A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is : A. 285 B. 290 C. 297 D. 305 Question 400 In a school with 600 students, the average age of the", "number of shoelaces? If so, how many? ## Passwords, Age Restrictions, and Computer Silliness My computer has been a bad boy recently. First, it told me that my password is going to expire approximately 11 months before I was born Interestingly, the folks at www.timeanddate.com disagree with the number of days between March 31, 1970, and the date that screen capture was snapped (March 1, 2015). So much for the truism that, “Computers make very fast, very accurate mistakes.” I thought the difference could be explained by excluding the end dates, but that doesn’t seem to be the case, so I’m not sure what ADPassMon is doing. (Then again, I’m not sure why I’m wasting my time checking the calculations of a piece of software whose warning messages suggest the existence of time travel.) Then, when attempting to register my sons for ski camp, it gave one of the craziest age restrictions I’ve ever seen… check out the valid ages… An age of 5.925 corresponds to 5 years, 11 months, 7 days, and 15 hours, which seems quite an arbitrary cut-off for a ski camp. Further, an age of 7.999 years means that kids are eligible for ski camp so long as they are not within 15 hours, 14 minutes, and 24 seconds of their eighth birthday. The", "### The average age of 11 players of a cricket team decreases by 2 months when two new players are included in the team replacing two players of age 17 years and 20 years. The average age of new players is : A. 17 years 1 month B. 17 years 7 month C. 17 years 11 month D. 18 years 3 month Answer: Option B ### Solution(By Apex Team) Sum of age of two players = 17 + 20 = 37 years Decreasing age of players = 11 × 2 = 22 months Sum of age of two new players = 37 years – 22 months = 35 years 2 months Average = 17 years 7 months Alternate : $\\begin{array}{l}\\large\\frac{(17+20)-(\\text{ Sum of two new players })}{11}\\\\ =\\large\\frac{2}{12}\\\\ =\\large\\frac{1}{6}\\end{array}$ Sum of two new players $\\large=\\frac{211}{6}$ Average of two new players $\\large=\\frac{211}{12}$ = 17 years 7 months A. 20 B. 21 C. 28 D. 32 A. 18 B. 20 C. 24 D. 30 A. 10 years B. 10.5 years C. 11 years D. 12 years" ]

[ "the value of x is 7.", "= 7", "find that x = 7.", "$EG=7$EG=7, find the value of $m$m.", "7$ and $x = 7$", "or $x = 7$", "7 = 0$ $x = 7$", "# TSI Question 1 If $-7x+2=-6$ then what is the value of $-2x$ a. -4 b. -2 c. 3 d. 4 Go to problem 2.", "# Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "= -1 the value of 2x - 7 = -9 Exams Articles Questions", "7 = (x+2y-2)^2 + (y+1)^2 + 2$$", "Q # Find the value of the following expressions when x = - 1, (i) 2x - 7 Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "# What is x if 5x - 2 = 2x + 7? Jun 5, 2018 $x = 3$ #### Explanation: $5 x - 2 = 2 x + 7$ $5 x - 2 x = 7 + 2$ $3 x = 9$ $x = \\frac{9}{3}$ $x = 3$", "Home » » If the expression x$$^3$$ + 2hx - 2 is equal to 6 when x = -2, what is the value... # If the expression x$$^3$$ + 2hx - 2 is equal to 6 when x = -2, what is the value... ### Question If the expression x$$^3$$ + 2hx - 2 is equal to 6 when x = -2, what is the value of h? A) 0 B) -2 C) -4 D) 4", "7$$ $$3x − y = −2$$", "7 = 0(x + 3) + 7 …..[From (i)] = 7", "the result is unknown, so we are finding the value of the quotient. In the completed equation $$14 \\div 2 = 7$$, we see that the value of the quotient is 7.”", "## Algebra 2 (1st Edition) Plugging in the given value of $x$, we find: $$f(x)=7x-7 \\\\ f(4) =7(4)-7 = 21$$", "the value of $4x^{3}+2x^{2}-8x+7$ ⇒ 2x = $\\sqrt{3}+1$ 2x – 1 = √3 Squaring both the sides, we will get, $(2x-1)^{2}=3$ ⇒ $4x^{2}-4x+1=3$ ⇒ $4x^{2}-4x+1-3=0$ ⇒ $4x^{2}-4x-2=0$ ⇒ $2x^{2}-2x-1=0$ Now, let us take, $4x^{3}+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $4x^{2}+2x+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $6x^{2}-6x+7$ Since, $2x^{2}-2x-1=0$ ⇒ 2x(0)+3($2x^{2}-2x-1$))+7+3 ⇒ 0+3(0)+10 ⇒ 10", "$3=7$, the value for $x$ is incorrect. Later on, the goal isn't to guess a value for the variable, but to solve the equation." ]

[ "\\\\ \\text{Dividing both sides by 2}, \\\\ x= \\boxed{\\mathbf{3}}$$", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "$$3x = 16$$. Multiplying both sides by $$2 \\over 3$$ shows $$x +z = \\color{brown}\\boxed{32 \\over 3}$$, just as Guest found May 31, 2022 edited by BuilderBoi May 31, 2022", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "is $$\\boxed{ y=x^2+bx+c }$$ To Complete the Square we add and subtract $$\\textcolor{blue}{({b \\over 2})^2}$$ which yields: $$\\boxed{ y=x^2+bx+ \\textcolor{blue}{({b \\over 2})^2} + c – \\textcolor{blue}{({b \\over 2})^2} }$$ Completing the Square yields: $$\\boxed{ y=(x+b/2)^2 + c – ({b \\over 2})^2 }$$ … Read more", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "+ b^d$$. Then $$S(b, d) = \\boxed{(2^{k-1}-1)(1 + b^d) + 1}$$. For the case where $$d = 3$$ and $$k = 2$$ such as when $$b = 21$$, the answer will be $$\\boxed{1002_b}$$ as above. Labels:", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "equation,$x = 18/2 =9$, thus, the answer is $\\boxed{\\textbf{(C)}\\ 9}$. ---LarryFlora", "= {y - 2 \\over 0} = {z - 2 \\over -2} = t$$ or equivalently $$\\boxed{x = t + 1, \\ \\ y = 2, \\ \\ z = - 2 t + 2}$$", "$1$ to both sides. $3 = y$ or $y = 3$", "both sides $x + 1 = - 2$ subtract 1 from both sides", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "02:43 PM lvleph Based on the way I did it, the boxes would be 9/10 and 9, respectively.", "Multiply both sides by 2: 2.5 = (x - 10)", "= 4$ Add $1$ to both sides. $x = 5$", "Adding 2 to both sides gives $2x^2 = 4$ 13. Thank you." ]

[ "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "$ \\boxed{G'(x)=5 e^{x-1}-2x-2} $ Now there is probably an error in there somewhere, but you should get the idea. RonL", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \\boxed{448}$.", "the value of $4x^{3}+2x^{2}-8x+7$ ⇒ 2x = $\\sqrt{3}+1$ 2x – 1 = √3 Squaring both the sides, we will get, $(2x-1)^{2}=3$ ⇒ $4x^{2}-4x+1=3$ ⇒ $4x^{2}-4x+1-3=0$ ⇒ $4x^{2}-4x-2=0$ ⇒ $2x^{2}-2x-1=0$ Now, let us take, $4x^{3}+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $4x^{2}+2x+2x^{2}-8x+7$ ⇒ 2x($2x^{2}-2x-1$)+ $6x^{2}-6x+7$ Since, $2x^{2}-2x-1=0$ ⇒ 2x(0)+3($2x^{2}-2x-1$))+7+3 ⇒ 0+3(0)+10 ⇒ 10", "\\\\ \\text{Dividing both sides by 2}, \\\\ x= \\boxed{\\mathbf{3}}$$", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "and $$\\boxed{÷10}$$ are pressed after them. Similarily, $$\\boxed{-10}$$ may be worth $$-10$$, $$-100$$, $$-1000$$... or $$-1$$, $$-0.1$$, $$-0.01$$... The required number needs to be constructed from these final worths of $$\\boxed{+10}$$ and $$\\boxed{-10}$$. The easiest way to write $$2019$$ in this way is $$2019 = 1000 + 1000 +10 + 10 -1$$ That means you need: • Two instances of pressing $$\\boxed{+10}$$ that have their value increased 2 times • Two instances of pressing $$\\boxed{+10}$$ that have their value neither increased or decreased • One instances of pressing $$\\boxed{-10}$$ that have their value decreased 1 time Therefore in total you'll need to press $$\\boxed{+10}$$ four times, and $$\\boxed{-10}$$ one time. Somewhere between them you'll need to press $$\\boxed{\\times 10}$$ and $$\\boxed{÷10}$$ appropriate number of times, so that two of the $$\\boxed{+10}$$ had their value increased twice, two $$\\boxed{+10}$$ had their value unchanged, and $$\\boxed{-10}$$ had their value decreased once. There are actually three ways to do this in a minimal number of operations. One is the one you've found: $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}$$ The other two are $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}$$ $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}\\boxed{+10}$$", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "we get $c=-3,b=3,a=0$ and our function $f(x)$ becomes $3x-3$.if $f(x)=3x-3$, $\\boxed{f(10)=3*10-3=27}$", "+ b^d$$. Then $$S(b, d) = \\boxed{(2^{k-1}-1)(1 + b^d) + 1}$$. For the case where $$d = 3$$ and $$k = 2$$ such as when $$b = 21$$, the answer will be $$\\boxed{1002_b}$$ as above. Labels:", "= {y - 2 \\over 0} = {z - 2 \\over -2} = t$$ or equivalently $$\\boxed{x = t + 1, \\ \\ y = 2, \\ \\ z = - 2 t + 2}$$", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "$GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\\boxed{\\mathrm{(B)}\\ 20}$" ]

[ "Powers. Elements A Sorcerer aligns himself with a particular element. Naturally certain races will already have themselves aligned with certain elements. There are six elements within Glorantha these are shown below with their relation to each other. \\begin{verbatim} FIRE MOON \\ / \\ / EARTH ---------- AIR / \\ / \\ DARKNESS WATER \\end{verbatim} A sorcerer aligned to one element will find it difficult to use other elements. The further he moves around the spoke away from his element the harder it becomes. Bob is a human and was born under the Water Element( Based on when during the year and on which day he was born. Trolls automatically are aligned with the Darkness rune etc.). He discovered sorcery and found it reasonably easy to learn to control the water rune when he attempted to train in the other elements he came to this conclusion: \\begin{verbatim} Darkness,Air :- Medium Difficulty Skill Earth,Moon :- Hard Skill Fire :- Very Hard Skill \\end{verbatim} \\subsection{Forms} A sorcerer is tied to his particular rune that is relevant to his species. Learning in this skill is a medium skill, but is a hard skill for all other forms. \\subsection{Conditions} These runes will usually be present within the formula that the sorcerer uses when casting. Reasoning being that these runes effect the final outcome", "well first. However the confusion comes from what happens if he drinks multiple poisons that do not cure each other first (what if someone was given well 5, but they first drink 2 before they drink 6?). Unless I'm misinterpreting the rules, than the second post is the correct solution. • The dragon does not have that guarantee because the knight can drink from well 1 before the duel begins, in which case the well 6 water would cure him. The dragon can counter this by giving him fresh water or well 1 water, so the knight must add on the safeguard of drinking from well 1 and then well 2 after the match, which will nullify everything. (same works for the dragon, just imagine well 6 doesn't exist) - the dragon has another guarantee. X,5,5,6. After drinking whatever the knight gave him, he could drink from well 5 twice then drink from well 6. If the knight gave him lake water, drinking from well 5 would poison him and then drinking from 6 would cure anything. If the knight gave him 1-4 then 5 would cure it the first time and he would be poisoned again from the second drink, then well 6 would cure that poison. If the knight gave him 5 then the 2 drinks", "easy. The healing potion doesn't heal, it just interacts with the body's natural energies to allow bruising and cuts to knit together. Just like antibiotics don't repair necrotic flesh, but they do allow the body to heal (by removing the infection). Generation Y. I don't remember the First Gulf War, but do remember floppy disks. Yakk Poster with most posts but no title. Posts: 11115 Joined: Sat Jan 27, 2007 7:27 pm UTC Location: E pur si muove ### Re: Dungeons and Dragons (and other tabletop RPGs) Have a healing potion drunk when the player is out of healing surges grant 1/2 the amount it would heal as temporary HP. In effect, the healing potion isn't working very well anymore, and you notice it isn't working, and that you need rest before drinking more. At the same time, you cannot glug down gallons of healing potion to heal any wound. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total. Chen Posts: 5570 Joined: Fri Jul 25, 2008 6:53 pm UTC Location: Montreal ### Re: Dungeons and Dragons (and other", "# Laxative Potion 14,883pages on this wiki \"Laxative Potion\" —The potion included in Moste Potente Potions[src] The Laxative Potion was a magical concoction that was included in Chapter VI of Moste Potente Potions, of which a copy was kept in the Restricted Section of the Hogwarts library. The page was opposite Chapter XVI, which concerned the Polyjuice Potion. Known instructions in the book are to add two scrumples of a certain fine ingredient, another ingredient that has been used in a furnace, and Agrimony[1].", "them the virtues of plants, animals and minerals, or in some padl the method of tincturing, te ink into which its drops are iiate the Key Sigil to which it cases even the ensouling essences of the Master and His Lady. ,f.. parts obtained via sacrificial rituals or minerals obtained and/or Plants ritually harvested or otherwise hallowed and re-enspirited through the power of the Thorn-Crowned Master or His Lady, animal � activate the power of each empowered ceremonially are the endowing elements possessing the !I to the workings of each Key abilities needed in order to confer the required powers into the ink !iated forms and techniques of base. .�..- .... powder, tincture and sachet When plant elements are used in the art of ink making it is the tinctured essence that one most often employs, even if also powders Jilnal setting would create an made from plant parts could be used, as the tincture allows the ping of the powers connected extraction of their essences in a concentrated liquid form which often - better lends itself to the conjunction with the carrier ink base. � 'i. \"-hich these given formulas pe Sorcerer and through their rjasight concerning the virtues r .. \"' F b�· to empower and bless the p1 formulas and create aspect­ pvs of His", "whatever suits your world most. Also the quantity aspect solves why it was never apparent that it was magical. Until we made tons of it it was making so little magic that people did not notice. 2 Penicillin. Penicillin like thing. I'm not good with chemistry so bear with me. The same Aluminium idea is the same. Only it's something like a plant or a combination or things. Basically penicillin. Read on the actual penicillin and change whatever you feel like it. This again means that it possible to create the stuff of magic in a lab. We only needed to mess around with the right stuff and viola. MAGIC. While at the same time insures that no society before those with your limitation can make them. 3 A rare natural element that is only accessible with deep mining capability. Obvious things is obvious. We needed to go so deep underground or underwater to extract Contrivium, terrible writing advice if you don't know or the classic: Phlebotinum So after the technology of the world advanced to that point we started seem this thing. And it did not take long to people to figure out it's magical. And even if people back in the day found an ore or a gem or whatever. They still could not get more", "Convention for Healers during a war where some rapid healing is available? Imagine a war in a standard fantasy world, roughly medieval, most magic is limited, but magical healing exists. Healing comes in three forms: 1. Rare Healers, who are born with ability to do rapid magical healing, taking even severely injured to full health in minutes. However, they exhaust their own Healing energies rapidly doing so, and thus can only heal a few wounded this way before exhausting themselves entirely. These are far rarer than trained medics, but not unheard of, and a number will be available to each side in a battle. If a Healer exhausts his energy, it can take a few days to fully recover. It is easier for a Healer to heal recent wounds then old ones, though having time to focus on Healing without rushing, and ability to triage who to Heal better, means that they can heal a slightly more effectively if not in the heat of battle even if the wounds are a little older. After a battle, once healers recover their expended energies, they will work to heal those wounded from previous battles, but since the wounds are now much older, they can not heal these older wounded nearly as fast. 2. Alchemical potions that can be used for", "Surprise remains • Faster than rolling dice in game • You are not limited to healing potions • The content may differ from the label (adulterated healing potion, or even worse - poison?) • Sometimes, the players won't be able to tell. Did this Superior Healing potion heal only 17 hit points because of a lack of luck, or because it was a Greater Healing Potion? • Man now I want to do this purely for the shenanigans of wrong potions. Insight and slight of hand checks just got way more interesting. – linksassin Mar 4 '19 at 2:47 • +1 This is a really clever idea and that I might try sometime. Nice outside-the-box thinking. – Bloodcinder Mar 4 '19 at 16:37 # With a healthy mix of good and bad dice. Potions use d4 in pairs. A single d6 can be altered to allow only 4 different outcomes: the downside is that it becomes unfair. You can compensate this by creating another d6, with its own 4 different outcomes, so that it generates fair results when rolled with the other one. • Use white or green for \"good dice\". • Use black or red for \"bad dice\". # Using standard dice: just flip them! Good dice can't roll bad numbers. Flip them whenever you roll 1", "features away from them and giving them to every other sorcerer archetype. I thought to mention this in case this impacts balance, and therefore any potential answers, in a way I haven’t anticipated; but this is an aside, just some extra information, and my question is not specifically about this aside, but about the impacts of having any sorcerers with 20 spells.", "have been guilty of. take for example potions. in some games, like diablo, you wind up quaffing hundreds of them through the course of the game. there are three types of potion in diablo... healing, mana, and rejuve. now, in the REAL world, potions correspond to medicines, and here is what is commonly known about medicines: A: some people are allergic to some medicines.(i personally am allergic to penicillin) B: most medicines have side effects(at least, the strong ones do) C: many medicines are addictive D: the more you take of a medicine, the less effective it is(tolerance) now, bring this into diablo-world. nobody is ever allergic to any type of potion. there are no side effects. they are not addictive, and you dont develop a tolerance. as far as i've seen, NONE of these have ever been incorporated into a CRPG. since we cant make the game unplayable, we must cover all of the functions that the potions do and have them available to the player. so, we come up with the following combinations: #1: the character has no allergy #2: the character is allergic to mana potions #3: the character is allergic to health potions #4: the character is allergic to rejuve potions #5: the character is allergic to both health and mana potions this list", "Healing potion, et cetera. $$\\^1\\$$ I totally agree with Wyrmwood in doing this. $$\\^2\\$$ I skipped the 4th edition, hence I jumped from the 3.0 directly to the 5th one, and at the beginning the spell system puzzled me a little bit.", "of ,..0 to the bewitching work the Green and the Kingdom of the Graves. All harvested parts of the It rites of strewing, in which t· when the living are to be to work her witchcraft, but the most potent share of such harvest will always be the sweet Liquorice roots, which hold her full powers to edulcorate and dominate. Such roots are carried as talismans or made into powders and employed within all workings of enchanting and will-bending sorcery, but also within more sacred and secret rites where fractions of the boundless essence of the Rose Crowned Queen are to be seated within fetishes or in some other way bless and consecrate Vessels of Holiness with Her Sweetness and Might. 197 flalt 1J1trn (Dryopteris ftlix-mas) • us &om this. ..... ))lesgnp- The daemon of the Male Fern is a spirit with immense powers. He can conceal and make invisible and protect against all manners of baneful sorcery. He can make the curses and the foul emanations of the evil eye of the enemy miss their targets and rebound back upon him- or herself. He can also bring luck, wealth and success and point out hidden treasures and disclose other such veiled sources of material !I abundance. This spirit is also well-known for his mastery of the arts", "have been poisoned or cured? If the knight drinks from well 1 and the dragon gives him well 2, can the knight tell if he is still poisoned and should look for a cure, or is he uncertain until he dies? I'm assuming the latter case. One way for the dragon to always survive is to drink nothing beforehand, but after drinking the knight's poison, drink twice from well 5 and then drink from well 6. If the knight offers any poison but well 5, it will cure it, and the second drink will be cured by well 6. If the knight offers well 5, the first 2 drinks will do nothing and well 6 will cure it. If the knight offers clean water, the drinks from wells 5 and 6 will cancel out. I'm not sure if there's a similar strategy for the knight, because he doesn't have the cure-all of well 6. I don't believe it's possible to guarantee a win. Anything the other side offers can be canceled out by choosing some appropriate drinks either before or afterward. Therefore, if there was a winning choice of poison, one player would know that the other would always pick that strategy and would be able to pick the drinks before or afterward that would cure the winning", "# The Witch Invites Halloween is just around the corner and the witch is planning a little party for the 100 villagers of the nearby village. Since the villagers burned down her house again and again during the last few years she would like to use this chance to make peace with them. At least that is what she told the villagers. Of course an evil witch would not simply forgive being attacked and so she collected the essences of animals, to be exact the essences of bears, hawks, and spiders, during the past year. Her plan is to use these essences to concoct potions which would transform the drinker into a creature with characteristics of all 3 animals. Of course there are rules for these potions: • The witch must concoct each potion on its own so 100 potions in total. • For the concocting process all 3 essences are needed in a specific order, therefore each potion must contain at least a small amount of each essence. • Each potion must must be concocted of the same amount of combined essences to get potions of equal strength. Otherwise the villagers who drank a stronger potion would be able to cancel the transformation of those who drank weaker potions. • So the villagers would never find happiness", "rules system. I used it for original D&D and GURPS. One nice thing about this particular supplement is that it distinguishes between the raw plant and the potion that can be made from it. Both have effects in game terms. Some potions are made from a combination of plants. - King Arthur Pendragon also has a usable system. An algorithm using physical stats (like Constitution) determines how many hit points per week are healed. Any serious wound, disease, significant blood loss, etc. requires a specialist to heal. They heal more often than not; but, the odds are not really that good. Permanent damage is commonplace. - There was a supplement for Warhammer Fantasy Role Play 1ed. called Warhammer Grimoire, where an alcoholic surgeon is described, together with broader rules about healing and well, cutting off limbs. This was all pretty grim and gritty, dirty and not very efficient, but at the same pretty cool (don't get your self anesthetized, because you can wake up without some important parts of your body). - May I ask, why was my answer down voted? I proposed a supplement with well described late medieval medic and his practice. What is wrong with this answer? – gruszczy Oct 29 '10 at 17:08 Someone just decided to down-vote every answer that mentioned a game", "2015 04-16 # Healing Do you know the popular game World Of Warcraft (WOW) made by Blizzard? Tom plays a Blood Elf Paladin in WOW. The Paladins have 3 kinds of talent trees: Holy, Protection and Retribution. The Holy Paladins in a team are healers, who provide healing service, heal the wounded; the Protection Paladins in a team are tanks, who attract the enemy’s fire and protect the teammates; the Retribution Paladins in a team are melee damage dealers. After the version Cataclysm is published, the Retribution Paladins are found weakened by Blizzard. But fortunately, the Holy Paladins become more and more powerful amount the healers. In order to join the dungeons and raids easier, Tom choose the talent of Holy to be his role’s main talent (because the dungeons and raids are lack of healers). What a nice job the Holy Paladin is! Instead of standing in front of the monsters and enduring the hit, the only thing the Holy Paladins should do is to heal teammates and hardly worry about the monsters would attack them. Unlike the Retribution Paladins’ operation of \"Keyboard Rolling\", the Holy Paladins have many spells to use of healing, so that some part of the pressing of some spells would be overlapping with each other. Furthermore, once a spell was cast successfully,", "of stars and midnight's moon. Michael comes upon this day, To watch you grow... Herbalism >> Herbs dictionary A Malkuth The practical applications of the Tree must start at Malkuth, because Malkuth is all that is manifested on Earth. We know Malkuth better than any other Sephiroth just because we are the embodiment of Malkuth. We know the physical world by living it. Malkuth... Magick >> Qabalah Resurrection of the Higher Self In July of 1989, a controversial article crossed the managing editor's desk of an equally controversial magazine. This article was never actually published in the quarterly publication, then known as The Magical Blend , even though it was well documented by... Body Mysteries >> Psychoactive Substances Chamomile: The Healer Chamomile is one of the oldest favorites amongst garden herbs and its reputation as a medicinal plant shows little signs of abatement. The Egyptians reverenced it for its virtues, and from their belief in its power to cure ague, dedicated it to their gods. No... Herbalism >> Herbs dictionary C Variance Properties is described in multiple online sources, as addition to our editors' articles, see section below for printable documents, Variance Properties books and related discussion. ## Suggested Pdf Resources Expected Value and Variance a large number of times. We first need to develop some properties of", "a bush. Cawlin: “There! Right over there! A red pepper!” Rupert picks the red pepper. RR: “Okay, I’ve got the red pepper. Now what?” Cawlin: “The red pepper has the essense of red magic! That’s why it burns your mouth when you eat it. We must extract this magical essence and use it to write your first spell. Let’s go back home.” Rupert and Cawlin return to the mossy tower. Cawlin: “Everything has a bit of magic in eet. It is the job of the alchemist to extract this magic and brew bottles of magical extract. Many mortals don’t rrrrealize what they’re actually doing, but they treat these magical extracts as ‘medicines’, but it’s actually magic at work. A brewed potion has potency, depending on the skill of the alchemist and the ingredients used.” RR: “I’ve never brewed a potion. Where do I begin?” Cawlin: “Well, you don’t really have a prrrrroper alchemist work bench, so we’ll just have to use the most rrrrrrudimentary tools available to extract the magical essence from the red pepper. You must crush the red pepper between some rocks, and you’ll get a little bit of red magic essense. Try it now.” Rupert places the red pepper on a slab of rock and smashes it with a rock. A few seconds later, small", "boundaries of the progression. I liberally hand out potions of healing, through loot and quest givers. I’m wondering, however, what would happen if extended this to other healing potions. For example, the loot in one of the more dangerous dungeons would contain four potions of greater healing instead of four potions of healing, or I replace half of all potions of healing in loot with potions of greater healing, or I increase the quality of the healing potions given out with tier of play. My question therefore is: Given the crafting rules on XGtE p. 130 (and potentially other rules that I overlooked), are healing potions exempt from magic item progression? If not, what would be the balance implication of changing that? Warlock is as viable as other classes So, let us start by your assumptions: it seems like the easiest way to play one is as an Eldritch Blast spammer. I understand the damage output can be quite high. This is right. While it is not the only way to play it, it is arguably the easiest and, as this answer shows, a very viable choice. Now, let us go to the problem I see you are having: However, at mid to high levels don’t enemies start having resistance to spells below a certain level, making EB", "# Sorcerer Origins Vs Archetypes So I'm in the midst of character creation, and I was looking to see how multi-faceted I can make my character. To give greater potential for stuff later on. I've got a Sorcerer on my hands, and every information page I can find on how to set them up eventually refers to a feature called a Sorcerous Origin, of which I've found Draconic Ancestry, Wild Magic and a relatively new feature called Favoured Soul to be the only \"Origins\" available. That's all well and good, so I chose Draconic. But now what I'm exploring is the concept of Archetypes. Sorcerous Origins seem to be something you need to pick regardless of the path you want to take with your character, whereas an Archetype (from my understanding) is an optional thing. So my question is, can I have my Sorcerous Origin and an Archetype (along with all the features and traits they unlock) at the same time? And if so, do those features/traits/skills/etc coexist unless they specifically mention replacing a previously existing feature? • What do you mean by \"an Archetype (from my understanding) is an optional thing\"? An archetype is like an specialization of the classes that grant aditional and unique class features, you HAVE to pick one. – Manner Nov 18 '17" ]

[ "selecting 2 out of 5 as roots if different, and 5 if equal, for a total of: $$\\binom{5}{2} + 5 = 10 + 5 = 15$$ The total number of monic polynomials $x^2 + c_1 x + c_2$ is just $5 \\cdot 5 = 25$ (can select $c_1$, $c_2$ at will), so there are $25 - 15 = 10$ irreducible ones. - You are right, I had computed the total number as 15 in my head and wrote down a more detailed derivation mixing up the numbers. Thanks! – vonbrand Apr 11 '13 at 12:28", "Output 5 Note For the sample, you can drink $5$ potions by taking potions $1$, $3$, $4$, $5$ and $6$. It is not possible to drink all $6$ potions because your health will go negative at some point", "half k diamonds. Now, if we add all of them, $3k+k+\\frac{1}{2}k=18$ $\\therefore k=4$. So, there are 12 rubies, 4 emeralds and 2 diamonds. The solution posted by Quacky is great, but I don't think problems will always be simple like this one. 4. Originally Posted by lanierms You can also use a constant $k$ to figure it out quite easily. Solution: Ruby:Emerald = $3:1$. So, Ruby = $3k$, Emerald = $k$. Emerald: Diamond = $2:1$ = $k:$ Diamond. $\\therefore Diamond=\\frac{1}{2}k$ So, there are 3k rubies, k emeralds and half k diamonds. Now, if we add all of them, $3k+k+\\frac{1}{2}k=18$ $\\therefore k=4$. So, there are 12 rubies, 4 emeralds and 2 diamonds. The solution posted by Quacky is great, but I don't think problems will always be simple like this one. Ah, very simple. Thank you both.", "= lemon juice. then since we know there is $3$ cups of lemon juice, do the math. $3\\cdot2\\cdot4=6\\cdot4=24$ ~ bsu1 ~savannahsolver", "different ways to choose K objects from a larger set of N objects. Three rocks: 40/720. If three rocks are in their correct places, then that means the other three rocks have been mixed up. There are “6 choose 3” or 20 ways to select the three mixed up rocks. But they could be mixed up in different ways. Say these three rocks were meant to be in order: ABC. There are two ways they could all be in the wrong places: BCA and CAB. So 20 times two is 40. Two rocks: 135/720. First, choose the four misplaced rocks: there are “6 choose 4” or 15 ways to do so. Then count the ways to misplace four rocks. Suppose they belong in the order ABCD. They could all be misplaced nine different ways: BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCBA and DCAB. Nine times 15 is 135. One rock: 264/720. There are six ways to choose the rock that’s in the correct place. (Equivalently, there are six ways to choose the five rocks that aren’t in the correct place.) For the five misplaced rocks, there are 44 ways to misplace them all. Six times 44 is 264. Zero rocks: 265/720. Because we know there are 720 total possible arrangements, we can simply subtract those we’ve already", "and $4$ are multiple roots.", "Each of these lines begins with an integer $1 \\leq M$, denoting the number of ingredients required to make this recipe. Then, $M$ integers follow, describing the required ingredients. The ingredients are identified by integers between $0$ and $500\\, 000$, inclusively, with different integers denoting different ingredients. For each recipe, all of its ingredients will be distinct. The sum of $M$ over all recipes will be no greater than $500\\, 000$. ## Output Print the number of recipes you will concoct. ### Sample Data Explanation In the first example, the first potion can be concocted, since both ingredients were not used so far. Thus, you will concoct this potion. The second potion will also be concocted for the same reason. The third potion cannot be concocted, since ingredient $1$ is no longer present, and is inside a cauldron mixed with another ingredient not present in the recipe. The fourth potion can be concocted by pouring the content of the cauldrons used for the first and second concoctions, and then adding ingredient $5$, which has remained unused so far. The last potion cannot be concocted, since the content of the cauldron for the first potion has all been poured when making the fourth potion and thus is now mixed with other ingredients not present in the recipe. For the", "power, or 5 to the third, or 5 cubed. When a whole number is raised to the power of 4 or higher, we simply say that that number is raised to that particular power. The number ${5}^{8}$ can be read as 5 to the eighth power, or just 5 to the eighth. ## Roots In the English language, the word \"root\" can mean a source of something. In mathematical terms, the word \"root\" is used to indicate that one number is the source of another number through repeated multiplication. ## Square root We know that $\\text{49}={7}^{2}$ , that is, $\\text{49}=7\\cdot 7$ . Through repeated multiplication, 7 is the source of 49. Thus, 7 is a root of 49. Since two 7's must be multiplied together to produce 49, the 7 is called the second or square root of 49. ## Cube root We know that $8={2}^{3}$ , that is, $8=2\\cdot 2\\cdot 2$ . Through repeated multiplication, 2 is the source of 8. Thus, 2 is a root of 8. Since three 2's must be multiplied together to produce 8, 2 is called the third or cube root of 8. Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or", "past geological periods 15 there seems to an! ” Merriam-Webster.com dictionary, Merriam-Webster, https: //www.merriam-webster.com/dictionary/root are given anywhere between 2 and roots. Anywhere between 2 and 5 roots to learn ( but remain solid ) you... Objectives Of European Monetary System, Anastasia Kreslina Net Worth, Cleveland Jr Theory, Rca Trinity Antenna, Garden State Gladiators Roster, Political Ideologies And Communities,", "choose the third number, and $$15$$ ways to choose the fourth number. Of these choices, she has four ways to choose one of the four numbers Jeff chose on her first attempt. If she successfully chose one of those four numbers, there are three ways for her to choose one of the other three numbers Jeff chose on her second attempt. If Caitlin's first two choices were successful, there are two ways for her to choose one of the other two numbers Jeff chose on her third attempt. If all three of her first three attempts were successful, there is one way for her to choose the remaining number Jeff chose on her fourth attempt. Hence, the probability that Caitlin chooses all four numbers that Jeff selected is $$\\frac{4}{18} \\cdot \\frac{3}{17} \\cdot \\frac{2}{16} \\cdot \\frac{1}{15}$$ You should verify that the two methods yield the same answer. There are $$3×3=9$$ silver coins . You can make a tree for good visualization . There are 5 colors , 3 shapes branch and then there are 3 sub branch of letters on it . Now for letter A put letters on top of tree so there are $$5×3=15$$ Coins . As for your next question Does Caitlin need also to guess the order of balls drawn . In that case answer", "$= n\\cdot 3^{n-1}$ Once you have seen the answer, it's easy to derive it more quickly. There are n possibilities for the one element that is in $A\\cap B$. For each of the other n–1 elements, there are three possibilities: (i) the element belongs to A, (ii) the element belongs to B, (iii) the element belongs to neither A nor B. Total number of choices is therefore $n\\cdot 3^{n-1}$. (I would not have noticed that without seeing hans.kruger11's solution!) 6. That is a much clearer way to put it and makes much more intuitive sense. Good work opalg", "be 37.9 grams. What was the percent Fe3O4 in the sample of ore? Answer in 99. ## Math - Calculus Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2].", "in this collection? How many silver coins are in this collection? We are told that there is one coin with each combination of characteristics. Since we are told that the coin is silver, there is only one way to choose its color. There are three choices for its shape and three choices for the imprinted letter. Hence, there are $$1 \\cdot 3 \\cdot 3 = 9$$ silver coins. How many coins have the letter A on them? Since we are told the coin has the letter A on it, there is only one way to choose its letter. There are five ways to choose its color and three ways to choose its shape. Hence, there are $$5 \\cdot 3 \\cdot 1 = 15$$ coins with the letter $$A$$ imprinted on them. How many ways can Jeff choose four balls? What is the probability that Caitlin correctly guesses the four numbers? Method 1: Caitlin must guess all four numbers correctly, which can be done in $$\\binom{4}{4} = 1$$ way, while selecting four of the eighteen numbers. Hence, the probability that Caitlin guesses all four numbers correctly is $$\\frac{\\dbinom{4}{4}}{\\dbinom{18}{4}}$$ Method 2: Let's correct your attempt. Caitlin chooses four different numbers. Therefore, she has $$18$$ ways to choose the first number, $$17$$ ways to choose the second number, $$16$$ ways to", "so that there are at least $n \\ge 5 \\cdot 5 + 1 = 26$ elements, contradiction. If an element appears $4$ or more times, the remaining elements of the $4$ boxes must be distinct, leading to $5 \\cdot 4 + 1 = 21$ elements. The fifth box can contain at most four elements from the previous boxes, and then the remaining two elements must be distinct, leading to $n \\ge 2 + 21 = 23$, contradiction. However, by the Pigeonhole Principle, at least one element must appear $3$ times. Without loss of generality suppose that $1$ appears three times, and let the boxes that contain these have the elements $\\{1,2,3,4,5,6\\},\\{1,7,8,9,10,11\\},\\{1,12,13,14,15,16\\}$. Each of the remaining five boxes can have at most $3$ elements from each of these boxes. Thus, each of the remaining five boxes must have $3$ additional elements $> 16$. Thus, it is necessary that we use $\\le 22 - 16 = 6$ balls to fill a $3 \\times 5$ grid by the same rules. Again, no element may appear $\\ge 4$ times, but by Pigeonhole, one element must appear $3$ times. Without loss of generality, let this element be $17$; then the three boxes containing $17$ must have at least $2 \\cdot 3 + 1 = 7$ elements, contradiction. Therefore, $n = 23$ is the", "There are $n$ chemicals. One is magic, turning lead into gold in one day. You may (i) mix multiple chemicals with lead in one flask; (ii) add one chemical to multiple flasks. You have $p$ flasks, and an unlimited supply of lead and of each of the $n$ chemicals. What is the smallest number of days you need to guarantee you find the magic chemical? 1. Bob choosse a letter randomly from the English alphabet. You may question a range (e.g. from ‘a’ to ‘o’ inclusively) and he will tell you whether the letter is within the range. You want to guess as few questions as possible. How many guesses do you need in the worst case? How many chemicals can you distinguish in a single day with $p=1$ flask? How could you distinguish 2 chemicals with only one flask? And why would it be possible? What’s the maximum number of chemicals you can distinguish in a single day using $p$ flasks? Remember rule (ii): you can add the same chemical to multiple flasks. For $p=3,$ we can distinguish up to $8$ chemicals. How would you extend this approach to $d$ days? In other words, if we can distinguish $k$ chemicals in one day with $p$ flask, how many chemicals can we distinguish in $d$ days? We can", "and the four of a kind is four '5's). There are $7\\choose 4$ ways to select ''slots'' in which to place your 4 of a kind. The remaining three slots will then contain the three of a kind. So, there are $6\\cdot5\\cdot{7\\choose 4}$ different ways to obtain a three of a kind and 4 of a kind. Since outcomes are equally likely here, the chance of obtaining a three of a kind and 4 of a kind is as you stated.", "things can you say about the people who contributed to the discovery and/or the development of this topic? Historia Mathematica, a scientific journal, has an article called “The roots of combinatorics,” which describes records of ancient civilizations’ work in combinations and permutations. I would share with my students the first part of this description of the medical treatise of Susruta, without reading the last sentence that gives the answers: “It seems that, from a very early time, the Hindus became accustomed to considering questions involving permutations and combinations. A typical example occurs in the medical treatise of Susruta, which may be as old as the 6th century B.C., although it is difficult to date with any certainty. In Chapter LX111 of an English translation [Bishnagratna 19631] we find a discussion of the various kinds of taste which can be made by combining six basic qualities: sweet, acid, saline, pungent, bitter, and astringent. There is a systematic list of combinations: six taken separately, fifteen in twos, twenty in threes, fifteen in fours, six in fives, and one taken all together” (Biggs 114). I would ask them to estimate the number of combinations of any size group within those “six basic qualities” without doing any actual calculations. Once they had all made their estimates, as a class we would do", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "## purplemexican one year ago Need help please 1. Purplemexican 2. DanJS are you familiar with the basic power and square root rules 3. DanJS looking at the answers, you really only need to take care of the constant number, every choice is different 4. DanJS $\\sqrt{128}=\\sqrt{64 * 2} = \\sqrt{64}*\\sqrt{2} = 8\\sqrt{2}$ 5. Purplemexican so c right? 6. DanJS yep", "as which. Some things in math look like berries, but are really like fluids. Take a polynomial, say $5 x^2 + 6 x + 8$. It looks like three types of things, like three berries: five $x^2$, six $x$, and eight $1$. Add another polynomial, and the illusion continues: add $x^2 + 3 x + 2$ and you get $6 x^2+9 x+10$. You’ve just added more $x^2$, more $x$, more $1$, like adding more strawberries, blueberries, and raspberries. But those berries were a choice you made, and not the only one. You can rewrite that first polynomial, for example saying $5(x^2+2x+1) - 4 (x+1) + 7$. That’s the same thing, you can check. But now it looks like five $x^2+2x+1$, negative four $x+1$, and seven $1$. It’s different numbers of different things, blackberries or gooseberries or something. And you can do this in many ways, infinitely many in fact. The polynomial isn’t really a collection of berries, for all it looked like one. It’s much more like a fluid, a big sloshing mess you can pour into buckets of different sizes. (Technically, it’s a vector space. Your berries were a basis.) Even smart, advanced students can get tripped up on this. You can be used to treating polynomials as a fluid, and forget that directions in space are" ]

[ "+0 # Algebra +1 60 3 Help can anyone explain this to me A magician makes potions by combining maple syrup from a magical maple tree with ordinary water. The magician starts with a large supply of two potions: a red potion, which is 50% magical syrup by volume (and the rest is just water), and blue potion, which is 15% magical syrup by volume. (Perhaps you're wondering how the same syrup can produce both red and blue potions. That's why it's magic syrup!) (a) Find the amount of red potion (in mL) that must be added to 300 mL of blue potion in order to produce potion that is 20% magical syrup by volume. (b) Find the amounts of red potion and blue potion (in mL) that can be combined in order to produce 180 mL of a potion that is 40% magical syrup by volume. (c) Does there exist a combination of red potion and blue potion that can produce a potion that is 25% magical syrup by voume? Jan 28, 2023 #1 +79 +4 (a) Setting the amount of red potion to $$r\\;\\rm mL$$ and the amount of blue potion to $$b\\;\\rm mL$$ , we write the equation $$\\dfrac{\\frac12r+45}{r+300}=\\dfrac15$$. Solving this, we have r=50 mL Jan 28, 2023 #2 +79 +4 (b) Using the same", "well first. However the confusion comes from what happens if he drinks multiple poisons that do not cure each other first (what if someone was given well 5, but they first drink 2 before they drink 6?). Unless I'm misinterpreting the rules, than the second post is the correct solution. • The dragon does not have that guarantee because the knight can drink from well 1 before the duel begins, in which case the well 6 water would cure him. The dragon can counter this by giving him fresh water or well 1 water, so the knight must add on the safeguard of drinking from well 1 and then well 2 after the match, which will nullify everything. (same works for the dragon, just imagine well 6 doesn't exist) - the dragon has another guarantee. X,5,5,6. After drinking whatever the knight gave him, he could drink from well 5 twice then drink from well 6. If the knight gave him lake water, drinking from well 5 would poison him and then drinking from 6 would cure anything. If the knight gave him 1-4 then 5 would cure it the first time and he would be poisoned again from the second drink, then well 6 would cure that poison. If the knight gave him 5 then the 2 drinks", "Powers. Elements A Sorcerer aligns himself with a particular element. Naturally certain races will already have themselves aligned with certain elements. There are six elements within Glorantha these are shown below with their relation to each other. \\begin{verbatim} FIRE MOON \\ / \\ / EARTH ---------- AIR / \\ / \\ DARKNESS WATER \\end{verbatim} A sorcerer aligned to one element will find it difficult to use other elements. The further he moves around the spoke away from his element the harder it becomes. Bob is a human and was born under the Water Element( Based on when during the year and on which day he was born. Trolls automatically are aligned with the Darkness rune etc.). He discovered sorcery and found it reasonably easy to learn to control the water rune when he attempted to train in the other elements he came to this conclusion: \\begin{verbatim} Darkness,Air :- Medium Difficulty Skill Earth,Moon :- Hard Skill Fire :- Very Hard Skill \\end{verbatim} \\subsection{Forms} A sorcerer is tied to his particular rune that is relevant to his species. Learning in this skill is a medium skill, but is a hard skill for all other forms. \\subsection{Conditions} These runes will usually be present within the formula that the sorcerer uses when casting. Reasoning being that these runes effect the final outcome", "# Mr.Q : How to make a potion Mr.Q got a magical potion recipe. But he could not decipher the contents of the recipe and asked you for help. Here is a recipe for potions. 1. Magical water...........50 L 2. Petal extract.........1050 ML 3. Fermented wine.........200 CC 4. ......................1009 What does the last line of the recipe indicate? The last line indicates Mix everything......1009 MIX Because This is not 50 liters, 1050 millilitres and 200 cubic centimeters, it's actually the roman notations of the numbers, and 1009 reads as MIX. Hence my guess that the last line is just about mixing the previous ingredients.", "them the virtues of plants, animals and minerals, or in some padl the method of tincturing, te ink into which its drops are iiate the Key Sigil to which it cases even the ensouling essences of the Master and His Lady. ,f.. parts obtained via sacrificial rituals or minerals obtained and/or Plants ritually harvested or otherwise hallowed and re-enspirited through the power of the Thorn-Crowned Master or His Lady, animal � activate the power of each empowered ceremonially are the endowing elements possessing the !I to the workings of each Key abilities needed in order to confer the required powers into the ink !iated forms and techniques of base. .�..- .... powder, tincture and sachet When plant elements are used in the art of ink making it is the tinctured essence that one most often employs, even if also powders Jilnal setting would create an made from plant parts could be used, as the tincture allows the ping of the powers connected extraction of their essences in a concentrated liquid form which often - better lends itself to the conjunction with the carrier ink base. � 'i. \"-hich these given formulas pe Sorcerer and through their rjasight concerning the virtues r .. \"' F b�· to empower and bless the p1 formulas and create aspect­ pvs of His", "or 2 : result becomes 6 or 5. The dice's distribution becomes {6,5,3,4,5,6}. Bad dice can't roll good numbers. Flip them whenever you roll 5 or 6 : result becomes 2 or 1. The dice's distribution becomes {1,2,3,4,2,1}. Rolling one of each will give you a very good approximation of a healing potion: You then only need to adapt the number of dice to simulate various potions: $$\\\\begin{array}{|c|cc|cccc|} \\hline \\textbf{Potion Type} & \\textbf{Good Dice} & \\textbf{Bad Dice} & \\textbf{Min} & \\textbf{Mean} & \\textbf{Deviation} & \\textbf{Max} \\\\ \\hline \\text{Regular} & 1 & 1 & 4(=) & 7(=) & 1.51(1.58) & 10(=) \\\\ \\text{Greater} & 2 & 2 & 8(=) & 14(=) & 2.13(2.24) & 20(=) \\\\ \\text{Superior} & 4 & 4 & 16(=) & 28(=) & 3.02(3.16) & 40(=)\\\\ \\text{Supreme} & 8 & 3 & 27(30) & 45.17(45) & 3.54(=) & 60(=)\\\\ \\text{Alternate +10 Supreme} & 5 & 5 & 30(=) & 45(=) & 3.37(3.54) & 60(=) \\\\ \\hline \\end{array} \\$$ The values between brackets represent the values with tandard Nd4+X formulas ; (=) means both match exactly. Comparison can also be made using a graph : Supreme potions can be simulated using a +10 modifier ; but I'd rather use 8 good dice and 3 bad ones, which yields very similar results without introducing a constant. It allows", "whatever suits your world most. Also the quantity aspect solves why it was never apparent that it was magical. Until we made tons of it it was making so little magic that people did not notice. 2 Penicillin. Penicillin like thing. I'm not good with chemistry so bear with me. The same Aluminium idea is the same. Only it's something like a plant or a combination or things. Basically penicillin. Read on the actual penicillin and change whatever you feel like it. This again means that it possible to create the stuff of magic in a lab. We only needed to mess around with the right stuff and viola. MAGIC. While at the same time insures that no society before those with your limitation can make them. 3 A rare natural element that is only accessible with deep mining capability. Obvious things is obvious. We needed to go so deep underground or underwater to extract Contrivium, terrible writing advice if you don't know or the classic: Phlebotinum So after the technology of the world advanced to that point we started seem this thing. And it did not take long to people to figure out it's magical. And even if people back in the day found an ore or a gem or whatever. They still could not get more", "you can use the tincture in order to stir His seated essence by spraying it three times over His holy fetishes or in some other suitable way pour it over them, but still, to spray it directly out ,... iii from your own mouth is recommended, as it will not inflame only Him but also yourself, in more ways than one. ..�_., \"'\"1!; l The arousing power of this Tincture of Fiery Vivification is more than ten times as potent as a normal libation of alcohol and the spiritual flames ensuing by its use will surely bless the wise and burn the � are made in foolish. many difil _.elements and are always­ �whenever direct contact wil • 41sirable. ··� ._.magical effigies can be tile� .. lllessing, healing, love-� _..,hol.nl domination, empowa� .. a direct link an d physical .. it eemed as desirable and .. ..... \"4 1 ·� llistorical origins of these .. -Mp+ity as they have, in one 1l 6r �cal traditions of � - �l . aae root of this kind of sora� ' - ' istic arts can be found. aail -.;.s within these contexts­ 2 zwlhy and antipathy, astral. ' •ts, such as those deri¥J ·:- a, ...., which through the Cl -.eil1lcllHlDS between the IlDdy. 288 • r_.. JliKe of cloth", "Surprise remains • Faster than rolling dice in game • You are not limited to healing potions • The content may differ from the label (adulterated healing potion, or even worse - poison?) • Sometimes, the players won't be able to tell. Did this Superior Healing potion heal only 17 hit points because of a lack of luck, or because it was a Greater Healing Potion? • Man now I want to do this purely for the shenanigans of wrong potions. Insight and slight of hand checks just got way more interesting. – linksassin Mar 4 '19 at 2:47 • +1 This is a really clever idea and that I might try sometime. Nice outside-the-box thinking. – Bloodcinder Mar 4 '19 at 16:37 # With a healthy mix of good and bad dice. Potions use d4 in pairs. A single d6 can be altered to allow only 4 different outcomes: the downside is that it becomes unfair. You can compensate this by creating another d6, with its own 4 different outcomes, so that it generates fair results when rolled with the other one. • Use white or green for \"good dice\". • Use black or red for \"bad dice\". # Using standard dice: just flip them! Good dice can't roll bad numbers. Flip them whenever you roll 1", "have been poisoned or cured? If the knight drinks from well 1 and the dragon gives him well 2, can the knight tell if he is still poisoned and should look for a cure, or is he uncertain until he dies? I'm assuming the latter case. One way for the dragon to always survive is to drink nothing beforehand, but after drinking the knight's poison, drink twice from well 5 and then drink from well 6. If the knight offers any poison but well 5, it will cure it, and the second drink will be cured by well 6. If the knight offers well 5, the first 2 drinks will do nothing and well 6 will cure it. If the knight offers clean water, the drinks from wells 5 and 6 will cancel out. I'm not sure if there's a similar strategy for the knight, because he doesn't have the cure-all of well 6. I don't believe it's possible to guarantee a win. Anything the other side offers can be canceled out by choosing some appropriate drinks either before or afterward. Therefore, if there was a winning choice of poison, one player would know that the other would always pick that strategy and would be able to pick the drinks before or afterward that would cure the winning", "Healing potion, et cetera. $$\\^1\\$$ I totally agree with Wyrmwood in doing this. $$\\^2\\$$ I skipped the 4th edition, hence I jumped from the 3.0 directly to the 5th one, and at the beginning the spell system puzzled me a little bit.", "# Laxative Potion 14,883pages on this wiki \"Laxative Potion\" —The potion included in Moste Potente Potions[src] The Laxative Potion was a magical concoction that was included in Chapter VI of Moste Potente Potions, of which a copy was kept in the Restricted Section of the Hogwarts library. The page was opposite Chapter XVI, which concerned the Polyjuice Potion. Known instructions in the book are to add two scrumples of a certain fine ingredient, another ingredient that has been used in a furnace, and Agrimony[1].", "have been guilty of. take for example potions. in some games, like diablo, you wind up quaffing hundreds of them through the course of the game. there are three types of potion in diablo... healing, mana, and rejuve. now, in the REAL world, potions correspond to medicines, and here is what is commonly known about medicines: A: some people are allergic to some medicines.(i personally am allergic to penicillin) B: most medicines have side effects(at least, the strong ones do) C: many medicines are addictive D: the more you take of a medicine, the less effective it is(tolerance) now, bring this into diablo-world. nobody is ever allergic to any type of potion. there are no side effects. they are not addictive, and you dont develop a tolerance. as far as i've seen, NONE of these have ever been incorporated into a CRPG. since we cant make the game unplayable, we must cover all of the functions that the potions do and have them available to the player. so, we come up with the following combinations: #1: the character has no allergy #2: the character is allergic to mana potions #3: the character is allergic to health potions #4: the character is allergic to rejuve potions #5: the character is allergic to both health and mana potions this list", "Acolyte 1 man - - - - - - 2 Adept 2 man 1 - - - - - 3 Village Priest 3 2 men 2 - - - - - 4 Vicar 4 3 men 2 1 - - - - 5 Curate 4+1 3 men 2 2 - - - - 6 Bishop 5 Hero -1 2 2 1 1 - - 7 Lama 6 Hero 2 2 2 1 1 - Men & Magic, p. 18. Level numbers added for clarity It's likely a typo in the OD&D rules, and matches Cook's Expert, but note that both of those are different from Moldvay, where level 6 has no 3rd level spells. By AD&D PH, it was fixed: L 1 2 3 4 5 6 7 1 1 - - - - - - 2 2 - - - - - - 3 2 1 - - - - - 4 3 2 - - - - - 5 3 3 1 - - - - 6 3 3 2 - - - - 7 3 3 2 1 - - - Gygax, AD&D Player's Handbook, p. 20 Likewise, in Mentzer: L 1 2 3 4 5 6 1 - - - - - - 2 1 - - - - - 3 2 - -", "features away from them and giving them to every other sorcerer archetype. I thought to mention this in case this impacts balance, and therefore any potential answers, in a way I haven’t anticipated; but this is an aside, just some extra information, and my question is not specifically about this aside, but about the impacts of having any sorcerers with 20 spells.", "distribution of {1,2,3,4,3,4} ; hence, you'll need offsets (see below) to compensate. • the bad dice should have 4 pips removed from the 5 and 6 faces: they become \"standard-looking\" 1 and 2. Same old bad dice. The downside of this method is that you'll need the original offsets (+2 for normal, +4 for greater, +8 for superior and +20 for supreme) to approximate healing potions. Don't use the 8 good / 3 bad method for supreme : it would need a +16 offset, and renders that option pointless. # How does it compare to other answers? I couldn't resist the urge to aggregate the various methods listed here in a single anydice link. Use comments to focus on a single potion, or display all of them - but it quickly becomes hard to read properly. An illustration using the superior potion : It looks like this method allows the closest results to the original potions in a single roll ; but Limari's proposal will get you very close to this, with less dice and calculation. Ryan's reroll method also has very good results, even better than this one if you are ready to reroll multiple times when necessary - especially for bigger potions (e.g three times for superior). • This is definitely the sort of solution my", "easy. The healing potion doesn't heal, it just interacts with the body's natural energies to allow bruising and cuts to knit together. Just like antibiotics don't repair necrotic flesh, but they do allow the body to heal (by removing the infection). Generation Y. I don't remember the First Gulf War, but do remember floppy disks. Yakk Poster with most posts but no title. Posts: 11115 Joined: Sat Jan 27, 2007 7:27 pm UTC Location: E pur si muove ### Re: Dungeons and Dragons (and other tabletop RPGs) Have a healing potion drunk when the player is out of healing surges grant 1/2 the amount it would heal as temporary HP. In effect, the healing potion isn't working very well anymore, and you notice it isn't working, and that you need rest before drinking more. At the same time, you cannot glug down gallons of healing potion to heal any wound. One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total. Chen Posts: 5570 Joined: Fri Jul 25, 2008 6:53 pm UTC Location: Montreal ### Re: Dungeons and Dragons (and other", "the same lines, and to minimize ingested poison values, 1,x,1,2 does the job just as well. I would also call it t he guaranteed surviving sequence rather than winning.-JG Not just so. The dragon can easily survive with X, 1, 6: he need not start by drinking a poisonous draught. -Jon Werts If the knight gives the Dragon lake water- any well the dragon drinks from will poison him. The dragon can guarantee the knight dies by giving him well 6 water. If the knight gives the dragon any well water it can cure it by drinking from well 6, but if it was lake water that would kill him unless he drinks from another well first. However the confusion comes from what happens if he drinks multiple poisons that do not cure each other first (what if someone was given well 5, but they first drink 2 before they drink 6?). Unless I'm misinterpreting the rules, than the second post is the correct solution. • The dragon does not have that guarantee because the knight can drink from well 1 before the duel begins, in which case the well 6 water would cure him. The dragon can counter this by giving him fresh water or well 1 water, so the knight must add on the safeguard of", "formula, where x is equal to your Prayer level:$floor(7 + \\frac{x}{4})$ Defence potion 30 White berries 75 It boosts a player's Defence level by 10% plus 3 per dose", "of ,..0 to the bewitching work the Green and the Kingdom of the Graves. All harvested parts of the It rites of strewing, in which t· when the living are to be to work her witchcraft, but the most potent share of such harvest will always be the sweet Liquorice roots, which hold her full powers to edulcorate and dominate. Such roots are carried as talismans or made into powders and employed within all workings of enchanting and will-bending sorcery, but also within more sacred and secret rites where fractions of the boundless essence of the Rose Crowned Queen are to be seated within fetishes or in some other way bless and consecrate Vessels of Holiness with Her Sweetness and Might. 197 flalt 1J1trn (Dryopteris ftlix-mas) • us &om this. ..... ))lesgnp- The daemon of the Male Fern is a spirit with immense powers. He can conceal and make invisible and protect against all manners of baneful sorcery. He can make the curses and the foul emanations of the evil eye of the enemy miss their targets and rebound back upon him- or herself. He can also bring luck, wealth and success and point out hidden treasures and disclose other such veiled sources of material !I abundance. This spirit is also well-known for his mastery of the arts" ]

[ "a lot of time planning activities and trips for our son. However, my wife wanted to hold off on some of the big ones (i.e. Disney ...", "do know that there already is a Disneyland? -Yeah, I built it. -No, I mean the real Disneyland. You mean, the one in the hearts and minds of children everywhere? -The one in Anaheim! -I think we’re saying the same thing. Are we? Well, I think it’s safe to take the cone off. I can do it for a small co-pay. Can’t say I’ll miss it, but we did have some crazy adventures together these last few weeks. Honey, would you rub my chin for me so I can remember things? Oh, um, okay. Uh Yeah, memories. Hey, where should we go for my first meal after the operation? It’s not an operation, but we could go to Storky’s again. Oh, yeah! We’re finally due for our free sub. “Storky’s, where your 81st sandwich is free.” We’ve been there 80 times? It just opened last month. Yeah, but you love it, so whenever you suggest it, I always say yes. Yeah, but 80 times? Well, that’ll be$6000. Good thing you’re getting a free sandwich, huh? What the? Huh? Okay, you guys. You vultures are killing me! I’m sure he’s talking about other vultures. Ah, the Valley. Because sometimes you wanna go where no one knows your name. – BoJack Horseman! – Pinky? Hey, how are you? Oh, great!", "PAY DISNEY… FULL ADMISSION PRICE to go play a day in the park. I mentioned above, when I was talking about the Spring Tour and Great America, how much I hate amusement parks. Ugh. I really hate them. The weekend after Disneyland was really unique… it’s going to be memorable for years to come. You see, we have an English final project for Villalobos. My group — or rather, I myself, am rather ambitious. Our presentation was going to rock the socks off all our classmates… in theory. We met all of Saturday evening, I think, and met Sunday at Hank’s house (Hank, clean up your yard…). And they we met… ALL OF MEMORIAL DAY… from 10 in the morning, through lunch, through dinner, through midnight… and they left at 6am in the morning on Tuesday. Naturally, I skipped school on Tuesday to finish editing our horrible, crappy video, which wasn’t even my job. The thing is, nobody in my group except me can edit video, so… Our presentation was on Wednesday. We first showed the video, which I typed up subtitles for the day before. Sound effects were added, but unfortunately Windows Movie Maker seems to be incapable of outputting playable video without considerable geekery, so we ended up throwing away many hours of work that we", "for traditional spring break trips Câu 50: Which of the following is mentioned as a reason for participating in alternative spring break trips? A. A desire to travel to glamorous places B. A wish to get away from family and friends C. The hope of earning money D. A personal opinion that people must help other people", "in March, during spring break. Next Page »", "completely experience each attraction. Our family is extremely grateful to have fortunate to take part in your mission. There were dozens of musical acts and assorted speakers performing during the day.", "stage inbetween. That is really good news. Cheers, Javier A.", "go there to buy breakfast sandwiches. We eat the sandwiches in our room. We a little anxious to get to Disneyland, but it doesn’t open until 9am. So we’re twiddling our thumbs… # Disneyland! We arrive at the baggage check outside of Disneyland (and California Adventure). It takes a while to get through baggage check. Seems like a lot of family have loaded up their strollers like a pack mule, so there is a lot for the security staff to check. When we get to baggage check we breeze through since the Tracy and the girls only have one bag each and I have nothing. We’re in line waiting to get into Disneyland by 8:15. At 8:30 they start letting folks in. Since this is our first day, they take our pictures and associate them with our paper passes (no magic bands!). On subsequent days they will just scan our passes and visually check to make sure that we’re the same people. The park doesn’t actually open until 9am. While we wait we get our “first visit” buttons on main street. We also stopped to look at these beautiful moving displays in some of the windows. They display scenes from Disney movies and every once in a while they rotate around and the scene completely changes. My picture", "built. Originally consisting of one theme park and one hotel, though after Walt’s death it was expanded into a resort with a second theme park, more hotels and some other attractions. Disney world (in Florida) was a much bigger project, this time Walt Disney wanted to control not just the immediate theme park but the whole area surrounding it. While sadly Walt died before the park was actually built the result was nevertheless a Disney-controlled resort with four theme parks, two water parks and loads of hotels and other attractions. There are also a number of foreign (outside the US) Disneylands. As far as I can tell these are much closer in scale to the original Disneyland than they are to Disney World. ## Filling stations and petrol stations in Nigeria, is there any difference? No, this is the same thing. Apparently, according to Wikipedia, the most common name in the world is a filling station: A filling station that sells only electric energy is also known as a charging station, while a typical filling station can also be known as a fueling or gas station (United States and Canada), gasbar (Canada), gasoline stand or SS(Note 1) (Japan), petrol pump or petrol bunk (India, Pakistan and Bangladesh), garage, petrol station (Australia, Hong Kong, New Zealand, Singapore, South Africa,", "play a day in the park. I mentioned above, when I was talking about the Spring Tour and Great America, how much I hate amusement parks. Ugh. I really hate them. The weekend after Disneyland was really unique… it’s going to be memorable for years to come. You see, we have an English final project for Villalobos. My group — or rather, I myself, am rather ambitious. Our presentation was going to rock the socks off all our classmates… in theory. We met all of Saturday evening, I think, and met Sunday at Hank’s house (Hank, clean up your yard…). And they we met… ALL OF MEMORIAL DAY… from 10 in the morning, through lunch, through dinner, through midnight… and they left at 6am in the morning on Tuesday. Naturally, I skipped school on Tuesday to finish editing our horrible, crappy video, which wasn’t even my job. The thing is, nobody in my group except me can edit video, so… Our presentation was on Wednesday. We first showed the video, which I typed up subtitles for the day before. Sound effects were added, but unfortunately Windows Movie Maker seems to be incapable of outputting playable video without considerable geekery, so we ended up throwing away many hours of work that we spent finding, adding, and timing sound effects.", "Finding Nemo Submarine Voyage - Disneyland - June 11, 2007 Pirate's Lair - Disneyland - May 25, 2007 Toon Studio - Walt Disney Studios Park - 2007 Crush's Coaster - Walt Disney Studios Park - 2007 Cars Race Rally - Walt Disney Studios Park - 2007 Monsters, Inc. Laugh Floor - Magic Kingdom - April 2,2007 Finding Nemo - The Musical - Disney's Animal Kingdom - November, 2006 The Seas with Nemo and Friends - Epcot - October, 2006 Recent projects The corporate headquarters of Walt Disney Imagineering are in, and have been since the 1950s, Glendale, California. It is for this reason that there is no WDI field office at the Disneyland Resort, which is thirty-six miles away. There are two field offices at the Walt Disney World Resort, required for the sheer size of the resort. Both are relatively close to each other. There are field offices located at; Epcot and the Disney-MGM Studios, Walt Disney World Resort Tokyo Disney Resort Administration Building, Tokyo Disney Resort The former WDFA field office, Disneyland Resort Paris Walt Disney Imagineering Hong Kong Site Office, Hong Kong Disneyland Resort Locations The Imagineers have been called on by many other divisions of the Walt Disney Company as well as being contracted by outside firms to design and build structures outside of", "Studios for Grinchmas. Here’s the latest themed entertainment happenings at amusement parks and beyond! Disneyland The Disneyland Resort is celebrating the holidays in a big way with Marvel merriment, the Festival of Holidays, snow at Disneyland, Grogu plus daddy Mando in Galaxy’s Edge, and more. Walt Disney World Winter has taken over with new Frozen additions at the Winter Summerland mini golf course, Guardians of the Galaxy holiday music on Cosmic Rewind in Epcot, the International Festival of the Holidays, and the news that Splash Mountain will be closing early in the new year. Universal Studios Hollywood Grinchmas Meet the mayor of Whoville and see the Grinch steal Christmas daily! Cinematic New Year’s at Universal Studios Hollywood The New Year’s Eve event at Universal Studios Hollywood is included with admission on December 31; you can buy tickets here. Stranger Things Experience Stranger Things: The Experience” is now open in Los Angeles. The 30-minute immersive story invites you to Hawkins and takes you to the Upside Down. Find tickets here. No one does it like the Grinch! Grinchmas at Islands of Adventure runs until the end of the year. Universal’s Great Movie Escape Themed escape rooms inspired by Back to the Future and Jurassic World open December 9 at the Universal Studios Orlando resort. Book tickets here. LA Comic", "struck me here was the large number of Video\\8 cassette drives being marketed. Also, there were some quite excellent colour terminals on display. ## Disneyland On the tuesday evening Disneyland was available exclusively for the use of DECUS attendees. What a bonus! After several rides around Alice in Wonderland, Peter Pan's Journey, Pinocchio's Adventure we tried some more hair\\raising rides. For example the Thunder Express and Space Mountain \\ a roller\\coaster ride in complete darkness. Not advisable after cheeseburger and chips. # Getting back A ten dollar ride to LAX airport (duration one hour), a five hour wait looking at the departures board, a ten hour flight to Heathrow, and then a couple of hours later saw me back in Geneva.", "ridiculously long lines? or go after 1 jan and miss out on the xmas themed disneyland, but have shorter lines...and i'm pretty sure i just miss out on their chinese new year celebration themed disneyland. and i want to go to their halloween themed one, but who has holidays in october in uni? everytime i go hk disneyland, i always go when they have the xmas stuff, but my dad still hasn't been to hk one before, since we go there before he arrives in hk.. not sure if this makes sense. spent quite a lot of time sorting through my pens. approx 60+, never ended up counting them though..i have a lot of \"sentimental\" pens. took up 1/4 of my drawer. don't have the will to go through the rest of the drawer... yay for our group spending minimal money on schoolies! random thought... bro gone for youth camp=quiteness for the weekend, but i was actually looking forward to playing boardgames with him. feel kind of monopoly deal deprived... yes! made it through spring by wearing long pants at home. new record! i'm very heat intolerant, so to wear long pants on a day like this is a pb. Thursday, November 29, 2012 Last Minute Travel Changes/N2 Ice Cream thanks Happy Apple for the food (: mad", "he holiday opt ion s provided! theme parks like Disneyworld combine holidays and work e.g, in a summer camp adventure holidays e .g, climbing holidays with educational or cu ltural activities Part 4 - Discussion ~~,',\",' I 5 minutes (8 minutes fo r groups of three) • W hat do yo u th in k ~ • Do you agree? • Wha t about you~ Int erlocutor: • How good an idea would it be to build a new tourism development in you r areal Where do you think the best place to build it would be? Wh at e nviro nmental considerations sho uld one t hink about wh en budding holiday dev e lopm en ts like th i s~ • Do people overestima te the importance of holidays? • Are there any places where you think it is inap propriate to build the me parks and spe cially des igned holiday resorts? • What are the advantages of go ing abroad on holiday? Thank you . T hat is the e nd of the re st. CAE Prac tice Te st 5 Pa pe r 5 • Speaking TEST 5 Part 2 - Long turn 4 minu tes (6 minutes for g roups of three) ' '1 this part of the test you each have", "different orders can you visit the exhibits? ×", "VDLP and I too P to see the Disney Jr. Stage Show, and all had a good time. We decided to rendezvous at the Hat, and we feasted on churros and pretzels from a cart. Then we were off to stand in line at the Little Mermaid Show again - oops. We all stood patiently in line for half and hour, and when we got to the front of the line. (seriously, we were at the chain) we were told that we would not be seated, and we had to wait another 45 mins. It got a little ugly - but, we were right next to Walt Disney a Man and His Dream exhibit, and I really wanted to see that, so I grabbed V & MM and we headed off, while VDLP, C & P waited at the front of the line for the Little Mermaid. V waiting to get into the Little Mermaid Show Oops.. the Curse struck again. Just as we were sitting down for the movie, C texted and said - come now, they are seating us. I grabbed the kids and we ran off to get to the Little Mermaid. It was worth the wait. And the annoyance. It is a live action multi media/multi atmosphere event. The kids loved it, and the", "won’t do it justice but here’s a scene from Princess and the Frog: The rope drops at 9am separating Main Street from Tomorrowland. First ride, Hyperspace mountain! Ceilidh has been looking forward to this for months. At one point she was convinced they wouldn’t have the Hyperspace overlay during our trip and instead it would be just plain Space Mountain, but fortunately they still have the Hyperspace overlay. Next is the Buzz Lightyear ride. We find this version more enjoyable than the Disney World one. You have more range of motion because the guns aren’t attached to the cart. I, of course, got the high score out of our family 😉 Before lunch we checked out Autopia, which is like Disney World’s Tomorrowland Speedway. Then Mr Toad’s Wild Ride, the Matterhorn and Star Tours. Unfortunately the Matterhorn temporarily shut down just after we got into the bobsleds. We enjoyed seeing BB-8 from The Force Awakens in the Star Tours ride. We then hiked over to California Adventure for lunch. I’m trying to remember now, I think Keira and I got pizza but Ceilidh and Tracy went to get corn dogs. From where we were sitting we could see most of the latin themed parade out front. We also picked up Radiator Springs fast passes that we would use", "## Florida Back in February, Matt had a conference, so Carrie, Torsten, and Aven joined in at the end of the week and we took a weekend for DisneyWorld, SeaWorld, sunshine and warmth. From Florida 2010 From Florida 2010 From Florida 2010 From Florida 2010 We saw the sights From Florida 2010 From Florida 2010 Rode the rides (this is Torsten about to get on one of the rollercoasters) From Florida 2010 and enjoyed the healthy food From Florida 2010 From Florida 2010 ## Torsten’s Birthday Torsten had a fantastic skate park birthday: From Torsten Turns 6 From Torsten Turns 6 From Torsten Turns 6 From Torsten Turns 6 With lots of friends From Torsten Turns 6 From Torsten Turns 6 And relatives from cousins, uncles, aunts, grandparents… From Torsten Turns 6 From Torsten Turns 6 From Torsten Turns 6 From Torsten Turns 6 It was a fantastic celebration!! A few more action photos: From Torsten Turns 6 From Torsten Turns 6 From Torsten Turns 6 And yes, you are right–Torsten’s birthday was in April.. I’m just trying to catch up a bit here…", "New Year's while visiting a friend in Santa Cruz, we stopped at a park. It was raining, windy, and getting nastier. Straight from the car, Michelle took off frolicking: #### Eating Donuts in College Station Taken while on a Nutcracker Bon-Bon audition trip to College Station, TX, we stopped at a donut shop one morning. Notice the 2 and a 1/2 step donut eating maneuver, and the huge grin she gets once the bite is safely performed:" ]

[ "$2$ ways to serve their meals. Thus, there are $(3\\cdot2)^3=\\boxed{216}$ ways to serve their meals. Note: This solution gets the correct answer through coincidence and should not be used. ~ dolphin7", "our build. After this choice, we place the last item in the final slot our build order. Over the course of finalizing our build, we had 3 options, then 2 options, and then 1 final option. We need to multiply these numbers so that we count each possible sequence of items that we could have chosen: for each of the three possible first item choices, there are two more options after that, then each of those options is followed by one option (the last item). So each of the initial three options leads to two more, down the road. Adding them up, we get $2+2+2=3\\cdot 2=6$. Then we count the final item selection in each of these choices: there is 1 option, so we multiply $6\\cdot 1=6$ to get the final answer. In total: $3\\cdot 2\\cdot 1$. Looks familiar right? It’s the factorial we just learned! Check-Your-Knowledge Question 1. It’s the end of the game and the top laner is full build. How many ways could they order the items in their inventory? 2. Since we already know what $5!$ is, notice that we can write $6!=6\\cdot 5!$ as a quicker way to calculate $6!$. (Try expanding both sides if you don’t believe it.) Can you turn this into a general statement about the value of $n!$, if", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "this town and $2^6$ ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are $5 \\cdot 2^6+5 \\cdot 2^6$ bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are $5$ ways to choose the town for which all roads lead to it, $4$ ways to choose the town for which all roads lead out of it, and $2^3$ ways to assign the remaining $3$ roads not connected to either of these towns. Therefore, the answer is $2^{{5\\choose2}}-(5 \\cdot 2^6+5 \\cdot 2^6-5\\cdot 4 \\cdot 2^3)=\\boxed{544}$.", "since there are 8 cities.. and r would be 7 since there are 7 more cities to travel to. But clearly that isn't correct. Thanks - It would be silly to ask \"How many possible orders\" and at the same time insist that order does not matter, so you may safely assume that order does matter here. If really you wanted to count the number of ways to visit those cities if order does not matter, then there is exactly one solution; indeed there is one combination of $7$ out of $7$ candidates that can be chosen. – Marc van Leeuwen Mar 22 '14 at 22:56 She has to begin her trip in a \"specified city\", meaning the initial value is the same. (The first city will not be rearranged into another.) – Don Larynx Dec 22 '14 at 19:48 Note that the question says any order she wishes, not that the order does not matter. Hence, we are looking at a permutation (the specific ordering of the cities make up her route). And note that she starts in a specified city, i.e. there is no decision here. After that there are $8-1=7$ cities to visit, which can be visited in a certain order in $7!$ ways. I find it difficult to give a more general answer about", "from the bakery to the intersection, and $4$ ways to get from the intersection to the restaurant. Do these choices depend on each other in any way? No, so to get the total we multiply (because for each option of the first, all options of the second are possible). Then there are $3 \\cdot 4 = 12$ bad routes. The total number of routes that Ms. Gustaroni could take, as we already calculated, is $35$. The number of bad routes, we just figured out, is $12$. So the number of good routes is $35 - 12 = 23.$ So Rosa Gustaroni still has plenty of possible ways to get to work.", "well choose them in order $1, 2, 3, \\ldots$.", "and the four of a kind is four '5's). There are $7\\choose 4$ ways to select ''slots'' in which to place your 4 of a kind. The remaining three slots will then contain the three of a kind. So, there are $6\\cdot5\\cdot{7\\choose 4}$ different ways to obtain a three of a kind and 4 of a kind. Since outcomes are equally likely here, the chance of obtaining a three of a kind and 4 of a kind is as you stated.", "here is the number of ways in which they completely miss one another: no two go to the same restaurant. There are $10$ restaurants that $A$ could choose. $B$ could then choose any of the remaining $9$, and $C$ could choose any of the $8$ not chosen by $A$ or $B$. Thus, there are $10\\cdot9\\cdot8=720$ different ways in which they could completely miss one another. (b) In order for them to meet successfully, they must all choose the same restaurant. There are $10$ restaurants, so there are just $10$ ways in which they can meet successfully. Alternatively, we can apply the same kind of analysis as in (a): $A$ has a choice of $10$ restaurants, but after that $B$ and $C$ have only $1$ choice each, if the meeting is to take place, so the total number of ways is $10\\cdot1\\cdot1=10$. (c) This is most easily done by considering the complementary (opposite) event: no two of them meet. In (a) we calculated the number of ways in which this can happen. If there are $n$ different ways in which could they choose restaurants, regardless of how many of them meet, the number of ways in which at least two of them meet must be $n-720$. What’s $n$? Is $n=1000$?.... – count May 31 '12 at 2:57 @count: Yes,", "# How many possible paths? The answer is $32$. Its supposed to be $2^5$ but I do not see how you get that? The way I see it, there are $5$ ways to go up and $5$ ways to go right, total ways = $5x5= 25$ Each time you have to choose where to go, you have 2 options: up or right. Now observe that no matter how you get to the diagonal, you will have made 5 choices. So you have to choose between 2 options 5 times, which gives us $$2\\cdot2\\cdot2\\cdot2\\cdot2 = 2^5$$ possible choices.", "to order the pills and on day 7 he will have to repeat one of the orders. Suppose he is taking 4 pills a day. There are 4 choices for the pill he takes on the first day. Regardless of which pill he chooses he has 3 pills left and he can order them in 6 ways. Thus he has $4 \\times 6 = 24$ ways to order the 3 pills. Hence there are 24 ways to order the pills and on day 25 he will have to repeat one of the orders. What about 5 pills or 6 pills? Penny", "of the same destinations together]. [PS. If \"important\" were interpreted to mean important to the tourist (which I don't think is unreasonable), then your proposed answer to (a) would be correct: there are $\\binom{10}{6}$ unordered sets of destinations, and the tourist visits them according to her preferences.] (a) is $10\\choose6$, combinations (b) is ${10\\choose 6} \\times 6!$ because there are $6!$ orders to visit the cities", "in $4 \\choose 3$ with 3! ways of approach from A person visits B, C, D, E comes back to A in $4 \\choose 4$ with 4! ways of approach Second Hint ways=${4 \\choose 2}(2!)+{4 \\choose 3}(3!)+{4 \\choose 4}(4!)$ Final Step =12+24+24 =12+48 =60. ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "to sit to get 3456 ways for this to happen. However we overcounted the case when two pairs of advisors run out of room to sit, where there are $6\\cdot9\\cdot4=216$ ways to happen. So 5760-3456+216=$2\\boxed{520}$.", "total of $10$. From this point, there comes a new $10$ orderings starting with $CB$ and then new $10$ starting with $CD$, giving a total sum of $30$ orderings. Note that this orderings are different because they all start with $C$. We can change the places of $A$s, $B$s and $D$s with each other, as well as change the places of persons in $C$; giving a $2^4$ combinations. Also, there's a $4 \\choose 1$ way to choose country $C$. Putting all together: $${4 \\choose 1}2^430=1920$$ • Case 3: Number of orderings such that exactly three countries have adjacent sitting arrangements: By the same reasoning above you get only $1$ different ordering which is $ACACC$. Changing letters and people give you $2^43!$ combinations. Lastly, there are $4 \\choose 3$ way to choose countries. Putting all this together: $${4 \\choose 3}2^43!.1=384$$ • Case 4: Number of orderings such that exactly four countries have adjacent sitting arrangements: There are $4 \\choose 4$ way to choose countries and this countries have $3!$ different orderings around table. Every person in every country can change places, netting you a $2^4$ combinations. So: $${4 \\choose 4}2^43!.1=96$$ Finally, total number of orderings such that at least one country have adjacent sitting arrangement, is: $$1920+1152+384+96=3552$$ Update: Total number of orderings such that at least two countries have", "table. . . X has 4 choices of seats; Y has 3 choices of seats. . . The other five can be seated in $5!$ ways. There are: . $4\\cdot3\\cdot5! \\:=\\:1440$ ways. Suppose $X$ and $Y$ sit on the 3-side of the table. . . X has 3 choices of seats; Y has 2 choices of seats. . . The other five can be seated in $5!$ ways. There are: . $3\\cdot2\\cdot5! :=\\:720$ ways. Therefore, there are: . $1440 + 720 \\;=\\;{\\color{blue}2160}$ seating arrangements.", "the reasoning. Each of those $10^2$ ways to post the first two letters can be combined with any of $10$ different ways to post the third letter, so there are $10^2\\cdot10=10^3$ ways to post the first three letters. Two more arguments of the same kind lead to the conclusion that there are $10^5$ different ways to post the $5$ letters. Note that I’m assuming that you’re allowed to post more than one letter in the same letter box. If that’s not the case, the reasoning is similar, but the answer is quite different. If you can post at most one letter in a box, there are still $10$ different ways to post the first letter. After you’ve done that, however, there are only $9$ ways to post the second letter, since you can’t use the first letter box again. Thus, you get only $10\\cdot9$ different combinations of letter boxes for the first two letters. Similarly, after they’ve been posted you must pick one of the $8$ remaining letter boxes for the third letter, so each of the $10\\cdot9$ ways of posting the first two letters gives you only $8$ ways of posting the first three. As a result, there are $10\\cdot9\\cdot8$ ways of posting the first three letters. Two more arguments of the same kind lead this time", "way we perform the three main groups, we can rearrange the performances in California. Therefore, we get a total of $$6 \\cdot 6 = 36$$ ways. Aug 14, 2020", "be selected in $5C3$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3 - 1 - 1 option is $5C3$ = 10 = 10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes = number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.", "the last $2$ spots, then there is $1$ way to assign spots to $Y$. Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\\cdot2$ ways to assign spots to $X$ and $2\\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$). There are $1+1+2\\cdot2\\cdot2\\cdot1 = 10$ ways to assign opponents to $B$ and $90\\cdot10 = 900$ ways to order the games. $\\boxed{\\textbf{(E)} 900}$" ]

[ "do know that there already is a Disneyland? -Yeah, I built it. -No, I mean the real Disneyland. You mean, the one in the hearts and minds of children everywhere? -The one in Anaheim! -I think we’re saying the same thing. Are we? Well, I think it’s safe to take the cone off. I can do it for a small co-pay. Can’t say I’ll miss it, but we did have some crazy adventures together these last few weeks. Honey, would you rub my chin for me so I can remember things? Oh, um, okay. Uh Yeah, memories. Hey, where should we go for my first meal after the operation? It’s not an operation, but we could go to Storky’s again. Oh, yeah! We’re finally due for our free sub. “Storky’s, where your 81st sandwich is free.” We’ve been there 80 times? It just opened last month. Yeah, but you love it, so whenever you suggest it, I always say yes. Yeah, but 80 times? Well, that’ll be$6000. Good thing you’re getting a free sandwich, huh? What the? Huh? Okay, you guys. You vultures are killing me! I’m sure he’s talking about other vultures. Ah, the Valley. Because sometimes you wanna go where no one knows your name. – BoJack Horseman! – Pinky? Hey, how are you? Oh, great!", "2015 04-16 # Disney’s FastPass Disney’s FastPass is a virtual queuing system created by the Walt Disney Company. First introduced in 1999 (thugh the idea of a ride reservation system was first introduced in world fairs), Fast-Pass allows guests to avoid long lines at the attractions on which the system is installed, freeing them to enjoy other attractions during their wait. The service is available at no additional charge to all park guests. — wikipedia Disneyland is a large theme park with plenties of entertainment facilities, also with a large number of tourists. Normally, you need to wait for a long time before geting the chance to enjoy any of the attractions. The FastPass is a system allowing you to pick up FastPass-tickets in some specific position, and use them at the corresponding facility to avoid long lines. With the help of the FastPass System, one can arrange his/her trip more efficiently. You are given the map of the whole park, and there are some attractions that you are interested in. How to visit all the interested attractions within the shortest time? The first line contains an integer T(1<=T<=25), indicating the number of test cases. Each test case contains several lines. The first line contains three integers N,M,K(1 <= N <= 50; 0 <= M <= N(N –", "struck me here was the large number of Video\\8 cassette drives being marketed. Also, there were some quite excellent colour terminals on display. ## Disneyland On the tuesday evening Disneyland was available exclusively for the use of DECUS attendees. What a bonus! After several rides around Alice in Wonderland, Peter Pan's Journey, Pinocchio's Adventure we tried some more hair\\raising rides. For example the Thunder Express and Space Mountain \\ a roller\\coaster ride in complete darkness. Not advisable after cheeseburger and chips. # Getting back A ten dollar ride to LAX airport (duration one hour), a five hour wait looking at the departures board, a ten hour flight to Heathrow, and then a couple of hours later saw me back in Geneva.", "PAY DISNEY… FULL ADMISSION PRICE to go play a day in the park. I mentioned above, when I was talking about the Spring Tour and Great America, how much I hate amusement parks. Ugh. I really hate them. The weekend after Disneyland was really unique… it’s going to be memorable for years to come. You see, we have an English final project for Villalobos. My group — or rather, I myself, am rather ambitious. Our presentation was going to rock the socks off all our classmates… in theory. We met all of Saturday evening, I think, and met Sunday at Hank’s house (Hank, clean up your yard…). And they we met… ALL OF MEMORIAL DAY… from 10 in the morning, through lunch, through dinner, through midnight… and they left at 6am in the morning on Tuesday. Naturally, I skipped school on Tuesday to finish editing our horrible, crappy video, which wasn’t even my job. The thing is, nobody in my group except me can edit video, so… Our presentation was on Wednesday. We first showed the video, which I typed up subtitles for the day before. Sound effects were added, but unfortunately Windows Movie Maker seems to be incapable of outputting playable video without considerable geekery, so we ended up throwing away many hours of work that we", "to all attractions within the park. A wristband or pass is then shown at the attraction entrance to gain admission. Disneyland opened in 1955 using the pay-as-you-go format. Initially, guests paid the ride admission fees at the attractions. Within a short time, the problems of handling such large amounts of coins led to the development of a ticket system that, while now out of use, is still part of the amusement-park lexicon. In this new format, guests purchased ticket books that contained a number of tickets, labeled “A,” “B” and “C.” Rides and attractions using an “A-ticket” were generally simple, with “B-tickets” and “C-tickets” used for the larger, more popular rides. Later, the “D-ticket” was added, then finally the now-famous “E-ticket,” which was used on the biggest and most elaborate rides, like Space Mountain. Smaller tickets could be traded up for use on larger rides (i.e., two or three A-tickets would equal a single B-ticket). – guests pay for only what they choose to experience – attraction costs can be changed easily to encourage use or capitalize on popularity – guests may get tired of spending money almost continuously – guests may not spend as much on food or souvenirs Pay-one-price An amusement park using the pay-one-price format will charge guests a single, large admission fee. The guest", "choice and my ideas were never any of the choices. There are two Innoventions - Innoventions East, and Innoventions West. I strongly recommend visiting Innoventions East and do \"Sum of all Thrills\". You design your own ride and use physics principles of mechanics/energy conservation. By this time I was sure that I had just lit a match to$300. However things got better. Most of Epcot is a kind of World's Fair. There are about a dozen exhibits dedicated to various countries, Mexico, Norway, China, etc. Each of these has souvenir shops, restaurants, and a small museum-like display concerning that country. In addition, some of the countries have an attraction like a ride or a movie. There were lines, but the longest one lasted a half hour and some were less than 10 minutes. We spent most of our time here and found it quite pleasant. The food is Americanized and of course, overpriced. This is generally true. However, there are a few exceptions. One of them is the sit-down restaurant at Morocco called Marakesh. It's an amazing restaurant. Also, if you go to the bakery at Norway, there are several authentic (based on reviews given my a Norwegian friend) items there. By early evening we had seen all of the country exhibits and were about to leave. However,", "ride each ride, 35 cents to ride each ride. And then '71, Magic Mountain opens, and they say, \"Screw that, one price, pay five bucks, get in, ride everything you want.\" It's simpler, and it has a magic feeling. Like, \"Oh my God, this is great! I make a sacrifice to enter utopia, that feels like a naturally human thing to do, and I can do whatever I want.\" It's a magic feeling. Disneyland realized they should do the same thing, and they did it, and it worked really well for them. Did that mean that the free-to-play model went away for [amusement parks]? It didn't, because if you look at county fairs, it's still what they do. Because a county fair can't afford to put up a big paywall and say you can't come in here if you don't pay, because they're a lower-end thing. So I think the same thing is going to happen with games. We're going to have these lower-end experiences where you can do that if you want. I don't think free-to-play is going to kill the \"pay one price to get in\" games. But I think what we're going to see is a gradual separation. I think there's going to be more on the ends and a little less in the middle.", "is then entitled to use almost all of the attractions in the park as often as they wish during their visit. The park might have some attractions that are not included in the admission charge; these are called “up-charge attractions” and can include bungee jumping or go-kart tracks or games of skill. However, the majority of the park’s attractions are included in the admission cost. The “pay-one-price” ticket was first used by George Tilyou at Steeplechase Park, Coney Island in 1897. The entrance fee was 25 cents for entrance to the 15-acre (61,000 m2) park and visitors could enjoy all of the attractions as much as they wanted. When Angus Wynne, founder of Six Flags Over Texas, first visited Disneyland in 1959, he noted that park’s pay-as-you-go format as a reason to make his park pay-one-price. He felt that a family would be more likely to visit his park if they knew, up front, how much it would cost to attend. – guests can more easily budget their visit – guests may be more likely to experience an attraction they’ve already paid for – guests may be willing to spend more on food and souvenirs Rides and attractions Mechanized thrill machines are what makes an amusement park out of a pastoral, relaxing picnic grove or retreat. Earliest rides", "?. example: • <eat name=\"Solstice Restaurant and Lounge\" alt=\"\" address=\"2801 California St\" directions=\"at Divisadero St\" phone=\"+1 415 359-1222\" email \"[email protected]\" fax=\"+1 415 359-1242\" url=\"http://www.solsticelounge.com/\" hours=\"5PM-midnight daily\" price=\"\\$8-\\$17\">Modern well decorated restaurant and lounge that serves up tasty, yet affordable treats, like \"gorgonzola mac-n-cheese\" and \"tempura battered fish tacos.\" It gets very packed in the evenings...a good sign!</eat> (WT-en) Asterix 16:54, 31 August 2008 (EDT) That's definitely happening, and I don't think it was that way before. I let the tech team know. --(WT-en) Peter Talk 23:53, 31 August 2008 (EDT) Archived from the Pub: Hi, would it be possible to implement the ability to have wiki links inside of listing tags in any of the fields? This would really help to make a proper listing out of a location when the location has an own article and one just wants to outline it's there and why the travellers should go see it. --(WT-en) FreakRob 05:29, 8 December 2008 (EST) Uhh, that's not what listings are for, really. Can you give an example? Why not just use a plain old paragraph? (WT-en) Jpatokal 05:49, 8 December 2008 (EST) I'm guessing he's looking for something like: <see name=\"[[Walt Disney World Resort]]\"></see>. The other fields in the see tag could be useful and would allow the listing for WDW to look the", "he holiday opt ion s provided! theme parks like Disneyworld combine holidays and work e.g, in a summer camp adventure holidays e .g, climbing holidays with educational or cu ltural activities Part 4 - Discussion ~~,',\",' I 5 minutes (8 minutes fo r groups of three) • W hat do yo u th in k ~ • Do you agree? • Wha t about you~ Int erlocutor: • How good an idea would it be to build a new tourism development in you r areal Where do you think the best place to build it would be? Wh at e nviro nmental considerations sho uld one t hink about wh en budding holiday dev e lopm en ts like th i s~ • Do people overestima te the importance of holidays? • Are there any places where you think it is inap propriate to build the me parks and spe cially des igned holiday resorts? • What are the advantages of go ing abroad on holiday? Thank you . T hat is the e nd of the re st. CAE Prac tice Te st 5 Pa pe r 5 • Speaking TEST 5 Part 2 - Long turn 4 minu tes (6 minutes for g roups of three) ' '1 this part of the test you each have", "the bracket is worth 1000 points to be divided among all who correctly guess the winner, and the overall winner is the one with the most points. Now that all of my colleagues know that the upsets enumerated above have already been taken by me their best responses are to pick the favorites and sure they will be correct with high probability on each, but they will split the 1000 points with everyone else and I will get the full 1000 on the inevitable one or two upsets that will come from that group. The Magic Kingdom of Data: The Walt Disney Co. recently announced its intention to “evolve” the experience of its theme park guests with the ultimate goal of giving everyone an RFID-enabled bracelet to transform their every move through the company’s parks and hotels into a steady stream of information for the company’s databases. …Tracking the flow through the parks will come next. Right now, the park prints out pieces of paper called “FastPasses” to let people get reservations to ride. The wristbands and golden orbs will replace these slips of paper and most of everything else. Every reservation, every purchase, every ride on Dumbo, and maybe every step is waiting to be noticed, recorded, and stored away in a vast database. If you add", "# P1Z2S - Editorial Author: Ke Bi Tester: Kevin Atienza Translators: Vasya Antoniuk (Russian), Team VNOI (Vietnamese) and Hu Zecong (Mandarin) Editorialist: Kevin Atienza Cakewalk ### PROBLEM: There are n Chefs, and the i th Chef wants to visit Phantasialand t_i times. The park offers a voucher which allows you to visit another time. The procedure for visiting the park and availing the offer is as follows. • Buy a ticket at the entrance of the park, and possibly avail a voucher. • Enter the theme park. While returning, sign your name in the voucher. Unsigned vouchers will not be allowed to be taken out of the park. • After the visit, the ticket counter takes back your ticket. • You can use vouchers signed with your name to visit the park again. This procedure has a flaw: the Chefs can exchange unsigned vouchers inside the park. Chefs thought of exploiting this flaw. What is the minimum number of tickets they should buy? ### QUICK EXPLANATION: Let s be the sum of the $t_i$s. Then the answer is \\max(n, \\lceil s/2 \\rceil), where \\lceil x \\rceil denotes the ceiling function. ### EXPLANATION: Let’s weaken the restrictions of the problem a bit, and say we allow to exchange vouchers anywhere, anytime, even the signed ones. What is the minimum", "trip to California. ### Problem Set 1.4 #### Exercises 1) Use the formula for combinations to find the value of each expression. Use a calculator to verify each answer. a) $_5 C _5$ b) $_6 C _4$ c) $_3 C _0$ d) $_7 C _3$ 2) In how many ways can 3 cards be selected from a standard deck of 52 cards? 3) In how many ways can three bracelets be selected from a box of ten bracelets? 4) In how many ways can a student select five questions to answer from an exam containing nine questions? 5) In how many ways can a student select five questions to answer from an exam containing nine questions if the student is required to answer the first and the last question? 6) The general manager of a fast-food restaurant chain must select 6 restaurants from 11 for a promotional program. In how many different possible ways can this selection be done? 7) There are 7 women and 5 men in a department. In how many ways can a committee of 4 people be selected? 8) For a fundraiser, a travel agency has donated 5 free vacations to Mexico as grand prizes in a raffle. Suppose that 220 people paid for raffle tickets. In how many different ways can the vacation", "Hide Easter Holidays Scandinavians often make vacation during the Easter holidays in the largest ski resort Åre. Åre provides fantastic ski conditions, many ski lifts and slopes of various difficulty profiles. However, some lifts go faster than others, and some are so popular that a queue forms at the bottom. Per is a beginner skier and he is afraid of lifts, even though he wants to ski as much as possible. Now he sees that he can take several different lifts and then many different slopes or some other lifts, and this freedom of choice is starting to be too puzzling... He would like to make a ski journey that: • starts at the bottom of some lift and ends at that same spot • has only two phases: in the first phase, he takes one or more lifts up, in the second phase, he will ski all the way down back to where he started • is least scary, that is the ratio of the time spent on the slopes to the time spent on the lifts or waiting for the lifts is the largest possible. Can you help Per find the least scary ski journey? A ski resort contains $n$ places, $m$ slopes, and $k$ lifts ($2 \\le n \\le 1000$, $1 \\le m \\le 1000$,", "• March 3rd 2008, 04:35 PM NICOLETTE Max went to the fair with 5 friends.some paid \\$2 each to ride the bumper cars.the rest paid \\$3 each to ride the roller coaster.they spent \\$16 in all.how many rode each ride ? if angel plants the same number of pepper plants in each of 5 rows, he will have none left over.if he plants 1 in the first row and then 1 more in each row than he did in the previous row, he will plant peppers in only 4 rows.how many plants does he have ? paul has 30 postcards.he has 4 times as many large postcards as small postcards.how many small postcards does he have ? how many large postcards ? can anyone help ? i've helped my daughter with about 20 of these questions, but now my brain has died (Crying) • March 3rd 2008, 04:48 PM Aryth Max went to the fair with 5 friends.some paid \\$2 each to ride the bumper cars.the rest paid \\$3 each to ride the roller coaster.they spent \\$16 in all.how many rode each ride ? I have 2 questions, one of them should reveal a problem in the problem: 1. Are you sure there are 5 friends? 2. Are you sure they spent \\$16 dollars? Because, as the problem", "## Thursday, October 14, 2010 ### Go San Francisco Card When you want to visit more than one attraction in a city, one way to save money is to buy an attraction pass. There can also be more than one attraction pass to choose from - here's what Smart Destinations offers: The Go San Francisco Card is an all access pass offering you the ultimate in convenience, cost savings, and flexibility for all the best things to do in the Bay Area. Choose 1, 2, 3, 5, or 7-day cards at a low, pre-paid price. The card provides free access to any and all of the 50 included attractions, saving up to 40% compared to buying separate tickets. Check the map to see where everything is located: View Larger Map To find other attraction passes, click on the attraction pass label in our side bar, or click on the name of the city or area you plan to visit and see what we've written about. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -", "+0 0 57 1 The Futurists, a rock group, are planning a concert tour with performances to begiven in five cities: San Francisco, Los Angeles, San Diego, Denver, and Las Vegas. How many ways can they arrange their itinerary if a.There are no restrictions? b.The tree performances in California must be given consecutively?​ Aug 14, 2020 #1 0 a. If there are no restrictions, they can go to any of the five cities first, then any of the remaining four, then any of the remaining three, etc. Therefore, the answer would be $$5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$$ ways. b. If they must give the performances in San Francisco, Los Angeles, and San Diego must be given consecutively, we can group the three together. They can either perform in one of the two other cities or in California, and then they can perform in the two other options they haven't done yet, and then the final option. We can order this in $$3 \\cdot 2 \\cdot 1 = 6$$ ways. However, we can also order the performances in California. They can perform in any of the three cities, then any of the remaining two, then the final option. From these, we also have $$3 \\cdot 2 \\cdot 1 = 6$$ ways. For every", "Probability # Rule of Sum A shopping mall has three parking lots. The first parking lot has $6$ empty parking spaces, the second has $80$ and the third has $310.$ If a car comes to the mall, how many choices are there for the car to park in a parking space? A pet store has $12$ green frogs, $10$ black and white dogs, $8$ brown cats, $7$ green iguanas, and $11$ yellow snakes. If you want a pet that is green or brown, how many choices do you have? On Calvin's street, each house is painted in a single color. There are 6 houses that are painted red, 9 houses that are painted blue, and 12 houses that are painted neither red nor blue. How many houses are there in total? Carlos went to the climbing gym, which has different types of climbing routes on the various walls. There are 10 top-roping routes on one wall, 6 lead-climbing routes on a second wall, and 9 bouldering routes on a third wall. If Carlos only has time to climb one route before he goes to work, how many choices does he have? The new books section of the library contains $7$ different nonfiction books, $4$ different romance softcovers, $6$ different mystery hardcovers, and $5$ different comics. If Anna would", "YOUR SECRETES. I WON’T CHANGE. I DON’T GIVE A FUCK WHAT YOU THINK. AT THE END OF THE DAY, ALL THAT MATTERS IS MY LIFE. IT IS SUCH A LIBERATING AND EMPOWERING CONVICTION. IT GIVES CLARITY AND FOCUS. # WRITING 160: PLAN THE CATALIA TRIP Hi, C. and J., 1. Time: For the trip, since C. returns on Friday, Sunday seems better. We can meet around 8:15am at the Long Beach ferry port, and take the ferry (\\$74) at 8:30am. The ride is about an hour. Then we can have breakfast/brunch on the island. We can return at either 3:45pm or 6pm. http://www.catalinaexpress.com/images/pdf/schedules/Catalina_Express_Schedule_Winter2015.pdf 2. Things to do: Zipline (\\$109, 2hrs) and snorkeling (2hrs) are very popular. If snorkeling is too much trouble, we can try zipline. If that is also too much, we can always find other stuff to do. http://www.visitcatalinaisland.com/specials-packages/activities/zip-ship http://www.yelp.com/biz/catalina-zip-line-eco-tour-avalon This paper studies a contract choice problem in the cross-sales setting. There are two competing manufacturers who sell substitutable products through two retailers to consumers. The demand is linear with uncertain intercept (potential). The retailers are better informed about the demand. There are two contract choices to organize the selling. The first is the wholesale price contract without information sharing, in which manufacturers decide based on the prior distribution of the demand, while each retailer $j$", "PF Gold P: 638 Jimmy - Now you know better. NEVER visit a Disney attraction without talking to Zz first... P: 2,179 Quote by Tsu Jimmy - Now you know better. NEVER visit a Disney attraction without talking to Zz first... I'm pretty much resolved to take this advice. Emeritus Sci Advisor PF Gold P: 29,239 In the meantime, for people who missed this the first time around, this is my own version of things one should/must not miss at WDW, including a few tips and hints. http://docs.google.com/View?id=df5w5j9q_2ch3cbn It is such a huge place with so many things to do. I seriously do not recommend going there (especially for first-timers and people who don't know it that well) without any kind of knowledge or some planning. Zz. Emeritus Sci Advisor PF Gold P: 29,239 I mentioned earlier about the \"Sum of All Thrills\" at Innoventions East at Epcot. This is something that I highly recommend people to do. It at least tries to incorporate basic mechanics into the design of your ride, so one would hope that while having fun, people will actually learn something, even something minuscule. This is what you see when you approach the attraction. You go into a briefing room where they tell you something about how to design your ride. You can choose" ]

[ "M14 Collapsible Stock, Missouri High School Baseball Recruits,", "students play tennis, 24% of the students play baseball, and 58% of the students playing neither tennis or baseball, if you pick a student at random what is the probability that the student plays both tennis and baseball In school 40% of the students play tennis, 24% of the students play baseball, and 58% of the students playing neither tennis or baseball, if you pick a student at random what is the probability that the student plays both tennis and baseball...", "all get a little giddy the first time we spark a social media dustup that gets all sorts of people talking about us and our pet opinions. Doesn't make him any more right in his opinions, but whatever. What really interests me is his choice of a sports-analogy that is more apt than he realizes. In the state of Texas, tens of thousands of young kids begin competing in organized football during elementary school. The enterprise is highly inclusive and exceedingly diverse. By the time these kids get to high school, they know a lot about the sport and have begun to develop skills. In high school, however, a weeding-out process begins. Not all kids make the junior varsity and varsity teams, and not all kids — even if they make the team — are apportioned equal playing time. As things progress to college, the weeding-out process becomes all the more acute. Playing on Friday nights as a high-school athlete in Texas is lots of fun with broad participation. Playing on Saturdays as a college athlete may be equally fun, but only the most competitive kids are on the field. The final weeding-out step comes when players are drafted by the National Football League — 32 teams sport 53-man rosters, meaning that only 1,696 young men are eligible", "sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball? In a certain Algebra 2 class of 26 students, 21 of them play basketball and 7 of them play baseball. There are 2 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?...", "? Free Version Easy # Venn Diagram Baseball Team GMAT-KP7EEL There are 22 players on an intramural baseball team. If fifteen players can play the 7 positions in the outfield, and ten players can play positions in the infield, how many players can play in both the infield and outfield? A $2$ B $3$ C $10$ D $15$ E $25$", "# Math Help - Should I use permutation or combination in here? 1. ## Should I use permutation or combination in here? A certain baseball team has 24 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, short stop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position. How may ways are there to select nine of the 24 players to be on the field (without regard to which position each player will have)? My reasoning is: 24x23x22x21x20x19x18x17x16 =474467051520 I don't get why this is wrong though. Any help would be appreciated! 2. I think it's 9*(24 choose 9) 3. Originally Posted by Felipe A certain baseball team has 24 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, short stop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position. How may ways are there to select nine of the 24 players to be on the field (without regard to", "Statistics At a certain high school, there are three sports: baseball, basketball, and football. Some athletes at this school play two of these three, but no athlete plays in all three. At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8. How many athletes at this school play baseball? Statement (1): 40 athletes play both baseball and football, and 75 play football only and no other sport Statement (2): 60 athletes play only baseball and no other sport For a discussion of ratios & proportions on the GMAT, further practice problems, and a solution of this problem, see: http://magoosh.com/gmat/2013/gmat-quant ... oportions/ Mike _________________ Mike McGarry Magoosh Test Prep Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939) Intern Joined: 31 Jan 2013 Posts: 17 Schools: ISB '15 WE: Consulting (Energy and Utilities) Re: At a certain high school, there are three sports: baseball, [#permalink] ### Show Tags 10 Sep 2013, 20:57 2 2 Given, ratio of Base B: Basket B: Foot B = 15:12:8 and non of them play all three games. Each form of game will have players who plays only that game and two games. 1) ((Only Base B)+(Base B+Basket B)+ 40 ) : (", "# The scatterplot shows the number of athletes on the field hockey and tennis teams at southern high school The scatterplot shows the number of athletes on the field hockey and tennis teams at southern high school during seven different years. which statement is best supported by the line of best fit? when there are 65 field hockey players, there will be about 40 tennis players. when there are 20 field hockey players, there will be about 20 tennis players. when there are 10 field hockey players, there will be about 35 tennis players. when there are 40 field hockey players, there will be about 25 tennis players. ## This Post Has 4 Comments 1. sarbjit879 says: b.When there are 20 field hockey players, there will be about 20 tennis players, Step-by-step explanation: 2. jazminpagan15 says: b Step-by-step explanation: because what it seems the rate is going down 3. salehabajwa101 says: Based on the graph: tennis players 40 35 30 25 20 15 10 hockey players 0 10 20 30 40 50 60 the number of tennis players decreases as the number of field hockey players increases. a decrease of 5 tennis players result to an increase of 10 hockey players. the statement that best supports the line of best fit is: c - when there are 10", "0 like 0 dislike There are 12 players on a baseball team. In how many different ways can the coach choose players for left field, right field,and center field? A) 36 B) 220 C) 1320 D) 1728 0 like 0 dislike 1320 (choice C) Explanation: There are 12 choices for left field. Then there are 12-1 = 11 choices for right field. Lastly, there are 11-1 = 10 choices for center field. In total we have 12*11*10 = 1320 different permutations or orderings. by", "75 + 40 +(Foot B+Basket B)) = 15:8 from the provided information, we can't deduct the total number of base ball players. Hence, insufficient.(Options A and D can be eliminated) 2) from the provided information, we can't deduct the total number of base ball players. Hence, insufficient. (Option B can be eliminated) Both 1 & 2 : (60 +(Base B+Basket B)+ 40 ) : ( 75 + 40 +(Foot B+Basket B)) = 15:8 From the provided information, we still can't deduct the total number of base ball players. Hence, insufficient. (Option C also can be eliminated) /SW GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 544 At a certain high school, there are three sports: baseball, [#permalink] ### Show Tags 31 Oct 2018, 13:47 1 Quote: At a certain high school, there are three sports: baseball, basketball, and football. Some athletes at this school play two of these three, but no athlete plays in all three. At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:18. How many athletes at this school play baseball? Statement (1): 40 athletes play both baseball and football, and 75 play football only and no other sport Statement (2): 60 athletes play only baseball and no other sport Obs.: my solution is", "they know a lot about the sport and have begun to develop skills. In high school, however, a weeding-out process begins. Not all kids make the junior varsity and varsity teams, and not all kids — even if they make the team — are apportioned equal playing time. As things progress to college, the weeding-out process becomes all the more acute. Playing on Friday nights as a high-school athlete in Texas is lots of fun with broad participation. Playing on Saturdays as a college athlete may be equally fun, but only the most competitive kids are on the field. The final weeding-out step comes when players are drafted by the National Football League — 32 teams sport 53-man rosters, meaning that only 1,696 young men are eligible to suit up for Sunday football. These are the best of the best athletes and are rewarded accordingly....I think of science in this same way. Emphasis added. Naturally this is just a re-hash of the baseball player analogy that Comradde PhysioProffe loves to deploy on these pages and it has a lot of truth in it. I wrote a post once upon a time that is relevant to this issue. HIGHLY relevant. Hmm. You know, I once watched a Rose Bowl in which an undersized mediocre looking, but nevertheless competent, quarterback", "are left = 180. Explanation: Number of bats each crate of baseball bats contains = 30. Number of crates of bats the store sells = 2. Number of crates of bats the store has = 8. Total number of bats the store has = Number of bats each crate of baseball bats contains × Number of crates of bats the store has = 30 × 8 = (10 × 8) + (10 × 8) + (10 × 8) = 80 + 80 + 80 = 160 + 80 = 240. Total number of bats the store sells = Number of bats each crate of baseball bats contains × Number of crates of bats the store sells = 30 × 2 = (10 × 2) + (10 × 2) + (10 × 2) = 20 + 20 + 20 = 40 + 20 = 60. Number of bats are left = Total number of bats the store has – Total number of bats the store sells = 240 – 60 = 10(24 – 6) = 10 × 18 = 180. Texas Test Math Lesson Check Question 8. A school cafeteria has 30 tables. Each table seats 6 students. How many students can be seated? (A) 120 (B) 150 (C) 180 (D) 306 Number of students can be seated =", "list of the eighteen player names was leaked. so... ## Who Plays For Which Team? • What is the order in which players are counted? TT1, SPS1, TT2 or TT1, TT2, TT3? Edit: Also, are the 18 given names their first names for the selection, or no? – Braegh Apr 12 at 22:24 • Partway through solving this, and unless I'm missing something, it'll be impossible to differentiate between players with the same initial: I can't tell which player is Morton and which is Miller or Myers. Also, I second Braegh's question. – Exal Apr 12 at 22:25 • @Braegh, yes first name, as stated in the puzzle text. Thank you for looking. – John S. Apr 12 at 23:39 As written in the comments, we are missi g a key informations : how are the players counted? This answers suppose a player from Tampa is counted, then a player from St-Petes. The players from Tampa are Babcock, Brown, Flynn, Jenkins, Lucas, RandalThor, Timmons, Taylor, Peters. All those who have a letter corresponding to an even number. The players from St-Petes are Adams, Carver, Gerson, Miller, Myers, Morton, Smith, Sawyer and Young. All those who have a letter corresponding to an odd number. • Aha, I was missing something--the fact he wasn't asking for the lineup in-order, just", "# Why do you suppose baseball means so much to Troy? Why do you suppose baseball means so much to Troy?", "high school baseball is one of the most popular levels of baseball in the United States. Unlike college athletics, there is no competitive national championship on the high school level. And, unlike college athletics, umpires on the high school level are not administered by a national organization (such as the NCAA). Rather, high school baseball is administered at the state level (usually by a statewide high school athletic association) and the qualifications for becoming and remaining a high school umpire are usually set by the entity overseeing high school baseball in each individual state. For example, the Florida High School Athletic Association sets forth minimum requirements for being a high school umpire in Florida.[10] Many statewide high school athletic associations contract with multiple local umpire associations throughout their state in which the local associations agree to train and provide umpires for high school games in each association's geographic area of the state in return for a \"booking fee\" being paid to the local associations by either the statewide high school association or by individual high schools. The local associations also promise to train their umpires to meet the state high school association's minimum requirements for umpires. For example, in Florida the Jacksonville Umpires Association trains and provides umpires for high school games in the Jacksonville area.[11] The specific", "# Math Help - How many Athletes? 1. ## How many Athletes? This High School has the following numbers of athletes on four of its boys' sports teams: 41 football players; 15 basketball players; 21 baseball players; 32 wrestlers No wrestler is also a basketball player. 9 boys play both football and baseball. 5 boys wrestle and play baseball. 12 boys wrestle and play football. 6 boys play basketball and football. 7 boys play basketball and baseball. 3 boys are on all of the teams except basketball. 4 boys are on all of the teams except wrestling. How many boys are involved in these four sports? 2. ## Re: How many Athletes? Are you familiar with a concept known as the Inclusion-Exclusion principle? Let A, B, C, and D be the sets of all football players, basketball players, baseball players, and wrestlers respectively. You can think of them like lists with the player's names written on them. When I write $|A|$, I mean the number of names on that list. When I write $A\\cap B$, that means a new list of names where we only write down the names that appear on both the football list and the basketball list. When I write $|A\\cap B|$, I mean the number of names on that list. The Inclusion-Exclusion principle states", "# Pico CTF We participated as Daemons of Khorne • We got 7th place out of 8013 High school teams • We got 10th place out of 12593 teams total (including College and beyond)", "extremely challenging; there are 30 teams, 40 standards rounds plus other picks. Furthermore, among those players left, many sign as amateur free agents post-draft. You're left with players from lower divisions, very small schools, 23-year-old seniors, bad bodies, soft tossers, poor defenders, etc. But, still, there may be players who aren't good MLB prospects, but who could still perform well as part of an independent league team. Looking at top framing college catchers was a bust; this is a premium defensive position and very little is overlooked. Among the undrafted senior hitters and pitchers there were several potential prospects, many of whom you'll read about in the book. The most important fact to keep in mind is that these are real people with real lives, real families and real hopes and dreams, and playing independent ball isn't nearly lucrative enough to pay the bills. Harsh reality will limit your pool even more, and those who choose to pursue it will face the additional stress of financial strain. That being said, was Ben and Sam's experiment a success? You'll have to read the book, but absolutely, some talent was found. ### A Bayes' Solution to Monty Hall For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem . Bayes' Theorem says that for two events", "# Question There are 750 players on the active rosters of the 30 Major League Baseball teams. A random sample of 15 players is to be selected and tested for use of illegal drugs. a. If 5% of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test? b. If 10% of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test? c. If 20% of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test? Sales0 Views30 • CreatedAugust 27, 2015 • Files Included", "is extremely challenging; there are 30 teams, 40 standards rounds plus other picks. Furthermore, among those players left, many sign as amateur free agents post-draft. You're left with players from lower divisions, very small schools, 23-year-old seniors, bad bodies, soft tossers, poor defenders, etc. But, still, there may be players who aren't good MLB prospects, but who could still perform well as part of an independent league team. Looking at top framing college catchers was a bust; this is a premium defensive position and very little is overlooked. Among the undrafted senior hitters and pitchers there were several potential prospects, many of whom you'll read about in the book. The most important fact to keep in mind is that these are real people with real lives, real families and real hopes and dreams, and playing independent ball isn't nearly lucrative enough to pay the bills. Harsh reality will limit your pool even more, and those who choose to pursue it will face the additional stress of financial strain. That being said, was Ben and Sam's experiment a success? You'll have to read the book, but absolutely, some talent was found. ### A Bayes' Solution to Monty Hall For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events" ]

[ "## College Algebra (10th Edition) $8$ students The number of surveyed students who plan to take only Biology and Chemistry during their first year can be found by looking for the region/s that represents the intersection of Biology and Chemistry. The diagram shows that the region common to both chemistry and biology contains the cardinalities $8$ and $2$. Note, however, that the cardinality $2$ represents the region common to all three subjects. Since the one we are looking for is the number of students who will take only Chemistry and Biology (and not any other subject), then we should not include $2$ in our count. Therefore, only 8 students plan to take only Chemistry and Biology during their first year.", "and each of them must study at least one of Biology, Physics, or Chemistry. This implies $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=90$ There are $36$ students who study Biology, $42$ who study Physics, and $40$ who study Chemistry. This implies $a_{1}+a_2+a_4+a_5=36$; $a_{2}+a_3+a_5+a_6=42$; $a_{4}+a_5+a_6+a_7=40$. Moreover, $9$ study Biology and Physics, $8$ study Biology and Chemistry, and $7$ study Physics and Chemistry. We will have $a_2+a_5=9$; $a_4+a_5=8$; $a_5+a_6=7$. We have this set of equations: ${\\begin{cases}a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=90\\\\a_{1}+a_2+a_4+a_5=36\\\\a_{2}+a_3+a_5+a_6=42\\\\a_{4}+a_5+a_6+a_7=40\\\\a_2+a_5=9\\\\a_4+a_5=8\\\\a_5+a_6=7\\end{cases}}$ Take the first equation and substract it with all other equations, we will have: ${\\begin{cases}a_3+a_6+a_7=54\\\\a_1+a_4+a_7=48\\\\a_1+a_2+a_3=50\\\\a_1+a_3+a_4+a_6+a_7=81\\\\a_1+a_2+a_3+a_6+a_7=82\\\\a_1+a_2+a_3+a_4+a_7=83\\end{cases}}$ • $6$th equation substract by $3$rd equation: $a_4+a_7=33$, combine with $2$nd equation we will have $a_1=15$. • $5$th equation substract by $3$rd equation: $a_6+a_7=32$, combine with $1$st equation we will have $a_3=22$, combine with $a_1=15$ and the $3$rd equation, we will have $a_2=13$. Combine with $a_2+a_5=9$ (first set of equaions), we have $a_5=-4$, so you are doing nothing wrong and the problem is incorrect.", "is $10+10+12+a=35$ . So, $a=35-10-10-12=3$ . Also it is known that 30 students study technical drawing. That is $10+11+a+b=30$ . So, $b=30-10-11-a=30-10-11-3=6$ . There are 60 students. Therefore, $10+10+11+12+a+b+c=60$ . We have $c=60-10-10-11-12-a-b=60-10-10-11-12-3-6=8$ . We can find the number of students that study biology. That is $12+a+b+c=12+3+6+8=29$ . Answer: 3 students study all the 3 subjects; 6 students study biology and technical drawing but not chemistry; 8 students study biology only; 29 students study biology.", "is below this banner. Can't find a solution anywhere? NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT? You will get a detailed answer to your question or assignment in the shortest time possible. ## Here's the Solution to this Question Using Venn diagram, we have: Let $a$ be the number of students that study all the 3 subjects, $b$ be the number of students that study biology and technical drawing but not chemistry, $c$ be the number of students that study biology only. We know that 35 students study chemistry. It is $10+10+12+a=35$ . So, $a=35-10-10-12=3$ . Also it is known that 30 students study technical drawing. That is $10+11+a+b=30$ . So, $b=30-10-11-a=30-10-11-3=6$ . There are 60 students. Therefore, $10+10+11+12+a+b+c=60$ . We have $c=60-10-10-11-12-a-b=60-10-10-11-12-3-6=8$ . We can find the number of students that study biology. That is $12+a+b+c=12+3+6+8=29$ . Answer: 3 students study all the 3 subjects; 6 students study biology and technical drawing but not chemistry; 8 students study biology only; 29 students study biology. Using Venn diagram, we have: Let $a$ be the number of students that study all the 3 subjects, $b$ be the number of students that study biology and technical drawing but not chemistry, $c$ be the number of students that study biology only. We know that 35 students study chemistry. It", "# Problem on number of elements in sets This question in my text book chapter named \"Permutation, Combination and Probability\". But I am stuck with Permutation, Combination and probability. All things are seems same to me. As I am new in these. Question is: There are 53 students taking Chemistry and Physics or both, 38 students taking chemistry, and 40 students taking physics. (i) How many students are taking both Chemistry and Physics? (ii) How many students are taking Chemistry but not Physics? (iii) How many students are taking Physics but not Chemistry? Can any one help me out. • \"Chemistry and Physics or both\" doesn't make any sense. Perhaps the question was supposed to read, \"53 students taking Chemistry OR Physics or both\". – Gerry Myerson Jun 23 '14 at 7:12 • (i) $38+40-53=25$. (ii) $38-25=13$. (iii) $40-25=15$. – barak manos Jun 23 '14 at 7:28 • Yes the correct answers. – user159627 Jun 23 '14 at 7:29 This is an elementary set theoretic problem. Let $P$ be the set of students taking Physics and $C$ be the set of students taking Chemistry. Then $P\\cup C$ is the set of students taking either Physics or Chemistry or both. And $P\\cap C$ is the set of students taking both Physics and chemistry. Also $P\\setminus C$ is the set", "of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 Math Expert Joined: 02 Sep 2009 Posts: 49303 Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 29 Dec 2012, 07:03 20 1 22 Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 $${Total} = {Chemistry } + {Biology} - {Both} + {Neither}$$; $$200 = 130 + 150 - {Both} + {Neither}$$; $${Neither} = {Both} - 80$$. Since given that $${Neither} \\geq 30$$, then $${Both} - 80 \\geq 30$$ --> $${Both} \\geq 110$$. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring in either of sciences, then $$110 \\leq {Both} \\leq 130$$. _________________ SVP", "the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 Math Expert Joined: 02 Sep 2009 Posts: 46283 Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 29 Dec 2012, 07:03 19 21 Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 $${Total} = {Chemistry } + {Biology} - {Both} + {Neither}$$; $$200 = 130 + 150 - {Both} + {Neither}$$; $${Neither} = {Both} - 80$$. Since given that $${Neither} \\geq 30$$, then $${Both} - 80 \\geq 30$$ --> $${Both} \\geq 110$$. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring", "students taking Mathematics, Physics and Chemistry, no student takes one subject only. The number of students taking Mathematics and Chemistry only, equals to four times the number taking Mathematics and Physics only. If the number of students taking Physics and Chemistry only equals to three times the number of students taking all three subjects, then the number of students taking all three subjects is A. $0$ B. $5$ C. $2$ D. $4$ E. $1$ ## Problem 24 The number of divisors of the number $2006$ is A. $3$ B. $4$ C. $8$ D. $5$ E. $6$", "# How many students studied chemistry, given that they all also took physics? Q:A class of 25 students has options as to which science subjects they study. The options are biology, chemistry, geology and physics. 15 of the 25 students took only two of these sciences, whilst the other 10 students took three. 5 students studied geology and 8 studied biology, but nobody took both. How many students studied chemistry, given that they all also took physics? I know that n(G)=5,n(B)=8,n(BnG)=0,n(BnCnGnP)=0, there is no one who takes only 1 subject and no one who takes all 4 subjects. Students who take chemistry will also take physics.I also know that n(CnP)=n(BnCnP)+n(PnCnG)=10. So how can I get the n(C)? Let's state Biology as B, Chemistry as C, Geology as G and Physics as P. What is given - $$N(T) = 25, N(B) = 8, N(G) = 5, N(P) = 25, N(B \\cap G) = 0$$, $$T$$ is total students. We also know $$10$$ students take $$3$$ of these $$4$$ subjects and $$15$$ take $$2$$ subjects. So all $$12 \\, ( = 25-8-5) \\,$$ students who take neither G nor B must have taken both P and C. $$N(P \\cap C \\cap \\neg(B \\cup G)) = 12$$ As $$10$$ students take $$3$$ subjects and nobody takes both B and G, those", "majoring in any subject then the students who take atleast one subject = 200 - 30 = 170 We know that n(A$\\cup$B) = n(A) + n(B) - n(A$\\cap$B) $\\Rightarrow$ 170 = 130 + 150 - n(A$\\cap$B) Solving we get n(A$\\cap$B) = 110. i.e., Students who can take both subjects are 110 But If more than 30 students are not taking any subject, what can be the maximum number of students who can take both the subjects? As there are 130 students are majoring in chemistry, assume these students are taking biology also. So maximum students who can take both the subjects is 130 So the number of students who can take both subjects can be any number from 110 to 130. 9) Kelly and Chris are moving into a new city. Both of them love books and thus packed several boxes with books. If Chris packed 60% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number of boxes Chris packed? Simple questions. If chris packs 60% of the boxes, kelly packs remaining 40% So Kelly : Chris = 40% : 60% = 2 : 3 10) Among a group of 2500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent", "for overlapping sets $$Total=A+B+C−(sum of 2−group overlaps)+(all three)+Neither$$ Let students who enrolled in none of the courses be x $$75 = 17 + 28 + 39 - (5 + 7 + 6) + 4 + x$$ $$75 = 70 + x$$ $$x = 5$$ Answer: D Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6806 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Out of 75 students, 17 students enrolled in a Physics class, 28 studen [#permalink] ### Show Tags 15 Mar 2018, 23:54 => Attachment: 777.png [ 10.71 KiB | Viewed 454 times ] The total number of students is given by $$a + b + c + d + e + f + g + h = 75.$$ Of these, $$a + d + f + g = 17$$ are enrolled in Physics, $$b + d + e + g = 28$$ are enrolled in Chemistry, and $$c + e + f + g = 39$$ are enrolled in Biology. Adding these three equations gives $$( a + b + c ) + 2( d + e + f ) + 3g = 84.$$ We also know that $$d + g = 5$$ study both Physics and Chemistry, $$e + g = 7$$ study both Chemistry and Biology, and $$f + g = 6$$ study", "P: 14 I got $$2,21 \\times 10^{-7}$$, which seems reasonable I guess. Maybe a bit low?! P: 76 You don't have the answer? It should be reasonable right? because its the equilibrum concentration right? P: 14 Nope dont have answer (yet) :(, but it seems kinda right.. Probably is equilibrum concentration.. P: 1,521 Quote by sveioen I got $$2,21 \\times 10^{-7}$$, which seems reasonable I guess. Maybe a bit low?! I got $$2,21 \\times 10^{-6}$$ instead. Related Discussions Biology, Chemistry & Other Homework 3 Biology, Chemistry & Other Homework 6 Biology, Chemistry & Other Homework 2 Biology, Chemistry & Other Homework 3 Chemistry 6", "in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 $${Total} = {Chemistry } + {Biology} - {Both} + {Neither}$$; $$200 = 130 + 150 - {Both} + {Neither}$$; $${Neither} = {Both} - 80$$. Since given that $${Neither} \\geq 30$$, then $${Both} - 80 \\geq 30$$ --> $${Both} \\geq 110$$. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring in either of sciences, then $$110 \\leq {Both} \\leq 130$$. Answer: D. _________________ ##### Most Helpful Community Reply SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1827 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 02 Oct 2014, 02:22 12 1 Refer diagram below: Attachment: sqa.png [ 4.78 KiB | Viewed 24937 times ] 130 + 150-x + 30 = 200 x = 110 Least of Chemistry(130) & Biology(150) is Chemistry =", "major. Now, the group of biology major can sit in 3! different ways, the group of computer science major in 4! different way, etc. Thus the total come up to 4!3!4!4!2! • I gave this a try and it still isn't correct (the answer for your equation was: 165888 ways). I did go ahead and try instead 4(3!4!4!2!) = 27648 ways, which was also incorrect. – ProjectDefy Apr 4 '17 at 5:00 • I ended up finding the answer, which you can see posted if you were curious to find the answer. – ProjectDefy Apr 4 '17 at 5:19", "+0 0 170 2 $$\\text{There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? }$$ Sep 8, 2019 #1 +6046 +1 $$\\text{There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them.~\\\\ 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and ~\\\\ 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? }$$ . Sep 9, 2019 #2 +23893 +1 There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? $$\\begin{array}{|c|c|c|rl|} \\hline \\text{calculus} & \\text{physics} & \\text{chemistry} & \\\\ \\hline x & x & x & 15 & \\text{ take calculus, physics, and chemistry} \\\\ \\hline x", "QUESTION # In a college, 60 students enrolled in chemistry, 40 in physics, 30 in biology, 15 in chemistry and physics, 10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects. Hint: In this question, find the number of students enrolled in at least one of the subjects first work out the sum of students with all three subjects, with two subjects only and with only one subject. Second, work out the no of students enrolled in a single subject only and finally work out the different values of each subject. Complete step-by-step solution - Let C, P and B are the students enrolled in the subjects Chemistry, Physics and Biology respectively. Number of students enrolled in the subject Chemistry = $n(C)=60$ Number of students enrolled in the subject Physics = $n(P)=40$ Number of students enrolled in the subject Biology = $n(B)=30$ Number of students enrolled in the subjects chemistry and Physics = $n(C\\bigcap P)=15$ Number of students enrolled in the subjects Physics and Biology = $n(P\\bigcap B)=10$ Number of students enrolled in the subjects Biology and chemistry = $n(B\\bigcap C)=5$ Number of students enrolled in the subjects Physics Biology and Chemistry = $n(P\\bigcap B\\bigcap C)=0$ Therefore, the number", "then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry or Biology. So our relevant total is 170. Out of these 170, 130 take Chemistry. Let's say that the rest 40 take Biology but then, the leftover 110 of Biology must overlap with Chemistry. So the overlap must be at least 110. The other side of the situation is that all 130 of Chemistry take Biology too so the overlap in that case will be 130. This gives us that the range of students majoring in both could be anywhere from 110 to 130. VeritasPrepKarishma i am bit confised with you solution cause we could have second option as well - see below the remake can you advise whats wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at", "which is length of $AD$. 5. (Probability and Statistics)The following chart gives the graduation ages of 10 students. What is the mean age of the graduating students? 1. $17$ 2. $17.1$ 3. $17.2$ 4. $17.3$ 5. $17.4$ EXPLANATION — According to the graph, there are: Two $16$ year-olds Five $17$ year-olds Three $18$ year-olds That means their ages, added up, will be $(16 + 16) + (17 + 17 + 17 + 17 + 17) + (18 + 18 + 18) = 171$. Note that because there are $10$ elements (the number of students), we must now divide the total by $10$. $171$ divided by $10 = 17.1$. 6. (Probability and Statistics) — A high school of $350$ students offers biology and chemistry courses. $213$ students are enrolled in biology while $155$ students are studying chemistry. $78$ students are studying both subjects. How many students are not studying either subject? 1. $18$ 2. $60$ 3. $70$ 4. $50$ 5. $78$ EXPLANATION — The correct answer is B. $60$ students. To solve this problem, first find the total number of students in the combined biology and chemistry courses $[213+155=368]$ Subtract from this total the number of students taking both courses, as they should only be counted once $[368-78=290]$. This leaves you with the number of students enrolled in biology", "110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry or Biology. So our relevant total is 170. Out of these 170, 130 take Chemistry. Let's say that the rest 40 take Biology but then, the leftover 110 of Biology must overlap with Chemistry. So the overlap must be at least 110. The other side of the situation is that all 130 of Chemistry take Biology too so the overlap in that case will be 130. This gives us that the range of students majoring in both could be anywhere from 110 to 130. VeritasPrepKarishma i am bit confised with you solution cause we could have second option as well - see below the remake can you advise whats wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at least 110. The other side of the situation is that all 150 of Biology take take Chemistry too so the overlap in that case will be 150 Note the", "chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry or Biology. So our relevant total is 170. Out of these 170, 130 take Chemistry. Let's say that the rest 40 take Biology but then, the leftover 110 of Biology must overlap with Chemistry. So the overlap must be at least 110. The other side of the situation is that all 130 of Chemistry take Biology too so the overlap in that case will be 130. This gives us that the range of students majoring in both could be anywhere from 110 to 130. VeritasPrepKarishma i am bit confised with you solution cause we could have second option as well - see below the remake can you advise whats wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at least 110. The other side of the situation" ]

[ "+0 # Probability 0 87 1 In a certain Algebra 2 class of 30 students, 7 of them play basketball and 11 of themplay baseball. There are 5 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball? Jun 12, 2022 #1 +2448 0 We know that in total, $$7 + 11 - 5 = 13$$ students play either basketball or baseball. This means that the probability is $$\\color{brown}\\boxed{13 \\over 30}$$ Jun 13, 2022", "## College Algebra (10th Edition) $8$ students The number of surveyed students who plan to take only Biology and Chemistry during their first year can be found by looking for the region/s that represents the intersection of Biology and Chemistry. The diagram shows that the region common to both chemistry and biology contains the cardinalities $8$ and $2$. Note, however, that the cardinality $2$ represents the region common to all three subjects. Since the one we are looking for is the number of students who will take only Chemistry and Biology (and not any other subject), then we should not include $2$ in our count. Therefore, only 8 students plan to take only Chemistry and Biology during their first year.", "and/or chemistry. The difference between this total of students and the number of students in the high school will be the number of students not studying either subject $[350-290=60]$. $60$ students are not enrolled in Biology and/or Chemistry. B is the correct answer. 1. $18$ is the difference between the actual number of students in the high school class and the total number of enrolled students in the combined science courses. This is a trap answer that faulty logic may lead one to choose. 2. Correct. 3. $70$ would be an answer resulting from faulty arithmetic or other human errors. 4. $50$ would be an answer resulting from faulty arithmetic or other human errors. 5. $78$ is the number of students studying both subjects. Feb 23, 2012 Jun 08, 2015", "# Combinations • May 19th 2009, 06:58 PM Nisar_0926 Combinations hey guys can anyone help me with question b) c) and d) i am really confused for some reason i am not getting the correct answer. 18) Twenty-Four girls try out for the senior softball team at St. Marcellinus. In how many ways can a team of 9 fielders (1 pitcher, 1 Catcher, 4 Infielders, 3 Outfieders) be selected given each of the following conditions? a) any girl can play any position. so for a) it is 24C9 = 1 307 504 b) Of the girls trying out are Pitchers, 5 are Catchers and the other trying out play any of the other 7 field positions. c) Emily must be pitcher and Jeannie must be the catcher, but the other girls can play any other position. d) Louise cannot be on the team b) 194 480 c) 170 544 d) 817 190 • May 20th 2009, 12:23 AM noob mathematician Quote: Originally Posted by Nisar_0926 b) Of the girls trying out are Pitchers, 5 are Catchers and the other trying out play any of the other 7 field positions. b) Is the number of pitchers given? c) ${22 \\choose 7}$ d) ${23 \\choose 9}$ • May 20th 2009, 03:34 AM Villa Incognito Quote: Originally Posted by noob mathematician", "Free Version Moderate # Permutations and Combinations 15 ALGEBR-MPWKCE A college has $10$ basketball players. A $5$-member team will be selected out of those $10$ players. Then $1$ player from the remaining team members will be chosen to serve as captain. How many different selections can be made in filling the roster and captain? A $120$ B $252$ C $1260$ D $151,200$", "+0 0 170 2 $$\\text{There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? }$$ Sep 8, 2019 #1 +6046 +1 $$\\text{There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them.~\\\\ 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and ~\\\\ 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? }$$ . Sep 9, 2019 #2 +23893 +1 There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? $$\\begin{array}{|c|c|c|rl|} \\hline \\text{calculus} & \\text{physics} & \\text{chemistry} & \\\\ \\hline x & x & x & 15 & \\text{ take calculus, physics, and chemistry} \\\\ \\hline x", "sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball? In a certain Algebra 2 class of 26 students, 21 of them play basketball and 7 of them play baseball. There are 2 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?...", "110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry or Biology. So our relevant total is 170. Out of these 170, 130 take Chemistry. Let's say that the rest 40 take Biology but then, the leftover 110 of Biology must overlap with Chemistry. So the overlap must be at least 110. The other side of the situation is that all 130 of Chemistry take Biology too so the overlap in that case will be 130. This gives us that the range of students majoring in both could be anywhere from 110 to 130. VeritasPrepKarishma i am bit confised with you solution cause we could have second option as well - see below the remake can you advise whats wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at least 110. The other side of the situation is that all 150 of Biology take take Chemistry too so the overlap in that case will be 150 Note the", "# Problem on number of elements in sets This question in my text book chapter named \"Permutation, Combination and Probability\". But I am stuck with Permutation, Combination and probability. All things are seems same to me. As I am new in these. Question is: There are 53 students taking Chemistry and Physics or both, 38 students taking chemistry, and 40 students taking physics. (i) How many students are taking both Chemistry and Physics? (ii) How many students are taking Chemistry but not Physics? (iii) How many students are taking Physics but not Chemistry? Can any one help me out. • \"Chemistry and Physics or both\" doesn't make any sense. Perhaps the question was supposed to read, \"53 students taking Chemistry OR Physics or both\". – Gerry Myerson Jun 23 '14 at 7:12 • (i) $38+40-53=25$. (ii) $38-25=13$. (iii) $40-25=15$. – barak manos Jun 23 '14 at 7:28 • Yes the correct answers. – user159627 Jun 23 '14 at 7:29 This is an elementary set theoretic problem. Let $P$ be the set of students taking Physics and $C$ be the set of students taking Chemistry. Then $P\\cup C$ is the set of students taking either Physics or Chemistry or both. And $P\\cap C$ is the set of students taking both Physics and chemistry. Also $P\\setminus C$ is the set", "are not involved with chemistry or biology. So that must mean the remaining 170 are involved with chemistry or biology to some extent. But comparing this 170 relevant students with the 130 chemistry majors and the 150 biology majors seems to show the numbers don't add up. The combined 130 chem and 150 bio majors = 280 majors, which is a lot more than the relevant students. So what exactly does this mean? Please view the video for further explanation on how to set this problem up and think through it. Track your OG progress here. Attachment: 222-small.jpg [ 44.94 KiB | Viewed 1246 times ] Re: Of the 200 students at College T majoring in one or more of the &nbs [#permalink] 19 Sep 2018, 22:22 Display posts from previous: Sort by", "would expect if we looked at Player A and Player B's numbers individually? A chemistry score of 0 indicates that the team performed no better or worse than expected when the two players played together. • Weighted Chemistry weights the Chemistry score by the number of possessions that were played by the pair of teammates to give a more accurate assessment of the teammate chemistry. The problem with the Chemistry score is that if a pair of teammates only played one possession together and outscored the opponent 2-0, they would have a ridiculously high chemistry score. The weighted chemistry gives more confidence in our assessment of chemistry to player pairs who were on the court a lot together. • When evaluating the effectiveness of a particular player, look at where his name appears in the Teammate Chemistry tab ranking when sorted by any of these three metrics. If his name appears in the top half of the ranking more often than the bottom half, this indicates that he was one of the more effective players on his team. 3. Individual Analysis • This tab takes things a step further by looking at one player at a time, seeing how the team performed when a player was paired with each of his teammates. When analyzing Player A, we can", "of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring in either of sciences, then 110<={Both}<=130. Bunuel, I have a question. I'm a bit confused about the wording: to me the current stem (....at least 30 of the students are not majoring in either chemistry or biology.....) sounds like \"at least 30 students are majoring in only one subject, chem or bio\". However, the only way to solve the problem is to apply that at least 30 students relate to the section \"neither\". Additionally, at the beginning of the stem it is provided that \"Of the 200 students at College T majoring in one or more of the sciences.....\". To me it sound like each of the students is majoring in at least one subject, bio or che. Am I wrong? Apologies if the question is stupid - I'm not a native speaker 1. At least 30 of the students are not majoring in either chemistry or biology, means that there are at least 30 of the students who are majoring in something else, maybe in math (so neither chemistry nor biology). 2. Of the 200 students at College T majoring in one or more of the sciences, means that all of them are majoring in at least one of", "then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry or Biology. So our relevant total is 170. Out of these 170, 130 take Chemistry. Let's say that the rest 40 take Biology but then, the leftover 110 of Biology must overlap with Chemistry. So the overlap must be at least 110. The other side of the situation is that all 130 of Chemistry take Biology too so the overlap in that case will be 130. This gives us that the range of students majoring in both could be anywhere from 110 to 130. VeritasPrepKarishma i am bit confised with you solution cause we could have second option as well - see below the remake can you advise whats wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at", "wrong with it ? Out of these 170, 150 take Biology. Let's say that the rest 20 take Chemistry but then, the leftover 110 of Chemistry must overlap with Biology. So the overlap must be at least 110. The other side of the situation is that all 150 of Biology take take Chemistry too so the overlap in that case will be 150 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8527 Location: Pune, India Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 25 Mar 2018, 03:30 1 dave13 wrote: VeritasPrepKarishma wrote: Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 This is how I would do this question: Total 200 students, Chemistry (130), Biology(150), Neither (30) So let's ignore 30 of the 200 and we only have to worry about the 170 students who do take Chemistry", "in one or more of the [#permalink] ### Show Tags 18 Sep 2014, 11:19 1 Maksym wrote: Bunuel wrote: Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 {Total} = {Chemistry } + {Biology} - {Both} + {Neither} 200 = 130 + 150 - {Both} + {Neither} --> {Neither} = {Both} - 80. Since given that {Neither}>=30, then {Both} - 80>=30 --> {Both}>=110. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring in either of sciences, then 110<={Both}<=130. Bunuel, I have a question. I'm a bit confused about the wording: to me the current stem (....at least 30 of the students are not majoring in either chemistry or biology.....) sounds like \"at least 30 students are majoring in only one subject, chem or bio\". However, the only way to solve the problem is to apply that at", "4) Did not major in biology (No Biology) We are given: Total = 200 Chemistry = 130 Biology = 150 No Chemistry and No Biology = at least 30 We can fill all this into our table. Remember, all the columns and rows sum together. Now that we have our chart initially filled in, we can determine the greatest and least number of students who majored in biology and chemistry. Because we are given that at least 30 students majored in neither biology or chemistry, let’s assume that this number is exactly 30 students. Using the value 30 as the \"No Chemistry-No Bio\" entry, we can solve for the remaining entries in the table by subtraction. We see that there are 110 students who majored in both biology and chemistry. Even though at this point we don’t know whether this is the maximum or the minimum number of students who could major in both sciences, we can guess it’s a minimum by looking at the answer choices given. Furthermore, if that is the case, the answer must be either choice D or E. Now, if 110 is indeed the minimum, how can we find the maximum? Recall that we used 30 as the number of students who majored in neither science, and 30 is the smallest number we", "the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 Math Expert Joined: 02 Sep 2009 Posts: 46283 Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 29 Dec 2012, 07:03 19 21 Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 $${Total} = {Chemistry } + {Biology} - {Both} + {Neither}$$; $$200 = 130 + 150 - {Both} + {Neither}$$; $${Neither} = {Both} - 80$$. Since given that $${Neither} \\geq 30$$, then $${Both} - 80 \\geq 30$$ --> $${Both} \\geq 110$$. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring", "My Math Forum Permutation/Combination July 8th, 2010, 06:04 PM #1 Newbie Joined: Jul 2010 Posts: 2 Thanks: 0 Permutation/Combination A certain baseball team has 24 players. Only nine can be on the field at a time. Each of the nine players on the field has a distinct field position: pitcher, catcher, first baseman, second baseman, third baseman, short stop, left field, right field, or center field. Assume for the moment that every player is qualified to play every position. How may ways are there to select nine of the 24 players to be on the field (without regard to which position each player will have)? My reasoning is: 24x23x22x21x20x19x18x17x16 =474467051520 I don't get why this is wrong though. Any help would be appreciated! July 8th, 2010, 06:10 PM #2 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Permutation/Combination 24 choose 9, which is 24! / (9! * *24-9)!). Your answer counts each team 9! times. July 9th, 2010, 06:41 AM #3 Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: Permutation/Combination Hello, Louis Felipe! Since it says \"without regard to which position each player will have\", [color=beige]. . [/color]the listing of the nine positions was unnecessary and", "of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 Math Expert Joined: 02 Sep 2009 Posts: 49303 Re: Of the 200 students at College T majoring in one or more of the [#permalink] ### Show Tags 29 Dec 2012, 07:03 20 1 22 Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 $${Total} = {Chemistry } + {Biology} - {Both} + {Neither}$$; $$200 = 130 + 150 - {Both} + {Neither}$$; $${Neither} = {Both} - 80$$. Since given that $${Neither} \\geq 30$$, then $${Both} - 80 \\geq 30$$ --> $${Both} \\geq 110$$. Now, as the number of students majoring in both chemistry and biology cannot possibly be more than the number of students majoring in either of sciences, then $$110 \\leq {Both} \\leq 130$$. _________________ SVP", "is $10+10+12+a=35$ . So, $a=35-10-10-12=3$ . Also it is known that 30 students study technical drawing. That is $10+11+a+b=30$ . So, $b=30-10-11-a=30-10-11-3=6$ . There are 60 students. Therefore, $10+10+11+12+a+b+c=60$ . We have $c=60-10-10-11-12-a-b=60-10-10-11-12-3-6=8$ . We can find the number of students that study biology. That is $12+a+b+c=12+3+6+8=29$ . Answer: 3 students study all the 3 subjects; 6 students study biology and technical drawing but not chemistry; 8 students study biology only; 29 students study biology." ]

[ "## Elementary Geometry for College Students (6th Edition) The altitude of an equilateral triangle splits the shape into two 30 60 90 triangles. If the hypotenuse is 10, then the short leg is 5 and therefore the long leg (the altitude) is 5$\\sqrt 3$. 5$\\sqrt 3$$\\approx$8.66", "Exercise condition: 4 Given $$PQR$$ is an equilateral triangle with each sides measures 14 $$cm$$. What is the length of altitude? The length of the altitude is $$cm$$. [Note: Insert the answer to the nearest whole number].", "altitude, and use 30-60-90 triangle properties. _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. If an equilateral triangle and a square have the same area [#permalink] 05 Sep 2017, 13:27 Display posts from previous: Sort by", "# Only Altitudes? Geometry Level 4 A triangle has altitudes of length $$15, 21$$ and $$35.$$ ts area is $$M\\sqrt3$$ square units.Find the value of $$M.$$ ×", "Free Version Difficult # Perimeter: Equilateral Triangle/Altitude Given GEOM-L2WE1K The altitude of an equilateral triangle is $8\\sqrt{3}$ meters. Find the perimeter of the triangle in meters. A $16$ B $48$ C $24\\sqrt{3}$ D $64\\sqrt{3}$ E $128\\sqrt{3}$", "### Exercise condition: 2 Calculate the value of $$SR$$ of an equilateral triangle $$PQR$$ if its each sides measures 492 $$mm$$ and $$PS$$ is the altitude of $$PQR$$. The length of $$SR$$ is $$mm$$.", "The difference between the areas of the circumcircle and incircle of an equilateral triangle is $147\\pi$ square units. Find the area of the triangle.", "# ABC is an equilateral triangle of side 2a. Question. ABC is an equilateral triangle of side 2a. Find each of its altitudes. Solution: Altitude of equilateral triangle $=\\frac{\\sqrt{3}}{2} \\times$ Side $=\\frac{\\sqrt{3}}{2} \\times 2 a=\\sqrt{3} a$", "deriving the area of an equilateral triangle given its perimeter. If the perimeter is $p$, the side is $\\frac p3$, the altitude is $\\frac {\\sqrt 3}6p$ – Ross Millikan Mar 4 '18 at 23:54", "picture is hard to read.) Finally, what is the height (altitude) of the equilateral triangle?, which you need to compute the area of the triangle, etc. Try again; I think you will be able to correct a mistake or two that you made. Last edited: Dec 1, 2016 3. Dec 3, 2016", "In an isosceles triangle the altitude is:", "Perimeter and Area of an Equilateral Triangle Geometry Level 1 The area of an equilateral triangle is $25\\sqrt{3}$ square rods. What is it’s perimeter in rods? ×", "an important theorem that we use in order to solve this problem. 1), The altitude of any isosceles triangle is the median, the angle bisector of top angle and the perpendicular bisector. 2), An equilateral triangle is a special case of an isosceles. This should tell you that the altitude, which lies on the $y$-axis in your case, will not only be perpendicular to the base, but it will also bisect the top angle. This forms a 30-60-90 triangle which you must have learned in geometry. So, if you want to find the height, just use the $1$-$2$-$\\sqrt{3}$ ratio. Does this help ?", "of an altitude of an equilateral triangle by observing that it divides the equilateral triangle into two right triangles have a side of length 1 as hypotenuse and half a side, of length 1/2, as a leg. Use the Pythagorean theorem to find the length of the other leg, the altitude. thanks a lot hallsofivy I have solved it to be sqrt 3/2", "Elementary Geometry for College Students (6th Edition) The height is equal to the altitude of the equilateral triangle in the center plus 2 radius (one on the bottom and one on the top.) Altitude=.5(4)$\\sqrt 3$ A=2$\\sqrt 3$ h=A+2r h=8+2$\\sqrt 3$ h$\\approx$11.46", "3/pi ) r 2 = 154 7/22... Pi ( x^2 ) / 12 circle inscribed in an equilateral triangle is x sqrt ( ). And simultaneously, a unique triangle and AD = h be the altitude CAT question is disucussed on EduRev Group...", "regular hexagon has sides of length 3 in. What is the area of the hexagon? ( Hint: the hexagon is made up a 6 equilateral triangles. ) 4. The area of an equilateral triangle is $36 \\sqrt{3}$ . What is the length of a side? 5. If a road has a grade of $30^\\circ$ , this means that its angle of elevation is $30^\\circ$ . If you travel 1.5 miles on this road, how much elevation have you gained in feet (5280 ft = 1 mile)?", "# An isosceles triangle has an altitude of 8cm. Its perimeter is 32cm. How can we find its area? I found: $48 c {m}^{2}$", "roads, and the side of the equilateral triangle has length $$1$$: You can build the store anywhere on or inside the triangle and the transportation cost will be $$\\sqrt{3}/2$$.", "Square $$AIME$$ has sides of length $$10$$ units. Isosceles triangle $$GEM$$ has base $$EM$$, and the area common to triangle $$GEM$$ and square $$AIME$$ is $$80$$ square units. Find the length of the altitude to $$EM$$ in $$\\triangle GEM$$. (第二十六届AIME1 2008 第2题)" ]

[ "$AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$. $[asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*\"24\", A--T, N); label(rotate(degrees(C-A))*\"24\", A--C, 2*NW); label(\"12\\sqrt 3\", C--D, E); label(\"12\\sqrt 3\", D--B, S); label(\"12\", A--D, S); pen p = fontsize(8)+red; MA(\"45^\\circ\", C,B,A,2); MA(\"30^\\circ\", B,A,T,2.5); MA(\"30^\\circ\", T,A,C,3.5); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, N); dot(\"T\", T, NE); dot(\"D\", D, S); [/asy]$ By 30-60-90 triangles, $AD=12$ and $CD=12\\sqrt{3}$. We also notice that $\\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\\sqrt{3}$. The base $AB$ is $12+12\\sqrt{3}$, and the altitude $CD=12\\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=\\boxed{291}$. -youyanli ## See also 2001 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"?\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr This is a 30-60-90 triangle. Thus, the shorter leg is half as long as the hypotenuse. BC = \\frac12 \\cdot AB = \\frac12 \\cdot (( 2 * BC ) + BCrs) = ( BC ) + BCrs 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ABs. How long is AC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", \"?\", ABs ); AC * AC * ACr This is a 30-60-90 triangle. Thus, the longer leg is \\frac{\\sqrt{3}}{2} times as long as the hypotenuse. AC = \\frac{\\sqrt{3}}{2} \\cdot AB = \\frac{\\sqrt{3}}{2} \\cdot (ABs) = AC + ACrs", "and multiplying these together proves the formula for isogonal lines. Hence, we have $$\\frac{BE}{15-BE}\\cdot \\frac{9}{6}=\\frac{169}{196}\\implies BE=\\frac{2535}{463}$$ so our desired answer is $\\boxed{463}.$ ## Solution 6 (Tangent subtraction formulas) Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle. $[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label(\"G\",G,N); label(\"F\",F,N); [/asy]$ Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\\angle DAG = \\alpha$. Now we know that $\\overline{DG} = \\frac{24}{5}$ and $\\overline{GC} = \\frac{18}{5}$. Therefore, $\\overline{AG} = \\frac{52}{5}$, so $\\tan{(\\alpha)} = \\frac{6}{13}$. Our goal now is to use tangent $\\angle EAG$ in triangle $AEG$. We set $\\overline{BE}$ to $x$, so $\\overline{ED} = 9 - x$ and $\\overline{EC} = 15 - x$, so $\\overline{EG} = \\frac{4}{5}(15-x)$ and $\\overline{GC} = \\frac{3}{5}(15-x)$ so $\\overline{AG} = \\frac{3x+25}{5}$. Now we just need tangent of $\\angle EAG$. We find this using $\\tan{(EAG)} =", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "$12 \\sqrt{2}$ is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it splits the square into two 45-45-90 triangles. $12 \\sqrt{2}$ would be the hypotenuse, or equal to $x \\sqrt{2}$. $12 \\sqrt{2} &= x \\sqrt{2}\\\\12 &= x$ b) Here, we are given the hypotenuse. Solve for $x$ in the ratio. $x \\sqrt{2} &= 16\\\\x &= \\frac{16}{\\sqrt{2}} \\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} = \\frac{16 \\sqrt{2}}{2} = 8 \\sqrt{2}$ In part b, we rationalized the denominator which we learned in the first section. ## 30-60-90 Triangles The second special right triangle is called a 30-60-90 triangle, after the three angles. To draw a 30-60-90 triangle, start with an equilateral triangle. Investigation 8-3: Properties of a 30-60-90 Triangle Tools Needed: Pencil, paper, ruler, compass 1. Construct an equilateral triangle with 2 inch sides. 2. Draw or construct the altitude from the top vertex to form two congruent triangles. 3. Find the measure of the two angles at the top vertex and the length of the shorter leg. The top angles are each $30^\\circ$ and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector. 4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical. $1^2 + b^2 &= 2^2\\\\1+b^2 &= 4\\\\b^2", "# Thread: Help with Special Triangle Problem 1. ## Help with Special Triangle Problem Would appreciate any help on the following problem, Thank You! 2. given $AB = 2$ BC is the hypotenuse of the 45-45-90 triangle ... $BC = (AB)\\sqrt{2} = 2\\sqrt{2}$ BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg. $BD = \\frac{2\\sqrt{2}}{\\sqrt{3}} = \\frac{2\\sqrt{6}}{3}$ from this point, you should be able to finish up. 3. Originally Posted by skeeter given $AB = 2$ BC is the hypotenuse of the 45-45-90 triangle ... $BC = (AB)\\sqrt{2} = 2\\sqrt{2}$ BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg. $BD = \\frac{2\\sqrt{2}}{\\sqrt{3}} = \\frac{2\\sqrt{6}}{3}$ from this point, you should be able to finish up. Thanks! So if you can just confirm for me: $CD = \\frac{2\\sqrt{42}}{3}$ (Used Pythagorean Theorem) and $x = \\frac{2\\sqrt{7}}{\\sqrt{3}}$ (Used Pythagorean Theorem again) Is this correct and in simplest form? 4. bump.... Anybody able to confirm if my previous answer was correct? Thanks!", "equilateral triangle area formula, $x=13\\sqrt{3}$, so $x^2$ is $\\boxed{507}$. ## Solution 2 Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle $BMN$ to solve for its area. ## Solution 3 Note that $AB=DB=16$ and $BE=BC=4$. Also, $\\angle ABE = \\angle DBC = 120^{\\circ}$. Thus, $\\triangle ABE \\cong \\triangle DBC$ by SAS. From this, it is clear that a $60^{\\circ}$ rotation about $B$ will map $\\triangle ABE$ to $\\triangle DBC$. This rotation also maps $M$ to $N$. Thus, $BM=BN$ and $\\angle MBN=60^{\\circ}$. Thus, $\\triangle BMN$ is equilateral. Using the Law of Cosines on $\\triangle ABE$, $$AE^2 = 16^2 + 4^2 - 2\\cdot 16\\cdot 4\\cdot\\left(-\\frac{1}{2}\\right)$$ $$AE = 4\\sqrt{21}$$ Thus, $AM=ME=2\\sqrt{21}$. Using Stewart's Theorem on $\\triangle ABE$, $$AM\\cdot ME\\cdot AE + AE\\cdot BM^2 = BE^2\\cdot AM + BA^2\\cdot ME$$ $$BM = 2\\sqrt{13}$$ Calculating the area of $\\triangle BMN$, $$[BMN] = \\frac{\\sqrt{3}}{4} BM^2$$ $$[BMN] = 13\\sqrt{3}$$ Thus, $x=13\\sqrt{3}$, so $x^2 = 507$. Our final answer is $\\boxed{507}$. Admittedly, this is much more tedious than the coordinate solutions. I also noticed that there are two more ways of showing that $\\triangle BMN$ is equilateral: One way is to show that $\\triangle ADB$, $\\triangle BMN$, and $\\triangle ECB$ are related by a spiral similarity centered at $B$. The other way is to use the Mean Geometry Theorem.", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 (mass points) We see that $EF:FB=3:7$ and $AF=FD$. We assign a mass of $7$ to $E$ and $3$ to $D$, making $F$ have mass $10$ and $A$ and $D$ each have mass 5.", "(opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using cosine: betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", AC + ACrs, \"x\" ); arc([0, 5*sqrt(3)/2], 0.8, 270, 300); label([-0.1, (5*sqrt(3)/2)-1], \"{30}^{\\\\circ}\", \"below right\"); Cosine is adjacent over hypotenuse (SOH CAH TOA), so \\cos {30}^{\\circ} = \\dfrac{ACdisp}{x}. We also know that \\cos{30}^{\\circ} = \\dfrac{\\sqrt{3}}{2}. Solving for x, we get \\qquad x \\cdot \\cos{30}^{\\circ} = ACdisp \\qquad x \\cdot \\dfrac{\\sqrt{3}}{2} = ACdisp \\qquad x = ACdisp \\cdot \\dfrac{2}{\\sqrt{3}} \\qquad x = ACdisp \\cdot \\dfrac{2\\cdot\\sqrt{3}}{3} So, x = ABs. randRange( 2, 6 ) randFromArray([ [1, \"\"], [3, \"\\\\sqrt{3}\"] ]) 2*BC + BCrs randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"x\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr We know the length of the hypotenuse of this triangle, and want to find the length of a leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine", "0 \\Rightarrow x = \\frac{-\\sqrt{3}\\pm 3\\sqrt{7}}{2}$. Since $x$ must be positive, $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow (C)$. ## Solution 2 (without trigonometry) Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$. $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label(\"$$E$$\", EE,SE); [/asy]$ As proved in the first solution, $\\angle OCD = 150^\\circ$. That makes $\\Delta OCE$ a $30-60-90$ triangle, so $OE = \\frac{x}{2}$ and $CE= \\frac{x\\sqrt 3}{2}$ Since $\\Delta ODE$ is a right triangle, $\\left({\\frac{x}{2}}\\right)^2 + \\left({\\frac{x\\sqrt 3}{2} +1}\\right)^2 = 4^2 \\Rightarrow x^2+x\\sqrt3-15 = 0$ Solving for $x$ gives $x =\\frac{3\\sqrt{7}-\\sqrt{3}}{2}\\Rightarrow (C)$ ## Solution 3 (simply Pythagorean Theorem) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$E$$\",(0,4.17)); label(\"$$F$$\",(0.75,4.15),W); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$ By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\\angle OED = 90^\\circ$. Since $OD = 4$ and $DE = \\frac{1}{2}$, $OE = \\sqrt{16-\\frac{1}{4}} = \\frac{\\sqrt{63}}{2} = \\frac{3\\sqrt{7}}{2}$. Since $\\triangle CED$ is $30-60-90$, $CE = \\frac{\\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\\triangle CED$). Thus $OC =", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "-- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$.", "their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\frac{1}{3}\\Rightarrow \\text{(C)}$", "# Help with Special Triangle Problem • May 22nd 2010, 05:21 AM mathguy20 Help with Special Triangle Problem Would appreciate any help on the following problem, Thank You! • May 22nd 2010, 05:48 AM skeeter given $AB = 2$ BC is the hypotenuse of the 45-45-90 triangle ... $BC = (AB)\\sqrt{2} = 2\\sqrt{2}$ BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg. $BD = \\frac{2\\sqrt{2}}{\\sqrt{3}} = \\frac{2\\sqrt{6}}{3}$ from this point, you should be able to finish up. • May 22nd 2010, 12:10 PM mathguy20 Quote: Originally Posted by skeeter given $AB = 2$ BC is the hypotenuse of the 45-45-90 triangle ... $BC = (AB)\\sqrt{2} = 2\\sqrt{2}$ BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg. $BD = \\frac{2\\sqrt{2}}{\\sqrt{3}} = \\frac{2\\sqrt{6}}{3}$ from this point, you should be able to finish up. Thanks! So if you can just confirm for me: $CD = \\frac{2\\sqrt{42}}{3}$ (Used Pythagorean Theorem) and $x = \\frac{2\\sqrt{7}}{\\sqrt{3}}$ (Used Pythagorean Theorem again) Is this correct and in simplest form? • May 27th 2010, 04:03 AM mathguy20 bump.... Anybody able to confirm if my previous answer was correct? Thanks!", "EDG}{\\text{shorter leg in} \\ \\triangle DFG} &= \\frac{\\text{hypotenuse in} \\ \\triangle EDG}{\\text{hypotenuse in} \\ \\triangle DFG}\\\\\\frac{6}{x} &= \\frac{10}{8}\\\\48 &= 10x\\\\4.8 &= x$ #### Example B Find the value of $x$ . Set up a proportion. $\\frac{shorter \\ leg \\ of \\ smallest \\ \\triangle}{shorter \\ leg \\ of \\ middle \\ \\triangle} &= \\frac{longer \\ leg \\ of \\ smallest \\ \\triangle}{longer \\ leg \\ of \\ middle \\ \\triangle}\\\\\\frac{9}{x} &= \\frac{x}{27}\\\\x^2 &= 243\\\\x &= \\sqrt{243} = 9 \\sqrt{3}$ #### Example C Find the values of $x$ and $y$ . Separate the triangles. Write a proportion for $x$ . $\\frac{20}{x} &= \\frac{x}{35}\\\\ x^2 &= 20 \\cdot 35\\\\x &= \\sqrt{20 \\cdot 35}\\\\x &= 10 \\sqrt{7}$ Set up a proportion for $y$ . Or, now that you know the value of $x$ you can use the Pythagorean Theorem to solve for $y$ . Use the method you feel most comfortable with. $\\frac{15}{y} &= \\frac{y}{35} && (10 \\sqrt{7})^2 + y^2 = 35^2\\\\y^2 &= 15 \\cdot 35 && \\qquad 700 + y^2 =1225\\\\y &= \\sqrt{15 \\cdot 35} && \\qquad \\qquad \\quad \\ y = \\sqrt{525} = 5 \\sqrt{21}\\\\y &= 5 \\sqrt{21}$ ### Guided Practice 1. Find the value of $x$ . 2. Now find the value of $y$ in $\\triangle RST$ above. 3. Write the similarity statement for the right triangles in", "{\\mathrm {adjacent} }{\\mathrm {opposite} }}}$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {b}{a}}}$ substitute $b = 1$, $a = 1$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {1}{1}}}$ $\\cot 45^{\\circ } = 1$ Case 2: Special angles 30o and 60o (from a 30o – 60o – 90o triangle) The following figure 7-3 represents an equilateral triangle with sides $a = 2$, $b = 2$, and $c =2$. Since equilateral triangle has congruent angles and the measure of angles in a triangle is $180^{\\circ }$, each angle measures $60^{\\circ }$. Let us draw an altitude from the vertex $B$. The altitude separates an equilateral triangle into two congruent right triangles. In Figure 7-4, ${\\displaystyle {\\overline {BD}}}$ is altitude, $ΔABD\\:≅\\:ΔCBD$, $∠BDA$ is a right angle, $m∠A=60^{\\circ }$, and $m∠ABD=30^{\\circ }$. We can determine the height h of these triangles by the Pythagorean theorem. $(AB)^{2}=(BD)^{2}+(AD)^{2}$ $(BD)^{2}=(AB)^{2} – (AD)^{2}$ Substitute $(BD) = h$, $AB = 2$, and $AD = 1$ in the formula $h^{2}=(2)^{2} – (1)^{2}$ $h^{2}= 3$ $h = \\sqrt{3}$ As the altitude $h$ splits the equilateral triangle into two congruent 30o – 60o – 90o triangles. Let us knock out one of those right triangles, let suppose $ABD$, and determine the values of trigonometric ratio for $30^{\\circ }$ and $60^{\\circ }$. When m B = 30o: The following Figure 7-5 represents the right-angled triangle", "# Difference between revisions of \"2009 AMC 10A Problems/Problem 10\" ## Problem Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\\triangle ABC$? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C)); dot(ps); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"3\",midpoint(A--D),NE); label(\"4\",midpoint(D--C),NE); [/asy]$ $\\mathrm{(A)}\\ 4\\sqrt3 \\qquad \\mathrm{(B)}\\ 7\\sqrt3 \\qquad \\mathrm{(C)}\\ 21 \\qquad \\mathrm{(D)}\\ 14\\sqrt3 \\qquad \\mathrm{(E)}\\ 42$ ## Solution It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\\cdot CD$. (See below for a proof.) Then $BD = \\sqrt{ 3\\cdot 4 } = 2\\sqrt 3$, and the area of the triangle $ABC$ is $\\frac{AC\\cdot BD}2 = 7\\sqrt3\\Rightarrow\\boxed{\\text{(B)}}$. Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get: 1. $AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \\cdot AD \\cdot DC$. 2. $AB^2 = AD^2 + BD^2$ 3. $BC^2 = BD^2 + CD^2$ Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD \\cdot DC$. Alternatively, note that $\\triangle ABD \\sim \\triangle BCD \\Longrightarrow \\frac{AD}{BD} = \\frac{BD}{CD}$." ]

[ "$12 \\sqrt{2}$ is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it splits the square into two 45-45-90 triangles. $12 \\sqrt{2}$ would be the hypotenuse, or equal to $x \\sqrt{2}$. $12 \\sqrt{2} &= x \\sqrt{2}\\\\12 &= x$ b) Here, we are given the hypotenuse. Solve for $x$ in the ratio. $x \\sqrt{2} &= 16\\\\x &= \\frac{16}{\\sqrt{2}} \\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} = \\frac{16 \\sqrt{2}}{2} = 8 \\sqrt{2}$ In part b, we rationalized the denominator which we learned in the first section. ## 30-60-90 Triangles The second special right triangle is called a 30-60-90 triangle, after the three angles. To draw a 30-60-90 triangle, start with an equilateral triangle. Investigation 8-3: Properties of a 30-60-90 Triangle Tools Needed: Pencil, paper, ruler, compass 1. Construct an equilateral triangle with 2 inch sides. 2. Draw or construct the altitude from the top vertex to form two congruent triangles. 3. Find the measure of the two angles at the top vertex and the length of the shorter leg. The top angles are each $30^\\circ$ and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector. 4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical. $1^2 + b^2 &= 2^2\\\\1+b^2 &= 4\\\\b^2", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "equilateral triangle area formula, $x=13\\sqrt{3}$, so $x^2$ is $\\boxed{507}$. ## Solution 2 Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle $BMN$ to solve for its area. ## Solution 3 Note that $AB=DB=16$ and $BE=BC=4$. Also, $\\angle ABE = \\angle DBC = 120^{\\circ}$. Thus, $\\triangle ABE \\cong \\triangle DBC$ by SAS. From this, it is clear that a $60^{\\circ}$ rotation about $B$ will map $\\triangle ABE$ to $\\triangle DBC$. This rotation also maps $M$ to $N$. Thus, $BM=BN$ and $\\angle MBN=60^{\\circ}$. Thus, $\\triangle BMN$ is equilateral. Using the Law of Cosines on $\\triangle ABE$, $$AE^2 = 16^2 + 4^2 - 2\\cdot 16\\cdot 4\\cdot\\left(-\\frac{1}{2}\\right)$$ $$AE = 4\\sqrt{21}$$ Thus, $AM=ME=2\\sqrt{21}$. Using Stewart's Theorem on $\\triangle ABE$, $$AM\\cdot ME\\cdot AE + AE\\cdot BM^2 = BE^2\\cdot AM + BA^2\\cdot ME$$ $$BM = 2\\sqrt{13}$$ Calculating the area of $\\triangle BMN$, $$[BMN] = \\frac{\\sqrt{3}}{4} BM^2$$ $$[BMN] = 13\\sqrt{3}$$ Thus, $x=13\\sqrt{3}$, so $x^2 = 507$. Our final answer is $\\boxed{507}$. Admittedly, this is much more tedious than the coordinate solutions. I also noticed that there are two more ways of showing that $\\triangle BMN$ is equilateral: One way is to show that $\\triangle ADB$, $\\triangle BMN$, and $\\triangle ECB$ are related by a spiral similarity centered at $B$. The other way is to use the Mean Geometry Theorem.", "$$16√3$$ C) $$24√3$$ D) 48 E) $$32√3$$ Attachment: equitriHfindA 09.16.18.jpg [ 37.97 KiB | Viewed 210 times ] Good question +1 to refresh a handy tool: the altitude of an equilateral triangle divides it into two congruent 30-60-90 triangles I. 30-60-90 triangles[/u] This method is as quick as the formula below and IMO easier to remember. In an equilateral triangle, each altitude = height = median Drop an altitude, BD, from ∠B to side AC The height, BD, divides ∆ABC into two congruent 30-60-90 triangles. • Altitude BD is a perpendicular bisector of vertex B and opposite side AC Resultant angles are 30, 60, and 90 degrees • Original angles A, B, and C = 60° • ∠ B is bisected into two 30° angles • At D, the altitude forms two 90° angles • Angles A and C = 60° • Both smaller triangles are 30-60-90 30-60-90 triangles have corresponding sides opposite those angles in the ratio $$x:x\\sqrt{3}:2x$$ Height of $$4\\sqrt{3}$$, opposite the 60° angle, corresponds with $$x\\sqrt{3}$$ $$4\\sqrt{3}=x\\sqrt{3}$$ $$x=4$$ Base AC = BC = $$2x=8$$ (In ∆BCD, side BC is opposite the 90° angle. Sides are equal.) Area, $$A=\\frac{b*h}{2}=\\frac{8*4\\sqrt{3}}{2}=16\\sqrt{3}$$ II. Formula For an equilateral triangle with side $$a$$, Height = $$\\frac{1}{2}a*\\sqrt{3}$$ Given: height = $$4\\sqrt{3}$$ Side length: $$4\\sqrt{3}=\\frac{1}{2}a*\\sqrt{3}$$ $$8\\sqrt{3}=a\\sqrt{3}$$ $$a=8$$ = base (all sides are equal)", "hexagon = $$(3\\sqrt{3} * side^2)$$ * $$\\frac{1}{2}$$ = $$(3\\sqrt{3} * 6^2)$$ * $$\\frac{1}{2}$$ = $$54\\sqrt{3}$$. Now if we donot know the formula then, please refer to the diagram below: Now we can write : y° + y° + y° + y° + y° + y° = 360° or y° = 60°. Now we can see the regular hexagon is divided into 6 triangle. So each triangle will be equilateral triangle since it is inscribed in a regular hexagon. Now area of Equilateral triangle = $$\\sqrt{3} * (side^2)$$ * $$\\frac{1}{4}$$ = $$\\sqrt{3} * (6^2)$$ * $$\\frac{1}{4}$$ =$$9\\sqrt{3}$$ Therefore for 6 triangles we have = 6* $$9\\sqrt{3}$$ = $$54\\sqrt{3}$$. Now if we donot know the area of equilateral triangle then we can always find out: Take any one triangle inscribed in the regular hexagon:- SO the area of th triangle = $$\\frac{1}{2} * base * altitude$$ Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result) So $$altitude^2 = 6^2 - 3^2$$ = 27 or altitude =", "NOT CALCULATED. \\begin{align} h=\\dfrac{2\\sqrt{s(s-a)(s-b)(s-c)}}{b} \\end{align}, \\begin{align} h=\\dfrac{2}{a} \\sqrt{\\dfrac{3a}{2}(\\dfrac{3a}{2}-a)(\\dfrac{3a}{2}-a)(\\dfrac{3a}{2}-a)} \\end{align}, \\begin{align} h=\\dfrac{2}{a}\\sqrt{\\dfrac{3a}{2}\\times \\dfrac{a}{2}\\times \\dfrac{a}{2}\\times \\dfrac{a}{2}} \\end{align}, \\begin{align} h=\\dfrac{2}{a} \\times \\dfrac{a^2\\sqrt{3}}{4} \\end{align}, \\begin{align} \\therefore h=\\dfrac{a\\sqrt{3}}{2} \\end{align}. This implies that the orthocenter is on the triangle at M, the vertex of the right angle of the triangle (a) Obtuse Triangle : Δ YPR is an obtuse triangle. Replace area in the formula with its equivalent in the area of a triangle formula: 1/2bh. It should be noted that an isosceles triangle is a triangle with two congruent sides and so, the altitude bisects the base and vertex. Altitude of an equilateral triangle is the perpendicular drawn from the vertex of the triangle to the opposite side is calculated using Altitude=(sqrt(3)*Side)/2.To calculate Altitude of an equilateral triangle, you need Side (s).With our tool, you need to enter the respective value … The perimeter of an isosceles triangle is 100 ft. This can be simplified to . For any triangle with sides a, b, c and semiperimeter s = (a+b+c) / 2, the altitude from side ais given by 1. Similar Triangle Construct. Let us represent $$AB$$ and $$AC$$ as $$a$$, $$BC$$ as $$b$$ and $$AD$$ as $$h$$. h a = 2 s ( s − a ) ( s − b ) ( s − c ) a . Learn Altitude of", "{\\mathrm {adjacent} }{\\mathrm {opposite} }}}$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {b}{a}}}$ substitute $b = 1$, $a = 1$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {1}{1}}}$ $\\cot 45^{\\circ } = 1$ Case 2: Special angles 30o and 60o (from a 30o – 60o – 90o triangle) The following figure 7-3 represents an equilateral triangle with sides $a = 2$, $b = 2$, and $c =2$. Since equilateral triangle has congruent angles and the measure of angles in a triangle is $180^{\\circ }$, each angle measures $60^{\\circ }$. Let us draw an altitude from the vertex $B$. The altitude separates an equilateral triangle into two congruent right triangles. In Figure 7-4, ${\\displaystyle {\\overline {BD}}}$ is altitude, $ΔABD\\:≅\\:ΔCBD$, $∠BDA$ is a right angle, $m∠A=60^{\\circ }$, and $m∠ABD=30^{\\circ }$. We can determine the height h of these triangles by the Pythagorean theorem. $(AB)^{2}=(BD)^{2}+(AD)^{2}$ $(BD)^{2}=(AB)^{2} – (AD)^{2}$ Substitute $(BD) = h$, $AB = 2$, and $AD = 1$ in the formula $h^{2}=(2)^{2} – (1)^{2}$ $h^{2}= 3$ $h = \\sqrt{3}$ As the altitude $h$ splits the equilateral triangle into two congruent 30o – 60o – 90o triangles. Let us knock out one of those right triangles, let suppose $ABD$, and determine the values of trigonometric ratio for $30^{\\circ }$ and $60^{\\circ }$. When m B = 30o: The following Figure 7-5 represents the right-angled triangle", "- sy√2 / (4) Area = [(2y^2) - sy√2] / (4) Area = [y(2y - s√2)] / (4) Difference in area: [s^2 / 4] - [y(2y - s√2)] / (4) [s^2 - y(2y - s√2)] / (4) But, that's unlike ANY answer choice You need to get y in terms of s to get the ratio. y - side of square s - side of equilateral triangle Note that the diagonal of a square is $$\\sqrt{2}*side = \\sqrt{2}*y$$ Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC Altitude of equilateral triangle is given by $$\\sqrt{3}/2 * Side = \\sqrt{3}/2 * s$$ Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2 So $$\\sqrt{2}*y = \\sqrt{3}/2 * s + s/2$$ So $$y = s * (\\sqrt{3} + 1)/(2\\sqrt{2})$$ Now, area of square = $$y^2 = s^2(\\sqrt{3} + 1)^2/8$$ Area of triangle ABC = $$(\\sqrt{3}/4) * s^2$$ Area of triangle BEC $$= s^2/4$$ Area of triangle ADB = Area of triangle AFC $$= 1/2 * (s^2(\\sqrt{3} + 1)^2/8 - (\\sqrt{3}/4) * s^2 - s^2/4) = s^2/8$$ Hence the required ratio is $$(s^2/4)/(s^2/8) = 2$$ SVP Joined: 06 Sep 2013 Posts: 1764 Concentration: Finance Re: An equilateral triangle ABC", "triangle that has an altitude of le [#permalink] Show Tags 26 Apr 2018, 13:58 Bunuel wrote: What is the area of an equilateral triangle that has an altitude of length 24? (A) $$24 \\sqrt{3}$$ (B) $$48 \\sqrt{3}$$ (C) $$96 \\sqrt{3}$$ (D) $$192 \\sqrt{3}$$ (E) $$384 \\sqrt{3}$$ An equilateral triangle of side 2 will have an altitude $$\\sqrt{3}$$ (because the altitude of the equilateral triangle will form a 30-60-90 triangle with the base) Since, the altitude of the equilateral triangle is 24, the side is $$\\frac{48}{\\sqrt{3}}$$ Formula used: Area(equilateral triangle) = $$\\frac{\\sqrt{3}}{4} * side^2$$ Therefore, the area of the equilateral triangle for side $$\\frac{48}{\\sqrt{3}}$$ is $$\\frac{\\sqrt{3}}{4}*48*\\frac{48}{3} = 192 \\sqrt{3}$$ (Option D) _________________ You've got what it takes, but it will take everything you've got SC Moderator Joined: 22 May 2016 Posts: 1902 What is the area of an equilateral triangle that has an altitude of le [#permalink] Show Tags 26 Apr 2018, 20:05 Bunuel wrote: What is the area of an equilateral triangle that has an altitude of length 24? (A) $$24 \\sqrt{3}$$ (B) $$48 \\sqrt{3}$$ (C) $$96 \\sqrt{3}$$ (D) $$192 \\sqrt{3}$$ (E) $$384 \\sqrt{3}$$ Attachment: equi4.26.18.png [ 62.33 KiB | Viewed 484 times ] • An equilateral triangle's altitude always creates creates two 30-60-90 triangles Each vertex = 60° Vertex B is bisected: two 30° angles Vertex C", "## Elementary Geometry for College Students (6th Edition) The altitude of an equilateral triangle splits the shape into two 30 60 90 triangles. If the hypotenuse is 10, then the short leg is 5 and therefore the long leg (the altitude) is 5$\\sqrt 3$. 5$\\sqrt 3$$\\approx$8.66", "+0 # HELP 0 137 2 In convex quadrilateral $ABCD$, $AB=8$, $BC=4$, $CD=DA=10$, and $\\angle CDA=60^\\circ$. If the area of $ABCD$ can be written in the form $\\sqrt{a}+b\\sqrt{c}$ where $a$ and $c$ have no perfect square factors (greater than 1), what is $a+b+c$? Jul 29, 2022 #1 +284 0 Hint: Draw a perpendicular line from point C to segment DA. This creates a 30-60-90 triangle. Jul 29, 2022 #2 +1155 +3 Drop an altitude from C to AD to form a 30-60-90 triangle. Call the foot of this altitude E. CE is $$5\\sqrt{3}$$ and ED is 5. This means AE is $$10-ED$$, which is 5. Thus, if we draw AC, we form another 30-60-90 triangle, and we find that AC is 10, and that triangle ACD is equilateral. An equilateral triangle with a side length of 10 has an area $$25\\sqrt{3}$$. Now, we just have to calculate the area of triangle BCA, then add the two areas. Triangle BCA has sides 4, 8, and 10. We drop another altitude, this one to AC, but unfortunately, this doesn't lead to any nice triangles. So, we resort to algebra. Call the foot of this altitude F. Let $$FC=x$$, so $$AF=10-x$$. Also, let $$BF=h$$. Using the Pythagorean theorem on both triangle BCF and ABF, we get the two equations $$x^2+h^2=16$$ and", "we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \\implies x = 4$. We have $AO_A = BD$ because of the rectangle, so $\\sqrt{11^2-r^2} = \\sqrt{13^2-r^2}-4$. Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \\cdot \\sqrt{169-r^2}$. Subtracting, we get $8 \\cdot \\sqrt{169-r^2} = 64 \\implies \\sqrt{169-r^2} = 8 \\implies 169-r^2 = 64 \\implies r^2 = 105$. We also notice that since we had $\\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$. We now look at our second diagram. $CO_C = \\sqrt{19^2-r^2} = \\sqrt{361 - 105} = \\sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \\sqrt{30^2 - 12^2} = \\sqrt{900-144} = \\sqrt{756}$. Therefore, $O_AE^2 = AC^2 = \\boxed{756}$ ~KingRavi ## Solution 2 Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. $a-11,b-13,c-19$. According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560$. After solving we have $b-a=4$, plug this back to $11^2-a^2=13^2-b^2; a=4,b=8,c=16$ The desired value is $(11+19)^2-(16-4)^2=\\boxed{756}$ ~bluesoul ## Solution 3", "4] - [y(2y - s√2)] / (4) [s^2 - y(2y - s√2)] / (4) But, that's unlike ANY answer choice You need to get y in terms of s to get the ratio. y - side of square s - side of equilateral triangle Note that the diagonal of a square is $$\\sqrt{2}*side = \\sqrt{2}*y$$ Also note that the diagonal is composed of 2 parts, the altitude of the equilateral triangle + the altitude of triangle BEC Altitude of equilateral triangle is given by $$\\sqrt{3}/2 * Side = \\sqrt{3}/2 * s$$ Altitude of 45-45-90 triangle is half of the hypotenuse (which is s here). So altitude of BEC = s/2 So $$\\sqrt{2}*y = \\sqrt{3}/2 * s + s/2$$ So $$y = s * (\\sqrt{3} + 1)/(2\\sqrt{2})$$ Now, area of square = $$y^2 = s^2(\\sqrt{3} + 1)^2/8$$ Area of triangle ABC = $$(\\sqrt{3}/4) * s^2$$ Area of triangle BEC $$= s^2/4$$ Area of triangle ADB = Area of triangle AFC $$= 1/2 * (s^2(\\sqrt{3} + 1)^2/8 - (\\sqrt{3}/4) * s^2 - s^2/4) = s^2/8$$ Hence the required ratio is $$(s^2/4)/(s^2/8) = 2$$ Karishma this is an awesome explanation for this tough problem, thanks a lot Just one small doubt, how do you know that ADB and AFC right triangles are equal? and both different from BCE, I have a", "a sketch, the answer would have been obvious. Code: C * /:\\ / : \\ a=12 / : \\ b=8 / : \\ / : \\ / : /61o : B * - - - + - - - * A D Draw altitude $\\,CD$ to side $\\,AB.$ In right triangle $CDB\\!:\\;\\sin61^o \\:=\\:\\dfrac{CD}{12} \\quad\\Rightarrow\\quad CD \\:=\\:12\\sin61^o \\;\\approx\\;10.5$ Side $\\,b$ is only 8 units long. There is no triangle.", "the angle bisector, median and altitude. Also, length of altitude of equilateral triangle ie HI and JF = $\\dfrac{\\sqrt{3} \\times x}{2}$ Length of diagonal HF = $\\sqrt{2} \\times y$ From the figure, we can express the length of diagonal HF as given : $\\sqrt{2} \\times y = \\dfrac{\\sqrt{3} \\times x}{2} + x + \\dfrac{\\sqrt{3} \\times x}{2}$ $\\Rightarrow \\sqrt{2} \\times y = (\\sqrt{3} \\times x) + x$ $\\Rightarrow \\sqrt{2} \\times y = x \\times (\\sqrt{3} + 1)$ $\\Rightarrow y = x \\times (\\dfrac{\\sqrt{3} + 1}{\\sqrt{2}})$ Ratio of areas asked = $(\\dfrac{y}{x})^{2} = (\\dfrac{\\sqrt{3} + 1}{\\sqrt{2}})^{2} = \\dfrac{4+2\\sqrt{3}}{2} = 2+\\sqrt{3}$ Alternate Explanation: Let AB=BC=CD=DA=x, then the area the area of square ABCD will be $x^2$ Since AHD, DGC, CFB, and BAE are equilateral triangles with side x, the sum of area of these traingles will be $4\\times\\ \\frac{\\sqrt{\\ 3}}{4}\\times\\ x^2=\\sqrt{\\ 3}x^2$ We know that angle HAD=60, angle DAB=90, and angle BAE=60, thus angle HAE=360-(60+90+60)=150 Area of triangles HAE+HDG+GCF+FBE=4* area of triangle HAE In HAE, AH=AE=x, and angle HAE=150, thus area will be $\\frac{1}{2}\\times\\ x\\times\\ x\\times\\ \\sin\\ 150$ sin 150=sin(180-150)=sin 30=0.5 Thus area of 4 triangles will be 4*0.5*x*x*0.5=$x^2$ Thus, total area= $x^2+\\sqrt{\\ 3\\ }\\ x^2+x^2$=$x^2\\left(2+\\sqrt{\\ 3}\\right)$ Thus area of EFGH/area of ABCD=$2+\\sqrt{\\ 3}$ We can see that T$_{2}$ is formed by using the mid points of T$_{1}$. Hence, we can", "side 1 unit long, then the hypotenuse must be 2 units long since the triangle can be thought of as half of an equilateral triangle. Call the length of the altitude $$a$$. By the Pythagorean Theorem, we can say $$a^2+1^2=2^2$$ so $$a=\\sqrt3$$. Now, consider another right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side $$y$$ units long. By the Angle-Angle Triangle Similarity Theorem, it must be similar to the right triangle with angles 30, 60, and 90 degrees and with sides 1, $$\\sqrt3$$, and 2 units long. The scale factor is $$y$$, so a triangle with angles 30, 60, and 90 degrees has side lengths $$y, y\\sqrt3,$$ and $$2y$$ units long. In triangle $$ABC, 2y=5$$ so $$y=\\frac 52$$. That means $$DB$$ is $$\\frac 52$$ units and $$DC$$ is $$\\frac 52 \\sqrt 3$$ units. In triangle $$EGH, y\\sqrt 3=4$$ so $$y=\\frac{4}{\\sqrt 3}$$. That means $$FG$$ is $$\\frac{4}{\\sqrt 3}$$ units and $$EG$$ is $$2 \\frac{4}{\\sqrt 3}$$ or $$\\frac{8}{\\sqrt 3}$$ units.", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "= AP - PY = 10\\sqrt{17} - 18$. Since $\\triangle APX$ and $\\triangle AQY$ are similar, we have $\\frac{AY}{AX} = \\frac{QY}{PX}$, or $\\frac{10\\sqrt{17} - 18}{AX} = \\frac{18}{10}$. Solving for $AX$, we get $AX = \\frac{9(10\\sqrt{17} - 18)}{5} = 18\\sqrt{17} - 36$. Therefore, the altitude from $P$ to $QR$ has length $MY + AY = 8 + (10\\sqrt{17} - 18) = 10\\sqrt{17} - 10$. Thus, the area of $\\triangle PQR$ is $\\frac{1}{2} \\cdot PQ \\cdot PR = \\frac{1}{2} \\cdot 36 \\cdot 10\\sqrt{17} = 180 \\sqrt{17}$. Feb 23, 2023", "is the only regular polygon that can be divided into congruent equilateral triangles. Connect the vertices. There are 6 equilateral triangles. An equilateral triangle has three 60° angles. Six sides of a hexagon divide 360° at center into six 60° angles at center The other two angles of each triangle also = 60°; both bisect a vertex of 120° = 60° each Given: each side of hexagon = 8 cm From above: the six triangles are equilateral. Each side of each triangle = 8 cm Find the area of one of these triangles. Multiply by 6 for the area of the hexagon. Multiply by 2 for the area of the parallelogram, which = two of those triangles Area of equilateral triangle* is a handy thing to know, s = side: $$A =\\frac{s^2\\sqrt{3}}{4}$$ $$A =\\frac{8^2\\sqrt{3}}{4}$$ $$A =16\\sqrt{3}$$ Area of hexagon (six equilateral triangles): $$6 * 16\\sqrt{3}= 96\\sqrt{3}$$ sq cm Area of parallelogram (two equilateral triangles) $$2 *16\\sqrt{3} = 32\\sqrt{3}$$ sq cm (Hexagon area) - (parallelogram area) =$$(96\\sqrt{3}- 32\\sqrt{3}) = 64\\sqrt{3}$$ sq cm *If you don't remember the formula for the area of an equilateral triangle, draw one. Drop an altitude, which is a perpendicular bisector of the opposite side and of the vertex. That altitude creates two congruent right 30-60-90 triangles with side lengths that correspond to 30-60-90, in", "that we can calculate area of the triangle in two ways: (1/2)*leg1*leg2 = (1/2)*Altitude*Hypotenuse $$6 * 6 = Altitude * \\sqrt{2} * 6$$ $$Altitude = 3\\sqrt{2} = r + \\sqrt{2}r$$ $$r = 3*\\sqrt{2}*(\\sqrt{2} - 1)$$ $$r = 3*1.4*.4 = 1.7$$ Area of circle $$= \\pi*r^2 = \\pi*1.7^2 = 3.89\\pi$$ approximately Hi VeritasPrepKarishma This is quite tough problem. Do you think is possible on test day to face such problem. I think even in 3 minutes is quite hard to get to the solution . Anyway I would like to point something that I found in your solution and of course correct me if I'm wrong. So you wrote $$1.7^2 = 3.89$$ but $$1.7^2 = 2.89$$ now can we round up 2.89 to 4, other than that we can see that the answers are quite far. Thanks a lot You are right. A much closer value is about $$2.94 * \\pi$$ and that would mean about $$3*\\pi$$. $$4*\\pi$$ is too much of a stretch but there are no options closer to this value. The likelihood of such a question in actual GMAT is not very high though it is based on very basic concepts. Also, this formula, Leg1 * Leg2 = Altitude*Hypotenuse comes in very handy. It is something we discuss in our book and it makes complicated questions" ]

[ "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "the area of $\\triangle{PQR}$ is $\\boxed{\\textbf{(C) }2\\sqrt{2}}$. Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\\overrightarrow{PQ}=[-2,2,0]$ and $\\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\\text{area} = \\frac{1}{2}\\left|\\overrightarrow{PQ} \\times \\overrightarrow{PR}\\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\\sqrt{2}$, giving the area as $2\\sqrt{2}$ like before. ## Solution 2 As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\\triangle PQR$ is perpendicular to the plane determined by $ABCD$. To see this, consider the bird's-eye view, looking down upon $P$, $Q$, and $R$ projected onto $ABCD$: $[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label(\"A\", (0,0), SW); label(\"B\", (3,0), SE); label(\"C\", (3,3), NE); label(\"D\", (0,3), NW); label(\"P\", (2,0), S); label(\"Q\", (0,2), W); label(\"R\", (1,1), NE); dot((2,0)); dot((0,2)); dot((1,1)); draw((0,2)--(2,0)); [/asy]$ Additionally, we know that $PQ$ is parallel to the plane determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\\overline{PQ} = 2\\sqrt{2}$, so the area of $\\triangle PQR$ is $\\frac{1}{2}", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "# 2007 IMO Problems/Problem 4 ## Problem In $\\triangle ABC$ the bisector of $\\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area. ## Diagram $[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),R=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair K = midpoint(B--C); pair L = midpoint(A--C); pair I=incenter(A,B,C); pair O = circumcenter(A,B,C); dot(O); dot(A^^B^^C^^R^^K^^L); draw(C--R); draw(circumcircle(A,B,R)); draw(A--C--B); draw(A--B); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"R\",R,S); label(\"K\",K,N); label(\"L\",L,S); label(\"O\",O,N); draw(K--O); draw(L--O); pair Q = intersectionpoint(L--O, C--R); dot(Q); label(\"Q\",Q,SW); pair E = midpoint(C--R); dot(E); label(\"E\",E,W); draw(O--E, dashed); pair P = intersectionpoint(O--(c/2-1.2,0), C--R); dot(P); label(\"P\",P,W); draw(O--P); draw(R--L, dashed); draw(R--K, dashed); [/asy]$ ~KingRavi ## Solution 1 (Efficient) $\\angle{RQL} = 90+\\angle{QCL} = 90+\\dfrac{C}{2}$, and similarly $\\angle{RPK} = 90+\\angle{PCK} = 90+\\dfrac{C}{2}$. Therefore, $\\angle{RQL} = \\angle{RPK}$. Using the triangle area formula $A = \\dfrac{1}{2}bc\\sin{\\angle{A}}$ yields $RQ \\cdot QL = RP \\cdot PK = \\dfrac{PK}{QL} = \\dfrac{RQ}{RP}$ after cancelling the sines and constant. Draw line $QD$ perpendicular to $BC$ that intersects $BC$ at $D$, then $QD=QL$ because the perpendicular bisectors are congruent, (or alternatively $\\triangle QDC\\cong\\triangle QLC$). This presents us $\\dfrac{PK}{QL}=\\dfrac{PK}{QD}=\\dfrac{PC}{QC}$ by similar triangles; now, we have only to prove $\\dfrac{PC}{QC}=\\dfrac{RQ}{RP}$, or $RQ \\cdot QC=RP \\cdot", "= (-1,0); pair C = (0,0); pair B = (2,0); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label(\"A\",A,W); label(\"C\",C,S); label(\"B\",B,E); label(\"P\",P,N); [/asy]$ Let $A_1$ equal to the area of $\\Delta ACP$ and $A_2$ equal to the area of $\\Delta CBP$. Heron's Formula states that the area of an triangle with sides $a$ $b$ and $c$ is $$\\sqrt{s(s-a)(s-b)(s-c)}$$ where $s$, or the semiperimeter, is half of the triangle's perimeter. The semiperimeter $s_1$ of $\\Delta ACP$ is $$\\frac{[(r + 2) + (3 - r) + 1]}{2} = \\frac{6}{2} = 3$$ Use Heron's Formula to obtain $$A_1 = \\sqrt{3(2)(3-2-r)(3-3+r)} = \\sqrt{6r(1-r)} = \\sqrt{6r-6r^2}$$ Using Heron's Formula again, find the area of $\\Delta CBP$ with sides $r+1$, $2$, and $3-r$. $$s_2 = \\frac{(r + 1) + 2 + (3 - r)}{2} = 3$$ $$A_2 = \\sqrt{3(3-2)(3-1-r)(3-3+r)} = \\sqrt{3(2r-r^2)} = \\sqrt{6r-3r^2}$$ Now, $$2 \\cdot A_1 = A_2$$ $$2\\sqrt{6r-6r^2} = \\sqrt{6r-3r^2}$$ $$4(6r-6r^2) = 6r-3r^2$$ $$24r-24r^2 = 6r-3r^2$$ $$18r = 21r^2$$ $$r = \\frac{18}{21} = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ ## See Also 2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 •", "# Difference between revisions of \"Circumradius\" The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ## Formula for a Triangle Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply $R=\\frac{abc}{4A}$. Also, $A=\\frac{abc}{4R}$ ## Proof $[asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label(\"A\", A, W); label(\"B\", B, N); label(\"C\", C, E); label(\"D\", D, S); label(\"O\", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label(\"E\", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]$ We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\\angle ADB = \\angle BCA$ because they both subtend arc $AB$. Therefore, $\\triangle BAD \\sim \\triangle BEC$ by AA similarity, so we have $$\\frac{BD}{BA} = \\frac{BC}{BE},$$ or $$\\frac {2R} c = \\frac ah.$$ However, remember that area $\\triangle ABC = \\frac {bh} 2$, so $h=\\frac{2 \\times \\text{Area}}b$. Substituting this in gives", "The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar. The ratio of their sides is $\\frac{AM}{AB} = \\frac 58$, hence the ratio of their areas is $\\left( \\frac 58 \\right)^2 = \\frac{25}{64}$. And as the area of triangle $ABC$ is $\\frac{6\\cdot 8}2 = 24$, the area of triangle $AME$ is $24\\cdot \\frac{25}{64} = \\boxed{ \\frac{75}8 }$. ## Solution 3 (Only Pythagorean Theorem) $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",Ep,S); label(\"M\",M,NW); [/asy]$ Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \\implies 25 + ME^2$, which means $AE = \\sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \\sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 +", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "# 1974 USAMO Problems/Problem 5 ## Problem Consider the two triangles $\\triangle ABC$ and $\\triangle PQR$ shown in Figure 1. In $\\triangle ABC$, $\\angle ADB = \\angle BDC = \\angle CDA = 120^\\circ$. Prove that $x=u+v+w$. $[asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label(\"A\",A,N); label(\"B\",B,ESE); label(\"C\",C,SW); label(\"D\",D,NE); label(\"a\",(B+C)/2,S); label(\"b\",(C+A)/2,WNW); label(\"c\",(A+B)/2,NE); label(\"u\",(A+D)/2,E); label(\"v\",(B+D)/2,N); label(\"w\",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NE); label(\"x\",(P+Q)/2,S); label(\"x\",(Q+R)/2,E); label(\"x\",(R+P)/2,W); label(\"a\",(P+M)/2,SE); label(\"b\",(Q+M)/2,N); label(\"c\",(R+M)/2,E); label(\"Figure 1\",P-(0,1),S); [/asy]$ ## Solutions ### Solution 1 We rotate figure $PRQM$ by a clockwise angle of $\\pi/3$ about $Q$ to obtain figure $RR'QM'$: $[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NW); label(\"R'\",RR,NE); label(\"M'\",MM,ESE); [/asy]$ Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then $$[ABC] + \\frac{b^2 \\sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .$$ Then by symmetry, $$3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2\\bigl( [PMQ] + [QMR] + [RMP] \\bigr) = 2 [PRQ] .$$ But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}$. Therefore, the area of $ABC$ is $$\\frac{(wv+vu+uw)\\sqrt{3}}{4} .$$ Also, by the Law", "# Difference between revisions of \"Circumradius\" The circumradius of a cyclic polygon is the radius of the cirumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ## Formula for a Triangle Let $a, b$ and $c$ denote the triangle's three sides, and let $A$ denote the area of the triangle. Then, the measure of the of the circumradius of the triangle is simply $R=\\frac{abc}{4A}$. Also, $A=\\frac{abc}{4R}$ ## Proof Proof: [asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label(\"$A$\", A, W); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$O$\", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label(\"$E$\", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy] We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\\angle ADB = \\angle BCA$ because they both subtend arc $AB$. Therefore, $\\triangle BAD \\sim \\triangle BEC$ by AA similarity, so we have $$\\frac{BD}{BA} = \\frac{BC}{BE},$$ or $$\\frac {2R} c = \\frac ah.$$ However, remember that area $\\triangle ABC = \\frac {bh} 2$, so $h=\\frac{2 \\times \\text{Area}}b$. Substituting", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$" ]

[ "to $BC$ and $AD$! After this point, note that $AP = PF$. It is easily derived that the circumradius of $\\triangle ABC$ is $\\frac{10}{\\sqrt{2}}$. Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\\boxed{077}$. ~awang11's sol", "This is a 30-60 right triangle. We know the ratios of ths sides . . . . $AC : BC : AB \\:=\\:1 : \\sqrt{3} : 2$ Hence, we have: . $AC : BC : AB \\;=\\;\\frac{40\\sqrt{3}}{3} : 40 : \\frac{80\\sqrt{3}}{3}$ The height of the tree is: . $AC + AB \\;=\\;\\frac{40\\sqrt{3}}{3} + \\frac{80\\sqrt{3}}{3} \\;=\\;\\frac{120\\sqrt{3}}{3} \\;=\\;\\boxed{40\\sqrt{3}\\text{ m}}$ • Jul 19th 2008, 07:18 PM aronfelizardo hai just find the distance and u can find the answer ..... its just too easy....... (Cool)", "# How do you find the other two side of a right triangle ABC, if ∠B=60 and AB=12 where AB is the hypotenuse? Nov 6, 2014 $m a t h b f \\left\\{{30}^{\\circ} \\text{-\"60^circ\"-} {90}^{\\circ}\\right\\}$ Triangle The ratios of three sides of a ${30}^{\\circ} \\text{-\"60^circ\"-} {90}^{\\circ}$ triangle are: $1 : \\sqrt{3} : 2$ Since the posted Triangle ABC is a ${30}^{\\circ} \\text{-\"60^circ\"-} {90}^{\\circ}$ triangle, we have the ratios: $B C : A C : A B = 1 : \\sqrt{3} : 2$ $A B = 12 \\implies B C = \\frac{A B}{2} = \\frac{12}{2} = 6$ $A C = B C \\cdot \\sqrt{3} = 6 \\sqrt{3}$ Hence, $B C = 6$ and $A C = 6 \\sqrt{3}$. I hope that this was helpful.", "\\times ED$=$\\frac {1}{2} area of \\triangle ABC$ i.e $\\frac{1}{2} \\times AC \\times ED$=60 i.e $\\frac{1}{2} \\times 17 \\times x$ =60 i.e $x=\\frac {120}{7}$", "Answer Comment Share Q) # In a triangle $ABC,C=90^{\\circ}$.Then $\\large\\frac{a^2-b^2}{a^2+b^2}=$ $\\begin {array} {1 1} (1)\\;\\sin (A+B) & \\quad (2)\\;\\sin (A-B) \\\\ (3)\\;\\cos (A+B) & \\quad (4)\\;\\cos (A-B) \\end {array}$", "$6$, $6.5$, $7$, and $7.5$. However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\\frac{32(6)}{(6)^2-32}=48$. Thus the perimeter of $\\triangle ABC$ must be $2(x+y) = {108}$.", "## Elementary Algebra We are told: $A = 5B \\\\ B+ C = A - 60$ Since it is a triangle, we know: $A + B + C =180$ Substituting $B+C = A -60$, we obtain: $A + A - 60 = 180 \\\\ A = 120$ Thus, we find: $A = 5B \\\\ B = \\frac{120}{5} = 24$ Finally, we find C: $C = 180 - A -B = 180 - 120 - 24 = 36$", "ABC. Code: 3 5 A * * * * * o * * * * * * * B * 60° * 3 * * * * 5 o * * o x * * * * x * C Law of Cosines: . $(x+5)^2 \\;=\\;8^2 + (x+3)^2 - 2(8)(x+3)\\cos60^o$ . . . . . . . . . $x^2 + 10x + 25 \\;=\\;x^2 + 6x + 9 + 64 - 8x - 24$ And we have: . $12x \\:=\\:24 \\quad\\Rightarrow \\quad \\boxed{x \\:=\\:2\\text{ cm}}$ The triangle looks like this: Code: 8 A o * * * o B * 60° * 5 * * 7 * * o C Formula: . $\\text{Area} \\;=\\;\\tfrac{1}{2}\\,bc\\sin A$ Therefore: . $A \\;=\\;\\tfrac{1}{2}(5)(8)\\sin60^o \\;=\\;\\boxed{10\\sqrt{3}\\text{ cm}^2}$ 3. Thank you!", "+0 # pls help, an explanation would be really appreciated! 0 33 2 In triangle ABC, AB = BC = 17 and AC = 16. Find the inradius of triangle ABC. May 15, 2020 #1 0 You can use the formula r = K/s for the inradius to get r = 12/5. May 15, 2020 #2 +1500 +2 Actually you forgot to multiply by two(but good job) $$r = \\frac{[ABC]}{s} = \\frac{120}{(17+17+16)/2} = \\boxed{\\frac{24}{5}}.$$", "+0 geometry 0 67 1 Let AB = 6, BC = 8, and AC = 8. What is the area of the circumcircle triangle ABC of minus the area of the incircle of triangle ABC? May 27, 2021 #1 +602 +1 Drop $C$ to $AB$ at $D$. Because of congruence, $AD=BD=3$. Then $CD=\\sqrt{8^2-3^2}=\\sqrt{55}$. So the area of the triangle is $3\\sqrt{55}$. Then the semiperimeter is 11. So $11r=3\\sqrt{55}\\implies r=\\frac{3\\sqrt{55}}{11}$, which is the inradius. The incircle area is then $\\frac{45}{11}\\pi$ The area is also $\\frac{abc}{4R}$. So $\\frac{96}{R}=3\\sqrt{55}$ so $R=\\frac{32}{\\sqrt{55}}$. The area of the circumcircle is then $\\frac{1024}{55}\\pi$. So the answer is $\\frac{1024}{55}\\pi - \\frac{45}{11}\\pi=\\boxed{\\frac{799}{55}\\pi}$. May 27, 2021", "$a = 20, \\ \\ b = 34$ and $c = 42$ and $s =$ $\\frac{a+b+c}{2}$ $=$ $\\frac{20+34+42}{2}$ $= 48$ Therefore area of triangle $= \\sqrt{s(s-a)(s-b)(s-c)}$ $= \\sqrt{48(48-20)(48-34)(48-42)}$ $= \\sqrt{20 \\times 28 \\times 14 \\times 6} = 336 \\ cm^2$ For $\\triangle ACD$ Here $a = 42, \\ \\ b = 20$ and $c = 34$ and $s =$ $\\frac{a+b+c}{2}$ $=$ $\\frac{42+20+34}{2}$ $= 48$ Therefore area of triangle $= \\sqrt{s(s-a)(s-b)(s-c)}$ $= \\sqrt{48(48-42)(48-20)(48-34)}$ $= \\sqrt{48 \\times 6 \\times 28 \\times 14} = 336 \\ cm^2$ Therefore Area of $ABCD = 336 + 336 = 672 \\ cm^2$ $\\\\$", "$\\triangle AEF\\cup H\\sim\\triangle ABC\\cup A'$, so $$\\frac{AD}{2R}=1-\\frac{A'D}{2R}=1-\\frac{HS}{HA}=1-\\cos A,$$ i.e.\\ $AD=2R(1-\\cos A)$. We are given $R=25$, and by the law of sines, $\\sin A=\\frac{49}{50}$, so $\\cos A=\\frac{3\\sqrt{11}}{50}$, and $AD=50-3\\sqrt{11}$, so $50+3+11=\\boxed{064}$.", "+0 # Precalc 0 122 1 In triangle ABC , we have A = 90 and sin B = 4/17. Find cos C. Jul 27, 2021 $$\\cos C=\\cos(90-B)=\\sin B=\\frac{4}{17}$$", "because I plan to use them while solving. $$\\triangle ABC$$ has a 30- and a 60-degree angle. The remaining angle of the triangle, $$m\\angle CAB=90^{\\circ}$$ . The triangle is therefore classified as a 30-60-90 triangle. 30-60-90 triangles have a special property: their side lengths are always at a constant ratio of $$1:\\sqrt{3}:2$$ . Since I know the length of $$\\overline{AB}$$ , 1, I can determine the lengths of the others by setting up a proportion. $$\\frac{AB}{BC}=\\frac{1}{2}$$ This is the proportion that I mentioned earlier. $$BC=2AB$$ I am solving for BC to make the proportion into a simple equation. Substitute 1 for AB. $$BC=2$$ Now we know that BC=2. In order to go further with this solution, you must know what a centroid is. A centroid is a point of concurrency where all three medians of a triangle intersect in the interior of the triangle. This means you must know what a median is. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side of that vertex. This information will help us. In the diagram above, $$\\overline{AE}$$ is a median. Therefore, point E is a midpoint of $$\\overline{BC}$$ , so point E bisects the segment. Therefore, $$BE=\\frac{BC}{2}=\\frac{2}{2}=1$$ . This means that we have identified two side lengths and the measure", "2. 549 3. 427 4. 361 Connect with: Question 13: The real root of the equation $2^{6 x}+2^{3 x+2}-21=0$ is 1. $\\frac{\\log _{2} 7}{3}$ 2. $\\log _{2} 9$ 3. $\\frac{\\log _{2} 3}{3}$ 4. $\\log _{2} 27$ Connect with: Question 14: Let $a_{1}, a_{2}, \\ldots$ be integers such that $a_{1}-a_{2}+a_{3}-a_{4}+\\cdots+(-1)^{n-1} a_{n}=n,$ for all $n \\geq 1$ Then $a_{51}+a_{52}+\\cdots+a_{1023}$ equals 1. -1 2. 10 3. 0 4. 1 Connect with: Question 15: Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at 1. $10: 25 \\mathrm{am}$ 2. $10: 18 \\mathrm{am}$ 3. $10: 27 \\mathrm{am}$ 4. $10: 45 \\mathrm{am}$ Connect with: Question 16: In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is 1. 68 2. 72 3. 78 4. 80 Connect with: Question 17: In 2010, a library contained a total of 11500 books in two categories - fiction and nonfiction. In 2015, the library contained a total of 12760 books in", "hypotenuse of triangle $ABC$. Notice that $h - \\frac {1}{21}h = \\sqrt {440}$. (The height of $ABC$ decreased by the corresponding height of $T_5$) Thus, $(AB)(h) = (AC)(BC) = 22\\cdot 21^2$. Because $AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\\cdot22^2$, $AC + BC = (21)(22) = \\boxed{462}$.", "same thing: $FAX$ is a 30-60-90 triangle, and $FX = \\frac{2}{2} = 1$, so $AX = \\sqrt{3}$. We can now find $DX$: \\begin{center} $DX = AX - AD = \\sqrt{3} - \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{2}$. \\end{center} Recalling why we were doing all this, we find the area of $\\triangle{FDX}$. It is $\\frac{FX \\times DX}{2} = \\frac{1 \\times \\frac{\\sqrt{3}}{2}}{2} = \\boxed{\\frac{\\sqrt{3}}{4}}$. Let's summarize: \\begin{center} The area of $\\triangle{ABC}$ is $\\boxed{\\frac{\\sqrt{3}}{8}}$ The area of $\\triangle{FDX}$ is $\\boxed{\\frac{\\sqrt{3}}{4}}$, which proves our initial claim. \\end{center} $\\square{}$", "how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense... because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90. and how do you know sqrt 3/3 is pi/6? • April 16th 2008, 03:53 AM topsquark Quote: Originally Posted by >_<SHY_GUY>_< i dont get how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense... because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90. and how do you know sqrt 3/3 is pi/6? For the $tan(\\pi / 6)$ part, this is just the same as $tan(30)$, so let's work on that. $tan(30) = \\frac{sin(30)}{cos(30)} = \\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}$ $= \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3}$ Does this help? -Dan", "means that we can say: $$a\\triangle ABC=a\\triangle AIB+a\\triangle BIC+a\\triangle CIA$$ $$a\\triangle ABC={5r\\over 2}+{6r\\over 2}+{7r\\over 2}$$ $$a\\triangle ABC =9r$$ Now, we also know that $$a\\triangle ABC=\\sqrt{s(s-AB)(s-AC)(s-BC)}$$ wherein $$s={AB+AC+BC\\over2}$$. This is called Heron's Formula. Substituting the values, we get: $$\\sqrt{9*4*3*2}=9r$$ $$6\\sqrt{6}=9r$$ $$r={2\\sqrt{6}\\over3}$$ Mathhemathh Aug 10, 2018 #3 +87629 +2 Nice approach, Mathehemathh !!!! CPhill Aug 10, 2018 #4 +7387 +2 Herons Formula, when so applied, is something completely new to me. Thank you, Mathhhemathh! ! asinus Aug 10, 2018", "$[ABC] = 3p = 6s \\Rightarrow rs = 6s \\Rightarrow r = 6$. The incircle of $ABC$ breaks the triangle's sides into segments such that $AB = x + y$, $BC = x + z$ and $AC = y + z$. Since ABC is a right triangle, one of $x$, $y$ and $z$ is equal to its radius, 6. Let's assume $z = 6$. The side lengths then become $AB = x + y$, $BC = x + 6$ and $AC = y + 6$. Plugging into Pythagorean's theorem: $(x + y)^2 = (x+6)^2 + (y + 6)^2$ $x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36$ $2xy - 12x - 12y = 72$ $xy - 6x - 6y = 36$ $(x - 6)(y - 6) - 36 = 36$ $(x - 6)(y - 6) = 72$ We can factor $72$ to arrive with $6$ pairs of solutions: $(7, 78), (8,42), (9, 30), (10, 24), (12, 18),$ and $(14, 15) \\Rightarrow \\mathrm{(A)}$." ]

[ "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "= (-1,0); pair C = (0,0); pair B = (2,0); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label(\"A\",A,W); label(\"C\",C,S); label(\"B\",B,E); label(\"P\",P,N); [/asy]$ Let $A_1$ equal to the area of $\\Delta ACP$ and $A_2$ equal to the area of $\\Delta CBP$. Heron's Formula states that the area of an triangle with sides $a$ $b$ and $c$ is $$\\sqrt{s(s-a)(s-b)(s-c)}$$ where $s$, or the semiperimeter, is half of the triangle's perimeter. The semiperimeter $s_1$ of $\\Delta ACP$ is $$\\frac{[(r + 2) + (3 - r) + 1]}{2} = \\frac{6}{2} = 3$$ Use Heron's Formula to obtain $$A_1 = \\sqrt{3(2)(3-2-r)(3-3+r)} = \\sqrt{6r(1-r)} = \\sqrt{6r-6r^2}$$ Using Heron's Formula again, find the area of $\\Delta CBP$ with sides $r+1$, $2$, and $3-r$. $$s_2 = \\frac{(r + 1) + 2 + (3 - r)}{2} = 3$$ $$A_2 = \\sqrt{3(3-2)(3-1-r)(3-3+r)} = \\sqrt{3(2r-r^2)} = \\sqrt{6r-3r^2}$$ Now, $$2 \\cdot A_1 = A_2$$ $$2\\sqrt{6r-6r^2} = \\sqrt{6r-3r^2}$$ $$4(6r-6r^2) = 6r-3r^2$$ $$24r-24r^2 = 6r-3r^2$$ $$18r = 21r^2$$ $$r = \\frac{18}{21} = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ ## See Also 2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 •", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "triangles of area 1 be $x$, then the base length of $\\Delta ADE=4x$. Notice, $\\big(\\frac{DE}{BC}\\big)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$, then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\big(\\frac{4}{\\sqrt{40}}\\big)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\big(1-\\frac{2}{5}\\big)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$ $[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label(\"B\",(0,0),SW); label(\"C\",(60,0),SE); label(\"E\",(50,10),E); label(\"D\",(10,10),W); label(\"A\",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$ Solution by ktong ## Solution 4 The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$. Therefore the area of the trapezoid is $40-16=\\boxed{24}$. ## Solution 5 You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$, so to find the area of such trapezoid $BCED$, we just take $40-16=\\boxed{24}$, like so. ∎ --anna0kear ## Solution 6 The combined area of the small triangles is $7$, and from the fact that each small triangle has an area of $1$, we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$, so will their areas have a proportion that is the square of the", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar. The ratio of their sides is $\\frac{AM}{AB} = \\frac 58$, hence the ratio of their areas is $\\left( \\frac 58 \\right)^2 = \\frac{25}{64}$. And as the area of triangle $ABC$ is $\\frac{6\\cdot 8}2 = 24$, the area of triangle $AME$ is $24\\cdot \\frac{25}{64} = \\boxed{ \\frac{75}8 }$. ## Solution 3 (Only Pythagorean Theorem) $[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); label(\"E\",Ep,S); label(\"M\",M,NW); [/asy]$ Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \\implies 25 + ME^2$, which means $AE = \\sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \\sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 +", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "area of $CMN$ in square inches is $[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label(\"A\", (0,0), S); label(\"B\", (1,0), S); label(\"C\", (1,1), N); label(\"D\", (0,1), N); label(\"M\", (0,.76), W); label(\"N\", (.82,0), S);[/asy]$ $\\mathrm{(A)\\ } 2\\sqrt{3}-3 \\qquad \\mathrm{(B) \\ }1-\\frac{\\sqrt{3}}{3} \\qquad \\mathrm{(C) \\ } \\frac{\\sqrt{3}}{4} \\qquad \\mathrm{(D) \\ } \\frac{\\sqrt{2}}{3} \\qquad \\mathrm{(E) \\ }4-2\\sqrt{3}$ ## Problem 20 Let $$T=\\frac{1}{3-\\sqrt{8}}-\\frac{1}{\\sqrt{8}-\\sqrt{7}}+\\frac{1}{\\sqrt{7}-\\sqrt{6}}-\\frac{1}{\\sqrt{6}-\\sqrt{5}}+\\frac{1}{\\sqrt{5}-2}.$$ Then $\\mathrm{(A)\\ } T<1 \\qquad \\mathrm{(B) \\ }T=1 \\qquad \\mathrm{(C) \\ } 12 \\qquad$ $\\mathrm{(E) \\ }T=\\frac{1}{(3-\\sqrt{8})(\\sqrt{8}-\\sqrt{7})(\\sqrt{7}-\\sqrt{6})(\\sqrt{6}-\\sqrt{5})(\\sqrt{5}-2)}$ ## Problem 21 In a geometric series of positive terms the difference between the fifth and fourth terms is $576$, and the difference between the second and first terms is $9$. What is the sum of the first five terms of this series? $\\mathrm{(A)\\ } 1061 \\qquad \\mathrm{(B) \\ }1023 \\qquad \\mathrm{(C) \\ } 1024 \\qquad \\mathrm{(D) \\ } 768 \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 22 The minimum of $\\sin\\frac{A}{2}-\\sqrt{3}\\cos\\frac{A}{2}$ is attained when $A$ is $\\mathrm{(A)\\ } -180^\\circ \\qquad \\mathrm{(B) \\ }60^\\circ \\qquad \\mathrm{(C) \\ } 120^\\circ \\qquad \\mathrm{(D) \\ } 0^\\circ \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 23 In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "the area of $\\triangle{PQR}$ is $\\boxed{\\textbf{(C) }2\\sqrt{2}}$. Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\\overrightarrow{PQ}=[-2,2,0]$ and $\\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\\text{area} = \\frac{1}{2}\\left|\\overrightarrow{PQ} \\times \\overrightarrow{PR}\\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\\sqrt{2}$, giving the area as $2\\sqrt{2}$ like before. ## Solution 2 As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\\triangle PQR$ is perpendicular to the plane determined by $ABCD$. To see this, consider the bird's-eye view, looking down upon $P$, $Q$, and $R$ projected onto $ABCD$: $[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label(\"A\", (0,0), SW); label(\"B\", (3,0), SE); label(\"C\", (3,3), NE); label(\"D\", (0,3), NW); label(\"P\", (2,0), S); label(\"Q\", (0,2), W); label(\"R\", (1,1), NE); dot((2,0)); dot((0,2)); dot((1,1)); draw((0,2)--(2,0)); [/asy]$ Additionally, we know that $PQ$ is parallel to the plane determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\\overline{PQ} = 2\\sqrt{2}$, so the area of $\\triangle PQR$ is $\\frac{1}{2}", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "# Problem #2136 2136 In the right triangle $\\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\\triangle DBF$ to that of $\\triangle ACE$? $\\mathrm{(A) \\ } \\frac{1}{4} \\qquad \\mathrm{(B) \\ } \\frac{9}{25} \\qquad \\mathrm{(C) \\ } \\frac{3}{8} \\qquad \\mathrm{(D) \\ } \\frac{11}{25} \\qquad \\mathrm{(E) \\ } \\frac{7}{16}$ $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",C--B,W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved." ]

[ "used? 9 tiles will cover 1 sq ft, 3ft x 5ft=15 sq ft, so 9 x 15 = 135 tiles will be needed. RonL", "How many sheets will be needed to cover it?", "SEARCH HOME Math Central Quandaries & Queries Question from Jill, how many 4\" x 4\" square tiles would I need for 50 square feet? Hi Jill, $3 \\times 4\" = 12\" = 1 \\mbox{ foot}$ and hence it takes $3 \\times 3 = 9,$ 4 inch square tiles to make a square foot. You need to cover 50 square feet so that will take at least $9 \\times 50 = 450$ tiles. Depending on the shape of the region you want to cover there may be some cutting required and hence some cutting. Hence you may need more than 450 tiles. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "# How Many SQ Feet Will a Bundle of Shingles Cover? A squarenof shingle will cover 100 square feet of root. Since a square requires threenbundles, one bundle of shingle covers about 33 square feet roof. Reference:", "land have the same area. Do you agree with Noah? Explain your reasoning. Exercise $$\\PageIndex{7}$$ A homeowner is deciding on the size of tiles to use to fully tile a rectangular wall in her bathroom that is 80 inches by 40 inches. The tiles are squares and come in three side lengths: 8 inches, 4 inches, and 2 inches. State if you agree with each statement about the tiles. Explain your reasoning. 1. Regardless of the size she chooses, she will need the same number of tiles. 2. Regardless of the size she chooses, the area of the wall that is being tiled is the same. 3. She will need two 2-inch tiles to cover the same area as one 4-inch tile. 4. She will need four 4-inch tiles to cover the same area as one 8-inch tile. 5. If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles. (From Unit 1.1.2)", "4 designer details to make your shower stand out 5 Steps to Calculate How Much Tile You Need Tiles, Tiles Kiln Ceramic Subway Tile Caspian Blue Modwalls Designer Brick Calculator Estimate the Bricks and Mortar Needed How Much Does 1 Cubic Foot of Concrete Weigh? Weight Pin on screenwall Pin en Esculturas de pared Glowing Pebble Wallpaper Family photos wall decor, Home How to keep liquor out of site? Wall pallet shelves aka my Marble Art 1.5 in. Hexagon Ceramic Mosaic in 2021", "the area of the water level of the pool, if after filling 25 m3 of water level by 10 cm? a) 25 m2 b) 250 m2 c) 2500 dm2 d) 25,000 cm2 • The book The book has 280 pages. Each side is a rectangle with sides 15 cm and 22 cm. a) how many leaves has the book? b) at least how many square meters of paper should be used to production this book? • Rectangle 45 The perimeter of a rectangle is 60cm. If the length of the rectangle is 20cm. a)find the width b)find the area. • Pool tiles The pool is 25m long, 10m wide, and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many tiles are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc? • Tiles The rectangular floor of the 6 x 1.8 m dimensions is to be covered with 50 cm square tiles. How many tiles will be needed? • Tiles How many 46 cm square tiles will cover a floor 22.08 m by 8.74 m? • Tiles How many square tiles with an area of 121 cm2 have to be ordered for the square room's", "'ll need to cover your Surface area packs you 'll need to know many... ) foot, but most people only know how many legs you need, for instance, the square of. The square footage of each board to let water drain I approach this of. Times the Width 're installing a wood picket fence, you need for your project say that working out many. Doing so, save yourself some money of siding on a 20′ wide wall will require 420 feet... Need 200 linear feet are the easiest way to calculate your needs project! Outcome you would want the Bingo caller in order to obtain the outcome... Multiple your Single board Coverage = ( Width + Expansion Gap along Width ) Stair! Possible to limit any material waste and in doing so, save yourself some money can,... Be ordered that amount to linear feet so, save yourself some.! The price per board foot, but I can purchase 20 board how many boards do i need calculator of siding, or 35.... With our easy loft Leg calculator need for your accent wall or floor tiles you will need know! Need help calculating how many pickets to purchase then enter the length of project. To calculate your needs help you work out how many pickets to purchase cost of $60", "each floor.", "the tile strip in millimeters if 12 tiles are kept in a line? Solution: Length of Square Tile = 120 cm Length of 12 Tile Strips in mm = 120*12 = 1440 mm Therefore, the Length of the Tile Strip in mm is 1440mm. Scroll to Top", "as quartzite); slate slabs (used mainly for outdoor applications like patios). Laminate Flooring: Laminate floors are made from layers of paper glued together under high pressure with plastic resin adhesive in between each one until it reaches 1–1/2 inches thick — then they’re pressed into sheets that can be used to cover old wooden floors or laid over subfloors like concrete slabs or plywood panels which won’t absorb moisture directly underneath them due to their impermeability factor due to its composition is mostly made up out of plastic resins rather than actual wood material used when creating real timber products such as solid hardwoods or engineered woods. ## How to estimate flooring cost Estimate the area of your floor by multiplying the length by the width. The length is usually easier to measure, but you can use either. For example, if you’re doing a square room (like a bedroom or living room), use diagonal measurements: if it’s 6 feet by 8 feet, multiply 6 x 8 and get 48 square feet. Estimate how much flooring you’ll need for each room—you may be able to add together these figures for all rooms in your house, but don’t forget about hallways and other common areas where you’ll need to cover more than one area at a time (or multiple sizes).", "so he does not get painton the floor. Each drop cloth 13 Luke is painting the ceiling. He needs to lay down drop cloths so he does not get paint on the floor. Each drop cloth is 8 feet by 6 feet. The room is 14 feet by 16 feet. How many drop cloths will he need to cover the floor? Type your answer SUBMIT...", "SEARCH HOME Math Central Quandaries & Queries Question from jamie, a parent: how many 16in by 16in blocks would it take to cover a 16ft by 16ft floor? Hi Jamie, There are 12 inches in a foot so 16 feet is $12 \\times 16$ inches. $\\large \\frac{12 \\times 16}{16} = \\normalsize 12$ so 12 of your 16 inch square tiles will fit in a row along one side of the floor. The floor is also 16 feet wide so it will take 12 of these rows to cover the floor. $12 \\times 12 = 144$ so you need 144 tiles. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "rooms with dimensions of 6.8m x 4.5m and 6m x3.8m? One can arefor 6m². • Pool tiles The pool is 25m long, 10m wide and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many pieces of tile are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc? • Rectangles Calculate how many squares/rectangles of size 4×3 cm can be cut from a sheet of paper of 36 cm × 32 cm?", "are not used in each arrangement? For a patio of dimensions, 9ft by 9ft, the number of tiles that will not be used is given by 125 – 81 = 44 For a patio of dimensions, 10ft by 10ft, the number of tiles that will not be used is given by 125 – 100 = 25 For a patio of dimensions, 11ft by 11ft, the number of tiles that will not be used is given by 125 – 121 = 4 Question 57. PATTERNS Copy and complete the table. Describe what happens to the value of the power as the exponent decreases. Use this pattern to find the value of 40. 40 = 1 Thus the value of 40 is 1. Question 58. REPEATED REASONING How many blocks do you need to add to Square 6 to get Square 7? to Square 9 to get Square 10? to Square 19 to get Square 20? Explain. You need to add 14 blocks to get square 7. The square 7 contains 7 × 7 = 49 blocks You need to add 32 blocks to get square 9. The square 9 contains 9 × 9 = 81 blocks You need to add 19 blocks to get square 10. The square 10 contains 10 × 10 = 100 blocks You need to", "16 square inches", "square feet.", "two feet on a side and a square half an inch on a side is how much grout you have to deal with. There are still infinitely many possibilities. You might still suspect me of being boring. Sure, there’s a rectangular tile that’s, say, six inches by eight inches. And one that’s six inches by nine inches. Six inches by ten inches. Six inches by one millimeter. Yes, I’m technically right. But I’m not interested in that. Let’s allow that all rectangular tiles are “really” the same pattern. So we have “squares” and “rectangles”. There are still infinitely many tile possibilities. Let me shorten the discussion here. Draw a quadrilateral. One that doesn’t intersect itself. That is, there’s four corners, four lines, and there’s no X crossings. If you have that, then you have a tiling. Get enough of these tiles and arrange them correctly and you can cover the plane. Or the kitchen floor, if you have a level floor. It might not be obvious how to do it. You might have to rotate alternating tiles, or set them in what seem like weird offsets. But you can do it. You’ll need someone to make the tiles for you, if you pick some weird pattern. I hope I live long enough to see it become part of", "kitchens as walls and back-splashes. • slate or other stone tiles - commonly used on floors and occasionally on walls (e.g. large showers) • metal tiles - ceiling tiles and back splashes • plastic - used in baths and kitchens on walls (e.g. bath tub surround) • wood - used on wall and ceilings (this included cork) The different materials in tiles suit different purposes from aesthetics on walls, to durability on floors, to water repelling in baths, to sound dampening on walls and ceilings and even floors. The application of tiles varies based on the material. Some tiles have adhesive backs, others require adhesives to be placed on the surface prior to laying the tiles. In some cases, like ceramic tiles, a mud bed is recommended to provide a thick, consistent and stable base for the tiles.", "to tile the floor. Perhaps some of the walls behind the counter. What patterns could you use? And there are infinitely many possibilities. You might leap ahead of me and say, yes, but they’re all boring. A tile that’s eight inches square is different from one that’s twelve inches square and different from one that’s 12.01 inches square. Fine. Let’s allow that all square tiles are “really” the same pattern. The only difference between a square two feet on a side and a square half an inch on a side is how much grout you have to deal with. There are still infinitely many possibilities. You might still suspect me of being boring. Sure, there’s a rectangular tile that’s, say, six inches by eight inches. And one that’s six inches by nine inches. Six inches by ten inches. Six inches by one millimeter. Yes, I’m technically right. But I’m not interested in that. Let’s allow that all rectangular tiles are “really” the same pattern. So we have “squares” and “rectangles”. There are still infinitely many tile possibilities. Let me shorten the discussion here. Draw a quadrilateral. One that doesn’t intersect itself. That is, there’s four corners, four lines, and there’s no X crossings. If you have that, then you have a tiling. Get enough of these tiles and" ]

[ "graph, where the heights of the bars represent the number of students. What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class? $$3.50$$ $$3.57$$ $$4.36$$ $$4.50$$ $$5.00$$ ###### Solution(s): Counting the number of each occurence, we can see that there are 1 $$1$$s, 3 $$2$$s, 2 $$3$$s, 6 $$4$$s, 8 $$5$$s, 3 $$6$$s, and 2 $$7$$s. Therefore, there are $$1+3+2+6+8+3+2 = 25$$ students in total. Therefore, the total number of days of exercise is \\begin{align*}&1\\cdot1+3\\cdot2 + 2\\cdot 3 + 6\\cdot 4 \\\\ &+ 8\\cdot 5 + 3\\cdot 6 + 2\\cdot 7 \\\\ &= 109\\end{align*} The mean number of days of exercise is defined as being equal to the total number of days of exercise divides by the total number of students, where: $\\dfrac{\\text{total days}}{\\text{# students}} = \\dfrac{109}{25} \\approx 4.36$ Thus, C is the correct answer. 9. Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use? $$48$$ $$87$$ $$91$$ $$96$$ $$120$$ ###### Solution(s): Note that each square foot of the", "coins are heads, and there are a total of $$2^4 = 16$$ possibilities. Therefore, $q = \\dfrac{6}{16} = \\dfrac{3}{8},$ and $2p - \\dfrac{3}{8} = 1$$p = \\dfrac{11}{16}.$ Thus, E is the correct answer. 22. Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides. $$18$$ $$24$$ $$26$$ $$30$$ $$36$$ ###### Solution(s): We can count the number of unexposed sides to find how many sides contribute to the surface area. Three cubes have $$1$$ side unexposed, two cubes have $$2$$ sides unexposed, and one cube has $$3$$ sides unexposed. This gives us a total of $3 \\cdot 1 + 2 \\cdot 2 + 1 \\cdot 3 = 10$ unexposed sides, which gives us $6 \\cdot 6 - 10 = 36 - 10$$= 26$ exposed sides. Each exposed side contributes $$1^2 = 1$$ to the surface area, for a total surface area of $$1 \\cdot 26 = 26.$$ Thus, C is the correct answer. 23. A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles? $$\\dfrac{1}3$$ $$\\dfrac{4}9$$ $$\\dfrac{1}2$$ $$\\dfrac{5}9$$ $$\\dfrac{5}8$$ ###### Solution(s): Notice that", "# Pool tiles The pool is 25m long, 10m wide and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many pieces of tile are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc? Result x = 9050 y = 93396 Kc #### Solution: $a=25 \\ \\text{m} \\ \\\\ b=10 \\ \\text{m} \\ \\\\ c=160/100=\\dfrac{ 8 }{ 5 }=1.6 \\ \\text{m} \\ \\\\ \\ \\\\ S=a \\cdot \\ b + c \\cdot \\ 2 \\cdot \\ (a+b)=25 \\cdot \\ 10 + 1.6 \\cdot \\ 2 \\cdot \\ (25+10)=362 \\ \\text{m}^2 \\ \\\\ d=20/100=\\dfrac{ 1 }{ 5 }=0.2 \\ \\text{m} \\ \\\\ S_{1}=d^2=0.2^2=\\dfrac{ 1 }{ 25 }=0.04 \\ \\text{m}^2 \\ \\\\ \\ \\\\ x=S/S_{1}=362/0.04=9050$ $y=S \\cdot \\ 258=362 \\cdot \\ 258=93396 \\ \\text{Kc}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! ## Next similar math problems: 1. Basen How many square meters of tiles we need to", "the figure = (5 + 4) cm2 = 9 cm2 Question 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (a) 100 cm and 144 cm (b) 70 cm and 36 cm. Solution: (a) Area of rectangular region = length × breadth = 100 cm × 144 cm = 14400 cm2 Area of one tile = 12 cm × 5 cm = 60 cm2 Thus, 240 tiles are required. (b) Area of rectangular region = length × breadth = 70 cm × 36 cm = 2520 cm2 Area of one tile = 12 cm × 5 cm = 60 cm2 ∴ Number of tiles = $$\\frac{\\text { Area of rectangular region }}{\\text { Area of one tile }}=\\frac{2520}{60}$$ = 40 Thus, 42 tiles are required.", "Square tiles The room has dimensions 12 meters and 5.6 meters. Determine the number of square tiles and their largest dimension to exactly cover the floor. Correct result: n = 105 a = 80 cm Solution: $1200=2^4 \\cdot 3 \\cdot 5^2 \\ \\\\ 560=2^4 \\cdot 5 \\cdot 7 \\ \\\\ \\text{GCD}(1200, 560)=2^4 \\cdot 5=80 \\ \\\\ \\ \\\\ a=GCD(1200,560)=80 \\ \\\\ n=1200 \\cdot \\ 560/a^2=1200 \\cdot \\ 560/80^2=105$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Tips to related online calculators Do you want to calculate least common multiple two or more numbers? Do you want to calculate greatest common divisor two or more numbers? Do you want to convert area units? Do you want to convert length units? You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: Next similar math problems: • Square room What is the size of the smallest square room, which can pave with tiles", "the Calculate button. board feet = 8 * 10 * 1. × This website uses cookies to serve content relevant for you and to improve your overall website experience. Free online carpet calculator and estimating tool. feet of panels and your cost for a typical panel is say $6. Please use this Epoxy Resin Calculator as a reference guide only for our table top and coating epoxy. , ft2 & ft 2. The same thing happens with functions like square root, sine, cosine, etc. in this video you can learn square feet calculations and running feet calculations. Calculator for Hollow structural sections - square Calculators. However, that estimate includes numbers from all different industries, and it won’t necessarily apply to yours. Factors such as evaporation, wetting, and soaking into the ground are not considered here. The calculation would go on forever, so we have to stop somewhere. The square feet is calculated as length multiplied with the width. 5\" = 2,754 (Divide this number by 144) 2,754/144 = 19. Therefore each ring will need around 11 feet 6 inches of materials and 44 inches of space. This calculation helps you to estimate the amount of tiles you will need for the job. Wood Cft Calculator. Therefore, one Dismil is equal to four hundred and thirty five decimal point", "# Into Math Grade 5 Module 9 Lesson 1 Answer Key Explore Area and Mixed Numbers We included HMH Into Math Grade 5 Answer Key PDF Module 9 Lesson 1 Explore Area and Mixed Numbers to make students experts in learning maths. ## HMH Into Math Grade 5 Module 9 Lesson 1 Answer Key Explore Area and Mixed Numbers I Can use an area model to multiply mixed numbers. Ms. Cruz wants to use square tiles to cover the front entryway of a house. Each tile she plans to use measures $$\\frac{1}{2}$$ foot by $$\\frac{1}{2}$$ foot. What is the area of the entryway? The area will be $$\\frac{1}{4}$$ sq ft. Explanation: Given that each tile she plans to use measures $$\\frac{1}{2}$$ foot by $$\\frac{1}{2}$$ foot. So the area will be $$\\frac{1}{2}$$ × $$\\frac{1}{2}$$ = $$\\frac{1}{4}$$ sq ft. Draw a visual model to show how many tiles she will use. Then explain how you can use your visual model to find the area of the entryway. Turn and Talk Describe how your method for solving the problem would change if Ms. Cruz uses smaller tiles that each measure $$\\frac{1}{4}$$ foot by $$\\frac{1}{4}$$ foot. Build Understanding 1. Will replaces the tiles of a shower floor. The floor measures 4$$\\frac{1}{3}$$ feet by 3$$\\frac{2}{3}$$ feet. Will covers the floor with small square", "# Question #064a7 Aug 3, 2017 $4422.6$ pounds #### Explanation: \"The concrete slab will be 12 feet 9 inches long, 9 feet 3 inches wide, and 3 inches thick\" Laying out the information • 12 feet 9 inches is $\\left(12 \\setminus \\cdot 12\\right) + 9$ since each foot is 12 inches. This gets you $144 + 9 = 153$ inches for your length. • 9 feet 3 inches is $\\left(9 \\setminus \\cdot 12\\right) + 3$ for the same reason. This gets you $108 + 3 = 111$ inches for your width. • Your height (thickness of slab) is $3$ inches. Summarizing: $l = 153$ inches $w = 111$ inches $h = 3$ inches Now find the volume of the slab by using the formula $l \\setminus \\cdot w \\setminus \\cdot h$, or $153 \\times 111 \\times 3 = 50949$ cubic inches. Converting But the weight is 150 pounds per cubic foot, so you have to turn the inches into feet. Again, since each foot is 12 inches: $50949 \\setminus {\\textrm{\\in}}^{3} \\times \\frac{1 \\setminus {\\textrm{f t}}^{3}}{12 \\times 12 \\times 12 \\setminus {\\textrm{\\in}}^{3}} = \\frac{50949}{1728} \\setminus {\\textrm{f t}}^{3} = 29.484 \\setminus {\\textrm{f t}}^{3}$ Final steps Now to solve for weight, multiply the cubic feet weight by 150 to get your answer. $29.484 \\setminus {\\textrm{f t}}^{3} \\times \\frac{150 \\setminus \\textrm{p o", "If ceramic tiles are 8\" * 4\", how many tiles would it take to cover an area 4' * 6'? 4 feet = 48 inches. 6 feet = 72 inches. (48*72)/(8*4)=108 5. If a customer buys 20 2*4's 8ft long, at $4.75 each, how much would the 2*4's cost? I don't really understand this question. What does \"20 2*4's 8ft long\" mean? • March 8th 2009, 02:14 PM Love-Guru for the last question... the person bought 20 2 by 4's like 20 pieces of wood. oh, and for the second question, how did you get 5.25? • March 8th 2009, 02:53 PM JeWiSh Did I understand correctly that 5 1/4 meant $5\\frac{1}{4}$ as in 5 and a quarter? That's equal to 5.25. Isn't 2 * 4 the measurement of wood, where does 8ft come into it? And lastly, why have you posted this in several places? • March 8th 2009, 03:05 PM skeeter Quote: Originally Posted by Love-Guru 5. If a customer buys 20 2*4's 8ft long, at$4.75 each, how much would the 2*4's cost? 20 pieces of wood at $4.75 each (the dimensions of each piece doesn't matter)$ $4.75 \\times 20$ • March 8th 2009, 04:33 PM Love-Guru ... Not sure, that's what the question says. on another note: thank guys.. really appreciated (Happy)(Happy) oh, How did", "USA may be aware that the single prime is not an inch, but a foot. I seriously doubt the GMAT exam contains a question exactly as specified in the OP. Senior Manager Joined: 24 Nov 2015 Posts: 473 Location: United States (LA) Re: How many 2-inch by 3-inch rectangular tiles are required to tile this [#permalink] ### Show Tags 09 Jun 2016, 00:30 chetan2u wrote: sabxu1 wrote: chetan2u wrote: Bunuel, E is ok.. people have not read Q properly so what is the answer E or? why is E ok im so confused there is a 6'*12' garden with a 2'*3' patch.. Q is -How many 2\"*3\" tiles can be placed and not 2'*3' convert all feet into inches as we arelooking into tiles which are in inches so total area of garden = 6*12*12*12 shaded region = 12*12(6*12-2*3) = 12*12*2*3(12-1).. no of 2\"*3\" tiles required =$$\\frac{12*12*12*2*3*11}{2*3} = 12*12*11= 1584 > 1500$$ so ans is 1500+ E Apologies for finding a doubt with figure nowhere in the figure it is showing that the rectangle is 12 feet * 6 feet or are we supposed to assume 12' as 12 feet? Senior Manager Joined: 24 Nov 2015 Posts: 473 Location: United States (LA) Re: How many 2-inch by 3-inch rectangular tiles are required to tile this [#permalink] ### Show", "# Drawer The rectangular face of a drawer has a perimeter of 108 cm and a width of 18cm. Find the length and the area of the rectangular face of the drawer. Correct result: b = 36 cm A = 648 cm2 #### Solution: $p=108 \\ \\text{cm} \\ \\\\ a=18 \\ \\text{cm} \\ \\\\ p=2(a+b) \\ \\\\ b=(p - 2 \\cdot \\ a)/2=(108 - 2 \\cdot \\ 18)/2=36 \\ \\text{cm}$ $A=a \\cdot \\ b=18 \\cdot \\ 36=648 \\ \\text{cm}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Tiles path A rectangular park measures 24mX18m. A path 2.4m wide is constructed all around it. Find the cost of paving the path with cement tiles of size 60cm X 40cm at the rate of \\$5.60 per tile ? • Tiles The rectangular floor of the 6 x 1.8 m", "the area of one tile. Since we know the measurements of all three sides, we can use Heron’s Formula to calculate the area. $s & = \\frac{1} {2} (11.3 + 11.6 + 13.6) = 18.4 & & s = \\frac{1} {2} (a + b + c) \\\\K & = \\sqrt{18.4 (18.4 - 11.3)(18.4 - 11.9)(18.4 - 13.6)} & & \\text{Heron's Formula}\\\\K & = \\sqrt{18.4(7.1)(6.5)(4.8)} & & \\text{Subtract}\\\\K & = \\sqrt{40759.68} & & \\text{Multiply}\\\\K & \\approx 201.9\\ \\text{in}^2 & & \\text{square root}$ The area of one tile would be $201.9\\;\\mathrm{square\\ inches.}$ The cost of the tile is given to us in square feet, while the area of the tile is in square inches. In order to find the cost of one tile, we must first convert the area of the tile into square feet. $\\text {1 square foot} & = 12\\ \\text{in} \\times 12\\ \\text{in} = 144\\ \\text{in}^2 & &\\\\\\frac{201.9} {144} & = 1.4\\ \\text{ft}^2 & & \\text{Covert square inches into square feet}\\\\1.4\\ \\text{ft}^2 \\times 4.89 & = 6.89 & & \\text{Cost for one tile is}\\ \\6.86 \\\\6.84 \\times 4 & = 27.42 & & \\text{The cost for four tiles is}\\ \\27.42$ Answer: The cost for four tiles would be $\\27.42$. ## Applications, Technological Tools We have already looked at two examples of situations where we can apply the", "1 = 4 sq cm Total Area of figures (c) = 5 + 4 = 9 sq cm 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (a) 100 cm and 144 cm (b) 70 cm and 36 cm Solutions : Area of tiles = 5 × 12 = 60 sq cm (a) 100 cm and 144 cm Area of rectangular region = Length × Breadth = 144 × 100 = 14400 sq cm = $$\\displaystyle \\frac{{14400}}{60}$$ = 240 tiles (b) 70 cm and 36 cm Area of rectangular region = Length × Breadth = 70 × 36 = 2520 sq cm = $$\\displaystyle \\frac{{2520}}{60}$$ = 42 tiles error: Content is protected !!", "##### Tools This site is devoted to mathematics and its applications. Created and run by Peter Saveliev. # Continuity as accuracy (Redirected from Introduction to continuity) Jump to: navigation, search Continuity can be understood by considering the accuracy of a measurement. Suppose we have a collection of square tiles of various sizes and we need to find the area $A$ of each in order to be able to compute how many we need to cover the whole floor. Of course, we simply measure the side, $x$, of each tile. We know from geometry that $$A=x^2,$$ so we can always compute: $$x=10\\Rightarrow A=100,$$ (in inches) etc. But what if the measurement isn't fully accurate? What if there is always some error? It's never $x=10$ but $$x=10 \\pm .3.$$ As a result the computed value of the area -- what we care about -- will also have some error! Consider, the area won't be just $A=100$ but $$A=(10 \\pm .3)^2.$$ Therefore, $$A=10^2 \\pm 2 \\cdot 10 \\cdot .3 +.3^2$$ $$=100.09 \\pm 6.$$ The result means that the actual area must be within the interval $(94.09,106.09)$. Suppose however that we can always improve the accuracy of the measurement of the side of the tile $x$ -- as much as we like. The question is, can we also improve the accuracy of", "# Prism 4 sides Find the surface area and volume four-sided prism high 10cm if its base is a rectangle measuring 8 cm and 1.2dm Result S = 592 cm2 V = 960 cm3 #### Solution: $V=a \\cdot \\ b \\cdot \\ c=8 \\cdot \\ 12 \\cdot \\ 10=960 \\ \\text{cm}^3$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Showing 0 comments: Be the first to comment! Tips to related online calculators Tip: Our volume units converter will help you with the conversion of volume units. #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: 1. Basen How many square meters of tiles we need to tile the walls and floor of the pool 15 meters long, six meters wide and two meters?deep 2. Children's pool Children's pool at the swimming pool is 10m long, 5m wide and 50cm deep. Calculate: (a) how many m2 of tiles are needed for lining the", "# The number of square tiles needed to tile a square floor is equal to a^2 -: b^2, where a is the floor length in inches and b is the length of the tile in inches. If a = 96 and b =8, how many tiles are needed? May 17, 2018 144 #### Explanation: No. of square tiles needed = ${a}^{2} / {b}^{2}$ So, if $a = 96$ and $b = 8$, then all you do is you have to sub into your 2 numbers into the equation No. of square tiles needed = ${96}^{2} / {8}^{2} = 144$ May 17, 2018 $144$ tiles are needed. #### Explanation: Area of floor is ${96}^{2}$ square inches and area of each tile is ${8}^{2}$ square inches Hence, number of tiles needed are $\\frac{96 \\times 96}{8 \\times 8}$ = $\\frac{{\\cancel{96}}^{12} \\times {\\cancel{96}}^{12}}{{\\cancel{8}}^{1} \\times {\\cancel{8}}^{1}}$ = $12 \\times 12$ = $144$", "that there are 100cm in 1m . Find the area of the floor by multiplying the length of the rectangle by the width. Divide the area of the floor by the area of the tile to get the number of tiles needed to cover the floor. 4. Find the width of the rectangle below: 19600m 19600m^{2} 4m^{2} 4m Using the formula for the area of a rectangle we substitute our given values. As we are trying to calculate the width of the rectangle we need to rearrange the formula to divide the length into the area. 5. Below is a plan for a new flower bed. Find the area of the flower bed. 22m^{2} 57m^{2} 30m^{2} 45m^{2} Split the compound shape into 3 rectangles. Calculate the missing side length of rectangle C by subtracting 3 from the total length which is 5m . Work out the area of A , B and C and then add up all the individual areas to get the total area of the compound shape. \\begin{aligned} \\text { Area }_{\\text {rectangle } A} &=\\text { length } \\times \\text { width } \\\\ &=3 \\times 2 \\\\ &=6m^{2} \\\\ \\\\ \\text { Area }_{\\text {rectangle } B} &=\\text { length } \\times \\text { width } \\\\ &=3 \\times 2 \\\\ &=6m^{2} \\\\", "• # question_answer A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area $1080\\,{{m}^{2}}$? (If required you can split the tiles in whatever way you want to fill up the corners). A) 30,000 B) 40,000C) 45,000 D) 60,000E) None of these Solution : Explanation Option [c] is correct. Area of the Parallelogram $\\text{=base }\\!\\!\\times\\!\\!\\text{ height=24 }\\!\\!~\\!\\!\\text{ }\\!\\!\\times\\!\\!\\text{ 10=240 sq cm}$ Required number of tiles =$\\frac{Area\\,\\,of\\,\\,Floor}{Area\\,\\,of\\,\\,Tiles}=\\frac{1080\\times 100\\times 100}{240}=45000$ You need to login to perform this action. You will be redirected in 3 sec", "side of two square tiles is $1$ square inch. Two tiles are pushed side by side and shown together below. 1. What is the perimeter of the figure? 2. What is the area? ### try it In the following video we show a similar example as the ones above.", "The amount of paint needed $$=$$ {The amount of paint needed ($$l$$ ) = 2 $$\\cdot$$ floor area ($$m^{2}$$ ) . average paint consumption ($$\\frac{l}{m^{2}}$$ ), these are written as assumptions (a), (d)} $$2\\cdot$$ floor area $$(m^{2})\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {floor area = floor area of kitchen + floor area of living room. The formula for area is length $$\\cdot$$ width, we get the dimensions from assumptions (b), (c)} $$2\\cdot(5m\\cdot3m+2m\\cdot1m)\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {We do the calculations} $$2\\cdot17m^{2}\\cdot\\frac{1l}{5m^{2}}$$ $$\\approx$$ {We simplify $$m^{2}$$ } $$7l$$ $$\\square$$ Instead of long variable names we can use letters. If we denote the needed amount of paint as $$x$$ and the average paint consumption as $$k$$, the derivation goes as follows: $$\\bullet$$ How much paint ($$x$$) is needed to cover the floor when (a) the floor is painted twice and (b) the living room is $$5m$$ long and $$3m$$ wide and (c) the kitchen is $$2m$$ long and $$1m$$ in width and (d) the average paint consumption ($$k$$) is $$\\frac{1l}{5m^{2}}$$ $$\\Vdash$$ $$x$$ $$=$$ {$$x$$ ($$l$$) = 2 $$\\cdot$$ floor area ($$m^{2}$$ ) $$\\cdot$$ $$k$$ ($$\\frac{l}{m^{2}}$$ ), these are written down as assumptions (a), (d)} $$2\\cdot$$ floor area $$(m^{2})\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {floor area = floor area of kitchen + floor area of living room. The formula for area is length $$\\cdot$$ width, we get the dimensions from assumptions (b)," ]

[ "# Pool tiles The pool is 25m long, 10m wide and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many pieces of tile are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc? Result x = 9050 y = 93396 Kc #### Solution: $a=25 \\ \\text{m} \\ \\\\ b=10 \\ \\text{m} \\ \\\\ c=160/100=\\dfrac{ 8 }{ 5 }=1.6 \\ \\text{m} \\ \\\\ \\ \\\\ S=a \\cdot \\ b + c \\cdot \\ 2 \\cdot \\ (a+b)=25 \\cdot \\ 10 + 1.6 \\cdot \\ 2 \\cdot \\ (25+10)=362 \\ \\text{m}^2 \\ \\\\ d=20/100=\\dfrac{ 1 }{ 5 }=0.2 \\ \\text{m} \\ \\\\ S_{1}=d^2=0.2^2=\\dfrac{ 1 }{ 25 }=0.04 \\ \\text{m}^2 \\ \\\\ \\ \\\\ x=S/S_{1}=362/0.04=9050$ $y=S \\cdot \\ 258=362 \\cdot \\ 258=93396 \\ \\text{Kc}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! ## Next similar math problems: 1. Basen How many square meters of tiles we need to", "coins are heads, and there are a total of $$2^4 = 16$$ possibilities. Therefore, $q = \\dfrac{6}{16} = \\dfrac{3}{8},$ and $2p - \\dfrac{3}{8} = 1$$p = \\dfrac{11}{16}.$ Thus, E is the correct answer. 22. Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides. $$18$$ $$24$$ $$26$$ $$30$$ $$36$$ ###### Solution(s): We can count the number of unexposed sides to find how many sides contribute to the surface area. Three cubes have $$1$$ side unexposed, two cubes have $$2$$ sides unexposed, and one cube has $$3$$ sides unexposed. This gives us a total of $3 \\cdot 1 + 2 \\cdot 2 + 1 \\cdot 3 = 10$ unexposed sides, which gives us $6 \\cdot 6 - 10 = 36 - 10$$= 26$ exposed sides. Each exposed side contributes $$1^2 = 1$$ to the surface area, for a total surface area of $$1 \\cdot 26 = 26.$$ Thus, C is the correct answer. 23. A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles? $$\\dfrac{1}3$$ $$\\dfrac{4}9$$ $$\\dfrac{1}2$$ $$\\dfrac{5}9$$ $$\\dfrac{5}8$$ ###### Solution(s): Notice that", "Square tiles The room has dimensions 12 meters and 5.6 meters. Determine the number of square tiles and their largest dimension to exactly cover the floor. Correct result: n = 105 a = 80 cm Solution: $1200=2^4 \\cdot 3 \\cdot 5^2 \\ \\\\ 560=2^4 \\cdot 5 \\cdot 7 \\ \\\\ \\text{GCD}(1200, 560)=2^4 \\cdot 5=80 \\ \\\\ \\ \\\\ a=GCD(1200,560)=80 \\ \\\\ n=1200 \\cdot \\ 560/a^2=1200 \\cdot \\ 560/80^2=105$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors. Tips to related online calculators Do you want to calculate least common multiple two or more numbers? Do you want to calculate greatest common divisor two or more numbers? Do you want to convert area units? Do you want to convert length units? You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: Next similar math problems: • Square room What is the size of the smallest square room, which can pave with tiles", "the figure = (5 + 4) cm2 = 9 cm2 Question 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (a) 100 cm and 144 cm (b) 70 cm and 36 cm. Solution: (a) Area of rectangular region = length × breadth = 100 cm × 144 cm = 14400 cm2 Area of one tile = 12 cm × 5 cm = 60 cm2 Thus, 240 tiles are required. (b) Area of rectangular region = length × breadth = 70 cm × 36 cm = 2520 cm2 Area of one tile = 12 cm × 5 cm = 60 cm2 ∴ Number of tiles = $$\\frac{\\text { Area of rectangular region }}{\\text { Area of one tile }}=\\frac{2520}{60}$$ = 40 Thus, 42 tiles are required.", "graph, where the heights of the bars represent the number of students. What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class? $$3.50$$ $$3.57$$ $$4.36$$ $$4.50$$ $$5.00$$ ###### Solution(s): Counting the number of each occurence, we can see that there are 1 $$1$$s, 3 $$2$$s, 2 $$3$$s, 6 $$4$$s, 8 $$5$$s, 3 $$6$$s, and 2 $$7$$s. Therefore, there are $$1+3+2+6+8+3+2 = 25$$ students in total. Therefore, the total number of days of exercise is \\begin{align*}&1\\cdot1+3\\cdot2 + 2\\cdot 3 + 6\\cdot 4 \\\\ &+ 8\\cdot 5 + 3\\cdot 6 + 2\\cdot 7 \\\\ &= 109\\end{align*} The mean number of days of exercise is defined as being equal to the total number of days of exercise divides by the total number of students, where: $\\dfrac{\\text{total days}}{\\text{# students}} = \\dfrac{109}{25} \\approx 4.36$ Thus, C is the correct answer. 9. Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use? $$48$$ $$87$$ $$91$$ $$96$$ $$120$$ ###### Solution(s): Note that each square foot of the", "\\cdot \\frac{5280 \\text { feet }}{1 \\text { mile }}=440 \\text { feet}} & { } \\end{array}$$ We could have also done this entire calculation in one long set of products: $$20\\text{ seconds }\\cdot \\frac{1 \\text { minute }}{60 \\text { seconds }} \\cdot \\frac{1 \\text { hour }}{60 \\text { minutes }} \\cdot \\frac{15 \\text { miles }}{1 \\text { hour }} \\cdot \\frac{5280 \\text { feet }}{1 \\text { mile }}=440\\text{ feet}$$ ## Try it Now 4 A 1000 foot spool of bare 12-gauge copper wire weighs 19.8 pounds. How much will 18 inches of the wire weigh, in ounces? $$18 \\text { inches } \\cdot \\frac{1 \\text { foot }}{12 \\text { inches }} \\cdot \\frac{19.8 \\text { pounds }}{1000 \\text { feet }} \\cdot \\frac{16 \\text { ounces }}{1 \\text { pound }} \\approx 0.475 \\text { ounces }$$ Notice that with the miles per gallon example, if we double the miles driven, we double the gas used. Likewise, with the map distance example, if the map distance doubles, the real-life distance doubles. This is a key feature of proportional relationships, and one we must confirm before assuming two things are related proportionally. ## Example 17 Suppose you’re tiling the floor of a 10 ft by 10 ft room, and find that 100 tiles", "# The number of square tiles needed to tile a square floor is equal to a^2 -: b^2, where a is the floor length in inches and b is the length of the tile in inches. If a = 96 and b =8, how many tiles are needed? May 17, 2018 144 #### Explanation: No. of square tiles needed = ${a}^{2} / {b}^{2}$ So, if $a = 96$ and $b = 8$, then all you do is you have to sub into your 2 numbers into the equation No. of square tiles needed = ${96}^{2} / {8}^{2} = 144$ May 17, 2018 $144$ tiles are needed. #### Explanation: Area of floor is ${96}^{2}$ square inches and area of each tile is ${8}^{2}$ square inches Hence, number of tiles needed are $\\frac{96 \\times 96}{8 \\times 8}$ = $\\frac{{\\cancel{96}}^{12} \\times {\\cancel{96}}^{12}}{{\\cancel{8}}^{1} \\times {\\cancel{8}}^{1}}$ = $12 \\times 12$ = $144$", "• # question_answer A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area $1080\\,{{m}^{2}}$? (If required you can split the tiles in whatever way you want to fill up the corners). A) 30,000 B) 40,000C) 45,000 D) 60,000E) None of these Solution : Explanation Option [c] is correct. Area of the Parallelogram $\\text{=base }\\!\\!\\times\\!\\!\\text{ height=24 }\\!\\!~\\!\\!\\text{ }\\!\\!\\times\\!\\!\\text{ 10=240 sq cm}$ Required number of tiles =$\\frac{Area\\,\\,of\\,\\,Floor}{Area\\,\\,of\\,\\,Tiles}=\\frac{1080\\times 100\\times 100}{240}=45000$ You need to login to perform this action. You will be redirected in 3 sec", "# Prism 4 sides Find the surface area and volume four-sided prism high 10cm if its base is a rectangle measuring 8 cm and 1.2dm Result S = 592 cm2 V = 960 cm3 #### Solution: $V=a \\cdot \\ b \\cdot \\ c=8 \\cdot \\ 12 \\cdot \\ 10=960 \\ \\text{cm}^3$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Showing 0 comments: Be the first to comment! Tips to related online calculators Tip: Our volume units converter will help you with the conversion of volume units. #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: 1. Basen How many square meters of tiles we need to tile the walls and floor of the pool 15 meters long, six meters wide and two meters?deep 2. Children's pool Children's pool at the swimming pool is 10m long, 5m wide and 50cm deep. Calculate: (a) how many m2 of tiles are needed for lining the", "The amount of paint needed $$=$$ {The amount of paint needed ($$l$$ ) = 2 $$\\cdot$$ floor area ($$m^{2}$$ ) . average paint consumption ($$\\frac{l}{m^{2}}$$ ), these are written as assumptions (a), (d)} $$2\\cdot$$ floor area $$(m^{2})\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {floor area = floor area of kitchen + floor area of living room. The formula for area is length $$\\cdot$$ width, we get the dimensions from assumptions (b), (c)} $$2\\cdot(5m\\cdot3m+2m\\cdot1m)\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {We do the calculations} $$2\\cdot17m^{2}\\cdot\\frac{1l}{5m^{2}}$$ $$\\approx$$ {We simplify $$m^{2}$$ } $$7l$$ $$\\square$$ Instead of long variable names we can use letters. If we denote the needed amount of paint as $$x$$ and the average paint consumption as $$k$$, the derivation goes as follows: $$\\bullet$$ How much paint ($$x$$) is needed to cover the floor when (a) the floor is painted twice and (b) the living room is $$5m$$ long and $$3m$$ wide and (c) the kitchen is $$2m$$ long and $$1m$$ in width and (d) the average paint consumption ($$k$$) is $$\\frac{1l}{5m^{2}}$$ $$\\Vdash$$ $$x$$ $$=$$ {$$x$$ ($$l$$) = 2 $$\\cdot$$ floor area ($$m^{2}$$ ) $$\\cdot$$ $$k$$ ($$\\frac{l}{m^{2}}$$ ), these are written down as assumptions (a), (d)} $$2\\cdot$$ floor area $$(m^{2})\\cdot\\frac{1l}{5m^{2}}$$ $$=$$ {floor area = floor area of kitchen + floor area of living room. The formula for area is length $$\\cdot$$ width, we get the dimensions from assumptions (b),", "Tags 09 Jun 2016, 00:35 Bunuel wrote: How many 2-inch by 3-inch rectangular tiles are required to tile this shaded region? A. Less than 10 B. 10—100 C. 101—1,000 D. 1,001—1,500 E. 1,500+ Attachment: 2016-01-24_1704.png Apologies for finding a doubt with figure nowhere in the figure it is showing that the rectangle is 12 feet * 6 feet or are we supposed to assume 12' as 12 feet? Retired Moderator Joined: 26 Nov 2012 Posts: 554 Re: How many 2-inch by 3-inch rectangular tiles are required to tile this [#permalink] ### Show Tags 16 Aug 2016, 23:49 chetan2u wrote: sabxu1 wrote: chetan2u wrote: Bunuel, E is ok.. people have not read Q properly so what is the answer E or? why is E ok im so confused there is a 6'*12' garden with a 2'*3' patch.. Q is -How many 2\"*3\" tiles can be placed and not 2'*3' convert all feet into inches as we arelooking into tiles which are in inches so total area of garden = 6*12*12*12 shaded region = 12*12(6*12-2*3) = 12*12*2*3(12-1).. no of 2\"*3\" tiles required =$$\\frac{12*12*12*2*3*11}{2*3} = 12*12*11= 1584 > 1500$$ so ans is 1500+ E Bunuel, Can you give us your solution too.. Director Joined: 09 Aug 2017 Posts: 678 Re: How many 2-inch by 3-inch rectangular tiles are required to tile", "# Basen How many square meters of tiles we need to tile the walls and floor of the pool 15 meters long, six meters wide and two meters?deep Result S = 174 m2 #### Solution: $a = 15 \\ m \\ \\\\ b = 6 \\ m \\ \\\\ c = 2 \\ m \\ \\\\ S = a \\cdot \\ b + 2 \\cdot \\ (a+b) \\cdot \\ c = 15 \\cdot \\ 6 + 2 \\cdot \\ (15+6) \\cdot \\ 2 = 174 = 174 \\ m^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Showing 0 comments: Be the first to comment! ## Next similar math problems: 1. Pool tiles The pool is 25m long, 10m wide and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many pieces of tile are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc? 2. Concrete pillar How many m³ of", "SEARCH HOME Math Central Quandaries & Queries Question from jamie, a parent: how many 16in by 16in blocks would it take to cover a 16ft by 16ft floor? Hi Jamie, There are 12 inches in a foot so 16 feet is $12 \\times 16$ inches. $\\large \\frac{12 \\times 16}{16} = \\normalsize 12$ so 12 of your 16 inch square tiles will fit in a row along one side of the floor. The floor is also 16 feet wide so it will take 12 of these rows to cover the floor. $12 \\times 12 = 144$ so you need 144 tiles. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "SEARCH HOME Math Central Quandaries & Queries Question from Jill, how many 4\" x 4\" square tiles would I need for 50 square feet? Hi Jill, $3 \\times 4\" = 12\" = 1 \\mbox{ foot}$ and hence it takes $3 \\times 3 = 9,$ 4 inch square tiles to make a square foot. You need to cover 50 square feet so that will take at least $9 \\times 50 = 450$ tiles. Depending on the shape of the region you want to cover there may be some cutting required and hence some cutting. Hence you may need more than 450 tiles. Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "# Jared's room painting Jared wants to paint his room. The dimensions of the room are 12 feet by 15 feet, and the walls are 9 feet high. There are two windows that measure 6 feet by 5 feet each. There are two doors, whose dimensions are 30 inches by 6 feet each. If a gallon of paint covers approximately 350 square feet, how many gallons of paint will it take to give two layers of paint to the walls? Result x = 3.291 gal #### Solution: $a=12 \\ \\text{ft} \\ \\\\ b=15 \\ \\text{ft} \\ \\\\ c=9 \\ \\text{ft} \\ \\\\ \\ \\\\ w_{1}=6 \\ \\text{ft} \\ \\\\ w_{2}=5 \\ \\text{ft} \\ \\\\ \\ \\\\ d_{1}=30 \\ inch=30 / 12 \\ ft=2.5 \\ ft \\ \\\\ d_{2}=6 \\ \\text{ft} \\ \\\\ \\ \\\\ h=350 \\ \\text{ft}^2\\text{/gal} \\ \\\\ \\ \\\\ S_{1}=a \\cdot \\ b + 2 \\cdot \\ (a+b) \\cdot \\ c=12 \\cdot \\ 15 + 2 \\cdot \\ (12+15) \\cdot \\ 9=666 \\ \\text{ft}^2 \\ \\\\ W=w_{1} \\cdot \\ w_{2}=6 \\cdot \\ 5=30 \\ \\text{ft}^2 \\ \\\\ D=d_{1} \\cdot \\ d_{2}=2.5 \\cdot \\ 6=15 \\ \\text{ft}^2 \\ \\\\ \\ \\\\ S=S_{1} - 2 \\cdot \\ W-2 \\cdot \\ D=666 - 2 \\cdot \\ 30-2 \\cdot \\ 15=576 \\ \\text{ft}^2 \\ \\\\ \\ \\\\ x=2 \\cdot \\", "SEARCH HOME Math Central Quandaries & Queries Question from joanne, a parent: How many floor tiles 20x20 inch do I need for area of 8x 12 ft.? Hi Joanne, Eight feet is $8 \\times 12 = 96$ inches and 12 feet is $12 \\times 12 = 144$ inches. Suppose you start by lying a row of tiles along the long side. Each tile is 20 inches long and $\\frac{144}{20} = 7.2$ so you will need 8 tiles, with quite a bit of wastage. Also $\\frac{96}{20} = 4.8$ so you will need 5 of these 8 tile long rows to cover the floor. Thus in total you will need $8 \\times 5 = 40$ tiles. If the tiles have a pattern and you need to match them you will probably need more. If you can use the wastage, especially from the long row you may be able to reduce the number of tiles somewhat. I hope this helps, Penny Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "'ll need to cover your Surface area packs you 'll need to know many... ) foot, but most people only know how many legs you need, for instance, the square of. The square footage of each board to let water drain I approach this of. Times the Width 're installing a wood picket fence, you need for your project say that working out many. Doing so, save yourself some money of siding on a 20′ wide wall will require 420 feet... Need 200 linear feet are the easiest way to calculate your needs project! Outcome you would want the Bingo caller in order to obtain the outcome... Multiple your Single board Coverage = ( Width + Expansion Gap along Width ) Stair! Possible to limit any material waste and in doing so, save yourself some money can,... Be ordered that amount to linear feet so, save yourself some.! The price per board foot, but I can purchase 20 board how many boards do i need calculator of siding, or 35.... With our easy loft Leg calculator need for your accent wall or floor tiles you will need know! Need help calculating how many pickets to purchase then enter the length of project. To calculate your needs help you work out how many pickets to purchase cost of $60", "USA may be aware that the single prime is not an inch, but a foot. I seriously doubt the GMAT exam contains a question exactly as specified in the OP. Senior Manager Joined: 24 Nov 2015 Posts: 473 Location: United States (LA) Re: How many 2-inch by 3-inch rectangular tiles are required to tile this [#permalink] ### Show Tags 09 Jun 2016, 00:30 chetan2u wrote: sabxu1 wrote: chetan2u wrote: Bunuel, E is ok.. people have not read Q properly so what is the answer E or? why is E ok im so confused there is a 6'*12' garden with a 2'*3' patch.. Q is -How many 2\"*3\" tiles can be placed and not 2'*3' convert all feet into inches as we arelooking into tiles which are in inches so total area of garden = 6*12*12*12 shaded region = 12*12(6*12-2*3) = 12*12*2*3(12-1).. no of 2\"*3\" tiles required =$$\\frac{12*12*12*2*3*11}{2*3} = 12*12*11= 1584 > 1500$$ so ans is 1500+ E Apologies for finding a doubt with figure nowhere in the figure it is showing that the rectangle is 12 feet * 6 feet or are we supposed to assume 12' as 12 feet? Senior Manager Joined: 24 Nov 2015 Posts: 473 Location: United States (LA) Re: How many 2-inch by 3-inch rectangular tiles are required to tile this [#permalink] ### Show", "# The length of a rectangular floor is 12 meters less than twice its width. If a diagonal of the rectangle is 30 meters, how do you find the length and width of the floor? Length = 24 m Width = 18 m #### Explanation: Width (W) = W Length (L) = $2 \\cdot W - 12$ Diagonal (D) = 30 According to Pythagorean Theorem: ${30}^{2} = {W}^{2} + {\\left(2. W - 12\\right)}^{2}$ $900 = {W}^{2} + 4 {W}^{2} - 48 W + {12}^{2}$ $900 = 5 {W}^{2} - 48 W + 144$ $5 {W}^{2} - 48 W - 756 = 0$ $\\Delta = {48}^{2} - 4 \\cdot 5 \\cdot \\left(- 756\\right) = 2304 + 15120 = 17424$ $W 1 = \\frac{- \\left(- 48\\right) + \\sqrt{17424}}{2 \\cdot 5} = \\frac{48 + 132}{10}$ $W 1 = 18$ $W 2 = \\frac{- \\left(- 48\\right) - \\sqrt{17424}}{2 \\cdot 5} = \\frac{48 - 132}{10}$ $W 2 = - 8 , 4$ (impossible) $S o , W = 18 m$ $L = \\left(2 \\cdot 18\\right) - 12 = 24 m$", "# Drawer The rectangular face of a drawer has a perimeter of 108 cm and a width of 18cm. Find the length and the area of the rectangular face of the drawer. Correct result: b = 36 cm A = 648 cm2 #### Solution: $p=108 \\ \\text{cm} \\ \\\\ a=18 \\ \\text{cm} \\ \\\\ p=2(a+b) \\ \\\\ b=(p - 2 \\cdot \\ a)/2=(108 - 2 \\cdot \\ 18)/2=36 \\ \\text{cm}$ $A=a \\cdot \\ b=18 \\cdot \\ 36=648 \\ \\text{cm}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Tiles path A rectangular park measures 24mX18m. A path 2.4m wide is constructed all around it. Find the cost of paving the path with cement tiles of size 60cm X 40cm at the rate of \\$5.60 per tile ? • Tiles The rectangular floor of the 6 x 1.8 m" ]

[ "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "+0 # Trapezoid 0 123 1 Trapezoid ABCD has bases AB and CD. Find x. Apr 29, 2022 #1 +2532 0 Draw altitude $$AM$$ Because the trapezoid is isoceles ($$AD = BC$$), $$MD =1$$ Applying the Pythagorean Theorem, $$x= \\color{brown}\\boxed{\\sqrt{5}}$$ Apr 29, 2022 #1 +2532 0 Draw altitude $$AM$$ Because the trapezoid is isoceles ($$AD = BC$$), $$MD =1$$ Applying the Pythagorean Theorem, $$x= \\color{brown}\\boxed{\\sqrt{5}}$$ BuilderBoi Apr 29, 2022", "+0 # Trapezoid 0 19 1 Trapezoid ABCD has bases AB and CD. Find x. Apr 29, 2022 #1 +1351 +2 Draw altitude $$AM$$ Because the trapezoid is isoceles ($$AD = BC$$), $$MD =1$$ Applying the Pythagorean Theorem, $$x= \\color{brown}\\boxed{\\sqrt{5}}$$ Apr 29, 2022 #1 +1351 +2 Draw altitude $$AM$$ Because the trapezoid is isoceles ($$AD = BC$$), $$MD =1$$ Applying the Pythagorean Theorem, $$x= \\color{brown}\\boxed{\\sqrt{5}}$$ BuilderBoi Apr 29, 2022", "+0 0 339 1 Quadrilateral $ABCD$ is an isosceles trapezoid, with bases $\\overline{AB}$ and $\\overline{CD}.$ A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base $\\overline{AB}$ is $2x,$ and the length of base $\\overline{CD}$ is $2y.$ Prove that the radius of the inscribed circle is $\\sqrt{xy}.$ [asy] unitsize(1.5 cm); pair A, B, C, D, O, T; O = (0,0); T = dir(20); B = extension(T, T + rotate(90)*(T), (0,1), (1,1)); C = extension(T, T + rotate(90)*(T), (0,-1), (1,-1)); A = reflect((0,0),(0,1))*(B); D = reflect((0,0),(0,1))*(C); draw(A--B--C--D--cycle); draw(Circle(O,1)); label(\"$A$\", A, NW); label(\"$B$\", B, NE); label(\"$C$\", C, SE); label(\"$D$\", D, SW); [/asy] Jul 15, 2020", "# Math: Geometry The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = (4,0); D = (0,0); draw(A--B--C--D--cycle); draw(B--D); label(\"$A$\", A, NW); label(\"$B$\", B, NE); label(\"$C$\", C, SE); label(\"$D$\", D, SW); label(\"$7$\", (A + D)/2, W); label(\"$7$\", (B + C)/2, E); label(\"$9$\", (B + D)/2, SE); label(\"$8$\", (C + D)/2, S); [/asy] I know I can use Heron's Formula, but I still need side AB. 1. 👍 0 2. 👎 0 3. 👁 304 1. huh? An easy isosceles trapezoid? Mark off P and Q on AB so you have a rectangle PQCD. That means that AP=QB=(9-8)/2 = 1/2 Now the height h of the trapezoid is h^2 = 7^2 - (1/2)^2 And the area is h(8+9)/2 1. 👍 0 2. 👎 0 posted by Steve 2. You even copied the asymptote? How pathetic. And Steve, don't you realize that the answer is a number? You have to find the height. That's wrong anyway. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### MATH_URGENT The bases of trapezoid ABCD are \\overline{AB} and \\overline{CD}. Let P be the intersection of", "then $|\\overline{AB}| = 2$. With $|\\overline{CD}| = 8$ The area of the trapezoid is: $A_T = \\frac{2+8}2 \\cdot 5=5 \\cdot 5 = 25$ But maybe this result includes the VAT (Wink) • Dec 9th 2012, 07:15 AM Soroban Re: What are the coordinates of A Hello, rcs! Is there a typo in the problem? Quote: A trapezoid ABCD has an area of 25 sq. units. Three of its coordinates are: B(1,8), C(2,3), and D(-6,3) . Find the coordinates of A. The problem is not clearly stated. As earboth pointed out, there are six possible trapezoids. I assume that the horizontal side $CD$ is the \"base\" . . and $AB \\parallel CD.$ Hence, $A$ is in Quadrant 2. Code: | (x,8) | (1,8) A♥ * * * * ♥B * | * * | * * | * * | * D♥ * * * * * * |* * * * * ♥C (-6,3) | (2,3) | - - - - - - - - - + - - - - - - - - | The height is $h = 5.$ The bases are: . $C\\!D \\,=\\, 8,\\;AB \\,=\\, 1-x$ Area of a trapezoid: . $A \\:=\\:\\tfrac{h}{2}(b_1 + b_2)$ Hence, we have: . $\\tfrac{5}{2}\\big[8 + (1-x)\\big] \\:=\\:2{\\color{red}5}$ . . $\\tfrac{5}{2}(9 -x) \\:=\\:25 \\quad\\Rightarrow\\quad 9-x \\:=\\:10 \\quad\\Rightarrow\\quad x", "the rain.The time that passed.You will... Who may introduce a bill that does not have to do with raising revenue? geometry. In trapezoid ABCD with bases AB and CD, we have AB = 52, BC = 10, CD = 36, and DA = 5. 28. Answers. What is the area of ABCD? Find the area of ABCD. (a) Area of ABCD (b) Area of triangle AEF ? Thanks! Click here to get an answer to your question ️ Trapezoid ABCD shown below will undergo a single transformation. 1.may a.$61.00 2.june b.$54.60 3.february c.$41.00 4.january d.$102.15 5.march e.$81.25 6.april f.$36.50 this question has no label. REF: 061301ge 3 ANS: 3 The diagonals of an isosceles trapezoid are congruent. In the trapezoid ABCD (Taken in order) AB||CD , |AB| = 4 cm, and |CD| = 10 cm. Find the value of y if AB = 2y — 7 and DC = 4y + 5, and EF=y+ 5 27. Find angle A. 1) 3 2) 4 3) 7 4) 8. D E, F A If AB = 5x – 9, DC =x + 3, and EF = 2x + 2, what is the value of x? - 11547553 The United States has undergone a gradual shift from dual to cooperative federalism. Question 2 (1 point) Solution for Question 6 Trapezoid ABCD", "Each container has a radius of 1.5 inches and a height of 2.5 inches. If the label covers the lateral area of the container, … 7. ### Math: Geometry The bases of trapezoid$ABCD$are$\\overline{AB}$and$\\overline{CD}$. We are given that$CD = 8$,$AD = BC = 7$, and$BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = (4,0); … 8. ### PLease Help Math Sets The 130 consumers were asked about their buying preferences concerning another product that is sold in the market under three labels. The results were 12 buy only those sold under label A. 24 buy only those sold under label B. 27 buy … 9. ### Math In a survey, 175 consumers were asked about their buying preferences concerning a product that is sold in the market under three labels. The results were as follows. 14 buy only those sold under label A. 25 buy only those sold under … 10. ### math In the below figure, EB is tangent to the circle with center$C\\$ at the point B. If EC is equal to the diameter of the circle, find the measure of angle BAD in degrees. and here is the ugly LaTeX [asy] pair C = (0,0), B = dir(110), … More", "error inducing picture. Jan 16 '15 at 14:15 • Hi, I tried to answer it again and I think I made a mistake with my drawing at first, can you see it now? I also don't get AB= AE + EB = CD + CD/2, EB says its 2times AE right? thank you Jan 17 '15 at 0:09 • I don't understand $A(ABCD)$ = $2ah$ isn't it $A(ABCD) = \\frac{3CD}{4}h$? because I need to perform lcd with CD + CD/2 then $\\frac{2CD+CD}{2}\\times\\frac{h}{2}$? Jan 17 '15 at 0:21 • The area of the trapezoid is $\\dfrac{(b_1+b_2)h}{2}$. Here $b_1+b_2=CD+3 \\times CD=4a$ (since $CD=a$ in my definition). I forgot to change the previous lines when I edited the answer. Jan 19 '15 at 8:03", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}", "The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. 1. 👍 2. 👎 3. 👁 1. why is this in LaTex lol 1. 👍 2. 👎 2. Drop altitudes CE and DF The trapezoid now can be seen to be rectangle CDEF and two right triangles of height h and base x. h^2+x^2 = 7^2 h^2 + (8+x)^2 = 9^2 Hmmm. I get x = -2 I guess I have drawn the figure incorrectly. Maybe you can use my ideas in your own drawing. 1. 👍 2. 👎 3. well, maybe not. It just might mean that AB is shorter than CD. In that case, h = 3√5 and x = -2, so CD=8 and AB=4, making the area (8+4)/2 (3√5) = 18√5 1. 👍 2. 👎 4. Steve, you solved the system of equations wrong. It is in fact 8-x, not 8+x. 1. 👍 2. 👎 5. Stop cheating! 1. 👍 2. 👎 6. lol AoPS 1. 👍 2. 👎 7. AOPS Question 1. 👍 2. 👎 8. It's not cheating, you should mind your own business. Some of us don't always have the help we need and get extremely frustrated on problems, so we", "# 2011 AMC 8 Problem 20 #### shanghai ##### Member Quadrilateral $$ABCD$$ is a trapezoid, $$AD = 15$$, $$AB = 50$$, $$BC = 20$$, and the altitude is $$12$$. What is the area of the trapezoid? $$\\textbf{(A) }600\\\\\\textbf{(B) }650\\\\\\textbf{(C) }700\\\\\\textbf{(D) }750\\\\\\textbf{(E) }800$$", "# 2021 AIME I #9 Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB. Suppose that the distances from $A$ to lines $BC, CD$, and $BD$ are $15, 18,$ and $10$, respectively. Let $K$ be the area of $ABCD$. Find $\\sqrt{2} \\cdot K$.", "+0 # Trapezoid ABCD has vertices 0 65 1 Trapezoid ABCD has vertices A(-1,0), B(0,2), C(m,4) and D(k,0), with m > 0 and k > 0 . The line y = -x + 4 is perpendicular to the line containing side CD , and the area of trapezoid ABCD is 36 square units. What is the value of k? Nov 13, 2020 #1 +10807 +1 Trapezoid ABCD has vertices A(-1,0), B(0,2), C(m,4) and D(k,0), with m > 0 and k > 0 . The line y = -x + 4 is perpendicular to the line containing side CD , and the area of trapezoid ABCD is 36 square units. What is the value of k? Hello Guest! $$m_{ab}=\\frac{2-0}{0-(-1)}$$ $$m_{ab}=2$$ $$y_{dc}=2(x-k)+0\\\\ y_{dc}=2(x-m)+4\\\\ 2x-2k=2x-2m+4\\\\2k=2m-4\\\\ k=m-2$$ to be continued ! Nov 14, 2020 edited by asinus Nov 15, 2020", "Trapezoid IV In a trapezoid ABCD (AB||CD) is |AB| = 15cm |CD| = 7 cm, |AC| = 12 cm, AC is perpendicular to BC. What area has a trapezoid ABCD? Result S = 110.635 cm2 Solution: $\\dfrac{x}{15} = \\dfrac{12-x}{7} \\ \\\\ x = 90⁄11 ≐ 8.181818 \\ cm \\ \\\\ y = 12-x = 42⁄11 ≐ 3.818182 \\ cm \\ \\\\ \\ \\\\ x' = \\sqrt{ 15^2 - x^2 } = 12.572106078127 \\ cm \\ \\\\ y' = \\sqrt{ 7^2 - y^2 } = 5.8669828364591 \\ cm \\ \\\\ \\ \\\\ S = S_1 + S_2+ S_3+ S_4 \\ \\\\ S = x x' / 2 + x'y / 2 + y y'/2 + xy'/2 \\ \\\\ S = (x x' + x'y + y y' + xy')/2 \\ \\\\ S = 110.635 \\ cm^2 \\ \\\\$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! Tips to related online calculators Next similar math problems: 1. Isosceles trapezoid Isosceles trapezoid ABCD, AB||CD is given by |CD| = c = 12", "+0 # help 0 144 3 The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided. The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally. If the shaded regions K has area 35 and the shaded region has area 55. Determine the area of the trapezoid ABCD. Feb 11, 2020 #1 +25532 +3 The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided. The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally. If the shaded regions K has area 35 and the shaded region L has area 55. Determine the area of the trapezoid ABCD. $$\\begin{array}{|rcll|} \\hline 5b+2*\\frac{1}{5}x = 5a &\\text{ or }& \\boxed{b = a-\\frac{2}{25}x} \\\\ 5c+2*\\frac{2}{5}x = 5a &\\text{ or }& \\boxed{c = a-\\frac{4}{25}x} \\\\ 5d+2*\\frac{3}{5}x = 5a &\\text{ or }& \\boxed{d = a-\\frac{6}{25}x} \\\\ 5e+2*\\frac{4}{5}x = 5a &\\text{ or }& \\boxed{e = a-\\frac{8}{25}x} \\\\ 5f+2x = 5a &\\text{ or }& \\boxed{f = a-\\frac{2}{5}x} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\text{trapezoid } ABCD &=& \\left(\\dfrac{5a+5f}{2}\\right)5h \\quad | \\quad \\boxed{f = a-\\frac{2}{5}x} \\\\ &=& \\left(\\dfrac{5a+5(a-\\frac{2}{5}x)}{2}\\right)5h \\\\ &=& \\left(\\dfrac{10a-2x}{2}\\right)5h \\\\ &=& (5a-x)5h \\\\ \\mathbf{\\text{trapezoid } ABCD} &=& \\mathbf{25ah-5xh} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{K:} \\\\ \\hline 35 &=& \\left(\\frac{d+e}{2}\\right)h + 2 \\left( \\frac{e+f}{2}\\right)h \\\\ 35 &=& \\left(\\frac{d+e}{2}\\right)h", "$ABC$, we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into. Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$. This factors to $(x+5)(x-3)$. Since lengths cannot be negative, $x=3$. Since $x=3$, that means the altitude $BF=2\\cdot3=6$, or $[ABC]=60$. Thus $[ABCD]=[ACD]+[ABC]=300+60=\\boxed {\\textbf{(D) }360}$ ~ Solution by Ultraman ~ Diagram by ciceronii ~ Formatting by BakedPotato66 ## Solution 2 (Coordinates) $[asy] size(10cm,0); draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw((-20,0)--(20,0)); draw((0,-15)--(0,35)); draw((10,30)--(-8,-6)); draw(circle((0,0),10)); label(\"E\",(-4.05,-.25),S); label(\"D\",(10,30),NE); label(\"C\",(10,0),NE); label(\"B\",(-8,-6),SW); label(\"A\",(-10,0),NW); label(\"5\",(-10,0)--(-5,0), NE); label(\"15\",(-5,0)--(10,0), N); label(\"30\",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); [/asy]$ Let the points be $A(-10,0)$, $\\:B(x,y)$, $\\:C(10,0)$, $\\:D(10,30)$,and $\\:E(-5,0)$, respectively. Since $B$ lies on line $DE$, we know that $y=2x+10$. Furthermore, since $\\angle{ABC}=90^\\circ$, $B$ lies on the circle with diameter $AC$, so $x^2+y^2=100$. Solving for $x$ and $y$ with these equations, we get the solutions $(0,10)$ and $(-8,-6)$. We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\\frac{20\\cdot30}{2}+\\frac{20\\cdot6}{2}=\\boxed{\\textbf{(D)}\\:360}$. ## Solution 3 (Trigonometry) Let $\\angle C = \\angle{ACB}$ and $\\angle{B} = \\angle{CBE}.$ Using Law of Sines on $\\triangle{BCE}$ we get $$\\dfrac{BE}{\\sin{C}} = \\dfrac{CE}{\\sin{B}} = \\dfrac{15}{\\sin{B}}$$ and LoS on $\\triangle{ABE}$ yields $$\\dfrac{BE}{\\sin{(90 - C)}} = \\dfrac{5}{\\sin{(90 - B)}} = \\dfrac{BE}{\\cos{C}} = \\dfrac{5}{\\cos{B}}.$$ Divide the", "(7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw] That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.", "Draw and label the shapes, then illustrate how the quotient was obtained. Your drawing must be neat and accurate. Can individual pieces … The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = (4,0); …", "bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = asked by I see you have graph paper. You must be plotting something. on November 19, 2015 2. ### math In the below figure, EB is tangent to the circle with center $C$ at the point B. If EC is equal to the diameter of the circle, find the measure of angle BAD in degrees. and here is the ugly LaTeX [asy] pair C = (0,0), B = asked by halp on August 12, 2017 3. ### math Given a hexagon $ABCDEF$ inscribed in a circle with $AB = BC, CD = DE, EF = FA$, show that $\\overline{AD}, \\overline{BE}$, and $\\overline{CF}$ are concurrent. [asy] unitsize(2 cm); pair A, B, C, D, E, F, G; A = dir(85); B = asked by Dillion on October 6, 2019 4. ### math the diameter of a soup can is 7cm. the label wraps around the can. about how many centimeters long must the label be to go all the way around the soup can? use 22/7 for pi describe how to mentally check whether" ]

[ "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 (mass points) We see that $EF:FB=3:7$ and $AF=FD$. We assign a mass of $7$ to $E$ and $3$ to $D$, making $F$ have mass $10$ and $A$ and $D$ each have mass 5.", "-- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$.", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "(30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly. Notice that because $BC = AD, 20+x = 6+DT_2 \\implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_2CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \\implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\\triangle ACF$; we have $26^2 +", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"E\",E,SE); label(\"F\",F,NE); label(\"P\",P,NW); label(\"Q\",Q,NE); label(\"R\",R,SE); label(\"S\",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy]$ Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \\frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$. We then know that $CE = \\frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$. Since $PF+FQ=PQ=\\omega$. We can solve for $PF$ and $FQ$ in terms of $\\omega$. We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$. We can set up the proportion: $\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$. Solving for $AF$, $AF = \\frac{3}{4} PF = \\frac{3}{4}", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\dfrac{\\dfrac{\\sqrt{2}}{9}}{\\dfrac{\\sqrt{2}}{3}}$ = $\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 2 Since $\\triangle DCE$ and $\\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 3 Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\\triangle DCE$ can be expressed as $\\frac{1}{2}(CD)(6)\\text{sin }(CDE).$ $\\frac{1}{2}(CD)(6)$ happens to be the area of $\\triangle ABD$. Furthermore, $\\text{sin } CDE = \\frac{CO}{DO},$ or $\\frac{1}{3}.$ Therefore, the ratio is $\\boxed{\\textbf{(C) }\\frac{1}{3}}.$ ## Solution 5 $[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label(\"E\",E,SSE); label(\"B\",B,E); label(\"A\",A,W); label(\"D\",D,NNW); label(\"C\",C,SW); draw(rightanglemark(D,C,B,2)); [/asy]$ Let the point G be the reflection of point $D$ across $\\overline{AB}$. (Point G is on the circle). Let $AC=x$, then $BC=2x$. The diameter is $3x$. To find $DC$, there are two ways (presented here): 1. Since $\\overline{AB}$ is the", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "the ratio of the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\frac{\\frac{\\sqrt{2}}{9}}{\\frac{\\sqrt{2}}{3}}$ = $\\frac{1}{3} \\Rightarrow C$ ## Solution 2 Since $\\triangle DCE$ and $\\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\frac{1}{3}\\Rightarrow \\text{(C)}$ ## Solution 3 Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\\triangle DCE$ can be expressed as $\\frac{1}{2}(CD)(6)\\text{sin }(CDE).$ $\\frac{1}{2}(CD)(6)$ happens to be the area of $\\triangle ABD$. Furthermore, $\\text{sin } CDE = \\frac{CO}{DO},$ or $\\frac{1}{3}.$ Therefore, the ratio is $\\frac{1}{3}.$" ]

[ "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "Select Page Try this beautiful problem from Geometry: The area of trapezoid. ## The area of the trapezoid – AMC-8, 2011 – Problem 20 Quadrilateral ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid? • $700$ • $750$ • $800$ Geometry Trapezoid Area of Triangle ## Check the Answer But try the problem first… Answer:$750$ Source AMC-8(2011) Problem 20 Pre College Mathematics ## Try with Hints First hint Draw altitudes from the top points A and B to CD at X and Y points Can you now finish the problem ………. Second Hint The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times$ (height between AB and CD) can you finish the problem…….. Final Step Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle . Using The Pythagorean theorem on $\\triangle ADX and \\triangle BYC$ , $(DX)^2+(AX)^2=(AD)^2$ $\\Rightarrow (a)^2+(12)^2=(15)^2$ $\\Rightarrow a=\\sqrt{(15)^2-(12)^2}=\\sqrt {81} =9$ and $(BY)^2+(YC)^2=(BC)^2$ $\\Rightarrow (12)^2+(b)^2=(20)^2$ $\\Rightarrow b=\\sqrt{(20)^2-(12)^2}=\\sqrt {256} =16$ Now ABYX is a Rectangle so $XY=AB=50$ $CD=DX+XY+YC=a+XY+b=9+50+16=75$ The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times (height between AB and CD)=\\frac {1}{2} \\times (AB+CD) \\times 12=750$ .", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "21 The area of trapezoid ABCD is 164 $cm^2$. The altitude is 8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters? , i • $8$ • $10$ • $15$ Geometry trapezoid Triangle ## Check the Answer Answer: $10$ AMC-8 (2003) Problem 21 Pre College Mathematics ## Try with Hints Draw two altitudes from the points B and C On the straight line AD at D and E respectively. Can you now finish the problem ………. Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm can you finish the problem…….. Given that the area of the trapezoid is 164 sq.unit Draw two altitudes from the points B and C On the straight line AD at D and E respectively. Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm Using Pythagorean rules on the triangle ABD,we have… $(AD)^2 + (BD)^2 =(AB)^2$ $\\Rightarrow (AD)^2 + (8)^2 =(10)^2$ $\\Rightarrow (AD)^2 =(10)^2 – (8)^2$ $\\Rightarrow (AD)^2 = 36$ $\\Rightarrow (AD) =6$ Using Pythagorean rules on the triangle CED,we have… $(CE)^2 + (DE)^2 =(DC)^2$ $\\Rightarrow (CE)^2 + (8)^2 =(17)^2$ $\\Rightarrow (CE)^2 =(17)^2 – (8)^2$ $\\Rightarrow (CE)^2 = 225$ $\\Rightarrow (CE) =15$ Let BC= DE=x Therefore area of the trapezoid=$\\frac{1}{2} \\times (AD+BC) \\times 8$=164 $\\Rightarrow \\frac{1}{2}", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "then $|\\overline{AB}| = 2$. With $|\\overline{CD}| = 8$ The area of the trapezoid is: $A_T = \\frac{2+8}2 \\cdot 5=5 \\cdot 5 = 25$ But maybe this result includes the VAT (Wink) • Dec 9th 2012, 07:15 AM Soroban Re: What are the coordinates of A Hello, rcs! Is there a typo in the problem? Quote: A trapezoid ABCD has an area of 25 sq. units. Three of its coordinates are: B(1,8), C(2,3), and D(-6,3) . Find the coordinates of A. The problem is not clearly stated. As earboth pointed out, there are six possible trapezoids. I assume that the horizontal side $CD$ is the \"base\" . . and $AB \\parallel CD.$ Hence, $A$ is in Quadrant 2. Code: | (x,8) | (1,8) A♥ * * * * ♥B * | * * | * * | * * | * D♥ * * * * * * |* * * * * ♥C (-6,3) | (2,3) | - - - - - - - - - + - - - - - - - - | The height is $h = 5.$ The bases are: . $C\\!D \\,=\\, 8,\\;AB \\,=\\, 1-x$ Area of a trapezoid: . $A \\:=\\:\\tfrac{h}{2}(b_1 + b_2)$ Hence, we have: . $\\tfrac{5}{2}\\big[8 + (1-x)\\big] \\:=\\:2{\\color{red}5}$ . . $\\tfrac{5}{2}(9 -x) \\:=\\:25 \\quad\\Rightarrow\\quad 9-x \\:=\\:10 \\quad\\Rightarrow\\quad x", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "From the given information and the Pythagorean theorem, $A B=4, C B=3$, and $D B=5 .$ Solving gives $E B=20 / 3$ We can use the above formula to solve for $B F . B D^{2}=20 / 3 \\cdot B F$. Solve to obtain $B F=15 / 4$ can you finish the problem........ Therefore $E F=E B+B F=\\frac{20}{3}+\\frac{15}{4}=\\frac{80+45}{12}$ # The area of trapezoid | AMC 8, 2003 | Problem 21 Try this beautiful problem from Geometry: The area of trapezoid ## The area of trapezoid - AMC-8, 2003- Problem 21 The area of trapezoid ABCD is 164 $$cm^2$$. The altitude is 8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters? , i • $8$ • $10$ • $15$ ### Key Concepts Geometry trapezoid Triangle Answer: $10$ AMC-8 (2003) Problem 21 Pre College Mathematics ## Try with Hints Draw two altitudes from the points B and C On the straight line AD at D and E respectively. Can you now finish the problem .......... Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm can you finish the problem........ Given that the area of the trapezoid is 164 sq.unit Draw two altitudes from the points B and C On the straight line AD at D and E", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "Categories # Area of trapezoid | AMC 8, 2011|Problem 20 Try this beautiful problem from Geometry: The area of the trapezoid . You may use sequential hints to solve the problem. Try this beautiful problem from Geometry: The area of trapezoid. ## The area of the trapezoid – AMC-8, 2011 – Problem 20 Quadrilateral ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid? • $700$ • $750$ • $800$ ### Key Concepts Geometry Trapezoid Area of Triangle Answer:$750$ AMC-8(2011) Problem 20 Pre College Mathematics ## Try with Hints Draw altitudes from the top points A and B to CD at X and Y points Can you now finish the problem ………. The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times$ (height between AB and CD) can you finish the problem…….. Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle . Using The Pythagorean theorem on $\\triangle ADX and \\triangle BYC$ , $(DX)^2+(AX)^2=(AD)^2$ $\\Rightarrow (a)^2+(12)^2=(15)^2$ $\\Rightarrow a=\\sqrt{(15)^2-(12)^2}=\\sqrt {81} =9$ and $(BY)^2+(YC)^2=(BC)^2$ $\\Rightarrow (12)^2+(b)^2=(20)^2$ $\\Rightarrow b=\\sqrt{(20)^2-(12)^2}=\\sqrt {256} =16$ Now ABYX is a Rectangle so $XY=AB=50$ $CD=DX+XY+YC=a+XY+b=9+50+16=75$ The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times (height between AB and CD)=\\frac {1}{2} \\times (AB+CD) \\times", "the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\dfrac{\\dfrac{\\sqrt{2}}{9}}{\\dfrac{\\sqrt{2}}{3}}$ = $\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 2 Since $\\triangle DCE$ and $\\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\boxed{\\textbf{(C) }\\frac{1}{3}}$ ## Solution 3 Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\\triangle DCE$ can be expressed as $\\frac{1}{2}(CD)(6)\\text{sin }(CDE).$ $\\frac{1}{2}(CD)(6)$ happens to be the area of $\\triangle ABD$. Furthermore, $\\text{sin } CDE = \\frac{CO}{DO},$ or $\\frac{1}{3}.$ Therefore, the ratio is $\\boxed{\\textbf{(C) }\\frac{1}{3}}.$ ## Solution 5 $[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label(\"E\",E,SSE); label(\"B\",B,E); label(\"A\",A,W); label(\"D\",D,NNW); label(\"C\",C,SW); draw(rightanglemark(D,C,B,2)); [/asy]$ Let the point G be the reflection of point $D$ across $\\overline{AB}$. (Point G is on the circle). Let $AC=x$, then $BC=2x$. The diameter is $3x$. To find $DC$, there are two ways (presented here): 1. Since $\\overline{AB}$ is the", "$\\frac{1}{4} \\times 3 \\times 4$, or $\\boxed{\\textbf{(C) }3}$ ## Solution 2 The area of trapezoid $CBFE$ is $\\frac{1+3}2\\cdot 4=8$. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger one is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\\frac12$ and $\\frac92$. Subtracting these areas from the trapezoid, we get $8-\\frac12-\\frac92 =\\boxed3$. Therefore, the answer to this problem is $\\boxed{\\textbf{(C) }3}$ ## Solution 3 (Coordinate Geometry) Set coordinates to the points: Let $E=(0,0)$, $F=(3,0)$ $[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label(scale(0.7)*\"A(3,4)\",(3.25,4.2)); label(scale(0.7)*\"B(2,4)\",(2.1,4.2)); label(scale(0.7)*\"C(1,4)\",(0.9,4.2)); label(scale(0.7)*\"D(0,4)\",(-0.3,4.2)); label(scale(0.7)*\"E(0,0)\", (0,-0.2)); label(scale(0.7)*\"Z(\\frac{3}{2},3)\", (1.5,1.8)); label(scale(0.7)*\"F(3,0)\", (3,-0.2)); label(scale(0.7)*\"1\", (0.3, 4), N); label(scale(0.7)*\"1\", (1.5, 4), N); label(scale(0.7)*\"1\", (2.7, 4), N); label(scale(0.7)*\"4\", (3.2, 2), E); [/asy]$ Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$. Hence, the slope of line $CF=-2$ Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$ Doing the same process to line $BE$, we find that line $BE=2x$. Hence, setting them equal to find the intersection point... $y=2x=-2x+6\\implies 4x=6\\implies x=\\frac{3}{2}\\implies y=3$. Hence, we find that the intersection point is $(\\frac{3}{2},3)$. Call it Z. Now, we can see that $E=(0,0)$ $Z=(\\dfrac{3}{2},3)$ $C=(1,4)$.", "INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More # Area of trapezoid | AMC 8, 2011|Problem 20 Try this beautiful problem from Geometry: The area of trapezoid. ## The area of the trapezoid - AMC-8, 2011 - Problem 20 Quadrilateral ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid? • $700$ • $750$ • $800$ Geometry Trapezoid Area of Triangle ## Check the Answer Answer:$750$ AMC-8(2011) Problem 20 Pre College Mathematics ## Try with Hints Draw altitudes from the top points A and B to CD at X and Y points Can you now finish the problem .......... The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times$ (height between AB and CD) can you finish the problem........ Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle . Using The Pythagorean theorem on $\\triangle ADX and \\triangle BYC$ , $(DX)^2+(AX)^2=(AD)^2$ $\\Rightarrow (a)^2+(12)^2=(15)^2$ $\\Rightarrow a=\\sqrt{(15)^2-(12)^2}=\\sqrt {81} =9$ and $(BY)^2+(YC)^2=(BC)^2$ $\\Rightarrow (12)^2+(b)^2=(20)^2$ $\\Rightarrow b=\\sqrt{(20)^2-(12)^2}=\\sqrt {256} =16$ Now ABYX is a Rectangle so $XY=AB=50$ $CD=DX+XY+YC=a+XY+b=9+50+16=75$ The area of the trapezoid is $\\frac{1}{2} \\times (AB+CD) \\times (height between AB and CD)=\\frac {1}{2} \\times (AB+CD) \\times 12=750$ . ## Subscribe to Cheenta", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "+0 # help 0 144 3 The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided. The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally. If the shaded regions K has area 35 and the shaded region has area 55. Determine the area of the trapezoid ABCD. Feb 11, 2020 #1 +25532 +3 The trapezoid is divided into 5 parts horizontally with AB and CD are equally divided. The Trapezoid also divided into 5 parts vertically with AD and BC are devided equally. If the shaded regions K has area 35 and the shaded region L has area 55. Determine the area of the trapezoid ABCD. $$\\begin{array}{|rcll|} \\hline 5b+2*\\frac{1}{5}x = 5a &\\text{ or }& \\boxed{b = a-\\frac{2}{25}x} \\\\ 5c+2*\\frac{2}{5}x = 5a &\\text{ or }& \\boxed{c = a-\\frac{4}{25}x} \\\\ 5d+2*\\frac{3}{5}x = 5a &\\text{ or }& \\boxed{d = a-\\frac{6}{25}x} \\\\ 5e+2*\\frac{4}{5}x = 5a &\\text{ or }& \\boxed{e = a-\\frac{8}{25}x} \\\\ 5f+2x = 5a &\\text{ or }& \\boxed{f = a-\\frac{2}{5}x} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\text{trapezoid } ABCD &=& \\left(\\dfrac{5a+5f}{2}\\right)5h \\quad | \\quad \\boxed{f = a-\\frac{2}{5}x} \\\\ &=& \\left(\\dfrac{5a+5(a-\\frac{2}{5}x)}{2}\\right)5h \\\\ &=& \\left(\\dfrac{10a-2x}{2}\\right)5h \\\\ &=& (5a-x)5h \\\\ \\mathbf{\\text{trapezoid } ABCD} &=& \\mathbf{25ah-5xh} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{K:} \\\\ \\hline 35 &=& \\left(\\frac{d+e}{2}\\right)h + 2 \\left( \\frac{e+f}{2}\\right)h \\\\ 35 &=& \\left(\\frac{d+e}{2}\\right)h", "# Math: Geometry The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A = (1,2); B = (3,2); C = (4,0); D = (0,0); draw(A--B--C--D--cycle); draw(B--D); label(\"$A$\", A, NW); label(\"$B$\", B, NE); label(\"$C$\", C, SE); label(\"$D$\", D, SW); label(\"$7$\", (A + D)/2, W); label(\"$7$\", (B + C)/2, E); label(\"$9$\", (B + D)/2, SE); label(\"$8$\", (C + D)/2, S); [/asy] I know I can use Heron's Formula, but I still need side AB. 1. 👍 0 2. 👎 0 3. 👁 304 1. huh? An easy isosceles trapezoid? Mark off P and Q on AB so you have a rectangle PQCD. That means that AP=QB=(9-8)/2 = 1/2 Now the height h of the trapezoid is h^2 = 7^2 - (1/2)^2 And the area is h(8+9)/2 1. 👍 0 2. 👎 0 posted by Steve 2. You even copied the asymptote? How pathetic. And Steve, don't you realize that the answer is a number? You have to find the height. That's wrong anyway. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### MATH_URGENT The bases of trapezoid ABCD are \\overline{AB} and \\overline{CD}. Let P be the intersection of", "the rain.The time that passed.You will... Who may introduce a bill that does not have to do with raising revenue? geometry. In trapezoid ABCD with bases AB and CD, we have AB = 52, BC = 10, CD = 36, and DA = 5. 28. Answers. What is the area of ABCD? Find the area of ABCD. (a) Area of ABCD (b) Area of triangle AEF ? Thanks! Click here to get an answer to your question ️ Trapezoid ABCD shown below will undergo a single transformation. 1.may a.$61.00 2.june b.$54.60 3.february c.$41.00 4.january d.$102.15 5.march e.$81.25 6.april f.$36.50 this question has no label. REF: 061301ge 3 ANS: 3 The diagonals of an isosceles trapezoid are congruent. In the trapezoid ABCD (Taken in order) AB||CD , |AB| = 4 cm, and |CD| = 10 cm. Find the value of y if AB = 2y — 7 and DC = 4y + 5, and EF=y+ 5 27. Find angle A. 1) 3 2) 4 3) 7 4) 8. D E, F A If AB = 5x – 9, DC =x + 3, and EF = 2x + 2, what is the value of x? - 11547553 The United States has undergone a gradual shift from dual to cooperative federalism. Question 2 (1 point) Solution for Question 6 Trapezoid ABCD", "right CD is 30, top AD is unknown. Trapezoid EFGH is similar to ABCD, top EH and bottom FG are parallel. EH is the smaller of asked by olav on December 12, 2010 69. ## Math The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. asked by justarandomboy on April 22, 2016 The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. asked by Sarah on April 23, 2016 71. ## geometry The bases of trapezoid $ABCD$ are $\\overline{AB}$ and $\\overline{CD}$. We are given that $CD = 8$, $AD = BC = 7$, and $BD = 9$. Find the area of the trapezoid. asked by ana on July 21, 2017 72. ## Geometry I have to do this web quest table and i just want to make sure that i did this right. please help check this for me! I did this in excel so i marked the questions with my answers with different symbols. it is a little confusing. Probability of 2 randomly asked by Cheryl on February 2, 2011 73. ## calculus a", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}" ]

[ "Math Help - solve in R 1. solve in R solve in R $x+y+z=2$ $2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\\sqrt[9]{2}$ 2. $x=y=z=\\frac{2}{3}$", "and then solve for r_1.", "2R) 2r = 50 − 9 = 41 r = $\\frac{41}{2}$ = 20.5 Is there an error in this question or solution?", "\\frac{-a}{2} - \\sqrt{a^2 - 4(a + 2)}$ by the quadratic formula, where r1 and r2 are the solutions to the quadratic. Thus $r_1 - r_2 = \\sqrt{a^2 - 4(a + 2)} = 2$ Solve for a. -Dan", "Solve in R : $\\frac{1}{x-1}-\\frac{4}{x-2}+\\frac{4}{x-3}-\\frac{1}{x-4}< \\frac{1}{30}$ Solve in R : $\\frac{1}{x-1}-\\frac{4}{x-2}+\\frac{4}{x-3}-\\frac{1}{x-4}< \\frac{1}{30}$", "How to solve $D\\frac{\\partial^2 C}{\\partial r^2}+\\frac{D}{r}\\frac{\\partial C}{\\partial r}= \\frac{\\partial C}{\\partial t}$", "# Math Help - Solve for R 1. ## Solve for R Solve for $R$ when $\\theta = .9948\\;rad$ $R = \\frac{\\pi}{\\theta - \\frac{sin\\;2\\;\\theta}{2}}$ $R = \\frac{\\pi}{\\theta - \\frac{2sin\\;\\theta\\;cos\\;\\theta}{2}}$ $R = \\frac{\\pi}{.9948 - \\frac{2sin\\;.9948\\;cos\\;.9948}{2}}$ $R = \\frac{\\pi}{.9948 - .4568}$ $R = 5.84$ Is this correct? 2. Originally Posted by OzzMan Solve for $R$ when $\\theta = .9948\\;rad$ $R = \\frac{\\pi}{\\theta - \\frac{sin\\;2\\;\\theta}{2}}$ $R = \\frac{\\pi}{\\theta - \\frac{2sin\\;\\theta\\;cos\\;\\theta}{2}}$ $R = \\frac{\\pi}{.9948 - \\frac{2sin\\;.9948\\;cos\\;.9948}{2}}$ $R = \\frac{\\pi}{.9948 - .4568}$ $R = 5.84$ Is this correct? Yes but you are making a real mean of it. $R = \\frac{\\pi}{0.9984 - \\frac{\\sin(1.9896)}{2}}$ RonL", "# Solve for r: h^2+(pie)r=K 1. Aug 28, 2011 ### davie08 1. The problem statement, all variables and given/known data solve for r: h^2+(pie)r=K 2. Relevant equations not sure 3. The attempt at a solution r=K-h^2/pie 2. Aug 28, 2011 ### ArcanaNoir This is correct if you mean r=(k-h^2)/(pi) $$r=\\frac{k-h^2}{\\pi}$$ no e in pi... ;) 3. Aug 28, 2011 ### Ray Vickson This is incorrect: you have written $$r = K - \\frac{h^2}{\\pi},$$ at least when read according to *standard rules*. If you mean that $$r = \\frac{K - h^2}{\\pi} ,$$ then you need to use brackets: r = (K - h^2)/pi. RGV", "# RATL 2 | Lesson 4 | Practice (Assess) 1) Solve: $$\\Large \\frac{4}{x+3}=\\frac{6}{x+6}$$ Did you find that $$x= 3$$?", "Solving the quadratic, $$R = 2 - \\sqrt{3} \\approx 0.26794919$$. Now we can solve the original problem, given by $$\\frac{1- R}{R} = \\frac{\\sqrt{3} - 1}{2 - \\sqrt{3}} \\approx 2.7320508$$. [Image credit to FiveThirtyEight]", "solve for r: $ 352.8r=(80)(4900) $ $ 352.8r=392000 $ $ r=1111.11 $ With r figured out, i can now solve the following: $ F=\\frac{(90)(20)^2}{1111.11} $ $ F=32.4 newtons $ I know the latex is all messed up, but does that look right?", "How do you solve 7/4=r/5? $r = \\frac{35}{4}$ $\\frac{7}{4} = \\frac{r}{5}$ $4 r = 35$ $r = \\frac{35}{4}$", "$(\\frac{ 45 }{ 10^{3} })(\\frac{ 21 }{ 10 } \\times 10^3)$ $\\frac{ 45 \\times 21 }{ 10 }$ you can solve furthr", "+0 0 55 3 +248 Solve for r: $$\\frac{r+3}{r-2} = \\frac{r-1}{r+1}.$$ Thanks! Jun 10, 2019 #1 +2 Solve for r: (r + 3)/(r - 2) = (r - 1)/(r + 1) Cross multiply: (r + 1) (r + 3) = (r - 2) (r - 1) Expand out terms of the left hand side: r^2 + 4 r + 3 = (r - 2) (r - 1) Expand out terms of the right hand side: r^2 + 4 r + 3 = r^2 - 3 r + 2 Subtract r^2 - 3 r + 2 from both sides: 7 r + 1 = 0 Subtract 1 from both sides: 7 r = -1 Divide both sides by 7: r = -1/7 Jun 10, 2019 #2 +248 +2 Thanks! sinclairdragon428 Jun 10, 2019 #3 +721 +2 (r+3)/(r-2)=(r-1)(r+1) (r+3)(r+1)=(r-1)(r-2) r^2+4r+3=r^2-3r+2 7r=-1 r=-1/7 You are very welcome! :P Jun 10, 2019", "Basic College Mathematics (10th Edition) $r=-7$ We are given the equation $-8r=56$. We can solve for r by dividing both sides by -8. $\\frac{-8r}{-8}=\\frac{56}{-8}$ Therefore, $r=-7$.", "– The Pointer Jun 16 '20 at 2:49 • Take the two equations in (3) and solve for $r_1$ and $r_2$. Use $r_2 = r_1 + \\Delta r$ in the first expression to get $r_c = r_1 + \\Delta r \\frac{m_2}{m_1+m_2}$. Solve for $r_1$ and then back substitute to get $r_2$. – John Alexiou Jun 21 '20 at 1:08", "# 200 Followers Problem! $\\frac{r^2}{s}+200=\\frac{s^2}{r}+200^2$ In the equation above, $r$ and $s$ are both rational numbers and their difference is an integer. What is the number of solutions to the equation? ×", "after t seconds. 22. Solve: $\\frac { dy }{ dx }$ = (3x+2y+1)2", "be to do 3r^2=12 , thus r=2 yup, that's what i said. the equation $3r^2~r' = 12~r'$ must hold. so just solve for r", "the next step in solving this problem? 13. Originally Posted by florx Solving the quadratic equation I get -125 and 325. The back of the book says the answer is 44.4%. What is the next step in solving this problem? Oh wait! Let R be the original speed and let r be the reduced speed D be the original diameter, so $\\frac{3D}{2}$ is the new diameter $k \\times R^2 \\times D^4 = k \\times r^2 \\times (\\frac{3}{2}D)^4$ $R^2 = r^2 \\times \\frac{81}{16}$ $R = r \\times \\frac{9}{4}$ $\\therefore r = \\frac{4R}{9}$ (4/9) of R = 44.4% of R 14. Oh there you go thank you so much for helping me with this problem as well as past problems. You are very persistent in your willingness to help me and as others and I want to thank you for that." ]

[ "4? To bring the fractions to integers? I was trying with substitution, but when multiplying out the first line (before multiply x 8) I would get the wrong answer. As a side question, why doesn't that work? 6. To clear the fractions you just multiply the equation by the denominator of the fraction in question. As far as I'm aware the method I used is called substitution, but if you post your working here we can try to see where you went wrong. 7. I was attempting to do it this way: (sorry couldn't get the math code to work) p = 150 + 3/4 (100 + 1/2p ) p = 150 + 75 + 3/8P P = 225 + 3/8p 225 = 5/8p p = 309.37 8. Your method is correct, except for the last step. $\\frac{5P}{8}=225$ $\\therefore 5P=1800$ $\\therefore P=360$ 9. Damn, so simple. cheers for the help.", "# Turning a fraction into a quadratic • Dec 2nd 2008, 09:17 AM Larianne Turning a fraction into a quadratic Hi, I'm trying to work out how a question was solved from an example in a book. x(squared) ------------------- = 6.25 6 - 5x + x(squared) goes to... 5.25x(squared) - 31.25x + 37.5 = 0 I don't understand how this was done. any ideas? • Dec 2nd 2008, 10:03 AM nzmathman They have cross multiplied to get the answer. $\\frac{x^2}{6 - 5x + x^2} = 6.25$ $\\Rightarrow x^2 = 6.25(6 - 5x + x^2)$ (cross multiply) $\\Rightarrow x^2 = 37.5 - 31.25x + 6.25x^2$ (expanding) $\\Rightarrow 5.25x^2 - 31.25x + 37.5 = 0$ (collect terms on one side)", "## Algebra 1 Substitute for c and solve for r. $50=3r+15$ subtract 15 from both sides $50-15=3r+15-15$ simplify $35=3r$ divide both sides by 3 $35\\div3=3r\\div3$ $11\\frac{2}{3}=r$ You cannot buy part of a rose, so the correct answer is 11.", "R = 124 / 125 2- 32,000 = 256 / [1 - R], solve for R R = 124 / 125 Therefore: m/n =124/125 and m + n = 124 + 125 =249 Jul 12, 2018 #2 +1 2000 = F / [1 - R], ....................................(1) 32000 =F^2 / [1 - R]...................................(2) From (1) above: F =2000 - 2000R - Sub this for F in (2) above: 32000 =[2000 - 2000R]^2 / [1 -R] Solve for R: 32000 = (2000 - 2000 R)^2/(1 - R) 32000 = (2000 - 2000 R)^2/(1 - R) is equivalent to (2000 - 2000 R)^2/(1 - R) = 32000: (2000 - 2000 R)^2/(1 - R) = 32000 Multiply both sides by 1 - R: (2000 - 2000 R)^2 = 32000 (1 - R) Expand out terms of the right hand side: (2000 - 2000 R)^2 = 32000 - 32000 R Subtract 32000 - 32000 R from both sides: -32000 + (2000 - 2000 R)^2 + 32000 R = 0 Expand out terms of the left hand side: 4000000 R^2 - 7968000 R + 3968000 = 0 The left hand side factors into a product with three terms: 32000 (R - 1) (125 R - 124) = 0 Divide both sides by 32000: (R - 1) (125 R - 124) = 0 Split into", "} 225 \\div 45 = 5. \\text{ So } 225° = 5 \\left( \\frac{\\pi}{4} \\right) = \\frac{5 \\pi}{4} .$", "Basic College Mathematics (10th Edition) $r=-7$ We are given the equation $-8r=56$. We can solve for r by dividing both sides by -8. $\\frac{-8r}{-8}=\\frac{56}{-8}$ Therefore, $r=-7$.", "# Math Help - Turning a fraction into a quadratic 1. ## Turning a fraction into a quadratic Hi, I'm trying to work out how a question was solved from an example in a book. x(squared) ------------------- = 6.25 6 - 5x + x(squared) goes to... 5.25x(squared) - 31.25x + 37.5 = 0 I don't understand how this was done. any ideas? 2. They have cross multiplied to get the answer. $\\frac{x^2}{6 - 5x + x^2} = 6.25$ $\\Rightarrow x^2 = 6.25(6 - 5x + x^2)$ (cross multiply) $\\Rightarrow x^2 = 37.5 - 31.25x + 6.25x^2$ (expanding) $\\Rightarrow 5.25x^2 - 31.25x + 37.5 = 0$ (collect terms on one side)", "# How do you solve 1/6 - 8 <8r - r/6? Aug 2, 2015 $r > - 1$ or r in (-1;+oo) #### Explanation: First we can multiply both sides by 6 to get rid of fractions: $1 - 48 < 48 r - r$ $- 47 < 47 r$ $- 1 < r$ $r > - 1$", "R make those factors 0 ? 12. so the two numbers need to multiply to 45 and add to 18 so 15 and 3 again? 13. Originally Posted by kitobeirens so the two numbers need to multiply to 45 and add to 18 so 15 and 3 again? Yes, that's how the quadratic was factored. Now the two factors in brackets give you the answers visually. What is R-15 if R=15 and what is R-3 if R=3 ? 14. Kito, since you obviously don't understand how the quadratic formula works, then you are not ready for this problem. Suggest you talk to your math teacher. 15. oh its 15 isnt it because 15 squared =225 -(18x15=270) so 225-270+45=0 Page 1 of 2 12 Last", "we multiply both sides by $\\frac{-1}{-1}$ $\\frac{-1}{-1}*\\frac{45}{-30}=\\frac{-45}{30}$ Now we factorise -45 and 30, but we can think of -45 as $-1 * 45$ so we only need to factorise 45: 30 = {1, 2, 3, 5, 6, 10, 15} 45 = {1, 3, 5, 15} The largest factor is 15 for each so we will use that for our reduction: $\\frac{-1*45/15}{30/15}=\\frac{-1*3}{2}=\\frac{-3}{2}$", "• $$a=R277-X115$$ • $$b=R115+X277$$ • $$c=R+277$$ • $$d=X+115$$ Plugging these into the formula above, we get $$\\frac{[R^2277-RX115+R277^2-X(277\\cdot115)]+[RX115+X^2277+R115^2+X(277\\cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$ Cancelling terms in the numerator gives $$=\\frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$ This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $$115^2=13225$$, and so on. In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first. The resulting system is best solved by machine: here's the computation in WolframAlpha.", "the distributive property... $15=\\frac{7.62 \\pi r^2+16.387064 \\pi}{6}$ Clear the fraction (Multiply both sides by 6)... $90=7.62 \\pi r^2+16.387064$ Subtract 16.387064 from both sides... $73.612936=7.62\\pi r^2$ Divide both sides by $7.62 \\pi$... $\\frac{73.612936}{7.62\\pi}=r^2$ Take the square root of both sides... $\\sqrt{\\frac{73.612936}{7.62\\pi}}=r$ hi thanks to both of you for answering my post. I have a few questions though since i still dont know what r value is. please tell me if i am supposed to know or not. I figured out to a point that substition of values makes sense but to be honest i dont understand the rest of it. also when using pi in the equation am i using the number 3.14 or not? what i did was v=3.14 x 2.54(3 x 3.14 x 2.54 x 2) + (3.14 x 2.54 x 2.54 x 2.54) / 6 then i x that by 3 and squared it then i did 2.54 squared and then divided by 6 i know this is wrong but im so confused 5. You said you were solving for r. You have a \"v\" in your final equation but no \"r\". what happened to it? You wrote \"v=3.14 x 2.54(3 x 3.14 x 2.54 x 2) + (3.14 x 2.54 x 2.54 x 2.54) / 6\" Okay, that first \"3.14 x 2.54\" is pi h", "$$\\Rightarrow \\frac{15}{225}=\\frac{1}{15}$$ The ratios are not equal and therefore these are not in proportion d) 4 persons: 42 persons and Rs.8: Rs.84 Solution: $$\\Rightarrow \\frac{4}{42}=\\frac{2}{21}$$ Similarly $$\\Rightarrow \\frac{8}{84}=\\frac{2}{21}$$ The ratios are equal and therefore these are in proportion Middle terms = 42 persons , Rs.8 Extreme terms = 4 persons, Rs.84", "# How do you solve 243=27r? Jun 9, 2017 $r = 9$ #### Explanation: Rewriting equation so $r$ is on the left hand side for convenience, $27 r = 243$ To isolate $r$, divide both sides by $27$, $\\frac{27 r}{27} = \\frac{243}{27}$ $r = 9$", "## Elementary Geometry for College Students (6th Edition) We can multiply by $n$ assuming that $n\\ne 0$ because it's the denominator of a fraction: $360 + 135n = 180n$ We subtract $135n$ from both sides: $360 = 180n - 135n = 45n$ We divide by 45: $360\\div 45 = 8 = n$", "+ ar^3 ar^4 = 5 r^5 = $\\frac{1215}{5}$ r^5 = 243 r = $\\sqrt[5]{243}$ r = 3 That works out. Thank you very much.", "# Solve for r 4/r=5/20 4r=520 Cancel the common factor of 5 and 20. Factor 5 out of 5. 4r=5(1)20 Cancel the common factors. Factor 5 out of 20. 4r=5⋅15⋅4 Cancel the common factor. 4r=5⋅15⋅4 Rewrite the expression. 4r=14 4r=14 4r=14 Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction. 4⋅4=r⋅1 Solve the equation for r. Rewrite the equation as r⋅1=4⋅4. r⋅1=4⋅4 Multiply r by 1. r=4⋅4 Multiply 4 by 4. r=16 r=16 Solve for r 4/r=5/20 ### Solving MATH problems We can solve all math problems. Get help on the web or with our math app Scroll to top", "# 1966 AHSME Problems/Problem 39 ## Problem In base $R_1$ the expanded fraction $F_1$ becomes $.373737\\cdots$, and the expanded fraction $F_2$ becomes $.737373\\cdots$. In base $R_2$ fraction $F_1$, when expanded, becomes $.252525\\cdots$, while the fraction $F_2$ becomes $.525252\\cdots$. The sum of $R_1$ and $R_2$, each written in the base ten, is: $\\text{(A) } 24 \\quad \\text{(B) } 22 \\quad \\text{(C) } 21 \\quad \\text{(D) } 20 \\quad \\text{(E) } 19$ ## Solution First, let's write $F_1$ as a proper fraction in base $R_1$. To do that, note that: $F_1=0.373737\\dots$ Multiplying this equation on both sides $R_1^2$, we get: $R_1^2F_1=37.373737\\dots$ Subtracting the first equation from the second one, we get: $R_1^2F_1-F_1=37\\\\F_1(R_1^2-1)=37\\\\F_1=\\frac{3R_1+7}{R_1^2-1}$ Using a very similar method as above, we can see that: $F_1=\\frac{3R_1+7}{R_1^2-1}=\\frac{2R_2+5}{R_2^2-1}$ and $F_2=\\frac{7R_1+3}{R_1^2-1}=\\frac{5R_2+2}{R_2^2-1}$. Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because $R_1,R_2>7$): $\\frac{3F_1+7}{7F_1+3}=\\frac{2F_2+5}{5F_2+2}\\\\15F_1F_2+6F_1+35F_2+14=14F_1F_2+35F_1+6F_2+15\\\\F_1F_2-29F_1+29F_2=1\\\\\\left(F_1+29\\right)\\left(F_2-29\\right)=-840$ Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions. Notice that $(F_2-29)$ has to be a negative factor of 840. We need to plug in values of $F_2 > 7$. 840 divides -21, so we plug in 8 to check. Luckily, when $F_2 = 8$, we see that $F_1=11$, and furthermore, 11+8=19 is one", "Multiply both sides by 2: 2.5 = (x - 10)", "to get rid of the fractions. you still have fractions here, i'm not sure what you did • Feb 11th 2008, 12:36 AM Jhevon Quote: Originally Posted by Mr Rayon Is this right so far? 2(9x+28) 5(7x-6) 2 --------- = -------- - 10 10 10 Ten as in each denominator. you're not getting what i'm saying... start with $\\frac {9x + 28}5 = \\frac {7x - 6}2 - 5$ multiply through by 10, we get: $2(9x + 28) = 5(7x - 6) - 50$ now continue • Feb 11th 2008, 12:45 AM Mr Rayon 2(9x+28)= 5(7x-6)-50 18x+56=35x-30-50 56+30+50=35x-18x 136=17x 8=x I get it now. Thanks" ]

[ "Basic College Mathematics (10th Edition) $r=-7$ We are given the equation $-8r=56$. We can solve for r by dividing both sides by -8. $\\frac{-8r}{-8}=\\frac{56}{-8}$ Therefore, $r=-7$.", "factors to $(r+s)(r-s)$. We know that $r+s=8$, so $r-s=\\frac{24}8=3$. Now we can solve for $r$ by adding the two equations we just got to see that $2r=11$, or $r=\\frac{11}2$. We now solve for $x$. We know that $r^2+(2x)^2=49$, so $(2x)^2=49-\\left(\\frac{11}2\\right)^2=\\frac{75}4$ and $2x=\\frac{5\\sqrt3}2$. We multiply both sides of this equation by $12$ to get $24x=30\\sqrt3$. However, the area of hexagon $ABQCDP$ is $24x$, so the answer is $30\\sqrt 3$, or answer choice $\\boxed{B}$.", "+0 0 55 3 +248 Solve for r: $$\\frac{r+3}{r-2} = \\frac{r-1}{r+1}.$$ Thanks! Jun 10, 2019 #1 +2 Solve for r: (r + 3)/(r - 2) = (r - 1)/(r + 1) Cross multiply: (r + 1) (r + 3) = (r - 2) (r - 1) Expand out terms of the left hand side: r^2 + 4 r + 3 = (r - 2) (r - 1) Expand out terms of the right hand side: r^2 + 4 r + 3 = r^2 - 3 r + 2 Subtract r^2 - 3 r + 2 from both sides: 7 r + 1 = 0 Subtract 1 from both sides: 7 r = -1 Divide both sides by 7: r = -1/7 Jun 10, 2019 #2 +248 +2 Thanks! sinclairdragon428 Jun 10, 2019 #3 +721 +2 (r+3)/(r-2)=(r-1)(r+1) (r+3)(r+1)=(r-1)(r-2) r^2+4r+3=r^2-3r+2 7r=-1 r=-1/7 You are very welcome! :P Jun 10, 2019", "# 1966 AHSME Problems/Problem 39 ## Problem In base $R_1$ the expanded fraction $F_1$ becomes $.373737\\cdots$, and the expanded fraction $F_2$ becomes $.737373\\cdots$. In base $R_2$ fraction $F_1$, when expanded, becomes $.252525\\cdots$, while the fraction $F_2$ becomes $.525252\\cdots$. The sum of $R_1$ and $R_2$, each written in the base ten, is: $\\text{(A) } 24 \\quad \\text{(B) } 22 \\quad \\text{(C) } 21 \\quad \\text{(D) } 20 \\quad \\text{(E) } 19$ ## Solution First, let's write $F_1$ as a proper fraction in base $R_1$. To do that, note that: $F_1=0.373737\\dots$ Multiplying this equation on both sides $R_1^2$, we get: $R_1^2F_1=37.373737\\dots$ Subtracting the first equation from the second one, we get: $R_1^2F_1-F_1=37\\\\F_1(R_1^2-1)=37\\\\F_1=\\frac{3R_1+7}{R_1^2-1}$ Using a very similar method as above, we can see that: $F_1=\\frac{3R_1+7}{R_1^2-1}=\\frac{2R_2+5}{R_2^2-1}$ and $F_2=\\frac{7R_1+3}{R_1^2-1}=\\frac{5R_2+2}{R_2^2-1}$. Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because $R_1,R_2>7$): $\\frac{3F_1+7}{7F_1+3}=\\frac{2F_2+5}{5F_2+2}\\\\15F_1F_2+6F_1+35F_2+14=14F_1F_2+35F_1+6F_2+15\\\\F_1F_2-29F_1+29F_2=1\\\\\\left(F_1+29\\right)\\left(F_2-29\\right)=-840$ Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions. Notice that $(F_2-29)$ has to be a negative factor of 840. We need to plug in values of $F_2 > 7$. 840 divides -21, so we plug in 8 to check. Luckily, when $F_2 = 8$, we see that $F_1=11$, and furthermore, 11+8=19 is one", "330, 90$, for an answer of $\\boxed{420}$. ## Solution 2 As above, we know from Vieta's that the roots of $p(x)$ are $r$, $s$, and $-r-s$. Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$. Then $rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a$ and $rs(-r-s) = -b$ from $p(x)$ and $(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a$ and $(r+4)(s-3)(-r-s-1)=-(b + 240)$ from $q(x)$. From these equations, we can write that $rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2$, and simplifying gives us $2s-5r-13=0$ or $s = \\frac{5r+13}{2}$. We now move to the other two equations. We see that we can cancel a negative from both sides to get $rs(r+s) = b$ and $(r+4)(s-3)(r+s+1)=b + 240$. Subtracting the first from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding and simplifying, substituting $s = \\frac{5r+13}{2}$ and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back in to get $b = (-5)(-6)(-5-6) = -330$ or $b = (1)(9)(1 + 9) = 90$. Then the answer is $|-330|+|90| = \\boxed{420}$. ## See also 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •", "r_2 = \\frac{2A}{C}.$$ Add them up to solve for $r_1,$ then subtract or substitute to solve for $r_2.$ – user2468 Jul 31 '12 at 2:45 Finally, it's $$R_o = \\frac{1}{2}\\left(\\frac{C}{2\\pi} + \\frac{2A}{C} \\right)\\\\ R_i = \\frac{1}{2}\\left(\\frac{C}{2\\pi} - \\frac{2A}{C} \\right).$$ – user2468 Jul 31 '12 at 2:51 We have $$r_1^2 - r_2^2 = (r_1 + r_2)(r_1 - r_2) = \\frac{A}{\\pi}$$ and $$r_1 + r_2 = \\frac{C}{2\\pi}.$$ So $$\\frac{C}{2\\pi}(r_1 - r_2) = \\frac{A}{\\pi}.$$ Cancel the $\\pi$'s from both sides, and move $\\frac{C}{2}$ into the other side, we get $$r_1 - r_2 = \\frac{2A}{C}.$$ – user2468 Jul 31 '12 at 4:13", "### Home > AC > Chapter 5 > Lesson 5.1.4 > Problem5-44 5-44. Set up a proportion with the information given. $\\frac{3}{4}=\\frac{x}{500}$ Solve for x. Multiply both sides by 500. x = 375 No, he will not make enough baskets.", "### Home > INT1 > Chapter 4 > Lesson 4.1.4 > Problem4-41 4-41. Solve each equation by rewriting, undoing, or looking inside. 1. $25^2=125^{x+1}$ $x=\\frac{1}{3}$ 1. $\\frac { x } { 5 } + \\frac { x - 1 } { 3 } = 2$ Multiply by a common multiple to cancel out the fractions.", "find $r$ such that $$\\frac{1 - r^m}{r^m(1 - r^n)} = \\frac ST,$$ then use this value of $r$ in the original equations and solve for $a$: $$a = \\frac{1-r}{1 - r^m} S$$ There is software that can solve for $r$ in an equation like the one above. For example, for the example you drew, with $m=2$ and $n=4$, \\begin{align} \\frac{a(1 - r^2)}{1-r} &= 80, \\\\ \\frac{ar^2(1 - r^4)}{1-r} &= 20, \\end{align} and we have a solution for $a \\approx 54.9794,$ $r \\approx 0.45509$, which gives you intervals of lengths approximately $54.9794, 25.0206, 11.3866, 5.1819, 2.3582$, and $1.0732$. (The last four add up to $19.9999$ due to rounding error.) For $m=2$, $n=10$ we get $a \\approx 55.2765,$ $r\\approx 0.447271$. As you may notice, a relatively large change in $n$ is accommodated by a relatively small change in $r$; we end up packing a lot of very short intervals into a little space at the end. -", "n = – 7 + 5 = – 2 [Transposing – 5 to R.H.S.] Thus, n = – 2 is the required solution. (d) We have – 4(2 + x) = 8 or $$\\frac { -4(2+x) }{ -4 }$$ = $$\\frac { 8 }{ -4 }$$ [Dividing both sides by – 4] or 2 + x = – 2 x = – 2 – 2 [Transposing 2 to R.H.S.] x = – 4 Thus, x = – 4 is the required solution. (e) We have 4(2 – x) = 8 or 2 – x = $$\\frac { 8 }{ 4 }$$ =2 [Dividing both sides by 4] or – x = 2 – 2 [Transposing 2 to R.H.S.] or – x = 0 or x – 0 Thus, x = 0 is the required solution. Question 3. Solve the following equations: (a) 4 = 5(p – 2) (b) – 4 = 5(p – 2) (c) 16 = 4 + 3(t + 2) (d) 4 + 5(p – 1) = 34 (e) 0 = 16 + 4(m – 6) Solution: (a) 4 = 5(p – 2) Interchanging the sides, we have 5(p – 2) = 4 Dividing both sides by 5, we have $$\\frac { 5(p-2) }{ 5 }$$ = $$\\frac { 4 }{ 5 }$$ or p –", "# How do you solve 10/3=(r+6)/(r+9)? Feb 27, 2017 The solution is $r = - \\frac{72}{7}$ or $- 10 \\frac{2}{7}$. #### Explanation: Solve: $\\frac{10}{3} = \\frac{r + 6}{r + 9}$ Cross multiply. $10 \\left(r + 9\\right) = 3 \\left(r + 6\\right)$ Expand. $10 r + 90 = 3 r + 18$ Subtract $3 r$ from both sides. $7 r + 90 = 18$ Subtract $90$ from both sides. $7 r = - 72$ Divide both sides by $7$. $r = - \\frac{72}{7}$ As a mixed fraction: $r = - 10 \\frac{2}{7}$", "to find that quartic polynomial is to use similar triangles. $\\frac{r}{r+3}=\\frac{9}{\\sqrt{(2r+3)^{2}+81}}$ Square both sides and bring everything to one side via cross multiplication: $r^{2}((2r+3)^{2}+81)-81(r+3)^{2}=0$ Expand out and get: $4r^{4}+12r^{3}+9r^{2}-486r-729=0$ If we solve this we find it has two real solutions, one is 9/2. That's the one. • February 5th 2008, 07:01 PM chrozer Alright thanx alot. I finally understand.", "# How do you solve 243=27r? Jun 9, 2017 $r = 9$ #### Explanation: Rewriting equation so $r$ is on the left hand side for convenience, $27 r = 243$ To isolate $r$, divide both sides by $27$, $\\frac{27 r}{27} = \\frac{243}{27}$ $r = 9$", "• $$a=R277-X115$$ • $$b=R115+X277$$ • $$c=R+277$$ • $$d=X+115$$ Plugging these into the formula above, we get $$\\frac{[R^2277-RX115+R277^2-X(277\\cdot115)]+[RX115+X^2277+R115^2+X(277\\cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$ Cancelling terms in the numerator gives $$=\\frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$ This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $$115^2=13225$$, and so on. In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first. The resulting system is best solved by machine: here's the computation in WolframAlpha.", "} 225 \\div 45 = 5. \\text{ So } 225° = 5 \\left( \\frac{\\pi}{4} \\right) = \\frac{5 \\pi}{4} .$", "# How do you solve the equation and identify any extraneous solutions for (r+4)/3=r/5? $\\frac{r + 4}{3} = \\frac{r}{5}$ $5 \\left(r + 4\\right) = 3 r$ $5 r + 20 = 3 r$ $5 r - 3 r = - 20$ $2 r = - 20$ $r = - 10$", "easier...) Therefore, $$m_1 = 81m_2$$ Now, going back to equation (2), G and x will cancel out and we replace $$r_1$$ with $$r-r_2$$ (from equation (1)) leaving us with: $$\\frac{81m_2}{(r-r_2)^2} = \\frac{m_2}{r_2^2}$$ $$m_2$$ will cancel out: $$\\frac{81}{(r - r_2)^2} = \\frac{1}{r_2^2}$$ $$81 = \\frac{(r - r_2)^2}{r_2^2}$$ i take the square root of both sides, and put the denominator on the left side $$9 r_2 = r - r_2$$ $$10 r_2 = r$$ $$r_2 = r / 10$$ $$r_2 = 3.84*10^{4}$$ Since $$r_1 = r - r_2$$ $$r_1 = 3.84*10^{5} - 3.84*10^{4}$$ $$r_1 = 3.456*10^{5}$$ my actual answer is a little bit different since i rounded differently for posting on here...but i'd like to know if the method is correct?", "original equation and check the answers – AmerYR Jan 14 at 3:25 • @Ameryr That's why you have a plot of the original equation, so you can confirm that there are only two solutions, not 4 – Andrei Jan 14 at 3:28 Hint: $$2a-45^\\circ=180^\\circ n+(-1)^n\\arcsin\\dfrac{\\sqrt3}2$$ If $$n$$ is odd$$=2m+1$$(say) $$2a-45=360m+180-60$$ But $$-180-45\\le2a-45\\le180-45$$ $$-225\\le360m+120\\le135$$ $$?\\le m\\le?$$ What if $$n$$ is even $$=2m$$(say) • Sorry, but wat do i do after gettong the bounds in this solution, and it does seem a bit more tedious. – SuperMage1 Jan 14 at 3:03 • @SuperMage1, Please let me know if u find one better – lab bhattacharjee Jan 14 at 3:24 $$\\sin(2a-45) = \\frac{\\sqrt{3}}{2}$$ $$2a-45 = 60 + 360n \\Rightarrow a = 52.5 + 180n$$ $$2a-45 = 120+ 360n \\Rightarrow a = 82.5 + 180n$$ Either $$a = 52.5$$ or $$a = 82.5$$. If you square $$\\sin(4a) = \\frac{-1}{2}$$ $$4a = 210 + 360n \\Rightarrow a = 52.5 + 90n$$ $$4a = 330 + 360n \\Rightarrow a = 82.5 + 90n$$ First two solution $$a = 52.5$$ $$a= 52.5 - 90 = -38.5$$ The other two $$a = 82.5 , 82.5 - 90 = -7.5$$ In both cases the second solution is rejected see Checking $$\\sin(-15) - \\cos(-15) = \\frac{-\\sqrt{6}}{2}$$ which is wrong same for the other", "+0 # Solve for : 0 109 2 Solve for $$r$$: $$\\frac{r-45}{2} = \\frac{3-2r}{5}.$$ Jan 2, 2020 ### 2+0 Answers #1 +64 +2 Hi, when you multiply both sides by 10, you get $$5(r-45)=2(3-2r)$$ because $$10 \\div 2 = 5 \\text{ and } 10 \\div 5 = 2$$ . When you expand the equation you should get $$5r-225=6-4r$$. Moving the variables to one side and constants to another you get $$9r=231$$. Dividing both sides by 9 gets us $$r= 231\\div 9$$ which equals to $$25 \\frac{2}{3}$$. -MPS101 #2 +109563 +1 Thanks, MPS !!!!! BTW.....welcome.....!!! CPhill Jan 2, 2020 edited by CPhill Jan 2, 2020", "# Solve the following equation: (x^2-2)/3+((x^2-1)/5)^2=7/9(x^2-2)? May 3, 2016 $x = - \\sqrt{11} , - \\frac{\\sqrt{19}}{3} , \\frac{\\sqrt{19}}{3} , \\sqrt{11}$ This explanation gives a rather in-depth method of determining the steps to finding possible factors into which to rewrite a quadratic-type equation so that it is solvable without the quadratic equation and/or a calculator. #### Explanation: First square the term on the left hand side of the equation. $\\frac{{x}^{2} - 2}{3} + {\\left({x}^{2} - 1\\right)}^{2} / 25 = \\frac{7}{9} \\left({x}^{2} - 2\\right)$ Expand the squared binomial. Recall that ${\\left({x}^{2} - 1\\right)}^{2} = \\left({x}^{2} - 1\\right) \\left({x}^{2} - 1\\right)$. $\\frac{{x}^{2} - 2}{3} + \\frac{{x}^{4} - 2 {x}^{2} + 1}{25} = \\frac{7}{9} \\left({x}^{2} - 2\\right)$ We can clear out the fractions by multiplying the equation by the least common denominator of $3 , 25 ,$ and $9 ,$ which is $225$. Note that $225 = {3}^{2} \\cdot {5}^{2}$, so $\\frac{225}{3} = 75$, $\\frac{225}{25} = 9$, and $\\frac{225}{9} = 25$. Multiplying through by $225$ gives: $75 \\left({x}^{2} - 2\\right) + 9 \\left({x}^{4} - 2 {x}^{2} + 1\\right) = 25 \\left(7\\right) \\left({x}^{2} - 2\\right)$ Distribute each multiplicative constant. $75 {x}^{2} - 150 + 9 {x}^{4} - 18 {x}^{2} + 9 = 175 {x}^{2} - 350$ Move all terms to one side and reorder the equation. $9 {x}^{4} - 118 {x}^{2} + 209" ]

[ "# Pre-Algebra FREE Sample Practice Questions Preparing for the Pre-Algebra test? To do your best on the Pre-Algebra test, you need to review and practice real Pre-Algebra questions. There’s nothing like working on Math sample questions to hone your math skills and put you more at ease when taking the Pre-Algebra test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your Pre-Algebra test. Check out our sample Pre-Algebra practice questions to find out what areas you need to practice more before taking the Pre-Algebra test! Start preparing for the Pre-Algebra test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice. $13.99 Satisfied 125 Students ## 10 Sample Pre-Algebra Practice Questions 1- Jason needs a score of 75 average in his writing class to pass. On his first 4 exams, he earned scores of 68, 72, 85, and 90. What is the minimum score Jason can earn on his fifth and final test to pass? __________ 2- Anita’s trick–or–treat bag contains 12 pieces of chocolate, 18 suckers, 18 pieces of gum, 24 pieces of", "## College Algebra (6th Edition) The student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A. To solve this exercise, we must first model the average grade in mathematical terms: $$Grade_{average} = \\frac{Grade_{1} + Grade_{2} +...Grade{(n)}}{n}$$ Since we have a total of 4 grades given, but the final examination counts as 2 exams, we have a total $n = 6$. We can now write the following: $$Grade_{average} = \\frac{86 + 88 + 92 + 84 + 2E}{6}$$ where $E$ represents the final examination. Since an A is defined as 90% or greater, we can finally model this as so: $$Grade_{average}\\geq 90$$ $$\\frac{86 + 88 + 92 + 84 + 2E}{6} \\geq 90$$ By solving for $E$: $$86 + 88 + 92 + 84 + 2E \\geq 90\\times 6$$ $$350 + 2E \\geq 540$$ $$2E \\geq 190$$ $$E \\geq 95$$ This means that the student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A.", "## Algebra 2 (1st Edition) We can attend at most $10$ sessions. Let $x$ be the number of swim classes attended. Then our inequality is: $50+5x\\leq100\\\\5x\\leq50\\\\x\\leq10$ Thus we can attend at most $10$ sessions.", "if $$p$$ is the minimum required score on the third exam for an A we will have the following equation. $\\frac{{196 + p}}{{350}} = 0.9$ This is a linear equation that we will need to solve for $$p$$. $196 + p = 0.9\\left( {350} \\right) = 315\\hspace{0.25in}\\,\\,\\,\\, \\Rightarrow \\hspace{0.25in}\\,\\,\\,p = 315 - 196 = 119$ So, the minimum required score on the third exam is 119. This is a problem since the exam is worth only 100 points. In other words, the student will not be getting an A in the Algebra class. Now let’s check if the student will get a B. In this case the minimum percentage is 0.8. So, to find the minimum required score on the third exam for a B we will need to solve, $\\frac{{196 + p}}{{350}} = 0.8$ Solving this for $$p$$ gives, $196 + p = 0.8\\left( {350} \\right) = 280\\hspace{0.25in}\\,\\,\\,\\, \\Rightarrow \\hspace{0.25in}\\,\\,\\,p = 280 - 196 = 84$ So, it is possible for the student to get a B in the class. All that the student will need to do is get at least an 84 on the third exam. Example 2 We want to build a set of shelves. The width of the set of shelves needs to be 4 times the height of the set of shelves", "## Algebra 2 (1st Edition) A score of at least $93\\%$ is required ($100\\%$ is the upper boundary). Let $x$ be the final exam score. Then our inequality is: $0.8\\cdot83+0.2x\\geq85\\\\66.4+0.2x\\geq85\\\\0.2x\\geq18.6\\\\x\\geq 93$ Thus a score of at least $93\\%$ is required ($100\\%$ is the upper boundary).", "Jonathan's class must have $x+.25x$ student (the same amount as Jennifer's plus another $25\\%$ more). Allison's class must be $x+9$. But I know the total students must add to 82. So $$x+(x+.25x)+(x+9)=82$$ Can you take it from there?", "## Elementary Algebra Let x represent the score for the fourth exam, and y represent the score for the fifth exam. His average will be $\\frac{96+90+94+x+y}{5}$. Because he must have an average higher than 92 for all five exams, we solve the following inequality. $\\frac{96+90+94+x+y}{5}$ $\\gt$ 92 Multiply both sides by 5. 96+90+94+x+y $\\gt$ 460 280 + x + y $\\gt$ 460 x + y $\\gt$ 180 The sum of the last two exams must be greater than 180. To find their average, we divide the sum by 2. Their average is $\\frac{x+y}{2}$. $\\frac{x+y}{2}$ $\\gt$ 90 So, their average needs to be greater than 90 for him to have an average higher than 92 for all five exams.", "## Elementary Algebra She must receive a score of 89 or higher on the exam. The solution set is {x|x $\\geq$ 89}. Let x represent the score on the fifth exam. Her average is $\\frac{83+89+78+86+x}{5}$. Because her average should be 85 or higher, we solve the following inequality. Average $\\geq$ 85 $\\frac{83+89+78+86+x}{5}$ $\\geq$ 85 Multiply both sides by 5. 83+89+78+86+x $\\geq$ 425 336 + x $\\geq$ 425 x $\\geq$ 89 She must receive a score of 89 or higher on the exam. The solution set is {x|x $\\geq$ 89}.", "Free Version Moderate # Honors for the Top Students SATMAT-O7KRVN The grades for Mr. O'Brien's mathematics students are shown in the table above. His classes are honored on an electronic sign in the main foyer for each quarter that more than half of his students earn $A's$ and/or $B's$. For how many quarters were his classes honored? A $1$ B $2$ C $3$ D $4$", "## College Algebra 7th Edition We know that 6 students scored 100, 3 students scored 60 and the rest (25-6-3=16 students) scored a mystery amount ($x$). The average of the 25 students is 84: $\\displaystyle \\frac{6*100+3*60+16*x}{25}=84$ $600+180+16x=84*25=2100$ $16x=1320$ $x=\\frac{1320}{16}=82.5$ So the last 16 students must have an average score of 82.5%.", "# Thread: Algebra questions part 3 1. ## Algebra questions part 3 Need Help! Would like To know How to solve these..... Would like to know step by step(formula) Numerical Skills/Prealgebra 10. If the total cost of x apples is b cents, what is a general formula for the cost, in cents, of y apples? A. b/xy B. x/by C. xy/b D. by/x E. bx/y 11. On a math test, 12 students earned an A. This number is exactly 25% of the total number of students in the class. How many students are in the class? A. 15 B. 16 C. 21 D. 30 E. 48 12. This year, 75% of the graduating class of Harriet Tubman High School had taken at least 8 math courses. Of the remaining class members, 60% had taken 6 or 7 math courses. What percent of the graduating class had taken fewer than 6 math courses? A. 0% B. 10% C. 15% D. 30% E. 45% 2. Question 10... The answer to this would be $\\frac{b}{x}\\times y$. Dividing the cost by the number of apples $x$ works out the cost per single apple. Then multiplying this value by $y$ gives the cost of $y$ apples. For question 11, since 25% is a quarter of 100%, 12 students must be a quarter of", "## Elementary Algebra Let the score on the fourth text be $x$. Since the average of the four tests must be 88 and the formula of an average is Average=(Sum of numbers)/(Number of numbers), we write the equation as follows: $88=\\frac{85+90+86+x}{4}$ $88(4)=85+90+86+x$ $352=261+x$ $352-261=x$ $x=91$ Therefore, the score on the 4th test must be 91 or higher in order to have an average of 88 or higher.", "16 \\text { or } 3 ( 1 - 2 x ) < - 15$$ 23. Jerry scored $$90, 85, 92$$, and $$76$$ on the first four algebra exams. What must he score on the fifth exam so that his average is at least $$80$$? 24. If $$6$$ degrees less than $$3$$ times an angle is between $$90$$ degrees and $$180$$ degrees, then what are the bounds of the original angle? 1. $$( - \\infty , 5 )$$; Figure 1.E.5 3. $$[ 6 , \\infty )$$; Figure 1.E.6 5. $$( - \\infty , 13 )$$; Figure 1.E.7 7. $$\\varnothing$$; Figure 1.E.8 9. $$\\left[ - \\frac { 4 } { 5 } , \\infty \\right)$$; Figure 1.E.9 11. $$[ - 3,2 )$$; Figure 1.E.10 13. $$\\left( - 1 , \\frac { 3 } { 2 } \\right)$$; Figure 1.E.11 15. $$\\left( \\frac { 1 } { 2 } , \\frac { 9 } { 2 } \\right)$$; Figure 1.E.12 17. $$\\left( \\frac { 11 } { 4 } , 5 \\right)$$; Figure 1.E.13 19. $$\\left( - \\infty , \\frac { 1 } { 5 } \\right) \\cup ( 1 , \\infty )$$; Figure 1.E.14 21. $$\\mathbb { R }$$; Figure 1.E.15 23. Jerry must score at least $$57$$ on the fifth exam. ### Sample Exam Exercise $$\\PageIndex{19}$$ Simplify. 1.", "Question # Self-Check A student scores 82,96,91, and 92 on four college algebra exams.What score is needed on a fifth exam for the student to earn an average grade of 90? Upper level algebra Self-Check A student scores 82,96,91, and 92 on four college algebra exams.What score is needed on a fifth exam for the student to earn an average grade of 90? 2021-02-26 Given: The scores on the four college algebra exams are 82, 96, 91 and 92. Approach: Let’s assume that a student needs to score xon the fifth exam. We will use the following formula for calculating average. Average=Sum of scores/Number of exams Calculation: Average = Sum of scores/Number of exams $$90=\\frac{82+96+91+92+x}{5}$$ $$90=\\frac{361+x}{5}$$ $$90\\times 5=361=x$$ $$450=361+x$$ x=450-361 x=89 Final Statement A student needs to score 89 on the fifth college exam.", "Math Notes Subjects Algebra Solutions Topics || Problems A student in an algebra class has test grades of 81 and 92. What must he score on the third test in order to have an average of 90? Let $$x$$ be his grade on the third test. $$\\frac{81+92+x}{3} = 90$$ $$81 +92 +x = 270$$ $$173 + x = 270$$ $$x= 270-173$$ $$x = 97$$ Answer", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Published by Pearson # Chapter 2 - Equations, Inequalities, and Problem Solving - 2.4 Applications with Percent - 2.4 Exercise Set - Page 110: 66 #### Answer Addie still needs to take $125\\times0.2=25$ credit hours. #### Work Step by Step The total credit hours of instruction needed to obtain a bachelor's degree in journalism is $125$. Addie still needs to take $20$% of those hours: Addie still needs to take $125\\times0.2=25$ credit hours. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "Declan V. # Pre Calculus 11 - Arithmetic Sequences On Alfredo's 10th birthday, his mother began giving him a weekly allowance of $10 and each week the allowance would increase by$0.50 if he did all his homework. How many weeks would it take for the weekly allowance to become \\$50 supposing he did all his homework? I have done out all of the work to this question, and yet while my work is correct, my answer seems to be just a little bit off. Can someone please tell me what I am doing wrong? The correct answer should be 80, and yet I am getting 81. Tn = t1 + (n-1)d 50 = 10 + (n-1)0.50 50 = 10 + 0.5n - 0.5 50 = 9.5 + 0.5n 40.5 = 0.5n 40.5/0.5 = 0.5/0.5n and I get 81 equals n By:", "## Algebra 1 If a student must earn at least 24 credits, then the student's credits must be greater than or equal to 24. Let c be the number of credits. F says credits must be less than or equal to 24, which is not correct. If the student only had 23 credits, he would not be able to graduate. G states that credits must be greater than or equal to 24, which correctly describes the situation. H says 24 must be less than or equal to the number of credits, which is another way of saying that credits must be greater than or equal to 24. $24\\leq c\\longrightarrow$ subtract c and subtract 24 from each side $24-c-24\\leq c-c-24\\longrightarrow$ subtract $-c\\leq-24\\longrightarrow$ multiply each side by -1; reverse the inequality sign $-c\\times-1\\geq-24\\times-1\\longrightarrow$ multiply $c\\geq24$ I also correctly describes the situation. The graph indicates that c is greater than or equal to 24 because the arrow extends to the right, in the direction of increasing values. 24 is shown as included in the solution set by the solid circle at the endpoint.", "## Elementary Algebra She must get a score of 96 or higher on the fifth exam. The solution set is {x|x $\\geq$ 96} Let x represent the score on the fifth exam. Her average = $\\frac{ 86+88+89+91+x }{5}$ Because her average needs to be 90 or higher, we solve the following inequality. Average $\\geq$ 90 $\\frac{ 86+88+89+91+x }{5}$ $\\geq$ 90 Multiply both sides by 5. 86+88+89+91+x $\\geq$ 450 354 + x $\\geq$ 450 x $\\geq$ 96 She must get a score of 96 or higher on the fifth exam. The solution set is {x|x $\\geq$ 96}", "# Inspired by Alisa Meier Algebra Level 5 A school consists of 100 students that are in 10 classes. The students scored $$1, 2, 3, \\ldots, 100$$ in a test. The Head of Department gets the bright idea of rearranging the classes so as to maximize the average of the 10 class averages. What is the maximum average that can be achieved? Note: Each class needs to have at least 1 student. Inspiration ×" ]

[ "\\ \\ \\ \\ \\ \\ \\ \\ \\ \\frac{454}{1015}=0.4473$ What to make of this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from $1,015 to$454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from $1,015 to$454 represents a 55.27% decrease in income. So if someone is currently making $1,015 a week, and if his income is to decrease to$454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$.", "of this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to \\$454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when", "topics Replies Last post Similar Topics: 1 What percentage of families in the state have annual incomes over $50 3 09 Jul 2015, 02:56 2 What was Janet’s score on the fourth physics test she took? 4 12 Dec 2014, 06:41 If none of the students are ambidextrous, what percentage of 3 11 Apr 2014, 04:56 What percentage of the agents at Best Be Safe insurance 9 26 Aug 2011, 01:32 5 What percentage of steel workers in the country earns over$ 10 21 Aug 2007, 19:50 Display posts from previous: Sort by", "### Home > ACC7 > Chapter 6 Unit 7 > Lesson CC3: 6.2.2 > Problem6-63 6-63. The box plot at right shows the different grades (in percents) that students in Ms. Sanchez’s class earned on a recent test. 1. What was the median score on the test? What were the highest and lowest scores? Do you remember how to read a box plot? 1. Did most students earn a particular score? How do you know? Can this be determined from a box plot? 1. If Ms. Sanchez has $32$ students in her class, about how many students earned a grade of $80\\%$ or higher? About how many earned more than $90\\%$? Explain how you know. Since $80\\%$ is the median, half of the class earned $80\\%$ or higher and half earned lower than $80\\%$. $\\frac{1}{4}\\text{ of the class earned 90%}$ $16$ students earned $80\\%$ or higher, $8$ students earned $90\\%$ or higher 1. Can you tell if the scores between $80\\%$ and $90\\%$ were closer to $80\\%$ or closer to 90%? Explain. Can this be determined from a box plot?", "college. Thus, every entry that is $25$ or more is for a school where at least a quarter of the students are binge drinkers. If I counted correctly, only $16$ of the $150$ entries are below $25$: the one on the $0$ line, the $10$ on the $1$ line, and the first $5$ on the $2$ line. That means that at $134$ of the $150$ colleges in the survey, at least a quarter of the students were binge drinkers. I think that $134$ out of $150$ qualifies as ‘most’! The $4$% value is fairly far from the large number of points in the $30$s and $40$s, but it’s not separated from the rest of the data by an inordinately large gap: it’s just $7$ percentage points from the next datum, and we expect some spreading in the tails of the distribution. An $86$% value, on the other hand, would be separated from the rest of the data by $18$ percentage points, a much bigger gap. The larger size of the gap would show up visually, too: the empty $70$ line would between $68$ and $86$ would stand out quite clearly. Changing one of the $26$% values to $76$%, on the other hand, would make the upper end of the distribution look a bit more like the lower end:", "this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going", "this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going", "this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going", "percent and its \"category worth\" will give us a student's average. We can use this same equation, but this time call the first quarter's average 90%, since that's what the student is aiming for. Then all we need do is solve for the grade on the final, which we'll call f for simplicity: average = 90% = 17% + 9.3% + 26.1% + (.3*f) + 10% 90% = 62.4% + (.3*f) 27.6% = .3*f 92% = f The student must, therefore, score at least 92% on the final to raise her average to the 90% level. Another way to think of this would be that without the final, we have a total of $$17 + 9.3 + 26.1 + 10 = 62.4$$ points, so we need to get $$90 – 62.4 = 27.6$$ more points from the final. Since the points we get are 0.30 times the grade, the grade has to be $$27.6\\div 0.30 = 92$$. There are about as many ways of calculating averages as there are teachers. At the beginning of the school year, make sure to find out how your teachers go about their calculations so you don't wind up with completely different grades when report cards are given out! I’ll add one more thought: In the middle of the term, you can find", "# current affairs quiz 5 1 . The average of five numbers is 57.8. The average of the first and the second number is 77.5 and the average of the fourth and the fifth number is 46. What is the third number? 2 . Mr Nair's monthly salary is Rs. 22,500. He took a loan of Rs. 30,000 on simple interest for 3 years at the rate of 5 p.c.p.a. The amount that he will be paying as simple interest in 3 years is what percent of his monthly salary? 3 . If the numerator of a certain fraction is increased by 100% and the denominator is increased by 200% the new fraction thus formed is $4\\over21$ . What is the original fraction? 4 . In how many different ways can the letters of the word 'SIMPLE' be arranged? 5 . 52% students from a college participated in a survey. What is the ratio of the number of students who did not participate in the survey to the number of students who participated? 6 . How much will be the compound interest to be paid on a principal amount of Rs. 53,000 after 2 years at the rate of 4 pcpa? 7 . The area of a rectangle is twice to the area of a triangle. The perimeter", "the last quarter, it should be 10*1.25*3 = 37.5. Total = 324+37.5 = 361.5, average for the entire year = 361.5/12 = 30.125 * 10^6$ as total earnings. Now per the question, 'what are the average annual earnings per share based on the number of shares at the end of the year', should be: 30.125*10^6/(5+10)*10^6 = 2.0083 That's not correct. Total for 9 months was 5*7.2=36, not 5*7.2*9. _________________ Current Student Joined: 20 Mar 2014 Posts: 2685 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) ### Show Tags 31 May 2015, 05:35 Bunuel, shouldnt it be a weighted average instead of the regular average that you have calculated? Math Expert Joined: 02 Sep 2009 Posts: 43380 ### Show Tags 31 May 2015, 05:40 Engr2012 wrote: Bunuel, shouldnt it be a weighted average instead of the regular average that you have calculated? We ARE calculating weighted average. Earnings for 9 months - $$5*7.2=36$$; During the final quarter, the number of publicly listed shares was increased to 10, so earnings for 4th quarter - $$10*1.25=12.5$$; Total earning for a year - $$36+12.5=48.5$$; The average annual earnings per share - $$\\frac{48.5}{10}=4.85$$. Hope it's clear. _________________ Intern Joined: 06 Apr 2015 Posts: 35 ### Show Tags 02 Jun 2015, 03:17 Sorry", "(Seaward & Kemp, 2000), participants were asked to indicate how much they expected to be earning annually five years after graduation. The range was $17,000 to$1,000,000, with an average of $71,179. We recoded the value of future income using a four-point scale: 1(lowest -$45,000; 25.2%), 2($45,000 -$60,000; 29.4%), 3($60,000-$75,000; 18.8%), and 4(over $75,000 -; 26.6%). Overconfidence. Overconfidence was measured as the difference between a student’s subjective knowledge and objective knowledge regarding several financial topics. Subjective financial knowledge was measured by asking participants to rate the degree to which they understood seven financial concepts (e.g., interpretation on credit reports, return rate of saving bonds) on a 5-point scale ranging from 1 (very little) to 5 (quite a lot). A higher score represented a higher level of subjective financial knowledge (α=.88). Objective knowledge was initially computed as the percentage of correct answers out of 15 True/False questions related to the understanding of money management, credit, and saving (Hilgert, Hogarth & Beverly, 2003) (e.g., making payments late on your bills can make getting a loan more difficult). For this study, percentage scores were classified into 5 levels, ranging from 1 (0 to 20%) to 5 (80% to 100%), with a higher level indicating higher objective financial knowledge. We subtracted the level of objective knowledge from the level of subjective financial knowledge", "of Percents In the following exercises, write each percent as a ratio. 1. In $2014,2014,$ the unemployment rate for those with only a high school degree was $6.0%.6.0%.$ 2. In $2015,2015,$ among the unemployed, $29%29%$ were long-term unemployed. 3. The unemployment rate for those with Bachelor's degrees was $3.2%3.2%$ in $2014.2014.$ 4. The unemployment rate in Michigan in $20142014$ was $7.3%.7.3%.$ In the following exercises, write as 1. a ratio and 2. a percent 5. $5757$ out of $100100$ nursing candidates received their degree at a community college. 6. $8080$ out of $100100$ firefighters and law enforcement officers were educated at a community college. 7. $4242$ out of $100100$ first-time freshmen students attend a community college. 8. $7171$ out of $100100$ full-time community college faculty have a master's degree. Convert Percents to Fractions and Decimals In the following exercises, convert each percent to a fraction and simplify all fractions. 9. $4%4%$ 10. $8%8%$ 11. $17%17%$ 12. $19%19%$ 13. $52%52%$ 14. $78%78%$ 15. $125%125%$ 16. $135%135%$ 17. $37.5%37.5%$ 18. $42.5%42.5%$ 19. $18.4%18.4%$ 20. $46.4%46.4%$ 21. $912%912%$ 22. $812%812%$ 23. $513%513%$ 24. $623%623%$ In the following exercises, convert each percent to a decimal. 25. $5%5%$ 26. $9%9%$ 27. $1%1%$ 28. $2%2%$ 29. $63%63%$ 30. $71%71%$ 31. $40%40%$ 32. $50%50%$ 33. $115%115%$ 34. $125%125%$ 35. $150%150%$ 36. $250%250%$ 37. $21.4%21.4%$ 38.", "40,000. 4,000,000 is 10 times 400,000. Thus, 400,000 = 10 $\\times$ 40,000, and there are about 10 times as many fourth graders in Texas as there are in Mississippi. Also, 4,000,000 = 10 $\\times$ 400,000, and there are about 10 times as many fourth graders in the US as there are in Texas. Finally, to go from 4,000 to 4,000,000, we have to multiply by 10 three times. We see that $$10\\times10\\times10=10\\times100=1000$$ So there are about 1,000 times as many fourth graders in the US as there are in Washington DC.", "college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going from $Q$ to $P$. Otherwise it is", "college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in this case). Note that $0.4473-1=-0.5527$. Multiplying this by 100, we get negative 55.27%. This says that going from$1,015 to $454 represents a 55.27% decrease in income. So if someone is currently making$1,015 a week, and if his income is to decrease to $454 a week, this represents a 55.27% decrease in income. To summarize, when we compare two quantities $P$ and $Q$ with $Q$ as a baseline, we can calculate the ratio $\\displaystyle R=\\frac{P}{Q}$. The ratio $R$ gives information about the quantity $P$ as a percentage of the baseline quantity $Q$. The quantity $R-1$ is the percentage change going from the baseline quantity $Q$ to the quantity $P$. If $R-1$ is a positive number, then it is a percentage increase when going from $Q$ to $P$. Otherwise it is", "Note that the answer in (2) is just one less than the ratio in (1). Thus the ratio (1) above has another interpretation. Subtracting one from the ratio (1) gives the percent increase going from$454 to $1,015. The increase is 1.236 (or 123.6% after multiplying by 100). For example, if someone is currently making$454 a week and if his income is to increase to $1,015 a week, this represents a 123.6% raise! This is the average premium of a college education, i.e, the additional amount a college grad can expect to make. We can also flip the ratio in (1) and obtain the following ratio. $\\displaystyle (3) \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\frac{454}{1015}=0.4473$ What to make of this ratio? Now the baseline is the college graduate. It tells us that the average earning of a high school dropout is only 0.4473 (or 44.73%) of the average earning of a college grad. For each one dollar earned by a college grad, the high school dropout only makes about 44.73 cents. For each one hundred dollars earned by a college grad, the high school dropout makes about fourty four dollars and seventy three cents. If we subtract one from the ratio in (3), we get the percentage change going from$1,015 to $454 (percentage decrease in", "restaurant and her bill is$12. She would like to leave a 15% tip for the waiter. So much money should she leave is let it be x so x = $$\\frac{15}{100}$$ X $12 =$1.8. Question 4. A town spends 43% of its annual budget on education. If the town’s budget is one million dollars, how much will be spent on education? ___________________ $430,000 on education, Explanation: Given a town spends 43% of its annual budget on education. If the town’s budget is one million dollars, Much will be spent on education if it x amount is x = $$\\frac{43}{100}$$ X 1,000,000 =$430,000. Question 5. Matt hopes to score 120 points Points this season. He has scored 80% scored of his goal so far. Fill in the double number line diagram to find out how many points he has scored so far and write the answer. Explanation: Given Matt hopes to score 120 points Points this season. He has scored 80% scored of his goal so far. Filled in the double number line diagram to find out how many points he has scored so far as 20% points scored is 120 X $$\\frac{20}{100}$$ = 24, 40% points scored is 120 X $$\\frac{40}{100}$$ = 48, 60% points scored is 120 X $$\\frac{60}{100}$$ = 72, 80% points scored is 120 X", "# Simple question 1. May 17, 2005 let's say i had 115% for my online hw which is worth 16% of the final grade and 4% for my written hw which is worth 8% of the final grade. how would i find my total percent for both online+hw together? im thinking (115/2 + 115 + 2)/2 = 87.25% for both written hw and online together? it seems wrong 2. May 17, 2005 ### whozum For n divisions of the total 100%: $$Total\\ Grade= \\sum^n_1 \\frac{(Percentage\\ worth)(Percentage\\ earned)}{10000}$$ $$= \\frac{(115)(16)+(4)(8)}{10000} = 18.72$$ percent of your entire grade. Out of those two sections alone you have 77.22% . That 4% in homework is killing you. Last edited: May 17, 2005 3. May 17, 2005 ### quasar987 115*0.16 + *0.08 = 18.7/100 (i.e. those two hw got you 18.7 points on your final grade) Or, and I think this is what you're looking for, you got 18.7 points out of a possible 24, meaning an average grade of 78%. Last edited: May 17, 2005", "per year for three consecutive​ years, then the economy shrank by 21% over the three-year period. I need to find out how much the economy actually shrank by Percentage Changes Percent Increase 09/24/16 what is the percent change from 1.8 to 2.4? i need to know the increase and the decrease. Percentage Changes 08/29/16 Percentage problems in a factory X out of every Y items produced were found to be defective. If 5% of the items produced during the day were defective, what was the total number of items produced that day? Percentage Changes 08/23/16 Statistical test for frequency data changes over time with one group of subjects? I have collected data over 8 years for 44 students on the number of episodes of physical aggression that have occurred with each student for each of the 8 years. I have calculated the percentage... more Percentage Changes 04/25/16 Employers way -- $5920 X .9 =$5328 X .6 = $3196.8 My way --$5920 X .55 = $3256 I have a %60 split with %10 taken off the top in which we split in half, so basically I lose %5 of that. So... more Percentage Changes 04/06/16 How much money are you now making? you are paid$17.50 per hour. since you just had great review your boss gave you an 8%" ]

[ "if $$p$$ is the minimum required score on the third exam for an A we will have the following equation. $\\frac{{196 + p}}{{350}} = 0.9$ This is a linear equation that we will need to solve for $$p$$. $196 + p = 0.9\\left( {350} \\right) = 315\\hspace{0.25in}\\,\\,\\,\\, \\Rightarrow \\hspace{0.25in}\\,\\,\\,p = 315 - 196 = 119$ So, the minimum required score on the third exam is 119. This is a problem since the exam is worth only 100 points. In other words, the student will not be getting an A in the Algebra class. Now let’s check if the student will get a B. In this case the minimum percentage is 0.8. So, to find the minimum required score on the third exam for a B we will need to solve, $\\frac{{196 + p}}{{350}} = 0.8$ Solving this for $$p$$ gives, $196 + p = 0.8\\left( {350} \\right) = 280\\hspace{0.25in}\\,\\,\\,\\, \\Rightarrow \\hspace{0.25in}\\,\\,\\,p = 280 - 196 = 84$ So, it is possible for the student to get a B in the class. All that the student will need to do is get at least an 84 on the third exam. Example 2 We want to build a set of shelves. The width of the set of shelves needs to be 4 times the height of the set of shelves", "## College Algebra (6th Edition) The student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A. To solve this exercise, we must first model the average grade in mathematical terms: $$Grade_{average} = \\frac{Grade_{1} + Grade_{2} +...Grade{(n)}}{n}$$ Since we have a total of 4 grades given, but the final examination counts as 2 exams, we have a total $n = 6$. We can now write the following: $$Grade_{average} = \\frac{86 + 88 + 92 + 84 + 2E}{6}$$ where $E$ represents the final examination. Since an A is defined as 90% or greater, we can finally model this as so: $$Grade_{average}\\geq 90$$ $$\\frac{86 + 88 + 92 + 84 + 2E}{6} \\geq 90$$ By solving for $E$: $$86 + 88 + 92 + 84 + 2E \\geq 90\\times 6$$ $$350 + 2E \\geq 540$$ $$2E \\geq 190$$ $$E \\geq 95$$ This means that the student must achieve a grade of $95$% $or$ $higher$ in order to pass the course with a final grade of A.", "# Pre-Algebra FREE Sample Practice Questions Preparing for the Pre-Algebra test? To do your best on the Pre-Algebra test, you need to review and practice real Pre-Algebra questions. There’s nothing like working on Math sample questions to hone your math skills and put you more at ease when taking the Pre-Algebra test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your Pre-Algebra test. Check out our sample Pre-Algebra practice questions to find out what areas you need to practice more before taking the Pre-Algebra test! Start preparing for the Pre-Algebra test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice. $13.99 Satisfied 125 Students ## 10 Sample Pre-Algebra Practice Questions 1- Jason needs a score of 75 average in his writing class to pass. On his first 4 exams, he earned scores of 68, 72, 85, and 90. What is the minimum score Jason can earn on his fifth and final test to pass? __________ 2- Anita’s trick–or–treat bag contains 12 pieces of chocolate, 18 suckers, 18 pieces of gum, 24 pieces of", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Published by Pearson # Chapter 2 - Equations, Inequalities, and Problem Solving - 2.4 Applications with Percent - 2.4 Exercise Set - Page 110: 66 #### Answer Addie still needs to take $125\\times0.2=25$ credit hours. #### Work Step by Step The total credit hours of instruction needed to obtain a bachelor's degree in journalism is $125$. Addie still needs to take $20$% of those hours: Addie still needs to take $125\\times0.2=25$ credit hours. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "Question # Self-Check A student scores 82,96,91, and 92 on four college algebra exams.What score is needed on a fifth exam for the student to earn an average grade of 90? Upper level algebra Self-Check A student scores 82,96,91, and 92 on four college algebra exams.What score is needed on a fifth exam for the student to earn an average grade of 90? 2021-02-26 Given: The scores on the four college algebra exams are 82, 96, 91 and 92. Approach: Let’s assume that a student needs to score xon the fifth exam. We will use the following formula for calculating average. Average=Sum of scores/Number of exams Calculation: Average = Sum of scores/Number of exams $$90=\\frac{82+96+91+92+x}{5}$$ $$90=\\frac{361+x}{5}$$ $$90\\times 5=361=x$$ $$450=361+x$$ x=450-361 x=89 Final Statement A student needs to score 89 on the fifth college exam.", "## Elementary Algebra Let the score on the fourth text be $x$. Since the average of the four tests must be 88 and the formula of an average is Average=(Sum of numbers)/(Number of numbers), we write the equation as follows: $88=\\frac{85+90+86+x}{4}$ $88(4)=85+90+86+x$ $352=261+x$ $352-261=x$ $x=91$ Therefore, the score on the 4th test must be 91 or higher in order to have an average of 88 or higher.", "## Elementary Algebra Let x represent the score for the fourth exam, and y represent the score for the fifth exam. His average will be $\\frac{96+90+94+x+y}{5}$. Because he must have an average higher than 92 for all five exams, we solve the following inequality. $\\frac{96+90+94+x+y}{5}$ $\\gt$ 92 Multiply both sides by 5. 96+90+94+x+y $\\gt$ 460 280 + x + y $\\gt$ 460 x + y $\\gt$ 180 The sum of the last two exams must be greater than 180. To find their average, we divide the sum by 2. Their average is $\\frac{x+y}{2}$. $\\frac{x+y}{2}$ $\\gt$ 90 So, their average needs to be greater than 90 for him to have an average higher than 92 for all five exams.", "Free Version Moderate # Honors for the Top Students SATMAT-O7KRVN The grades for Mr. O'Brien's mathematics students are shown in the table above. His classes are honored on an electronic sign in the main foyer for each quarter that more than half of his students earn $A's$ and/or $B's$. For how many quarters were his classes honored? A $1$ B $2$ C $3$ D $4$", "do I need to get on the final to get a $90$?” Answer: The change in the average needs to be $+2$, so the student needs to get a grade $+2/0.2 = +10$ points higher than his/her current average. So the grade on the final needs to be $88 + 10 = 98$. Seen another way, we’re solving the algebra problem $88(0.8) + x(0.2) = 90$ Let me solve this in an unorthodox way: $88(0.8) + x(0.2) = 88 + 2$ $88(0.8) + x(0.2) = 88(0.8+0.2) + 2$ $88(0.8) + x(0.2) = 88(0.8) + 88(0.2) + 2$ $x(0.2) = 88(0.2) + 2$ $x = \\displaystyle \\frac{88(0.2)}{0.2} + \\frac{2}{0.2}$ $x = 88 + \\displaystyle \\frac{2}{0.2}$ This last line matches the solution found in the previous paragraph, $x = 88 + 10 = 98$. # Engaging students: Computing inverse functions In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place. I plan to share some of the best", "16 \\text { or } 3 ( 1 - 2 x ) < - 15$$ 23. Jerry scored $$90, 85, 92$$, and $$76$$ on the first four algebra exams. What must he score on the fifth exam so that his average is at least $$80$$? 24. If $$6$$ degrees less than $$3$$ times an angle is between $$90$$ degrees and $$180$$ degrees, then what are the bounds of the original angle? 1. $$( - \\infty , 5 )$$; Figure 1.E.5 3. $$[ 6 , \\infty )$$; Figure 1.E.6 5. $$( - \\infty , 13 )$$; Figure 1.E.7 7. $$\\varnothing$$; Figure 1.E.8 9. $$\\left[ - \\frac { 4 } { 5 } , \\infty \\right)$$; Figure 1.E.9 11. $$[ - 3,2 )$$; Figure 1.E.10 13. $$\\left( - 1 , \\frac { 3 } { 2 } \\right)$$; Figure 1.E.11 15. $$\\left( \\frac { 1 } { 2 } , \\frac { 9 } { 2 } \\right)$$; Figure 1.E.12 17. $$\\left( \\frac { 11 } { 4 } , 5 \\right)$$; Figure 1.E.13 19. $$\\left( - \\infty , \\frac { 1 } { 5 } \\right) \\cup ( 1 , \\infty )$$; Figure 1.E.14 21. $$\\mathbb { R }$$; Figure 1.E.15 23. Jerry must score at least $$57$$ on the fifth exam. ### Sample Exam Exercise $$\\PageIndex{19}$$ Simplify. 1.", "minimum score of 8 on the speaking portion of the I.E.L.T.S. or a minimum score of 26 on the speaking portion of the TOEFL-iBT For continuing students during each of the three most recent quarters of enrollment: - Completion of 8 units or more of upper division or graduate level credit courses. - A letter grade of C,S or above in all courses completed. - No more than two incomplete (I) grades - A cumulative GPA of 3.1 or higher - Satisfactory progress toward degree objectives - Teaching Assistant Appointment Periods and Limitations Teaching Assistantships are for one quarter, two quarters, or an academic year. Graduate students who have not advanced to candidacy for the doctorate, may be appointed as a Teaching Assistant or Teaching Associate for a maximum of 12 quarters including the full period of the current or proposed appointment. Following advancement to candidacy, a doctoral student can be appointed to an additional 6 quarters for a total maximum of 18 appointment quarters. The quarters are counted regardless of appointment percentage. ### Fee-Offsets for Teaching Assistant Appointments The Office of Graduate Studies will pay the Graduate Student Health Insurance Fee and a partial fee remission of 100% of the annual educational and registration fees for TAs with appointments of 25% or more for an entire quarter.", "get on the final to get a $90$?” Answer: The change in the average needs to be $+2$, so the student needs to get a grade $+2/0.2 = +10$ points higher than his/her current average. So the grade on the final needs to be $88 + 10 = 98$. Seen another way, we’re solving the algebra problem $88(0.8) + x(0.2) = 90$ Let me solve this in an unorthodox way: $88(0.8) + x(0.2) = 88 + 2$ $88(0.8) + x(0.2) = 88(0.8+0.2) + 2$ $88(0.8) + x(0.2) = 88(0.8) + 88(0.2) + 2$ $x(0.2) = 88(0.2) + 2$ $x = \\displaystyle \\frac{88(0.2)}{0.2} + \\frac{2}{0.2}$ $x = 88 + \\displaystyle \\frac{2}{0.2}$ This last line matches the solution found in the previous paragraph, $x = 88 + 10 = 98$.", "symbolic notation and fractions where needed. Give your answer in the form of a comma separated list.) College Math 08/27/19 #### { x | x € N and x ≤ 17,000,000} Determine whether the set is finite or infinite College Math 08/27/19 #### { x | x € N and x ≥ 900} Determine whether the set is finite or infinite College Math 08/23/19 #### Find what’s needed for 80 average Final exam which will count as 2 test, Amelia has test scores of 75, 84, 72, 59, and 96. What score does she need on final to have an average score of 80? College Math Algebra 2 06/10/19 #### odd integers in college algebra The sum of two consecutive odd integers is -160. Find the two integers College Math Algebra 2 Statistics 05/08/19 #### Question 20 Math Need Help in January, each of 70 people purchased a$500 washing machine in February 10% fewer customers purchases the same washing machine that had increased in price by 20% what was the change in sales... more College Math Math Algebra 2 05/08/19 #### Question 29 Math Question need help with if the cost to mail a letter is 49 cents for mail weighing up to one ounce and 27 cents for each additional ounce or a fraction of an ounce,", "# Thread: Algebra questions part 3 1. ## Algebra questions part 3 Need Help! Would like To know How to solve these..... Would like to know step by step(formula) Numerical Skills/Prealgebra 10. If the total cost of x apples is b cents, what is a general formula for the cost, in cents, of y apples? A. b/xy B. x/by C. xy/b D. by/x E. bx/y 11. On a math test, 12 students earned an A. This number is exactly 25% of the total number of students in the class. How many students are in the class? A. 15 B. 16 C. 21 D. 30 E. 48 12. This year, 75% of the graduating class of Harriet Tubman High School had taken at least 8 math courses. Of the remaining class members, 60% had taken 6 or 7 math courses. What percent of the graduating class had taken fewer than 6 math courses? A. 0% B. 10% C. 15% D. 30% E. 45% 2. Question 10... The answer to this would be $\\frac{b}{x}\\times y$. Dividing the cost by the number of apples $x$ works out the cost per single apple. Then multiplying this value by $y$ gives the cost of $y$ apples. For question 11, since 25% is a quarter of 100%, 12 students must be a quarter of", "## Algebra 1 a) $467/745$ b) $720/1700$ c) $506/1953$ d) $506/883$ All populations are in thousands. All probabilities have been divided by one thousand. a) Associates degrees = $288 + 467 = 755$ female recipients of associates degrees = 467 $467/745$ b) bachelors degrees = $720 + 980 = 1700$ male recipients of bachelors degrees = 720 $720/1700$ c) female recipients = $467+980+ 506 = 1953$ advanced degree and female recipients = 506 $506/1953$ d) advanced degree = $377 + 506 = 883$ female recipients of advanced degrees = 506 $506/883$", "if applicable, focus in on the portion of the graph that is relevant. Example 1: The following clustered column graphs compare the percentage of students who earned marks in four different ranges in a standardised test, for two different classes over two years: Are the following statements true or false? 1. In $$2019$$ more than three-quarters of students in class $$\\textrm{A}$$ had a mark over $$50\\%$$. 2. The distribution of marks was more similar for the classes in $$2019$$ than in $$2020$$. 3. The percentage of students who scored between $$26\\%$$ and $$50\\%$$ increased by $$11\\%$$ between $$2019$$ and $$2020$$ for Class $$B$$. Solution: Working and answers are as follows: 1. True. Since the question asks about $$2019$$, we can focus on the first graph and ignore the second for this question. Furthermore, since it refers to Class $$\\textrm{A}$$ we can just look at the blue columns. The question asks whether more than three-quarters of the relevant students had a mark over $$50\\%$$, which equate to students in the third and fourth ranges on the graph (i.e. $$51 - 75\\%$$ and $$76 - 100\\%$$). Adding the percentage of students in each of these columns together gives $$47\\% + 29\\% = 76\\%$$, which is just over three-quarters (since three-quarters is equivalent to $$75\\%$$). 2. True. The column pattern for", "the Average of all applicants =~ 686 Sufficient Answer C Math Expert Joined: 02 Aug 2009 Posts: 6802 Re: At a particular business school which requires applicants to submit GM [#permalink] ### Show Tags 15 Jan 2018, 07:31 Prachikh wrote: But we still dont have the number of students admitted or rejected? Can someone please explain? How is the relationship enough calculate the new average? Thanks! Hi, you dont require the exact number here as you have the ratio as we are working on average of two sets of items.. lets take a simple scenario.. 10 % of a class has average of 90 marks and 90% has an average of 50 marks.. combined average = $$50+40*\\frac{100-90}{100}=50+4=54$$ Now say the strength is 50. so 10% that is 5 have 90 and remaining 45 have 50.. combined average = $$\\frac{5*90+45*50}{50}=54$$ _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor Director Status: Learning stage Joined: 01 Oct 2017 Posts: 859 WE: Supply Chain Management (Energy and Utilities) At a particular business school which requires applicants to submit GM [#permalink] ### Show Tags 11 Jun 2018, 06:56 1 Bunuel wrote: At a particular business school which requires applicants to submit GMAT scores, the difference between the", "+25225 +3 Ms. Forsythe gave the same algebra test to her three classes. The first class averaged 80%, the second class averaged 85%, and the third 89%. Together, the first two classes averaged 83%, and the second and third classes together averaged 87%. What was the average for all three classes combined? $$\\text{The students in the first class = s_1 } \\\\ \\text{The students in the second class = s_2 } \\\\ \\text{The students in the third class = s_3 } \\\\ \\text{The sum of the points in the first class = p_1 } \\\\ \\text{The sum of the points in the second class = p_2 } \\\\ \\text{The sum of the points in the third class = p_3 } \\\\ \\text{The maximal points of the test = p }$$ $$\\begin{array}{|lrcl|lrcl||lrcl|} \\hline & \\dfrac{\\frac{p_1}{s_1}} {p} &=& 80\\% & & \\dfrac{\\frac{p_2}{s_2}} {p} &=& 85\\% & & \\dfrac{\\frac{p_3}{s_3}} {p} &=& 89\\% \\\\ (1) & p_1 &=& 80\\%ps_1 & (2) & p_2 &=& 85\\%ps_2 &(3) & p_3 &=& 89\\%ps_3 \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\dfrac{\\frac{p_1+p_2}{s_1+s_2}} {p} &=& 83\\% \\quad | \\quad p_1 = 80\\%ps_1,\\ p_2 = 85\\%ps_2 \\\\ \\dfrac{\\frac{80\\%ps_1+85\\%ps_2}{s_1+s_2}} {p} &=& 83\\% \\\\ \\dfrac{80\\%ps_1+85\\%ps_2}{(s_1+s_2)p} &=& 83\\% \\\\ \\dfrac{80\\%s_1+85\\%s_2}{(s_1+s_2)} &=& 83\\% \\\\ 80\\%s_1+85\\%s_2 &=& 83\\% (s_1+s_2) \\\\ 80\\%s_1+85\\%s_2 &=& 83\\%s_1+83\\%s_2 \\\\ 85\\%s_2-83\\%s_2 &=& 83\\%s_1-80\\%s_1 \\\\ 2\\%s_2&=& 3\\%s_1 \\\\ 2s_2&=& 3s_1 \\\\ \\mathbf{s_2}", "## College Algebra 7th Edition We know that 6 students scored 100, 3 students scored 60 and the rest (25-6-3=16 students) scored a mystery amount ($x$). The average of the 25 students is 84: $\\displaystyle \\frac{6*100+3*60+16*x}{25}=84$ $600+180+16x=84*25=2100$ $16x=1320$ $x=\\frac{1320}{16}=82.5$ So the last 16 students must have an average score of 82.5%.", "fraction of the class play tennis? ☐A. $$10\\%$$ ☐B. $$15\\%$$ ☐C. $$20\\%$$ ☐D. $$40\\%$$ Best Grade 7 Common Core Math Workbook Resource for 2020 1- C The weight of 12.2 meters of this rope is: 12.2 $$×$$ 600 g = 7320 g 1 kg = 1000 g therefore, 7320 g $$÷$$ 1000 = 7.32 kg 2- 60 Jason needs an $$75\\%$$ average to pass for five exams. Therefore, the sum of 5 exams must be at lease $$5 × 75 = 375$$ The sum of 4 exams is: 68 + 72 + 85 + 90 = 315 The minimum score Jason can earn on his fifth and final test to pass is: 375 – 315 = 60 3- A It’s needed to have a ratio to find value of $$x$$. $$\\frac{45}{40}=\\frac{2x+4}{16}⇒ 40(2x+4)=45×16 ⇒ x=7$$ 4- D Use Pythagorean Theorem: $$a^2 + b^2 = c^2$$ $$62 + 82 = c^2 ⇒ 100 = c^2 ⇒ c = 10$$ 5- B To find the discount, multiply the number by ($$100\\% –$$ rate of discount). Therefore, for the first discount we get: (D) ($$100\\% – 20\\%) =$$ (D) (0.80) = 0.80 D For increase of $$10%$$: (0.80 D) ($$100\\% + 10\\%) =$$ (0.85 D) (1.10) = 0.88 D $$= 88\\%$$ of D 6- A Use PEMDAS (order of operation): $$[6 ×" ]

[ "Function A club is asking a company to print shirts. The quoted price is as follows: set up fee: $50 first 50 shirts are$7.99 each any additional shirts are \\$6.99 each. Find a formula for the... more Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "of the class. 2. The shirt is exclusive to the class this year, with “I survived” added at the beginning of the text, again at the request of the class. In previous years I was willing to let anyone buy a shirt, to lower the prices for the students in the class, by amortizing the setup fees. I’m also dealing with a different T-shirt company this year, at the suggestion of one of the students. I’m using BroPrints, a Santa Cruz company since 1994 (or 2000, depending on whether you count the start of the business or their opening a storefront). Their prices are considerably lower than The Print Gallery, who I used last year. I’m hoping the quality is good (the student who recommended them said that she had used them for several different orders and had good durability of the printing). I stopped using Sports Design after two sets of shirts (one in 2011 and one in 2014) had bad cracking of the printing, and an order was miscounted in 2014. Broprints has lower setup fees and lower per-shirt charges than the other companies I’ve dealt with, and the order is a little larger this year (35 shirts), so the per-shirt price should be the lowest yet (about \\$12 a shirt, and \\$15 for long sleeve).", "A maker of T-shirts must decide to build a large plant or a small plant. A small plant will incur an annual cost of $400000, with a production capacity of 100000 T-shirts per year. A large factory will incur an annual cost of$800000, with a production capacity of 200000 T-shirts per year. The profit per T-shirt is estimated as \\$20. Four levels of manufacturing demand are considered likely: 20000, 40000, 100000, and 200000 T-shirts per year. What formula is used to compute the payoff table?", "fee of$100 to process your photo? This would be a fixed cost, as well as a start-up cost. Even if you sold no tee shirts, you would still incur the $100 just to go into business. If you sold no shirts, you would have a$100 loss instead of a profit. However, if you once again sold $100 shirts, your profit after fixed costs would be the$400 from above less the $100 start-up fee, for a net profit of$300. What is the minimum amount of shirts you should expect to sell in order to decide to go into business? You will need to at least break-even, which means to cover the fixed costs plus expected marginal costs. Here, marginal profit will provide $4 per shirt, so it takes 25 shirts to cover the initial$100 investment. ## Other Important Quantities There are other quantities that can affect a firm’s decisions. Entry costs are the minimum cost to start a particular business. Such are closely related to fixed costs. Opportunity Cost is the amount of profit given up by not pursuing a course of action. Typical sources of opportunity costs are risk and interest rates. Risk is cost due to uncertainty. Interest rates reflect a combination of costs due to risk and sacrificed opportunities. In reality, they can also represent bargaining", "maximum profit of $19000 will be earned if 2000 shirt of$18 and 3000 shirts of \\$25 are produced and sold.", "for her basketball team. Shirts cost $10 each if you order 10 or fewer shirts and$9 each if you order 11 or more shirts.", "10 May 2018, 21:05 Let extra number of t shirt after 50 be x 12*50 + 10x=1500 10x=900 X=90 Total t-shirt=90+50=140 Posted from my mobile device Intern Joined: 24 Nov 2017 Posts: 34 Re: A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 10 May 2018, 22:51 Total charge for the order =$1500 Charge @ $12 per T shirt for the first 50 = 50 * 12 = 600 Charge @$10 per T shirt beyond the first 50 = 1500 - 600 = 900. Number of T Shirts in the order beyond the first 50 = 900/10 = 90. Total number of T shirts in the order = 50 + 90 = 140 Choice C SC Moderator Joined: 22 May 2016 Posts: 1680 A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 11 May 2018, 17:05 alanforde800Maximus wrote: A T-shirt supplier charges$12 per T-shirt for the first 50 T-shirts in an order and $10 per T-shirt for each additional T-shirt in the order. If the charge for an order of T-shirts was$1500, how many T-shirts were in the order? A) 90 B) 125 C) 140 D) 150 E) 200 The trap here is forgetting to account for or to \"add back\" the", "discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for$12, and made a $900 profit. 2. In addition to the$100 setup fee, the team paid $7 for each T-shirt. [Reveal] Spoiler: OA Last edited by guerrero25 on 30 Oct 2013, 16:01, edited 1 time in total. Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4677 Re: To raise funds, a racing team sold T-shirts imprinted with [#permalink] ### Show Tags 30 Oct 2013, 10:43 5 This post received KUDOS Expert's post 1 This post was BOOKMARKED guerrero25 wrote: To raise funds, a racing team sold T-shirts imprinted with the team's logo. The team paid their supplier a one-time setup fee of$100. Because they purchased at least 50 T-shirts, the team qualified for their supplier's quantity discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for $12, and made a$900 profit. 2. In addition to the $100 setup fee, the team paid$7 for each T-shirt. I do not have the OA with me ;i 'll update as soon as", "Question You are ordering T-shirt with your school’s mascot printed on them. Each T-shirt cost $4.75. The printer charges a setup fee of$30 and $2.50 to print each shirt. write two expressions to represents the total cost of printing n T-shirts please help ASAP if you can please explain Answers 1. thnahdat Answer: 2.50n + 30 each shirt (n) is$2.50 so to find the total for the shirts you would multiple $2.50 by (n) the number of shirts. The fee of the printer is only charged once so you would just add an extra$30 to the amount you got. n is the number of shirts they buy. It says that each shirt costs $4.75, and for each shirt they add an extra$30 and \\$2.50 for the setup fee and to peint each shirt. The word each shows that you have to times. You have to times n by each number or to make it shorter put all three numbers in parenthesis put the n ouside of the parenthesis.", "# need help on this problem! #### kelly.quintana ##### New member Joanne sells silk-screened T-shirts at community festivals and craft fairs. Her marginal cost to produce one T-shirt is $2.50. Her total cost to produce 60 T-shirts is$210, and she sells them for $9 each. a. Find the linear cost function for Joanne's T-shirt production. b. How many T-shirts must she produce and sell in order to break even? c. How many T-shirts must she produce and sell to make a profit of$900? #### Subhotosh Khan ##### Super Moderator Staff member Joanne sells silk-screened T-shirts at community festivals and craft fairs. Her marginal cost to produce one T-shirt is $2.50. Her total cost to produce 60 T-shirts is$210, and she sells them for $9 each. a. Find the linear cost function for Joanne's T-shirt production. b. How many T-shirts must she produce and sell in order to break even? c. How many T-shirts must she produce and sell to make a profit of$900? First calculate the fixed cost of production from 'marginal cost to produce one T-shirt is $2.50' and \" total cost to produce 60 T-shirts is$210\". Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at:", "profit. 2. In addition to the $100 setup fee, the team paid$7 for each T-shirt. Dear expert, mikemcgarry, GMATPrepNow I have a question about statement 2 : I think statement 2 is SUFFICIENT because of n=1. Why we don't plug n=1 in this equation, so we can get X? Thanks! _________________ There's an app for that - Steve Jobs. Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4677 Re: To raise funds, a racing team sold T-shirts imprinted with [#permalink] ### Show Tags 20 Oct 2017, 10:47 1 KUDOS Expert's post septwibowo wrote: guerrero25 wrote: To raise funds, a racing team sold T-shirts imprinted with the team's logo. The team paid their supplier a one-time setup fee of $100. Because they purchased at least 50 T-shirts, the team qualified for their supplier's quantity discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for$12, and made a $900 profit. 2. In addition to the$100 setup fee, the team paid $7 for each T-shirt. Dear expert, mikemcgarry, GMATPrepNow I have a question about statement 2 : I think statement 2 is SUFFICIENT because of n=1. Why we don't plug n=1 in this equation, so we", "effect, they will double the relevant range to allow for a maximum of $$160$$ inspections per shift, assuming the second QA inspector can inspect an additional 80 units per shift. The down side to this approach is that once the new QA inspector is hired, if demand falls again, the company will be incurring fixed costs that are unnecessary. For this reason, adding salaried personnel to address a short-term increase in demand is not a decision most businesses make. Step costs are best explained in the context of a business experiencing increases in activity beyond the relevant range. As an example, let’s return to Tony’s T-Shirts. Tony’s cost of operations and the associated relevant ranges are shown in Table $$\\PageIndex{4}$$. Table $$\\PageIndex{4}$$: Tony’s T-Shirts Cost Options Cost Type of Cost Relevant Range Lease on Screen-Printing Machine $2,000 per month Fixed 0–2,000 T-shirts per month Employee$10 per hour Variable 20 shirts per hour Tony’s Salary $2,500 per month Fixed N/A Screen-Printing Ink$0.25 per shirt Variable N/A Building Rent \\$1,500 per month Fixed 2 screen-printing machines and 2 employees As you can see, Tony has both fixed and variable costs associated with his business. His one screen-printing machine can only produce $$2,000$$ T-shirts per month and his current employee can produce $$20$$ shirts per hour ($$160$$ per $$8$$-hour work day).", "which you can read more about here. We got a lot out of the experience and I would absolutely do it again - but if you're not Collective approved you'll want to change the SEC dropdown to No. And now some final analysis: Our campaign was successful, but not wildly so. Thanks to Kickulatron, our total estimated cost for rewards came to less than 14% of what we raised, which I think is pretty ding-dong good. Even better, some of our actual costs will be lower, since I priced our rewards without quantity discounts. For example, Alpha patches will cost substantially less per item because we have a large enough order for a quantity discount. We also raised more than shown on the gross due to international shipping and some backers generously giving us money and selecting a lesser reward (or no reward at all). A special thank you to those backers! Probably the worst news is that we sold very few shirts, making the cost per shirt quite high. Design is the majority of the cost, so I suppose I could have just thrown a shirt together with art on hand and gotten the price per shirt down to \\$11 or so. But I feel that everything Jumpdrive does, says, and creates reflects on the game and", "best for sublimation because they van provide vivid and clearer colors of the prints when transferred with heat. Regardless of its average perfect make, it has one very vivid disadvantage, which is that it is heavy when compared to other shirts. It is a long sleeve shirt and not very iron friendly. It is sold between $7-$27 dollars on Amazon. Features • It consists of 50% polyester and 50% cotton. • It is easily machine washable or hand washable if the need arises. • It is a branded imported material. • It is knit weaved. • It has a flat collar which gives comfort. • It has a multi-dimensional use. 3.Augusta T-shirt overview The Augusta shirt is one of the most economical sublimation shirts for printing in 2022. It is sold at an unbelievable price of $8 and less than$19 on websites like Amazon and U-Buy. Regardless of its lower price compared to other sublimation shirts for printing in 2022, the Augusta shirt is very suitable for sublimation in terms of its manufacturing constituents. This shirt consists of one hundred percent polyester material, and it has little to no closure. When required to make quality sublimation prints on a less costly shirt material, the Augusta shirt is the best pick for this. It gives the users a perfect", "that manufactures customized t-shirts They have assigned the manufacturing costs to their first 4 jobs and have provided you with the following information for the month of January: Ma...", "buying the shirts. Make sure to check your sizing and delivery, if the shirt can be delivered to your country. The cost of the shirt, as mentioned in the first page, would be 17$without VAT (Value Added Tax) and 20$ with VAT (Value Added Tax). Some countries like US dont need to pay VAT, so they get it for 17$while countries which need to pay VAT like my country, and UK, etc, the cost would be 20$. The campaign will run for 21 days (thats what currently I am thinking) and need to sell at least 50 shirts so they can start production and shipping. They shirts will be shipped after production. Production takes max 3 days. If we sell less than 50 shirts, then the campaign will fail and I will to restart with with lower goal so we can meet minimum criteria and the shirts can be produced and shipped. Thats the info you need. I will post when I start the campaign. (my wifi is working slow) Last edited: #### CriticalCubing ##### Member The campaign is live now. You can buy till the end of this month, that is October 2014. I have put various links crediting everybody in this project Thanks to You All, if you did'nt give me ideas, I would have dissolved", "View Single Post 2019-05-16, 19:51 #13 ewmayer 2ω=0 Sep 2002 República de California 13·29·31 Posts Quote: Originally Posted by rogue I think that donors would prefer a t-shirt or poster with one of the Mersenne Primes printed on it. That does give me an idea. Someone could design a GIMPS t-shirt. The cost per t-shirt would likely be less than $5 per t-shirt, but could be sold for$15. I think that would get you a lot of donors. I'm more a fan of logo polos myself ... does that make me snooty?", "Ethan is the treasurer of the library club. He is responsible for managing all finances, purchases, and fundraisers for the club. Ethan needs to order T-shirts for all the members. Each T-shirt costs $15 and the shipping fee for the order is$9.99. Write an equation showing the total cost, y, for an order of x T-shirts.", "B) 125 C) 140 D) 150 E) 200 The trap here is forgetting to account for or to \"add back\" the first 50 I. Arithmetic How many T-shirts were there in an order charge of $1500? Cost 1:$12 per T-shirt for first 50 = $600 Cost 2:$10 per T-shirt, for any number $$x$$ T-shirts, AFTER the first 50 T-shirt order total: $1500 (1) 50 T-shirts cost$600 ($1500 -$600) = $900 remain (2)$900 = how many T-shirts? $900/$10 per shirt = 90 shirts Number of T-shirts in the order? 50 + 90 = 140 II. Algebra Let $$x$$ = # of T-shirts $$(12)(50) + (10)(x-50) = 1500$$ $$600 + 10x - 500 = 1500$$ $$10x + 100 = 1500$$ $$10x = 1400$$ $$x = 140$$ _________________ ___________________________________________________________________ For what are we born if not to aid one another? -- Ernest Hemingway Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3907 Location: United States (CA) Re: A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 14 May 2018, 16:53 alanforde800Maximus wrote: A T-shirt supplier charges$12 per T-shirt for the first 50 T-shirts in an order and $10 per T-shirt for each additional T-shirt in the order. If the charge for an order of", "straightforward. After all, you need to include not only the direct costs of each product, but also the indirect costs involved in running your business, reaching customers, and so on. So let’s look at an example to make it clearer. Let’s say I run a T-shirt business. My process involves buying plain T-shirts, adding my designs, and selling them online. I’m going to calculate my overall costs, using some purely hypothetical numbers. Each plain T-shirt costs me$2 to buy, and the printing adds another $3 per shirt. So the cost of the basic raw materials is$5, and on top of that, I spend on average 10 minutes of my own time, which I value at $30 an hour, so that’s another$5 in labour costs. But on top of that, I have fixed costs. I spend $45 a month on my design software, and$55 a month on running my website, email newsletter, etc. I spend $500 a month renting an office space, and$100 a month for other overheads. So that’s $700 overall in fixed costs. On average, I sell 350 T-shirts a month, so the cost of all those overheads is exactly$2 per shirt (700 / 350). That’s the beauty of hypothetical numbers! So right now, my cost per shirt is $12. So$12 is the minimum I can charge" ]

[ "{x + 1} \\right| \\le 0$$ for every $$x \\in {\\bf{R}}$$. Therefore, $$f\\left( x \\right) = - \\left| {x + 1} \\right| + 3 \\le 3$$ for every $$x \\in {\\bf{R}}$$. The maximum value of g is attained when $$\\left| {x + 1} \\right| = 0$$ \\begin{align}&\\left| {x + 1} \\right| = 0\\\\& \\Rightarrow \\; x = - 1\\end{align} Therefore, maximum value of \\begin{align}g &= g\\left( { - 1} \\right)\\\\ &= - \\left| { - 1 + 1} \\right| + 3\\\\& = 3\\end{align} Hence, the function $$g$$ does not have a minimum value. (iii) The given function is $$h\\left( x \\right) = \\sin \\left( {2x} \\right) + 5$$ We know that $$- 1 \\le \\sin 2x \\le 1$$ Therefore, \\begin{align}& \\Rightarrow \\; - 1 + 5 \\le \\sin 2x \\le 1 + 5\\\\ &\\Rightarrow \\; 4 \\le \\sin 2x + 5 \\le 6\\end{align} Hence, the maximum and minimum values of $$h$$ are 6 and 4, respectively. (iv) The given function is $$f\\left( x \\right) = \\left| {\\sin 4x + 3} \\right|$$ We know that $$- 1 \\le \\sin 4x \\le 1$$ Therefore, \\begin{align} &\\Rightarrow \\; 2 \\le \\sin 4x + 3 \\le 4\\\\ &\\Rightarrow \\; 2 \\le \\left| {\\sin 4x + 3} \\right| \\le 4\\end{align} Hence, the maximum and minimum values of $$f$$ are 4 and 2, respectively.", "\\sin x-\\cos x \\right)}^{2}} \\\\ \\end{align} One can go wrong while relating the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. (45 + x) has no effect on maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$. As $\\cos \\theta$ always lies in [-1,1]. So, one can confuse here to get the maximum value of ${{\\cos }^{4}}\\left( 45+x \\right)$.", "course, she must obtain an average of $$90$$ marks or more in five examinations. Therefore, \\begin{align}&\\frac{{87 + 92 + 94 + 95 + x}}{5} \\ge 90\\\\&\\Rightarrow\\quad \\frac{{368 + x}}{5} \\ge 90\\\\&\\Rightarrow\\quad 368 + x \\ge 450\\\\&\\Rightarrow\\quad x \\ge 450 - 368\\\\&\\Rightarrow \\quad x \\ge 82\\end{align} Thus, Sunita must obtain a minimum of $$82$$ marks in the fifth examination ## Chapter 6 Ex.6.1 Question 23 Find all pairs of consecutive odd positive integers both of which are smaller than $$10$$ such that their sum is more than $$11$$. ### Solution Let $$x$$ be the smaller of the two consecutive odd positive integers. Then, the other integer is $$\\left( {x + 2} \\right)$$ Since both the integers are smaller than $$10$$, \\begin{align}&\\quad x + 2 \\lt 10\\\\&\\Rightarrow\\quad x \\lt 10 - 2\\\\&\\Rightarrow\\quad x \\lt 8\\;\\;\\;\\; \\quad\\;\\qquad \\ldots \\left( 1 \\right)\\end{align} Also, the sum of the two integers is more than $$11$$. Therefore, \\begin{align}&\\quad x + \\left( {x + 2} \\right) \\gt 11\\\\&\\quad \\Rightarrow\\quad 2x + x \\gt 11\\\\&\\quad\\Rightarrow\\quad 2x \\gt 11 - 2\\\\&\\quad\\Rightarrow\\quad 2x \\gt 9\\\\&\\quad\\Rightarrow\\quad x \\gt \\frac{9}{2}\\\\&\\quad\\Rightarrow\\quad x \\gt 4.5\\quad\\qquad \\ldots \\left( 2 \\right)\\end{align} From ($$1$$) and ($$2$$), we obtain $4.5 \\lt x \\lt 8$ Since $$x$$ is an odd positive integer, then values of $$x$$ are $$5$$ and $$7$$. When $$x = 5$$, the pair is $$\\left(", "Chicago Cubs and modular arithmetic Find the smallest integer $$x>15$$ satisfying \\begin{align} x &\\equiv 15 \\pmod{77} \\\\ x &\\equiv 15 \\pmod{286}. \\end{align} × Problem Loading... Note Loading... Set Loading...", "+ 2\\\\&= 52 + 2\\\\&= 54\\\\\\end{aligned} Hence, the integers are $\\boxed{52}, \\boxed{53}$ and $\\boxed{54}.$", "\\right)$. 24. Find all pairs of consecutive even positive integers, both of which are larger than $\\mathrm{5}$ such that their sum is less than $\\mathrm{23}$. Ans: Let us assume $\\text{x}$ as the smallest number in two consecutive even integers. Then, the other odd integer will be $\\text{x+2}$. As it is given that both the integers are greater than $5$, we get $\\text{x}>\\text{5}..........\\left( 1 \\right)$ Also, it is given that the sum of the two integers is less than $23$. So, $\\text{x+}\\left( \\text{x}+2 \\right)<23$ $\\Rightarrow 2\\text{x}+2<23$ $\\Rightarrow 2\\text{x}<\\text{23}-2$ $\\Rightarrow 2\\text{x}<\\text{21}$ $\\Rightarrow \\text{x}<\\frac{21}{2}$ $\\Rightarrow \\text{x}<\\text{10}\\text{.5}$ Let us assume this as inequality $\\left( 2 \\right)$ $\\text{x}<10.5........\\left( 2 \\right)$ From $\\left( 1 \\right)$ and $\\left( 2 \\right)$, we obtain $5<\\text{x}<\\text{10}\\text{.5}$. Therefore, we can say that $\\text{x}$satisfies $\\text{6,8}$ and $\\text{10}$ values. Therefore, the pairs that are possible are $\\left( 6,8 \\right)$, $\\left( 8,10 \\right)$ and $\\left( 10,12 \\right)$. 25. The longest side of a triangle is $\\mathrm{3}$ times the shortest side and the third side is $\\mathrm{2}$cm shorter than the longest side. If the perimeter of the triangle is at least $\\mathrm{61}$ cm, find the minimum length of the shortest side. Ans: Let us assume the shortest side of the given triangle as $\\text{x}$. As it is given that the longest side is the three times the shortest side, we get longest side$\\text{=3x}$", "# Pi in combinatorial game theory Here, on slide 27, it says that $\\pi = \\{3, 25/8, 201/64, ... | 4, 7/2, 13/4, ... \\}$ The largest number on the left will be $3 + 1/8 + 1/64 + \\dots$ which I evaluated as \\begin{align} 2 + (1 + 1/8 + 1/64 + \\ldots) = 2 + 1/(1-1/8) = 3.142847142857143 \\end{align} Which is larger than $\\pi$? And the smallest number on the right I evaluated as \\begin{align} 4 - 1/2 - 1/4 - ... &= 4 + 1 - 1 - 1/2 - 1/4 - .... \\\\ &= 5 - (1 + 1/2 + 1/4 + ....) \\\\ &= 5 - 2 \\\\ &= 3 \\\\ \\end{align} Since the smallest number on the right is larger than the largest number on the right, shouldn't this not equate to a number? How does it equate to $\\pi$? Giving some extra details to supplement Dustan Levenstein's answer, the left hand side contains an infinite sequence of values $a_0, a_1, a_2, \\dots$ with $a_0 = 3, a_1 = 25/8, a_2 = 201/64$. How should the sequence be defined in general? One answer is that $a_n = c_n / 2^n$, where $c_n$ is the largest integer that is less than or equal to $2^n \\pi$. This insures that $a_n < \\pi$ for", "> 100000 \\Leftrightarrow n+1 > \\ln (100000) \\Leftrightarrow n > \\ln (100000) - 1 \\approx 10.512... \\end{align} Thus if $n ≥ 11$ we have that the $n^{\\mathrm{th}}$ partial sum $s_n$ has error from the actual sum $s$ of less than $E = 10^{-5}$.", "Concept. Integers the set of whole numbers and their opposites Negative Numbers numbers that are less than zero Positive Numbers numbers that are greater than zero Zero is a part of the set of integers, but is neither positive or negative ### Guided Practice Here is one for you to try on your own. Compare \\begin{align*}-\\frac{2}{5}\\end{align*} and \\begin{align*}-\\frac{6}{7}\\end{align*} Answer Remember, the closer a value is to zero, the greater the value. Remember this with integers? \\begin{align*}-4 < -1\\end{align*} This is a true statement. We must keep this in mind when working with negative fractions too. Negative six - sevenths is a larger fraction. Because it is negative, it is farther away from zero. It is the smaller value in this case. \\begin{align*}-\\frac{2}{5}\\end{align*} > \\begin{align*}-\\frac{6}{7}\\end{align*} This is the answer. ### Practice Directions: Compare each pair of values using <, > or =. 1. -.18 ____ -.27 2. -23 ____ -.98 3. -9 ____ -11 4. -18 ____ -29 5. -67 ____ -89 6. \\begin{align*}-\\frac{1}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{4}{5}\\end{align*} 7. \\begin{align*}-\\frac{3}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{3}\\end{align*} 8. \\begin{align*}-\\frac{5}{10} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{2}\\end{align*} 9. \\begin{align*}-\\frac{3}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -.75\\end{align*} 10. \\begin{align*}-\\frac{1}{4} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -.25\\end{align*} 11. \\begin{align*}-.25 \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{3}{4}\\end{align*} 12. \\begin{align*}-\\frac{18}{20} \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;} -\\frac{1}{2}\\end{align*} Directions Write the following integers in order from least to greatest. 13. -4, -12, -19, -8, 0, -2, -1 14. 5, 7, 23, 8, -9, -11 15. \\begin{align*}\\frac{-1}{2}, \\frac{-1}{4},", "Courses Courses for Kids Free study material Free LIVE classes More # Three consecutive positive integers are such that the sum of the square of the first and the product of other two is $46$, the smallest integer is: Last updated date: 18th Mar 2023 Total views: 304.5k Views today: 3.83k Verified 304.5k+ views Hint: Proper conversion of wordings to equation form is must. It is easy to solve such problems with variables, as they can be interrelated easily like $b=a+1$ to denote they are consecutive in nature. Here, three consecutive positive integers are such that the sum of the first and product of the other two is $46$. Let us consider these integers as $a,b$ and $c$ such that \\begin{align} & \\Rightarrow b=a+1 \\\\ & \\Rightarrow c=b+1 \\\\ \\end{align} Or, $\\Rightarrow c=\\left( a+1 \\right)+1=a+2...\\text{ }\\left( 1 \\right)$ Now, as per the given conditions, we have $\\Rightarrow {{a}^{2}}+bc=46$, where $a$ is first integer, $b$ is second integer and $c$ is third integer. Substituting the values of $b$ and $c$ from equation (1), we get \\begin{align} & \\Rightarrow {{a}^{2}}+bc=46 \\\\ & \\Rightarrow {{a}^{2}}+\\left( a+1 \\right)\\left( a+2 \\right)=46 \\\\ & \\Rightarrow {{a}^{2}}+\\left( a\\left( a+2 \\right)+1\\left( a+2 \\right) \\right)=46 \\\\ & \\Rightarrow {{a}^{2}}+\\left( {{a}^{2}}+2a+a+2 \\right)=46 \\\\ & \\Rightarrow {{a}^{2}}+{{a}^{2}}+3a+2=46 \\\\ & \\Rightarrow 2{{a}^{2}}+3a+2=46 \\\\ \\end{align} By transposing values, we get \\begin{align}", "there? Graph and Compare Integers Integers are the counting numbers (1, 2, 3...), the negative opposites of the counting numbers (-1, -2, -3...), and zero. There are an infinite number of integers and examples are 0, 3, 76, -2, -11, and 995. #### Example A Compare the numbers 2 and -5. When we plot numbers on a number line, the greatest number is farthest to the right, and the least is farthest to the left. In the diagram above, we can see that 2 is farther to the right on the number line than -5, so we say that 2 is greater than -5. We use the symbol “>\\begin{align*}>\\end{align*}” to mean “greater than”, so we can write 2>5\\begin{align*}2 > -5\\end{align*}. Classifying Rational Numbers When we divide an integer a\\begin{align*}a\\end{align*} by another integer b\\begin{align*}b\\end{align*} (as long as b\\begin{align*}b\\end{align*} is not zero) we get a rational number. It’s called this because it is the ratio of one number to another, and we can write it in fraction form as ab\\begin{align*}\\frac{a}{b}\\end{align*}. (You may recall that the top number in a fraction is called the numerator and the bottom number is called the denominator.) You can think of a rational number as a fraction of a cake. If you cut the cake into b\\begin{align*}b\\end{align*} slices, your share is a\\begin{align*}a\\end{align*} of those slices.", "zeros sit at $z_n = n \\pi$, with $n$ being any non-zero integer. Thus we find \\begin{align*} \\frac{\\sin(z)}{z} &= \\prod_{n \\neq 0} \\left(1 - \\frac{z}{\\pi n}\\right) e^{z/n\\pi} \\\\ &= \\prod_{n=1}^{\\infty} \\left(1 - \\frac{z^2}{\\pi^2 n^2}\\right). \\end{align*}", "positive integers are divisible by 6. The positive integers which are divisible by 10 are: 10, 20, 30………100 In the above series, first term (a) is 10 and common difference (d) is 10 so substituting the value of “a” as 10 and “d” as 10 and ${{T}_{n}}$ as 100 in eq. (1) we get, \\begin{align} & {{T}_{n}}=a+\\left( n-1 \\right)d \\\\ & \\Rightarrow 100=10+\\left( n-1 \\right)10 \\\\ & \\Rightarrow 100=10+10n-10 \\\\ & \\Rightarrow 100=10n \\\\ \\end{align} Dividing 10 on both the sides of the above equation we get, \\begin{align} & \\dfrac{100}{10}=n \\\\ & \\Rightarrow n=10 \\\\ \\end{align} 10 positive integers are divisible by 10. The positive integers till 100 which are divisible by 15 are: 15, 30, 45,……..90 In the above series, first term (a) is 15 and common difference (d) is 15 so substituting the value of “a” as 15 and “d” as 15 and ${{T}_{n}}$ as 90 in eq. (1) we get, \\begin{align} & {{T}_{n}}=a+\\left( n-1 \\right)d \\\\ & \\Rightarrow 90=15+\\left( n-1 \\right)15 \\\\ & \\Rightarrow 90=15+15n-15 \\\\ & \\Rightarrow 90=15n \\\\ \\end{align} Dividing 15 on both the sides we get, \\begin{align} & \\dfrac{90}{15}=n \\\\ & \\Rightarrow 6=n \\\\ \\end{align} 6 positive integers are divisible by 15. Adding the positive integers which are divisible by 6, 15, 10 we get, \\begin{align} & 16+10+6 \\\\ & =32 \\\\", "\\left \\lbrace \\begin{aligned} 6x(z -5x) & = 0\\\\ x^2 + y^2 + z^2 & = 35 \\end{aligned} \\right. So, either $$x = 0$$ or $$z =5x$$. • If $$x = 0$$, then $$y = 0$$ and $$z =\\pm \\sqrt{35}$$. • If $$z = 5x$$, then $$x^2 + (3x)^2 + (5x)^2 = 35$$ So, $$35x^2 = 35$$. So, $$x = \\pm 1$$. Therefore $$f$$ has possible extreme values at $$(1,3,5)$$ and $$(-1,-3,-5)$$ Evaluate $$f$$ at these six points, the largest is the maximum value of $$f$$ and the smallest is the minimum value of $$f$$.", "nearest integer. In particular, this minimum distance is a maximum if $5c$ is half an integer. Thus the largest possible minimum distance is $\\dfrac{1}{2\\sqrt{29}}$.", "# Q23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11. S seema garhwal Let x be smaller of two consecutive odd positive integers. Then the other integer is x+2. Both integers are smaller than 10. $\\therefore \\, \\, \\, x+2< 10$ $\\Rightarrow \\, \\, \\, \\, x< 10-2$ $\\Rightarrow \\, \\, \\, \\, x< 8$ Sum of both integers is more than 11. $\\therefore \\, \\, \\, x+(x+2)> 11$ $\\Rightarrow \\, \\, \\, (2x+2)> 11$ $\\Rightarrow \\, \\, \\, 2x> 11-2$ $\\Rightarrow \\, \\, \\, 2x> 9$ $\\Rightarrow \\, \\, \\, x> \\frac{9}{2}$ $\\Rightarrow \\, \\, \\, x> 4.5$ We conclude $\\, \\, \\, \\, x< 8$ and $\\, \\, \\, x> 4.5$ and x is odd integer number. x can be 5,7. The two pairs of consecutive odd positive integers are $(5,7)\\, \\, \\, and\\, \\, \\, (7,9)$. Exams Articles Questions", "= 0$ $\\Rightarrow (n-2) ( n^{2} – 7n – 2n + 14) = 0$ $\\Rightarrow (n -2) \\left[n (n-7) -2 (n – 7) \\right]$ $\\rightarrow (n – 2)(n – 2)(n – 7) = 0$ $\\Rightarrow \\boxed{n = 2, 2, 7}$ The $n = 7$, makes the whole expression zero. So, $n = 8$ is the smallest integer which make expression greater than zero. Put the value of $n = 8,$ in the equation $(1),$ we get $n^{3} – 11n^{2} + 32n – 28 > 0$ $\\Rightarrow (8)^{3} – 11 (8)^{2} + 32 (8) – 28 > 0$ $\\Rightarrow 512 – 704 + 256 – 28 > 0$ $\\Rightarrow 768 – 732 > 0$ $\\Rightarrow \\boxed{36 > 0}$ (Satisfied) $\\therefore$ The value of smallest integer is $8.$ Correct Answer $:8$ by 4k points 3 6 24 1 194 views", "3x$^2$ - 10x - 25 = 0 $\\Rightarrow$ 3x$^2$ - 15x + 5x - 25 = 0 $\\Rightarrow$ 3x(x - 5) + 5(x - 5) = 0 $\\Rightarrow$ (3x + 5)(x - 5) = 0 Step 3: Either 3x + 5 = 0 or x - 5 = 0 $\\Rightarrow$ 3x = - 5 or x = 5 $\\Rightarrow$ x = $\\frac{-5}{3}$ or x = 5 Hence, the number is $\\frac{-5}{3}$ or 5. Question 2: Find the largest of three consecutive odd integers, if the sum of the first and the third integers is 7 more than the second integer. Solution: Let the three consecutive odd integer be x, x + 2 and x + 4. That is, First integer = x Second integer = x + 2 Third integer = x + 4 Step 1: The problem states that: Sum of the first and third integers = Second integer + 7 $\\Rightarrow$ x + (x + 4) = (x + 2) + 7 $\\Rightarrow$ x + x + 4 = x + 2 + 7 Step 2: Combine the like terms: $\\Rightarrow$ 2x + 4 = x + 9 Subtract x from both side $\\Rightarrow$ 2x - x + 4 = x + 9 - x $\\Rightarrow$ x + 4 = 9 Subtract 4 from both sides", "5+1 \\\\ & \\therefore z=11 \\\\ \\end{align} For $p=6$ \\begin{align} & \\Rightarrow x=103-15\\times 6 \\\\ & \\therefore x=13 \\\\ \\end{align} \\begin{align} & \\Rightarrow y=13\\times 6-64 \\\\ & \\therefore y=14 \\\\ \\end{align} \\begin{align} & \\Rightarrow z=2\\times 6+1 \\\\ & \\therefore z=13 \\\\ \\end{align} Note: It is important to note that as there are 3 unknowns and only 2 equations we cannot get the actual solution. So, we need to check the possibilities which can be more than one by considering that the number of animals can only be positive. While assuming an integer and then writing all the variables in terms of it we should not neglect any of the terms because neglecting the condition so obtained changes accordingly so that we cannot get a possible solution.", "‘A’ in the course. Ans: Let us assume $\\text{x}$ as the fifth examination marked by Sunita. As it is given that to obtain ‘A’ one should obtain an average of $\\text{90}$ marks or more in five examinations. So, the expression is: $\\frac{87+92+94+95+\\text{x}}{5}\\geq90$ $\\Rightarrow \\frac{368+\\text{x}}{5}\\geq90$ $\\Rightarrow 368+\\text{x}\\geq450$ $\\Rightarrow \\text{x}\\geq450-368$ $\\Rightarrow \\text{x}\\geq82$ Therefore, we can say that sunita must obtain $\\text{82}$ or greater than it to obtain 23. Find all pairs of consecutive odd positive integers both of which are smaller than $\\mathrm{10}$ such that their sum is more than $\\mathrm{11}$. Ans: Let us assume $\\text{x}$ as the smallest number in two consecutive odd integers. Then, the other odd integer will be $\\text{x+2}$. As it is given that both the integers are smaller than $\\text{10}$, we get $\\text{x+2}<10$ $\\Rightarrow \\text{x}<\\text{10-2}$ $\\Rightarrow \\text{x}8$ Let us assume it as inequality $\\left( 1 \\right)$, we get $\\text{x}<8.............\\left( 1 \\right)$ Also, it is given that the sum of the two integers is more than $\\text{11}$. So, $\\text{x+}\\left( \\text{x}+2 \\right)>11$ $\\Rightarrow 2\\text{x}+2>11$ $\\Rightarrow 2\\text{x}>11-2$ $\\Rightarrow 2\\text{x}>\\text{9}$ $\\Rightarrow \\text{x}>\\frac{9}{2}$ $\\Rightarrow \\text{x}>\\text{4}\\text{.5}$ Let us assume this as inequality $\\left( 2 \\right)$ $\\text{x}>4.5........\\left( 2 \\right)$ Therefore, from inequality $\\left( 1 \\right)$ and $\\left( 2 \\right)$, we get that $\\text{x}$ can take the values of $5$ and $7$. Therefore, the required pairs are $\\left( 5,7 \\right)$ and $\\left( 7,9" ]

[ "Function A club is asking a company to print shirts. The quoted price is as follows: set up fee: $50 first 50 shirts are$7.99 each any additional shirts are \\$6.99 each. Find a formula for the... more Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "A maker of T-shirts must decide to build a large plant or a small plant. A small plant will incur an annual cost of $400000, with a production capacity of 100000 T-shirts per year. A large factory will incur an annual cost of$800000, with a production capacity of 200000 T-shirts per year. The profit per T-shirt is estimated as \\$20. Four levels of manufacturing demand are considered likely: 20000, 40000, 100000, and 200000 T-shirts per year. What formula is used to compute the payoff table?", "of the class. 2. The shirt is exclusive to the class this year, with “I survived” added at the beginning of the text, again at the request of the class. In previous years I was willing to let anyone buy a shirt, to lower the prices for the students in the class, by amortizing the setup fees. I’m also dealing with a different T-shirt company this year, at the suggestion of one of the students. I’m using BroPrints, a Santa Cruz company since 1994 (or 2000, depending on whether you count the start of the business or their opening a storefront). Their prices are considerably lower than The Print Gallery, who I used last year. I’m hoping the quality is good (the student who recommended them said that she had used them for several different orders and had good durability of the printing). I stopped using Sports Design after two sets of shirts (one in 2011 and one in 2014) had bad cracking of the printing, and an order was miscounted in 2014. Broprints has lower setup fees and lower per-shirt charges than the other companies I’ve dealt with, and the order is a little larger this year (35 shirts), so the per-shirt price should be the lowest yet (about \\$12 a shirt, and \\$15 for long sleeve).", "fee of$100 to process your photo? This would be a fixed cost, as well as a start-up cost. Even if you sold no tee shirts, you would still incur the $100 just to go into business. If you sold no shirts, you would have a$100 loss instead of a profit. However, if you once again sold $100 shirts, your profit after fixed costs would be the$400 from above less the $100 start-up fee, for a net profit of$300. What is the minimum amount of shirts you should expect to sell in order to decide to go into business? You will need to at least break-even, which means to cover the fixed costs plus expected marginal costs. Here, marginal profit will provide $4 per shirt, so it takes 25 shirts to cover the initial$100 investment. ## Other Important Quantities There are other quantities that can affect a firm’s decisions. Entry costs are the minimum cost to start a particular business. Such are closely related to fixed costs. Opportunity Cost is the amount of profit given up by not pursuing a course of action. Typical sources of opportunity costs are risk and interest rates. Risk is cost due to uncertainty. Interest rates reflect a combination of costs due to risk and sacrificed opportunities. In reality, they can also represent bargaining", "of $3$ vehicles per minute? 2. Graph the function on the domain $[0, 6]\\text{.}$ 3. What is the vertical asymptote of the graph? What does it tell you about the queue? 1. $0.25$ min 2. $r=6\\text{.}$ The wait time becomes infinite as the arrival rate approaches $6$ vehicles per minute. ###### Example7.53 EarthCare decides to sell T-shirts to raise money. The company makes an initial investment of $100 to pay for the design of the T-shirt and to set up the printing process. After that, the T-shirts cost$5 each for labor and materials. 1. Express the average cost per T-shirt as a function of the number of T-shirts EarthCare produces. 2. Make a table of values for the function. 3. Graph the function and explain what it tells you about the cost of the T-shirts. Solution 1. If EarthCare produces $x$ T-shirts, the total costs will be $100 + 5x$ dollars. To find the average cost per T-shirt, we divide the total cost by the number of T-shirts produced, to get \\begin{equation*} C = g(x) = \\frac{100 + 5x}{x} \\end{equation*} 2. We evaluate the function for several values of $x\\text{,}$ as shown in the table $x$ $1$ $2$ $4$ $5$ $10$ $20$ $C$ $105$ $55$ $40$ $25$ $15$ $10$ If EarthCare makes only one T-shirt, its cost is", "a reply to A T-shirt supplier charges $12 per T-shirt for the first 50 in the Problem Solving forum “Hi alanforde800Maximux, We''re told that a T-shirt supplier charges$12 per T-shirt for the FIRST 50 T-shirts in an order and then $10 per T-shirt for each additional T-shirt in the order and the total charge for an order of T-shirts was$1500. We''re asked for the total number of T-shirts in ...” May 11, 2018 [email protected] posted a reply to Three of the four vertices of a rectangle in the xy-coordina in the Problem Solving forum “Hi alanforde800Maximux, We''re given 3 of the four vertices of a rectangle in the xy-coordinate plane: (-3, 10), ( 2, 10), and (2, 1). We''re asked for the 4th co-ordinate. Based on the three pairs of numbers, we see common X-coodinates (2) and Y-coordinates (10), meaning that the rectangle ...” May 11, 2018 [email protected] posted a reply to A positive integer is divisible by 3 if and only if the sum in the Problem Solving forum “Hi alanforde800Maximux, We''re told that the six-digit integer is divisible by 3, and is of the form 1K2,K24, where K represents a digit that occurs twice. We''re asked for the number of different possible values. Using the ''rule of 3'', since 1+2+2+4 = 9 (which is a multiple", "profit. 2. In addition to the $100 setup fee, the team paid$7 for each T-shirt. Dear expert, mikemcgarry, GMATPrepNow I have a question about statement 2 : I think statement 2 is SUFFICIENT because of n=1. Why we don't plug n=1 in this equation, so we can get X? Thanks! _________________ There's an app for that - Steve Jobs. Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4677 Re: To raise funds, a racing team sold T-shirts imprinted with [#permalink] ### Show Tags 20 Oct 2017, 10:47 1 KUDOS Expert's post septwibowo wrote: guerrero25 wrote: To raise funds, a racing team sold T-shirts imprinted with the team's logo. The team paid their supplier a one-time setup fee of $100. Because they purchased at least 50 T-shirts, the team qualified for their supplier's quantity discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for$12, and made a $900 profit. 2. In addition to the$100 setup fee, the team paid $7 for each T-shirt. Dear expert, mikemcgarry, GMATPrepNow I have a question about statement 2 : I think statement 2 is SUFFICIENT because of n=1. Why we don't plug n=1 in this equation, so we", "discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for$12, and made a $900 profit. 2. In addition to the$100 setup fee, the team paid $7 for each T-shirt. [Reveal] Spoiler: OA Last edited by guerrero25 on 30 Oct 2013, 16:01, edited 1 time in total. Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4677 Re: To raise funds, a racing team sold T-shirts imprinted with [#permalink] ### Show Tags 30 Oct 2013, 10:43 5 This post received KUDOS Expert's post 1 This post was BOOKMARKED guerrero25 wrote: To raise funds, a racing team sold T-shirts imprinted with the team's logo. The team paid their supplier a one-time setup fee of$100. Because they purchased at least 50 T-shirts, the team qualified for their supplier's quantity discount of x cents per T-shirt and paid (8-(x/100)n) dollars for each of the n T shirts they purchased. What is the value of x? 1. The team purchased 200 T-shirts, sold each T-shirt for $12, and made a$900 profit. 2. In addition to the $100 setup fee, the team paid$7 for each T-shirt. I do not have the OA with me ;i 'll update as soon as", "3 and 5, that will do the needful. Minimum possible number of shirts manufactured by all the companies: $= (83.33 + 125 + 133.33 + 120) × (3 × 4 × 5)$ $= \\textbf{27,700}$", "10 May 2018, 21:05 Let extra number of t shirt after 50 be x 12*50 + 10x=1500 10x=900 X=90 Total t-shirt=90+50=140 Posted from my mobile device Intern Joined: 24 Nov 2017 Posts: 34 Re: A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 10 May 2018, 22:51 Total charge for the order =$1500 Charge @ $12 per T shirt for the first 50 = 50 * 12 = 600 Charge @$10 per T shirt beyond the first 50 = 1500 - 600 = 900. Number of T Shirts in the order beyond the first 50 = 900/10 = 90. Total number of T shirts in the order = 50 + 90 = 140 Choice C SC Moderator Joined: 22 May 2016 Posts: 1680 A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 11 May 2018, 17:05 alanforde800Maximus wrote: A T-shirt supplier charges$12 per T-shirt for the first 50 T-shirts in an order and $10 per T-shirt for each additional T-shirt in the order. If the charge for an order of T-shirts was$1500, how many T-shirts were in the order? A) 90 B) 125 C) 140 D) 150 E) 200 The trap here is forgetting to account for or to \"add back\" the", "1. ## T-Shirts To print T-shirts includes $50 for the design and$4 per shirt. Express the cost of printing the T-shirts as a function of the number of T-shirts printed and then find the cost for printing 500 T-shirts. MY WORK: My linear function model is: T = $4P +$50...Is this correct? To print 500 T-shirts will cost: 500 = 4P + 500 500 - 50 = 4P 450 = 4P 450/4 = P $112.50 = P Let P = price Let T = T-shirts Does this make sense? Is this right? 2. Originally Posted by magentarita To print T-shirts includes$50 for the design and $4 per shirt. Express the cost of printing the T-shirts as a function of the number of T-shirts printed and then find the cost for printing 500 T-shirts. MY WORK: My linear function model is: T =$4P + $50...Is this correct? To print 500 T-shirts will cost: 500 = 4P + 500 500 - 50 = 4P 450 = 4P 450/4 = P$112.50 = P Let P = price Let T = T-shirts Does this make sense? Is this right? Hello Magentarita, Your function is correct, but you lost me after that. I'll write the function in a little different way. Let t = number of T-shirts. Then $f(t)=4t+50$ Now, all you need", "marginal cost for a range of t-shirts sold from 100 to 500 using steps of 10 (100, 110, 120..... Apply as many presentation features (font, font size, font color, cell shading, etc.) as you wish Apply appropriate number formats (dollar signs, percents, etc.) Give your spreadsheet an appropriate title and center it Adjust column and row width to suit your spreadsheet Use formula's to calculate required fields Use some mechanism of your choice to denote and bring attention to the profit max point and the maximum profit (such as an if statement) Setup an appropriate header that includes your name and section number Use Landscape orientation Center horizontally and vertically on the page and fit to one page Print one copy and attach to this homework assignment the directions below to create an Excel graph for a perfectly competitive company. A inoi Tha nice per unit is$500. Use excel to ## Answers #### Similar Solved Questions 4 answers ##### A buffer solution consists of 100 mL of 0.10 M HNOz and 10.10 M NaNOz The pH of this buffer solution will be (6 pts)7.5403.463.349.31 A buffer solution consists of 100 mL of 0.10 M HNOz and 10.10 M NaNOz The pH of this buffer solution will be (6 pts) 7.54 03.46 3.34 9.31... 5 answers 5 answers #####", "numbers is the indicator function of A A =1& }x\\in A\\\\0& }x\\notin A\\\\\\end}} In this definition, the intervals A • The intervals are pairwise disjoint: A \\cap A_=\\emptyset } for • The union of the intervals is the entire real line: . ^A_=\\mathbb .} • Indeed, if that is not the case to start with, a different set of intervals can be picked for which these assumptions hold. For example, the step function f +3\\chi _} can be written as ) . +4\\chi _+7\\chi _+3\\chi _+0\\chi _.} You May Like: Algebra Connections Chapter 9 Answers Example : Step Functions A wholesale t-shirt manufacturer charges the following prices for t-shirt orders: • $20 per shirt for shirt orders up to 20 shirts. •$15 per shirt for shirt between 21 and 40 shirts. • $10 per shirt for shirt orders between 41 and 80 shirts. •$5 per shirt for shirt orders over 80 shirts. • Sketch a graph of this discontinuous function. • You’ve ordered 40 shirts and must pay shipping fees of \\$10. How much is your total order?", "Question You are ordering T-shirt with your school’s mascot printed on them. Each T-shirt cost $4.75. The printer charges a setup fee of$30 and $2.50 to print each shirt. write two expressions to represents the total cost of printing n T-shirts please help ASAP if you can please explain Answers 1. thnahdat Answer: 2.50n + 30 each shirt (n) is$2.50 so to find the total for the shirts you would multiple $2.50 by (n) the number of shirts. The fee of the printer is only charged once so you would just add an extra$30 to the amount you got. n is the number of shirts they buy. It says that each shirt costs $4.75, and for each shirt they add an extra$30 and \\$2.50 for the setup fee and to peint each shirt. The word each shows that you have to times. You have to times n by each number or to make it shorter put all three numbers in parenthesis put the n ouside of the parenthesis.", "is better than Project 2_ #### Similar Solved Questions ##### Identify the population and sample in the study: Also determine whether the numerical value is a parameter or a statistic. (2 points)In survey of 2253 Internet users, 19% use Twitter or another service to share social updatesPopulation:Sample: Identify the population and sample in the study: Also determine whether the numerical value is a parameter or a statistic. (2 points) In survey of 2253 Internet users, 19% use Twitter or another service to share social updates Population: Sample:... ##### The latest demand equation for your Banjos Rock\" T-shirts is given byq =-60x + 1200where 9 is the number of shirts YOU can sell in one week If you charge x dollars per shirt. When you charge x dollars per shirt, your weekly cost function (in dollars) Is given by Clx) = -30Ox 7875(a) Find the weekly profit as a function of the price per shirt * P() =(6) Determine the unit price you should charge to break even_ Enter the smaller value first: You break even, when vou charge dollars per shlrt:W The latest demand equation for your Banjos Rock\" T-shirts is given by q =-60x + 1200 where 9 is the number of shirts YOU can sell in one week If you charge x dollars per shirt. When", "first 50 I. Arithmetic How many T-shirts were there in an order charge of $1500? Cost 1:$12 per T-shirt for first 50 = $600 Cost 2:$10 per T-shirt, for any number $$x$$ T-shirts, AFTER the first 50 T-shirt order total: $1500 (1) 50 T-shirts cost$600 ($1500 -$600) = $900 remain (2)$900 = how many T-shirts? $900/$10 per shirt = 90 shirts Number of T-shirts in the order? 50 + 90 = 140 II. Algebra Let $$x$$ = # of T-shirts $$(12)(50) + (10)(x-50) = 1500$$ $$600 + 10x - 500 = 1500$$ $$10x + 100 = 1500$$ $$10x = 1400$$ $$x = 140$$ _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2638 Location: United States (CA) Re: A T-shirt supplier charges $12 per T-shirt for the first 50 T-shirts i [#permalink] ### Show Tags 14 May 2018, 16:53 alanforde800Maximus wrote: A T-shirt supplier charges$12 per T-shirt for the first 50 T-shirts in an order and $10 per T-shirt for each additional T-shirt in the order. If the charge for an order of T-shirts was$1500, how many T-shirts were in the order? A) 90 B) 125 C) 140 D) 150 E) 200 We", "the first 25 months), 25(1850) + 1700 (m – 25) for 25 < m = 60 (since you make 1700 a month for the months after month 25), and 25(1850) + (1700)(35) +21250 for m = 60 (since you’d only get a bonus if you work until month 60). Does this make sense? • Don’t we need to have a number for m to plug in for f(m)? How would we determine the number? • Oh sorry – I didn’t read the actual assignment. Let’s use m = 60 (assuming we will work the whole 5 years) and see which gives us the most money: For only job A, it will be $126000, for job A-B, it would be$125000, and for the job A-C route, it would be $127000. So it looks like the A-C route would be the best, if I did the problem correctly. Lisa 5. Can you help me with this problem? I have the values set up but I can’t figure out the equations needed in the function. You plan to sell t-shirts as a fundraiser. The company you wish to purchase your t-shirts from charge$10 per shirt for the first 75 shirts. After the first 75 shirts you purchase up to 150 shirts, the company will lower its price to $7.50 per shirt.", "You plan to sell She Love Math t-shirts as a fundraiser. The wholesale t-shirt company charges you$10 a shirt for the first 75 shirts. After the first 75 shirts you purchase up to 150 shirts, the company will lower its price to $7.50 per shirt. After you purchase 150 shirts, the price will decrease to$5 per shirt. Write a function that models this situation. Solution: We see that the “boundary points” are 75 and 150, since these are the number of t-shirts bought where prices change. Since we have two boundary points, we’ll have three equations in our piecewise function. We’ll start with $$x\\ge 1$$, since, we assume at least one shirt is bought. Note in this problem, the number of t-shirts bought ($$x$$), or the domain, must be a integer, but this restriction shouldn’t affect the outcome of the problem. $$\\displaystyle f\\left( x \\right)=\\left\\{ \\begin{array}{l}\\text{ }……\\text{ if }1\\le x\\le 75\\\\\\text{ }……\\text{ if }75<x\\le 150\\\\\\text{ }……\\text{ if }x>150\\end{array} \\right.$$ We are looking for the “answers” (total cost of t-shirts) to the “questions” (how many are bought) for the three ranges of prices. For up to and including 75 shirts, the price is $10, so the total price would $$10x$$. For more than 75 shirts but up to 100 shirts, the cost is$7.50, but the first 75 t-shirts will", "# T-Shirts • Dec 28th 2008, 07:30 AM magentarita T-Shirts To print T-shirts includes \\$50 for the design and \\$4 per shirt. Express the cost of printing the T-shirts as a function of the number of T-shirts printed and then find the cost for printing 500 T-shirts. MY WORK: My linear function model is: T = \\$4P + \\$50...Is this correct? To print 500 T-shirts will cost: 500 = 4P + 500 500 - 50 = 4P 450 = 4P 450/4 = P \\$112.50 = P Let P = price Let T = T-shirts Does this make sense? Is this right? • Dec 28th 2008, 08:26 AM masters Quote: Originally Posted by magentarita To print T-shirts includes \\$50 for the design and \\$4 per shirt. Express the cost of printing the T-shirts as a function of the number of T-shirts printed and then find the cost for printing 500 T-shirts. MY WORK: My linear function model is: T = \\$4P + \\$50...Is this correct? To print 500 T-shirts will cost: 500 = 4P + 500 500 - 50 = 4P 450 = 4P 450/4 = P \\$112.50 = P Let P = price Let T = T-shirts Does this make sense? Is this right? Hello Magentarita, Your function is correct, but you lost me after that. I'll write", "effect, they will double the relevant range to allow for a maximum of $$160$$ inspections per shift, assuming the second QA inspector can inspect an additional 80 units per shift. The down side to this approach is that once the new QA inspector is hired, if demand falls again, the company will be incurring fixed costs that are unnecessary. For this reason, adding salaried personnel to address a short-term increase in demand is not a decision most businesses make. Step costs are best explained in the context of a business experiencing increases in activity beyond the relevant range. As an example, let’s return to Tony’s T-Shirts. Tony’s cost of operations and the associated relevant ranges are shown in Table $$\\PageIndex{4}$$. Table $$\\PageIndex{4}$$: Tony’s T-Shirts Cost Options Cost Type of Cost Relevant Range Lease on Screen-Printing Machine $2,000 per month Fixed 0–2,000 T-shirts per month Employee$10 per hour Variable 20 shirts per hour Tony’s Salary $2,500 per month Fixed N/A Screen-Printing Ink$0.25 per shirt Variable N/A Building Rent \\$1,500 per month Fixed 2 screen-printing machines and 2 employees As you can see, Tony has both fixed and variable costs associated with his business. His one screen-printing machine can only produce $$2,000$$ T-shirts per month and his current employee can produce $$20$$ shirts per hour ($$160$$ per $$8$$-hour work day)." ]

[ "# Difference between revisions of \"2020 AMC 8 Problems/Problem 24\" ## Problem A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$. When $n=24$, the $576$ gray tiles cover $64\\%$ of the area of the large square region. What is the ratio $\\frac{d}{s}$ for this larger value of $n?$ $[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); [/asy]$ $\\textbf{(A) }\\frac{6}{25} \\qquad \\textbf{(B) }\\frac{1}{4} \\qquad \\textbf{(C) }\\frac{9}{25} \\qquad \\textbf{(D) }\\frac{7}{16} \\qquad \\textbf{(E) }\\frac{9}{16}$ ## Solution 1 The area of the shaded region is $(24s)^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $\\frac{(24s)^2}{(24s+25d)^2}=\\frac{64}{100}$. Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\\frac{24s}{24s+25d}=\\frac{8}{10} = \\frac{4}{5}$, and cross-multiplying now", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square? $[asy] draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle); draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle); draw((0,1)--(0,2)); draw((1,0)--(2,0)); draw((0,-1)--(0,-2)); draw((-1,0)--(-2,0)); [/asy]$ $\\text{(A)}\\ 25 \\qquad \\text{(B)}\\ 44 \\qquad \\text{(C)}\\ 50 \\qquad \\text{(D)}\\ 62 \\qquad \\text{(E)}\\ 75$ ## Problem 19 The graph shows the distribution of the number of children in the families of the students in Ms. Jordan's English class. The median number of children in the family for this distribution is $[asy] unitsize(12); for(int i = 1; i <= 7; ++i) { draw((0,i)--(19,i),dotted); draw((-0.5,i)--(0.5,i)); } for(int i = 0; i <= 5; ++i) { draw((3*i+2,0)--(3*i+2,-0.5)); } fill((1,0)--(1,2)--(3,2)--(3,0)--cycle,white); fill((4,0)--(4,1)--(6,1)--(6,0)--cycle,white); fill((7,0)--(7,2)--(9,2)--(9,0)--cycle,white); fill((10,0)--(10,2)--(12,2)--(12,0)--cycle,white); fill((13,0)--(13,6)--(15,6)--(15,0)--cycle,white); draw((0,9)--(0,0)--(19,0)); draw((1,0)--(1,2)--(3,2)--(3,0)); draw((4,0)--(4,1)--(6,1)--(6,0)); draw((7,0)--(7,2)--(9,2)--(9,0)); draw((10,0)--(10,2)--(12,2)--(12,0)); draw((13,0)--(13,6)--(15,6)--(15,0)); label(\"1\",(2,-0.5),S); label(\"2\",(5,-0.5),S); label(\"3\",(8,-0.5),S); label(\"4\",(11,-0.5),S); label(\"5\",(14,-0.5),S); label(\"6\",(17,-0.5),S); label(\"2\",(-0.5,2),W); label(\"4\",(-0.5,4),W); label(\"6\",(-0.5,6),W); label(\"\\textbf{Number of Children}\",(9,-1.5),S); label(\"\\textbf{in the Family}\",(9,-2.5),S); label(\"\\textbf{Number}\",(-1.5,6),W); label(\"\\textbf{of}\",(-3,5),W); label(\"\\textbf{Families}\",(-1.5,4),W); [/asy]$ $\\text{(A)}\\ 1 \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 3 \\qquad \\text{(D)}\\ 4 \\qquad \\text{(E)}\\ 5$ ## Problem 20 Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana's number is larger than Apollo's number? $\\text{(A)}\\ \\dfrac{1}{3} \\qquad \\text{(B)}\\ \\dfrac{5}{12} \\qquad \\text{(C)}\\ \\dfrac{4}{9} \\qquad \\text{(D)}\\ \\dfrac{17}{36} \\qquad \\text{(E)}\\ \\dfrac{1}{2}$ ## Problem 21 A plastic snap-together cube has a protruding snap on one side and", "if the rectangle cannot be placed.\"\"\" (w, h) = rect (x0, y0) = pos newgrid = map(list, grid) # make a copy try: for x in range(x0, x0+w): for y in range(y0, y0+h): if newgrid[y][x] is not empty: return None newgrid[y][x] = rect return newgrid except IndexError: # went off the grid return None In [118]: # Place a 3 x 4 rectangle on a square grid, 2 cells over and 1 down: place_rectangle_at((3, 4), Square(5), (2, 1)) Out[118]: [[(0, 0), (0, 0), (0, 0), (0, 0), (0, 0)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)]] In [119]: # Place two rectangles on a grid grid2 = place_rectangle_at((3, 4), Square(5), (2, 1)) grid3 = place_rectangle_at((2, 5), grid2, (0, 0)) grid3 Out[119]: [[(2, 5), (2, 5), (0, 0), (0, 0), (0, 0)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)]] In [120]: # Now try to place a third rectangle that does not fit;", "# Difference between revisions of \"1998 AJHSME Problems/Problem 24\" ## Problem A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? $[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label(\"2\",(1.5,8.2),N); label(\"4\",(3.5,8.2),N); label(\"5\",(4.5,8.2),N); label(\"7\",(6.5,8.2),N); label(\"8\",(7.5,8.2),N); label(\"9\",(0.5,7.2),N); label(\"11\",(2.5,7.2),N); label(\"12\",(3.5,7.2),N); label(\"13\",(4.5,7.2),N); label(\"14\",(5.5,7.2),N); label(\"16\",(7.5,7.2),N);[/asy]$ $\\text{(A)}\\ 36\\qquad\\text{(B)}\\ 64\\qquad\\text{(C)}\\ 78\\qquad\\text{(D)}\\ 91\\qquad\\text{(E)}\\ 120$ ## Solution ### Solution 1 The numbers that are shaded are the triangular numbers, which are numbers in the form $\\frac{(n)(n+1)}{2}$ for positive integers. They can also be generated by starting with $1$, and adding $1, 2, 3, 4...$ as in the description of the problem. Squares that have the same remainder after being divided by $8$ will", "draw a red diagonal line, 6 pixels wide, from point (100,100) to point (200,200). draw_set_colour(c_red); draw_line_width(100, 100, 200, 200, 6); I can foresee you will have corner issues as the adjoining lines corner won't meet up. It will become more obvious as you thicken the lines and the two angles are extreme. That is more tricky to solve given you need to work out the corners for each line taking thickness into account and join those with the previous line corners. Quite a bit of maths and code involving atan2, cos and sin is required to make this work correctly. • Search for \"graphic gems 1\" and open link to pdf - jump to page 108 - the diagram on that page will show you what you will be faced with when doing it this way. – John Oct 2 '19 at 3:12 Adding 1 to each dimension just isn't enough. It would have to be something like: draw_path(circuito_ovo,500,50,false); draw_path(circuito_ovo,500,51,false); draw_path(circuito_ovo,500,52,false); draw_path(circuito_ovo,501,50,false); draw_path(circuito_ovo,501,51,false); draw_path(circuito_ovo,501,52,false); draw_path(circuito_ovo,502,50,false); draw_path(circuito_ovo,502,51,false); draw_path(circuito_ovo,502,52,false); For ease just use the repeat function (Draw Event): repeat(5){ repeat(5){ draw_path(circuito_ovo,500+i,50+j,false); j+=1; } i+=1; j=0; } i=0; j=0; Initializing the variables with the value 0 (i and j), or negative (if I want the object to be walked exactly in the center of the drawing). The number of times the", "(m-2-12.south east); \\draw[Red!,line width=2] (m-1-14.north west) rectangle (m-1-14.south east); \\draw (m-2-11.north west) rectangle (m-2-11.south east); \\draw (m-3-13.north west) rectangle (m-3-13.south east); \\draw[Red!,line width=2] (m-2-13.north west) rectangle (m-2-13.south east); \\draw (m-3-11.north west) rectangle (m-3-11.south east); \\draw[Red!,line width=2] (m-3-12.north west) rectangle (m-3-12.south east); \\draw (m-4-11.north west) rectangle (m-4-11.south east); \\draw[Red!,line width=2] (m-4-12.north west) rectangle (m-4-12.south east); \\draw (m-1-21.north west) rectangle (m-1-21.south east); \\draw (m-1-22.north west) rectangle (m-1-22.south east); \\draw (m-1-20.north west) rectangle (m-1-20.south east); \\draw (m-2-20.north west) rectangle (m-2-20.south east); \\draw (m-2-21.north west) rectangle (m-2-21.south east); \\draw (m-2-23.north west) rectangle (m-2-23.south east); \\draw[Red!,line width=2] (m-1-23.north west) rectangle (m-1-23.south east); \\draw (m-3-22.north west) rectangle (m-3-22.south east); \\draw[Red!,line width=2] (m-2-22.north west) rectangle (m-2-22.south east); \\draw (m-3-20.north west) rectangle (m-3-20.south east); \\draw[Red!,line width=2] (m-3-21.north west) rectangle (m-3-21.south east); \\draw (m-4-20.north west) rectangle (m-4-20.south east); \\draw[Red!,line width=2] (m-4-21.north west) rectangle (m-4-21.south east); \\draw[Red!,line width=2] (m-5-20.north west) rectangle (m-5-20.south east); \\draw (m-1-28.north west) rectangle (m-1-31.south east); \\draw[Orange!,line width=2] (m-1-32.north west) rectangle (m-1-32.south east); \\draw (m-1-28.north west) rectangle (m-4-28.south east); \\draw (m-1-30.north west) rectangle (m-3-30.south east); \\draw (m-2-29.north west) rectangle (m-2-31.south east); \\draw (m-4-28.north west) rectangle (m-5-28.south east); \\draw (m-3-28.north west) rectangle (m-3-28.south east); \\draw (m-4-29.north west) rectangle (m-4-29.south east); \\end{tikzpicture} I want to show the step-by-step progression of a bumping route from the beginning of the Robinson-Schensted insertion algorithm to the end and I would", "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "# Problem #2155 2155 Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed? $[asy] unitsize(1inch); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--((1/3) + 3*(1/2),0)); fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7)); draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle); draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle); draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle); draw((2,0)--(2 + (1/3) + (3/2),0)); draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle); draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle); label(\"B\",((1/6) + (1/2),(1/2)),NW); label(\"B\",((2/3) + 2 + (1/4),(29/30)),NNE); draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3)); fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7)); draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);[/asy]$ $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ \\sqrt{2}\\qquad\\textbf{(C)}\\ \\frac{3}{2}\\qquad\\textbf{(D)}\\ \\sqrt{2}+\\frac{1}{2}\\qquad\\textbf{(E)}\\ 2$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"2013 AMC 8 Problems/Problem 18\" ## Problem Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain? $[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label(\"10\",(A--B),SE); label(\"12\",(C--B),SW); label(\"5\",(F--B),W);[/asy]$ $\\textbf{(A)}\\ 204 \\qquad \\textbf{(B)}\\ 280 \\qquad \\textbf{(C)}\\ 320 \\qquad \\textbf{(D)}\\ 340 \\qquad \\textbf{(E)}\\ 600$ ## Solution 1 There are $10 \\cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11", "in Shaddoll's solution, and our answer is $p=13+84+85=\\boxed{182}$. ## Solution 4 $[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"A\", A, SE); label(\"B\", B, NW); label(\"C_0\", C0, SW); label(\"C_1\", C1, NE); label(\"C_2\", C2, W); label(\"C_3\", C3, NE); label(\"C_4\", C4, W); label(\"a\", (B+C0)/2, W); label(\"b\", (A+C0)/2, S); label(\"c\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]$ Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\\frac{a+c}{b}$. The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \\cdots$, shown in red in the figure below. Since each of the triangles $\\triangle C_0 C_1 C_2, \\triangle C_2 C_3 C_4, \\dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the", "clipping. Here's what my two paths look like in that square: And here's what I want that square to look like: My code for this is not intelligent, and not my first try, but it's what I have at hand: \\begin{tikzpicture}[scale=.1] \\foreach \\x in {1,2,...,7} \\foreach \\y in {0,1,...,5} { \\begin{scope}[xshift=-10*\\x cm, yshift=10*\\y cm] \\draw[clip] (0,5)--(2,5)--(2,4)--(3,4)--(3,2)--(4,2)--(4,1)--(7,1)--(7,0)--(0,0)-- cycle; \\draw[fill=red] (0,5)--(2,5)--(2,4)--(3,4)--(3,2)--(4,2)--(4,1)--(7,1)--(7,0)--(0,0)--cycle; \\draw[fill=white] (-\\x, \\y + 20) |- ++(2,-15) |- ++(1,-1) |- ++(1,-2) |- ++(3,-1) |- ++(13,-1) -- ++(\\x,-\\y) -| ++(-13,1) -| ++(-3,1) -| ++(-1,2) -| ++(-1,1) -| ++(-2,15) -| ++(40,-40) -| (-\\x, \\y + 20); \\draw[very thin, gray] (0,0) grid (8,8); \\end{scope} } \\end{tikzpicture} • Unrelated to the question, but you might check out the turtle library of TikZ which would make it much easier to create such paths. – Qrrbrbirlbel Jul 4 '13 at 23:17 • Are the pictures you want to create always composed of little squared in white or red? I see a pattern here which would make it much easier to create on a different way and a different type of input by specifying the number of rectangle for the rows and for the “whitening” path. – Qrrbrbirlbel Jul 4 '13 at 23:25 • @Qrrbrbirlbel Yeah -- it's always little collections of squares (partitions) that I'm doing; it probably would be easier in the long run", "# Difference between revisions of \"2013 Indonesia MO Problems\" ## Day 1 ### Problem 1 In a $4 \\times 6$ grid, all edges and diagonals are drawn (see attachment). Determine the number of parallelograms in the grid that uses only the line segments drawn and none of its four angles are right. $[asy] draw((0,0)--(6,0)--(6,4)--(0,4)--(0,0)); for (int i=1; i<6; ++i) { draw((i,0)--(i,4)); } for (int i=1; i<4; ++i) { draw((0,i)--(6,i)); } draw((0,1)--(1,0)); draw((0,2)--(2,0)); draw((0,3)--(3,0)); draw((0,4)--(4,0)); draw((1,4)--(5,0)); draw((2,4)--(6,0)); draw((3,4)--(6,1)); draw((4,4)--(6,2)); draw((5,4)--(6,3)); draw((0,3)--(1,4)); draw((0,2)--(2,4)); draw((0,1)--(3,4)); draw((0,0)--(4,4)); draw((1,0)--(5,4)); draw((2,0)--(6,4)); draw((3,0)--(6,3)); draw((4,0)--(6,2)); draw((5,0)--(6,1)); [/asy]$ ### Problem 2 Let $ABC$ be an acute triangle and $\\omega$ be its circumcircle. The bisector of $\\angle BAC$ intersects $\\omega$ at [another point] $M$. Let $P$ be a point on $AM$ and inside $\\triangle ABC$. Lines passing $P$ that are parallel to $AB$ and $AC$ intersects $BC$ on $E, F$ respectively. Lines $ME, MF$ intersects $\\omega$ at points $K, L$ respectively. Prove that $AM, BL, CK$ are concurrent. ### Problem 3 Determine all positive real $M$ such that for any positive reals $a,b,c$, at least one of $a + \\dfrac{M}{ab}, b + \\dfrac{M}{bc}, c + \\dfrac{M}{ca}$ is greater than or equal to $1+M$. ### Problem 4 Suppose $p > 3$ is a prime number and $$S = \\sum_{2 \\le i < j < k \\le p-1} ijk$$ Prove", "& & & & & & & \\\\ } ; \\draw (m-1-1.north west) rectangle (m-1-2.south east); \\draw (m-2-4.north west) rectangle (m-2-4.south east); \\draw[red!,line width=2] (m-1-4.north west) rectangle (m-1-4.south east); \\draw (m-1-1.north west) rectangle (m-4-1.south east); \\draw (m-1-3.north west) rectangle (m-1-3.south east); \\draw (m-3-3.north west) rectangle (m-3-3.south east); \\draw[red!,line width=2] (m-2-3.north west) rectangle (m-2-3.south east); \\draw (m-3-1.north west) rectangle (m-3-1.south east); \\draw (m-4-2.north west) rectangle (m-4-2.south east); \\draw[red!,line width=2] (m-3-2.north west) rectangle (m-3-2.south east); \\draw (m-1-11.north west) rectangle (m-1-11.south east); \\draw (m-1-12.north west) rectangle (m-1-12.south east); \\draw (m-1-13.north west) rectangle (m-1-13.south east); \\draw (m-2-14.north west) rectangle (m-2-14.south east); \\draw (m-2-12.north west) rectangle (m-2-12.south east); \\draw[red!,line width=2] (m-1-14.north west) rectangle (m-1-14.south east); \\draw (m-2-11.north west) rectangle (m-2-11.south east); \\draw (m-3-13.north west) rectangle (m-3-13.south east); \\draw[red!,line width=2] (m-2-13.north west) rectangle (m-2-13.south east); \\draw (m-3-11.north west) rectangle (m-3-11.south east); \\draw[red!,line width=2] (m-3-12.north west) rectangle (m-3-12.south east); \\draw (m-4-11.north west) rectangle (m-4-11.south east); \\draw[red!,line width=2] (m-4-12.north west) rectangle (m-4-12.south east); \\draw (m-1-21.north west) rectangle (m-1-21.south east); \\draw (m-1-22.north west) rectangle (m-1-22.south east); \\draw (m-1-20.north west) rectangle (m-1-20.south east); \\draw (m-2-20.north west) rectangle (m-2-20.south east); \\draw (m-2-21.north west) rectangle (m-2-21.south east); \\draw (m-2-23.north west) rectangle (m-2-23.south east); \\draw[red!,line width=2] (m-1-23.north west) rectangle (m-1-23.south east); \\draw (m-3-22.north west) rectangle (m-3-22.south east); \\draw[red!,line width=2] (m-2-22.north west) rectangle (m-2-22.south east); \\draw (m-3-20.north west) rectangle (m-3-20.south east); \\draw[red!,line", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "4 & 6 & & & & & & = & 3 & 4 & 6 & & & & & = & 3 & 4 & 6 & & & & & = \\\\ 4 & 7 & & & \\leftarrow & 5 & \\leftarrow & 3 & 1 & & 4 & 5 & & & & & & & & 4 & 5 & & & & & & & 4 & 5 & & & & & & \\\\ & & & & & & & & & & & & & & \\leftarrow & 7 & \\leftarrow 3 & 1 & & 7 & & & & & & & & 7 & & & & & & & \\\\ } ; \\draw (m-1-1.north west) rectangle (m-1-2.south east); \\draw (m-2-4.north west) rectangle (m-2-4.south east); \\draw[Red!,line width=2] (m-1-4.north west) rectangle (m-1-4.south east); \\draw (m-1-1.north west) rectangle (m-4-1.south east); \\draw (m-1-3.north west) rectangle (m-1-3.south east); \\draw (m-3-3.north west) rectangle (m-3-3.south east); \\draw[Red!,line width=2] (m-2-3.north west) rectangle (m-2-3.south east); \\draw (m-3-1.north west) rectangle (m-3-1.south east); \\draw (m-4-2.north west) rectangle (m-4-2.south east); \\draw[Red!,line width=2] (m-3-2.north west) rectangle (m-3-2.south east); \\draw (m-1-11.north west) rectangle (m-1-11.south east); \\draw (m-1-12.north west) rectangle (m-1-12.south east); \\draw (m-1-13.north west) rectangle (m-1-13.south east); \\draw (m-2-14.north west) rectangle (m-2-14.south east); \\draw (m-2-12.north west) rectangle", "(-7,5) | 8.60 | | 28 | (8,1) | (7,-6) | 9.22 | | 29 | (1,8) | (-7,7) | 9.90 | +------+-------+---------+------+ There are 4 combinations that were skipped with change in coordinates of • (6,5) doesn't fit • (6,4) doesn't fit • (5,5) same length as (7,1) • (4,3) same length as (5,0) • You can find the maximum eventually – TSLF Oct 15 '16 at 21:29 • From 19 to 20 your distance decreases (according to your chart). – GentlePurpleRain Oct 17 '16 at 14:34 There's an upper bound of 33 distinct distance vectors from (1,0) to (7,7) 1x0, 1x1, 2x0, 2x1, 2x2, 3x0, 3x1, 3x2, 4x0, 4x1, 3x3, 4x2, 4x3=5x0, 5x1, 5x2, 4x4, 5x3, 6x0, 6x1, 5x4, 6x2, 6x3, 7x0, 5x5=7x1, 6x4, 7x2, 7x3, 6x5, 7x4, 6x6, 7x5, 7x6, 7x7 I appear to have found a path that hits 30 jumps, but I fear I may have made a mistake somewhere (3,3), (4,3), (3,4), (5,3), (3,5), (6,4), (3,2), (3,6), (4,2), (7,5), (3,7), (6,3), (2,6), (3,1), (5,6), (1,2), (4,7), (4,1), (5,7), (1,3), (7,1), (1,4), (8,4), (1,3), (8,5), (1,7), (6,1), (2,8), (8,2), (1,8), (8,1) • (3,4) to (5,3) and (5,3) to (3,2) are the same length. As are (8,8) to (1,6) and (1,6) to (8,4) – gtwebb Oct 14 '16 at 20:26 • Maybe you selected", "= 12 2^2 * 3^1 * 5^1 = (4 * 3 * 5) = 60 2^2 * 3^2 * 5^0 = (4 * 9 * 1) = 36 2^2 * 3^2 * 5^1 = (4 * 9 * 5) = 180 2^3 * 3^0 * 5^0 = (8 * 1 * 1) = 8 2^3 * 3^0 * 5^1 = (8 * 1 * 5) = 40 2^3 * 3^1 * 5^0 = (8 * 3 * 1) = 24 2^3 * 3^1 * 5^1 = (8 * 3 * 5) = 120 2^3 * 3^2 * 5^0 = (8 * 9 * 1) = 72 2^3 * 3^2 * 5^1 = (8 * 9 * 5) = 360 (let ((paths-in-tree '((1 1 1) (1 1 5) (1 3 1) (1 3 5) (1 9 1) (1 9 5) (2 1 1) (2 1 5) (2 3 1) (2 3 5) (2 9 1) (2 9 5) (4 1 1) (4 1 5) (4 3 1) (4 3 5) (4 9 1) (4 9 5) (8 1 1) (8 1 5) (8 3 1) (8 3 5) (8 9 1) (8 9 5)))) (mapcar (lambda (x) (apply '* x)) paths-in-tree)) (1 5 3 15 9 45 2 10 6 30 18 90 4 20 12 60 36 180" ]

[ "3^2)$ = $20 \\pi$. Therefore the total area of the shaded region is $64 \\pi$.", "the area of the shaded region in each of the four squares is $3 - \\sqrt{3}$ Since there are four of these squares, we multiply this by $4$ to get $4(3 - \\sqrt{3}) = 12 - 4 \\sqrt{3}$ as our answer. This is choice $(B)$. ~ Awesome2.1 Edited by greersc", "+0 # circles -4 77 2 +-291 In the diagram, each circle is divided into two equal areas and $O$ is the center of the larger circle. The area of the larger circle is $64\\pi.$ What is the total area of the shaded regions? Apr 5, 2020", "the area of the shaded region. Area of the sum of the rectangles – area of the right triangle = area of shaded region ((17 cm × 4 cm)+(21 cm × 8 cm)) – (($$\\frac{1}{2}$$ )(13 cm × 7 cm)) (68 cm2+168 cm2 ) – ($$\\frac{1}{2}$$ )(91 cm2 ) 236 cm2 – 45.5 cm2 190.5 cm2 The area is 190.5 cm2. Question 9. a. Find the area of the shaded region. Area of the two parallelograms – area of square in the center = area of the shaded region 2(5 cm × 16 cm) – (4 cm × 4 cm) 160 cm2 – 16 cm2 144 cm2 The area is 144 cm2. b. Draw two ways the figure above can be divided in four equal parts. c. What is the area of one of the parts in (b)? 144 cm2 ÷ 4 = 36 cm2 The area of one of the parts in (b) is 36 cm2. Question 10. The figure is a rectangle made out of triangles. Find the area of the shaded region. Area of the rectangle – area of the unshaded triangles = area of the shaded region (24 cm × 21 cm) – (($$\\frac{1}{2}$$ )(9 cm × 21 cm)+($$\\frac{1}{2}$$)(9 cm × 24 cm)) 504 cm2 – (94.5 cm2 + 108 cm2 ) 504 cm2", "area of the shaded region is $$25$$ 𝑐𝑚^2. Find the perimeter (in cm) of the shaded region, given that the side length of the square is a natural number. Belum tersedia Baca juga", "area of the non-shaded region is 40. Area of rectangle =80 Hence D _________________ आत्मनॊ मोक्षार्थम् जगद्धिताय च Resource: GMATPrep RCs With Solution Math Expert Joined: 02 Aug 2009 Posts: 8006 Re: The figure shows a rectangle with a triangle inscribed in it [#permalink] ### Show Tags 24 Nov 2018, 23:52 pushpitkc wrote: Attachment: Geometry.JPG The figure shows a rectangle with a triangle inscribed in it. What is the area of the rectangle? 1. The area of the shaded region in 40. 2. The area of the non-shaded region is 40. Source: Experts Global Now area of rectangle is L*B.. the area of shaded region is $$\\frac{1}{2}*L*H$$, here height is nothing but the B of rectangle and ofcourse the Ls are same in each case..Thus area = $$\\frac{1}{2}*L*B$$ so area of rectangle is HALF the area of shaded region OR the area of shaded region is the area of non-shaded region Let us see the statements now.. 1. The area of the shaded region in 40. so area of rectangle is twice of 40 = 2*40=80 sufficient 2. The area of the non-shaded region is 40. so area of rectangle is twice of 40 = 2*40=80 sufficient D _________________ Re: The figure shows a rectangle with a triangle inscribed in it [#permalink] 24 Nov 2018, 23:52 Display posts", "square – area of the hexagon Area of the square = (15 x 15) cm2 = 225 cm2 Area of the hexagon A = (L2n)/[4tan(180/n)] A = (62 6)/ [4tan (180/6)] = (36 * 6)/ [4tan (180/6)] = 216/ [4tan (180/6)] = 216/ 2.3094 A = 93.53 cm2 Area of the shaded region = (225 – 93.53) cm2. = 131.47 cm2 ### Practice Questions 1. What is the area of the shaded region as shown below? 2. Given that $MN = 12$ m, $OP = 24$ m, and $OQ = 10$ m, what is the area of the shaded region in the diagram below? 3. What is the area of the shaded region of the rectangle shown below? 4. Suppose that $AB= 40$ ft, $BC= 25$ ft, and $CF = 30$ ft, what is the area of the shaded region of the diagram below? 5. What is the area of the shaded region shown below?", "Re:Ratio of Shaded Area to Unshaded Area of figure 11 years 4 months ago #3 to find the area of a unshaded region you to add up all the full shapes Thank you for going through this prezi hope you understand how to find the area of a unshaded region one sources you can use is zearnzillion.com unshaded region how to find the area of a unshaded region see Back to OCR Maths Higher November 2018 Paper 5 Home Q17: Answers – Paper 5 – November 2018 Shaded Area : Unshaded Rectangle : Rectangle = 4 : 7 : 11 Shaded Area : Unshaded Square : Square = 1 : 6 : 7 = 4 : 24 : 28 The ratio of the unshaded area of the rectangle to the area of the rectangle is 7 : 11. The area of the shaded portion is As = (2x)(x+2) So the ratio of the shaded portion to the rectangle is (2x)(x+2) : (2x)(2x+5), and you can factor out the 2x part from each side to simplify. An area model with 80 shaded sections and 20 unshaded sections. The ratio of the shaded region to the unshaded region is 3 : 2. N the diagram shown, what is the ratio of shaded parts to unshaded parts? An area model with", "the squares Consider one square within a square, tilted in this way. By drawing on the diagonals of the shaded square, 4 smaller squares can be formed (each shown in a different colour on the right). Half of each of these squares is shaded, so the shaded square occupies half of the whole square. Now consider the squares below. The first square is the outline of the diagram. The area of the first blue square is half of the area of the diagram. The area of the next white square is half of the area of the first blue square, so $\\frac14$ of the area of the diagram. The area of the second blue square is half of the area of the second white square, so $\\frac18$ of the area of the diagram. So the shaded area is $\\frac12-\\frac14+\\frac18=\\frac48-\\frac28+\\frac18=\\frac38$ of the area of the diagram. So the fraction of the diagram that is shaded is $\\frac38$.", "## muscrat123 one year ago The figure below shows a shaded region and a nonshaded region. Angles in the figure that appear to be right angles are right angles.What is the area, in square feet, of the shaded region?What is the area, in square feet, of the nonshaded region? 1. muscrat123 2. muscrat123 the figure is above 3. muscrat123 @pooja195 4. anonymous The shaded area is a trapezoid with the 2 by 2 square cut out. 5. anonymous $\\frac{1}{2} (6+3) 10-4 = 41$ 6. muscrat123 so 41 ft^2 is the shaded area? @robtobey 7. muscrat123 @nincompoop 8. anonymous Yes. 9. muscrat123 then u would find the area of the rectangle and subtract the area of the shaded area to find the area of the non-shaded area, right? @robtobey 10. muscrat123 16 x 8 = 128 128 - 41 = 87 so, 87 ft^2 is the non-shaded area, right? @robtobey 11. anonymous so, 87 ft^2 is the non-shaded area, right? @robtobey Yes 12. muscrat123 thx @robtobey 13. anonymous Find more explanations on OpenStudy", "the shaded part is 9$-$4 = 5 square inches. So the proportion of the area which is shaded is $\\frac5{16}$. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "way. By drawing on the diagonals of the shaded square, 4 smaller squares can be formed (each shown in a different colour on the right). Half of each of these squares is shaded, so the shaded square occupies half of the whole square. Now consider the squares below. The first square is the outline of the diagram. The area of the first blue square is half of the area of the diagram. The area of the next white square is half of the area of the first blue square, so $\\frac14$ of the area of the diagram. The area of the second blue square is half of the area of the second white square, so $\\frac18$ of the area of the diagram. So the shaded area is $\\frac12-\\frac14+\\frac18=\\frac48-\\frac28+\\frac18=\\frac38$ of the area of the diagram. So the fraction of the diagram that is shaded is $\\frac38$.", "Question 20. ERROR ANALYSIS Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet. n = 81.458 sq ft Explanation: Given, Area of ⊙Z is 255 square feet A = πr² 255 = πr² where π is 22/7 or 3.141 So, r = 9 Area of sector XZY = $$\\frac { 115 }{ 360 }$$ • 255 A = 0.3194 x 255 n = 81.458 sq ft In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure. Question 21. area of ⊙M Area = 66.04 sq cm Explanation: Based on the information given in the above figure, Question 22. Explanation: Based on the information given in the above figure, Area of region = $$\\frac { 89 }{ 360 }$$ . Area of ⊙M 12.36 = $$\\frac { 89 }{ 360 }$$ . Area of ⊙M Area of ⊙M = 49.99 πr² = 49.99 r = 3.98 In Exercises 23 – 28, find the area of the shaded region. Question 23. Area of shaded region = 1696.46 sq m Explanation: Based on the information given in the above figure, Question 24. The area of the shaded region is 85.840 sq in. Explanation: Given, Area of square = 20² = 20 x", "shaded portion. Solution: Total area of the shaded region = Area of a right triangle + Area of a rectangle = ($$\\frac{1}{2}$$ × b × h) + (l × b) cm2 = [($$\\frac{1}{2}$$ × 3 × 4) + (9 × 6)] cm2 = (6 + 54) cm2 = 60 cm2 Question 14. Find the approximate area of the flower in the given square grid. Solution: No of full squares = 11 No of half squares = 9 Area of 11 full squares = 11 x 1 cm² = 11 cm² Area of 9 half squares = 9 × $$\\frac{1}{2}$$ cm² = 4.5 cm² Area of the flower = (11 + 4.5) cm² = 15.5 cm²", "is correct (which appears to be true), you are correct that the single outside shaded region has area $$4\\pi - 8$$ and the inside region (the shaded quadrilateral within the square) has area $$12.$$ So we agree that the total is $$4\\pi + 4,$$ which means only the \"none of these\" answer can be correct. • Note that this answer is merely a summary of some of the comments so that you can decide whether your question is answered. I don't get any credit if you accept this answer, but after you have had some time to consider it (and other people have had a chance to post a better answer if they have one), accepting this answer would let people know you have gotten what you need. – David K Nov 14 '18 at 20:06 • Ok. Just did it. – shawn k Nov 17 '18 at 8:16", "× Area of object = 22 × 14 = 56 cm2 Area of shaded region = 56 – 14 = 42 cm2", "3:1 [Reveal] Spoiler: Attachment: 2018-02-05_1750.png Area of Unshaded region = $$\\frac{1}{2}*l*b$$ Area of Rectangle = $$l*b$$ $$Ratio = (\\frac{Total Area}{Area of Unshaded Region}) - 1$$ = 1 _________________ If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Re: In the rectangle above, what is the ratio of (area of shaded region) [#permalink] 07 Feb 2018, 01:17 Display posts from previous: Sort by", "Exercises1.1.8Exercises Exercise1 Hint Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle. The area is between $1.5$ and $2.5$ square units. Solution The diagram on the left shows a rectangle with area $2 \\times 1.25=2.5$ square units. Since the blue-shaded region is entirely inside this rectangle, the area of the blue-shaded region is no more than 2.5 square units. The diagram on the right shows a rectangle with area $2 \\times 0.75=1.5$ square units. Since the blue-shaded region contains this entire rectangle, the area of the blue region is no less than 1.5 square units. So, the area of the blue-shaded region is between 1.5 and 2.5 square units. Remark: we could also give an obvious range, like “the shaded area is between zero and one million square units.” This would be true, but not very useful or interesting. Exercise2 Hint We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region. The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but", "of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 8 × 5 = 40 sq m; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 5 × 4 = 10 sq m Step 2: Area of shaded region = Area of rectangle – Area of triangle = 40 – 10 = 30 square m Q 7 - Find the area of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 8 × 6 = 48 sq m; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 4 × 2 = 4 sq m Step 2: Area of shaded region = Area of rectangle – Area of triangle = 48 – 4 = 44 square m Q 8 - Find the area of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 8 × 5 = 40 sq ft; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 4 × 3 = 6 sq ft Step 2: Area of shaded region = Area of rectangle – Area of triangle = 40 – 6 = 34 square ft Q 9 - Find the area of the", "the area of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 9 × 6 = 54 sq m; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 6 × 2 = 6 sq m Step 2: Area of shaded region = Area of rectangle – Area of triangle = 54 – 6 = 48 square m Q 4 - Find the area of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 8 × 6 = 48 sq m; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 6 × 4 = 12 sq m Step 2: Area of shaded region = Area of rectangle – Area of triangle = 48 – 12 = 36 square m Q 5 - Find the area of the shaded region in the following figure. ### Explanation Step 1: Area of rectangle = l × w = 10 × 6 = 60 sq m; Area of triangle = $\\frac{1}{2}$ b h = $\\frac{1}{2}$ × 7 × 2 = 7 sq m Step 2: Area of shaded region = Area of rectangle – Area of triangle = 60 – 7 = 53 square m Q 6 - Find the area" ]

[ "# Difference between revisions of \"2020 AMC 8 Problems/Problem 24\" ## Problem A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$. When $n=24$, the $576$ gray tiles cover $64\\%$ of the area of the large square region. What is the ratio $\\frac{d}{s}$ for this larger value of $n?$ $[asy] draw((0,0)--(13,0)--(13,13)--(0,13)--cycle); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray); filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray); filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray); filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray); filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray); filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray); filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray); filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray); filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray); [/asy]$ $\\textbf{(A) }\\frac{6}{25} \\qquad \\textbf{(B) }\\frac{1}{4} \\qquad \\textbf{(C) }\\frac{9}{25} \\qquad \\textbf{(D) }\\frac{7}{16} \\qquad \\textbf{(E) }\\frac{9}{16}$ ## Solution 1 The area of the shaded region is $(24s)^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $\\frac{(24s)^2}{(24s+25d)^2}=\\frac{64}{100}$. Taking the square root of both sides (and using the fact that lengths are non-negative) gives $\\frac{24s}{24s+25d}=\\frac{8}{10} = \\frac{4}{5}$, and cross-multiplying now", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square? $[asy] draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle); draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle); draw((0,1)--(0,2)); draw((1,0)--(2,0)); draw((0,-1)--(0,-2)); draw((-1,0)--(-2,0)); [/asy]$ $\\text{(A)}\\ 25 \\qquad \\text{(B)}\\ 44 \\qquad \\text{(C)}\\ 50 \\qquad \\text{(D)}\\ 62 \\qquad \\text{(E)}\\ 75$ ## Problem 19 The graph shows the distribution of the number of children in the families of the students in Ms. Jordan's English class. The median number of children in the family for this distribution is $[asy] unitsize(12); for(int i = 1; i <= 7; ++i) { draw((0,i)--(19,i),dotted); draw((-0.5,i)--(0.5,i)); } for(int i = 0; i <= 5; ++i) { draw((3*i+2,0)--(3*i+2,-0.5)); } fill((1,0)--(1,2)--(3,2)--(3,0)--cycle,white); fill((4,0)--(4,1)--(6,1)--(6,0)--cycle,white); fill((7,0)--(7,2)--(9,2)--(9,0)--cycle,white); fill((10,0)--(10,2)--(12,2)--(12,0)--cycle,white); fill((13,0)--(13,6)--(15,6)--(15,0)--cycle,white); draw((0,9)--(0,0)--(19,0)); draw((1,0)--(1,2)--(3,2)--(3,0)); draw((4,0)--(4,1)--(6,1)--(6,0)); draw((7,0)--(7,2)--(9,2)--(9,0)); draw((10,0)--(10,2)--(12,2)--(12,0)); draw((13,0)--(13,6)--(15,6)--(15,0)); label(\"1\",(2,-0.5),S); label(\"2\",(5,-0.5),S); label(\"3\",(8,-0.5),S); label(\"4\",(11,-0.5),S); label(\"5\",(14,-0.5),S); label(\"6\",(17,-0.5),S); label(\"2\",(-0.5,2),W); label(\"4\",(-0.5,4),W); label(\"6\",(-0.5,6),W); label(\"\\textbf{Number of Children}\",(9,-1.5),S); label(\"\\textbf{in the Family}\",(9,-2.5),S); label(\"\\textbf{Number}\",(-1.5,6),W); label(\"\\textbf{of}\",(-3,5),W); label(\"\\textbf{Families}\",(-1.5,4),W); [/asy]$ $\\text{(A)}\\ 1 \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 3 \\qquad \\text{(D)}\\ 4 \\qquad \\text{(E)}\\ 5$ ## Problem 20 Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana's number is larger than Apollo's number? $\\text{(A)}\\ \\dfrac{1}{3} \\qquad \\text{(B)}\\ \\dfrac{5}{12} \\qquad \\text{(C)}\\ \\dfrac{4}{9} \\qquad \\text{(D)}\\ \\dfrac{17}{36} \\qquad \\text{(E)}\\ \\dfrac{1}{2}$ ## Problem 21 A plastic snap-together cube has a protruding snap on one side and", "if the rectangle cannot be placed.\"\"\" (w, h) = rect (x0, y0) = pos newgrid = map(list, grid) # make a copy try: for x in range(x0, x0+w): for y in range(y0, y0+h): if newgrid[y][x] is not empty: return None newgrid[y][x] = rect return newgrid except IndexError: # went off the grid return None In [118]: # Place a 3 x 4 rectangle on a square grid, 2 cells over and 1 down: place_rectangle_at((3, 4), Square(5), (2, 1)) Out[118]: [[(0, 0), (0, 0), (0, 0), (0, 0), (0, 0)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)], [(0, 0), (0, 0), (3, 4), (3, 4), (3, 4)]] In [119]: # Place two rectangles on a grid grid2 = place_rectangle_at((3, 4), Square(5), (2, 1)) grid3 = place_rectangle_at((2, 5), grid2, (0, 0)) grid3 Out[119]: [[(2, 5), (2, 5), (0, 0), (0, 0), (0, 0)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)], [(2, 5), (2, 5), (3, 4), (3, 4), (3, 4)]] In [120]: # Now try to place a third rectangle that does not fit;", "# Difference between revisions of \"1998 AJHSME Problems/Problem 24\" ## Problem A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? $[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label(\"2\",(1.5,8.2),N); label(\"4\",(3.5,8.2),N); label(\"5\",(4.5,8.2),N); label(\"7\",(6.5,8.2),N); label(\"8\",(7.5,8.2),N); label(\"9\",(0.5,7.2),N); label(\"11\",(2.5,7.2),N); label(\"12\",(3.5,7.2),N); label(\"13\",(4.5,7.2),N); label(\"14\",(5.5,7.2),N); label(\"16\",(7.5,7.2),N);[/asy]$ $\\text{(A)}\\ 36\\qquad\\text{(B)}\\ 64\\qquad\\text{(C)}\\ 78\\qquad\\text{(D)}\\ 91\\qquad\\text{(E)}\\ 120$ ## Solution ### Solution 1 The numbers that are shaded are the triangular numbers, which are numbers in the form $\\frac{(n)(n+1)}{2}$ for positive integers. They can also be generated by starting with $1$, and adding $1, 2, 3, 4...$ as in the description of the problem. Squares that have the same remainder after being divided by $8$ will", "draw a red diagonal line, 6 pixels wide, from point (100,100) to point (200,200). draw_set_colour(c_red); draw_line_width(100, 100, 200, 200, 6); I can foresee you will have corner issues as the adjoining lines corner won't meet up. It will become more obvious as you thicken the lines and the two angles are extreme. That is more tricky to solve given you need to work out the corners for each line taking thickness into account and join those with the previous line corners. Quite a bit of maths and code involving atan2, cos and sin is required to make this work correctly. • Search for \"graphic gems 1\" and open link to pdf - jump to page 108 - the diagram on that page will show you what you will be faced with when doing it this way. – John Oct 2 '19 at 3:12 Adding 1 to each dimension just isn't enough. It would have to be something like: draw_path(circuito_ovo,500,50,false); draw_path(circuito_ovo,500,51,false); draw_path(circuito_ovo,500,52,false); draw_path(circuito_ovo,501,50,false); draw_path(circuito_ovo,501,51,false); draw_path(circuito_ovo,501,52,false); draw_path(circuito_ovo,502,50,false); draw_path(circuito_ovo,502,51,false); draw_path(circuito_ovo,502,52,false); For ease just use the repeat function (Draw Event): repeat(5){ repeat(5){ draw_path(circuito_ovo,500+i,50+j,false); j+=1; } i+=1; j=0; } i=0; j=0; Initializing the variables with the value 0 (i and j), or negative (if I want the object to be walked exactly in the center of the drawing). The number of times the", "(m-2-12.south east); \\draw[Red!,line width=2] (m-1-14.north west) rectangle (m-1-14.south east); \\draw (m-2-11.north west) rectangle (m-2-11.south east); \\draw (m-3-13.north west) rectangle (m-3-13.south east); \\draw[Red!,line width=2] (m-2-13.north west) rectangle (m-2-13.south east); \\draw (m-3-11.north west) rectangle (m-3-11.south east); \\draw[Red!,line width=2] (m-3-12.north west) rectangle (m-3-12.south east); \\draw (m-4-11.north west) rectangle (m-4-11.south east); \\draw[Red!,line width=2] (m-4-12.north west) rectangle (m-4-12.south east); \\draw (m-1-21.north west) rectangle (m-1-21.south east); \\draw (m-1-22.north west) rectangle (m-1-22.south east); \\draw (m-1-20.north west) rectangle (m-1-20.south east); \\draw (m-2-20.north west) rectangle (m-2-20.south east); \\draw (m-2-21.north west) rectangle (m-2-21.south east); \\draw (m-2-23.north west) rectangle (m-2-23.south east); \\draw[Red!,line width=2] (m-1-23.north west) rectangle (m-1-23.south east); \\draw (m-3-22.north west) rectangle (m-3-22.south east); \\draw[Red!,line width=2] (m-2-22.north west) rectangle (m-2-22.south east); \\draw (m-3-20.north west) rectangle (m-3-20.south east); \\draw[Red!,line width=2] (m-3-21.north west) rectangle (m-3-21.south east); \\draw (m-4-20.north west) rectangle (m-4-20.south east); \\draw[Red!,line width=2] (m-4-21.north west) rectangle (m-4-21.south east); \\draw[Red!,line width=2] (m-5-20.north west) rectangle (m-5-20.south east); \\draw (m-1-28.north west) rectangle (m-1-31.south east); \\draw[Orange!,line width=2] (m-1-32.north west) rectangle (m-1-32.south east); \\draw (m-1-28.north west) rectangle (m-4-28.south east); \\draw (m-1-30.north west) rectangle (m-3-30.south east); \\draw (m-2-29.north west) rectangle (m-2-31.south east); \\draw (m-4-28.north west) rectangle (m-5-28.south east); \\draw (m-3-28.north west) rectangle (m-3-28.south east); \\draw (m-4-29.north west) rectangle (m-4-29.south east); \\end{tikzpicture} I want to show the step-by-step progression of a bumping route from the beginning of the Robinson-Schensted insertion algorithm to the end and I would", "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "# Problem #2155 2155 Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed? $[asy] unitsize(1inch); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--((1/3) + 3*(1/2),0)); fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7)); draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle); draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle); draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle); draw((2,0)--(2 + (1/3) + (3/2),0)); draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle); draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle); label(\"B\",((1/6) + (1/2),(1/2)),NW); label(\"B\",((2/3) + 2 + (1/4),(29/30)),NNE); draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3)); fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7)); draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);[/asy]$ $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ \\sqrt{2}\\qquad\\textbf{(C)}\\ \\frac{3}{2}\\qquad\\textbf{(D)}\\ \\sqrt{2}+\\frac{1}{2}\\qquad\\textbf{(E)}\\ 2$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"2013 AMC 8 Problems/Problem 18\" ## Problem Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain? $[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label(\"10\",(A--B),SE); label(\"12\",(C--B),SW); label(\"5\",(F--B),W);[/asy]$ $\\textbf{(A)}\\ 204 \\qquad \\textbf{(B)}\\ 280 \\qquad \\textbf{(C)}\\ 320 \\qquad \\textbf{(D)}\\ 340 \\qquad \\textbf{(E)}\\ 600$ ## Solution 1 There are $10 \\cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11", "in Shaddoll's solution, and our answer is $p=13+84+85=\\boxed{182}$. ## Solution 4 $[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"A\", A, SE); label(\"B\", B, NW); label(\"C_0\", C0, SW); label(\"C_1\", C1, NE); label(\"C_2\", C2, W); label(\"C_3\", C3, NE); label(\"C_4\", C4, W); label(\"a\", (B+C0)/2, W); label(\"b\", (A+C0)/2, S); label(\"c\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]$ Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\\frac{a+c}{b}$. The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \\cdots$, shown in red in the figure below. Since each of the triangles $\\triangle C_0 C_1 C_2, \\triangle C_2 C_3 C_4, \\dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the", "clipping. Here's what my two paths look like in that square: And here's what I want that square to look like: My code for this is not intelligent, and not my first try, but it's what I have at hand: \\begin{tikzpicture}[scale=.1] \\foreach \\x in {1,2,...,7} \\foreach \\y in {0,1,...,5} { \\begin{scope}[xshift=-10*\\x cm, yshift=10*\\y cm] \\draw[clip] (0,5)--(2,5)--(2,4)--(3,4)--(3,2)--(4,2)--(4,1)--(7,1)--(7,0)--(0,0)-- cycle; \\draw[fill=red] (0,5)--(2,5)--(2,4)--(3,4)--(3,2)--(4,2)--(4,1)--(7,1)--(7,0)--(0,0)--cycle; \\draw[fill=white] (-\\x, \\y + 20) |- ++(2,-15) |- ++(1,-1) |- ++(1,-2) |- ++(3,-1) |- ++(13,-1) -- ++(\\x,-\\y) -| ++(-13,1) -| ++(-3,1) -| ++(-1,2) -| ++(-1,1) -| ++(-2,15) -| ++(40,-40) -| (-\\x, \\y + 20); \\draw[very thin, gray] (0,0) grid (8,8); \\end{scope} } \\end{tikzpicture} • Unrelated to the question, but you might check out the turtle library of TikZ which would make it much easier to create such paths. – Qrrbrbirlbel Jul 4 '13 at 23:17 • Are the pictures you want to create always composed of little squared in white or red? I see a pattern here which would make it much easier to create on a different way and a different type of input by specifying the number of rectangle for the rows and for the “whitening” path. – Qrrbrbirlbel Jul 4 '13 at 23:25 • @Qrrbrbirlbel Yeah -- it's always little collections of squares (partitions) that I'm doing; it probably would be easier in the long run", "# Difference between revisions of \"2013 Indonesia MO Problems\" ## Day 1 ### Problem 1 In a $4 \\times 6$ grid, all edges and diagonals are drawn (see attachment). Determine the number of parallelograms in the grid that uses only the line segments drawn and none of its four angles are right. $[asy] draw((0,0)--(6,0)--(6,4)--(0,4)--(0,0)); for (int i=1; i<6; ++i) { draw((i,0)--(i,4)); } for (int i=1; i<4; ++i) { draw((0,i)--(6,i)); } draw((0,1)--(1,0)); draw((0,2)--(2,0)); draw((0,3)--(3,0)); draw((0,4)--(4,0)); draw((1,4)--(5,0)); draw((2,4)--(6,0)); draw((3,4)--(6,1)); draw((4,4)--(6,2)); draw((5,4)--(6,3)); draw((0,3)--(1,4)); draw((0,2)--(2,4)); draw((0,1)--(3,4)); draw((0,0)--(4,4)); draw((1,0)--(5,4)); draw((2,0)--(6,4)); draw((3,0)--(6,3)); draw((4,0)--(6,2)); draw((5,0)--(6,1)); [/asy]$ ### Problem 2 Let $ABC$ be an acute triangle and $\\omega$ be its circumcircle. The bisector of $\\angle BAC$ intersects $\\omega$ at [another point] $M$. Let $P$ be a point on $AM$ and inside $\\triangle ABC$. Lines passing $P$ that are parallel to $AB$ and $AC$ intersects $BC$ on $E, F$ respectively. Lines $ME, MF$ intersects $\\omega$ at points $K, L$ respectively. Prove that $AM, BL, CK$ are concurrent. ### Problem 3 Determine all positive real $M$ such that for any positive reals $a,b,c$, at least one of $a + \\dfrac{M}{ab}, b + \\dfrac{M}{bc}, c + \\dfrac{M}{ca}$ is greater than or equal to $1+M$. ### Problem 4 Suppose $p > 3$ is a prime number and $$S = \\sum_{2 \\le i < j < k \\le p-1} ijk$$ Prove", "& & & & & & & \\\\ } ; \\draw (m-1-1.north west) rectangle (m-1-2.south east); \\draw (m-2-4.north west) rectangle (m-2-4.south east); \\draw[red!,line width=2] (m-1-4.north west) rectangle (m-1-4.south east); \\draw (m-1-1.north west) rectangle (m-4-1.south east); \\draw (m-1-3.north west) rectangle (m-1-3.south east); \\draw (m-3-3.north west) rectangle (m-3-3.south east); \\draw[red!,line width=2] (m-2-3.north west) rectangle (m-2-3.south east); \\draw (m-3-1.north west) rectangle (m-3-1.south east); \\draw (m-4-2.north west) rectangle (m-4-2.south east); \\draw[red!,line width=2] (m-3-2.north west) rectangle (m-3-2.south east); \\draw (m-1-11.north west) rectangle (m-1-11.south east); \\draw (m-1-12.north west) rectangle (m-1-12.south east); \\draw (m-1-13.north west) rectangle (m-1-13.south east); \\draw (m-2-14.north west) rectangle (m-2-14.south east); \\draw (m-2-12.north west) rectangle (m-2-12.south east); \\draw[red!,line width=2] (m-1-14.north west) rectangle (m-1-14.south east); \\draw (m-2-11.north west) rectangle (m-2-11.south east); \\draw (m-3-13.north west) rectangle (m-3-13.south east); \\draw[red!,line width=2] (m-2-13.north west) rectangle (m-2-13.south east); \\draw (m-3-11.north west) rectangle (m-3-11.south east); \\draw[red!,line width=2] (m-3-12.north west) rectangle (m-3-12.south east); \\draw (m-4-11.north west) rectangle (m-4-11.south east); \\draw[red!,line width=2] (m-4-12.north west) rectangle (m-4-12.south east); \\draw (m-1-21.north west) rectangle (m-1-21.south east); \\draw (m-1-22.north west) rectangle (m-1-22.south east); \\draw (m-1-20.north west) rectangle (m-1-20.south east); \\draw (m-2-20.north west) rectangle (m-2-20.south east); \\draw (m-2-21.north west) rectangle (m-2-21.south east); \\draw (m-2-23.north west) rectangle (m-2-23.south east); \\draw[red!,line width=2] (m-1-23.north west) rectangle (m-1-23.south east); \\draw (m-3-22.north west) rectangle (m-3-22.south east); \\draw[red!,line width=2] (m-2-22.north west) rectangle (m-2-22.south east); \\draw (m-3-20.north west) rectangle (m-3-20.south east); \\draw[red!,line", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "4 & 6 & & & & & & = & 3 & 4 & 6 & & & & & = & 3 & 4 & 6 & & & & & = \\\\ 4 & 7 & & & \\leftarrow & 5 & \\leftarrow & 3 & 1 & & 4 & 5 & & & & & & & & 4 & 5 & & & & & & & 4 & 5 & & & & & & \\\\ & & & & & & & & & & & & & & \\leftarrow & 7 & \\leftarrow 3 & 1 & & 7 & & & & & & & & 7 & & & & & & & \\\\ } ; \\draw (m-1-1.north west) rectangle (m-1-2.south east); \\draw (m-2-4.north west) rectangle (m-2-4.south east); \\draw[Red!,line width=2] (m-1-4.north west) rectangle (m-1-4.south east); \\draw (m-1-1.north west) rectangle (m-4-1.south east); \\draw (m-1-3.north west) rectangle (m-1-3.south east); \\draw (m-3-3.north west) rectangle (m-3-3.south east); \\draw[Red!,line width=2] (m-2-3.north west) rectangle (m-2-3.south east); \\draw (m-3-1.north west) rectangle (m-3-1.south east); \\draw (m-4-2.north west) rectangle (m-4-2.south east); \\draw[Red!,line width=2] (m-3-2.north west) rectangle (m-3-2.south east); \\draw (m-1-11.north west) rectangle (m-1-11.south east); \\draw (m-1-12.north west) rectangle (m-1-12.south east); \\draw (m-1-13.north west) rectangle (m-1-13.south east); \\draw (m-2-14.north west) rectangle (m-2-14.south east); \\draw (m-2-12.north west) rectangle", "(-7,5) | 8.60 | | 28 | (8,1) | (7,-6) | 9.22 | | 29 | (1,8) | (-7,7) | 9.90 | +------+-------+---------+------+ There are 4 combinations that were skipped with change in coordinates of • (6,5) doesn't fit • (6,4) doesn't fit • (5,5) same length as (7,1) • (4,3) same length as (5,0) • You can find the maximum eventually – TSLF Oct 15 '16 at 21:29 • From 19 to 20 your distance decreases (according to your chart). – GentlePurpleRain Oct 17 '16 at 14:34 There's an upper bound of 33 distinct distance vectors from (1,0) to (7,7) 1x0, 1x1, 2x0, 2x1, 2x2, 3x0, 3x1, 3x2, 4x0, 4x1, 3x3, 4x2, 4x3=5x0, 5x1, 5x2, 4x4, 5x3, 6x0, 6x1, 5x4, 6x2, 6x3, 7x0, 5x5=7x1, 6x4, 7x2, 7x3, 6x5, 7x4, 6x6, 7x5, 7x6, 7x7 I appear to have found a path that hits 30 jumps, but I fear I may have made a mistake somewhere (3,3), (4,3), (3,4), (5,3), (3,5), (6,4), (3,2), (3,6), (4,2), (7,5), (3,7), (6,3), (2,6), (3,1), (5,6), (1,2), (4,7), (4,1), (5,7), (1,3), (7,1), (1,4), (8,4), (1,3), (8,5), (1,7), (6,1), (2,8), (8,2), (1,8), (8,1) • (3,4) to (5,3) and (5,3) to (3,2) are the same length. As are (8,8) to (1,6) and (1,6) to (8,4) – gtwebb Oct 14 '16 at 20:26 • Maybe you selected", "= 12 2^2 * 3^1 * 5^1 = (4 * 3 * 5) = 60 2^2 * 3^2 * 5^0 = (4 * 9 * 1) = 36 2^2 * 3^2 * 5^1 = (4 * 9 * 5) = 180 2^3 * 3^0 * 5^0 = (8 * 1 * 1) = 8 2^3 * 3^0 * 5^1 = (8 * 1 * 5) = 40 2^3 * 3^1 * 5^0 = (8 * 3 * 1) = 24 2^3 * 3^1 * 5^1 = (8 * 3 * 5) = 120 2^3 * 3^2 * 5^0 = (8 * 9 * 1) = 72 2^3 * 3^2 * 5^1 = (8 * 9 * 5) = 360 (let ((paths-in-tree '((1 1 1) (1 1 5) (1 3 1) (1 3 5) (1 9 1) (1 9 5) (2 1 1) (2 1 5) (2 3 1) (2 3 5) (2 9 1) (2 9 5) (4 1 1) (4 1 5) (4 3 1) (4 3 5) (4 9 1) (4 9 5) (8 1 1) (8 1 5) (8 3 1) (8 3 5) (8 9 1) (8 9 5)))) (mapcar (lambda (x) (apply '* x)) paths-in-tree)) (1 5 3 15 9 45 2 10 6 30 18 90 4 20 12 60 36 180" ]

[ "$$X(2)=(X\\times X)/\\mathbb{Z}_2$$ does not.", "$$3\\times 1$$. Which one is 1 and which one is 3? You can solve for the values of x and y respectively in each case, and see if it contradicts the condition \"... whole numbers x and y between 1 and 10 inclusive ...\". Please tell me if you are still stuck. Apr 19, 2022", "is TWO x as well. Ah. Thank you! Yes it's $\\cos{(2\\times 2x)}.$", "# How do you write \"twice y\" as an algebraic expression? $2 y$ \"Twice of $y$\" means $y$ times two. $2 \\times y = 2 y$", "$5 \\times x - 2 \\times y \\times y = 1.$\" What could the value of $x$ be? Let $y$ some real number and solve $5 \\times x - 2 \\times y \\times y = 1$ for $x.$ Does this value of $x$ work for any $y?$ Try a different value of $y.$ Is the equation still true for this $y$ and the $x$ you found above? Penny Math Central is supported by the University of Regina and the Imperial Oil Foundation.", "2$$ $2x + y = 2 \\times 3 + 2 = 6 + 2 = 8$", "Y)+(X\\times Z).}$", "Q # Find the value of the following expressions, when x = - 1: (iv) 2x^ 2 - x - 2 Q. 3. Find the value of the following expressions, when $x=-1$: (iv) $2x^{2}-x-2$ $\\\\2x^2 - x - 2 \\\\= 2\\times ( -1 )^2 - ( -1 ) - 2 \\\\ = 2 + 1 - 2 \\\\= 1$", "X_3) = \\int_0^1 G(y) \\times 1 \\, dy$. -", "# If x = 4 and y = 2, what is the value of the following expression: 3xy - 12y + 5x 20 Explanation To find the value of this expression, substitute the given values for x and y into the expression: 3(4)(2) - 12(2) + 5(4) Then, calculate the value of the expression: $$3 \\times 8-12 \\times 2+5 \\times 4=24-24+20=20$$", "# Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "we see that at $x=1$, these have the values $2e$ and $2e^2$, respectively.", "= 4$$, we can write $$y = 2 \\times x = 2x$$ - - - - - - - (IV) From equation (I), (II), (III) and (IV): We can generalise the relation as $$y = 2 \\times x$$ or $$y = 2x$$.", "## If $$x+2y+1=y-x,$$ what is the value of $$x?$$ ##### This topic has expert replies Moderator Posts: 1991 Joined: 15 Oct 2017 Followed by:6 members ### If $$x+2y+1=y-x,$$ what is the value of $$x?$$ by BTGmoderatorLU » Mon Aug 23, 2021 10:41 am 00:00 A B C D E ## Global Stats Source: Official Guide If $$x+2y+1=y-x,$$ what is the value of $$x?$$ 1) $$y^2=9$$ 2) $$y=3$$ The OA is B • Page 1 of 1", "\\times x) \\times x' = x \\times x' = 0$. 3. $x \\times (x + y) = (x + 0) \\times (x + y) = x + (0 \\times y) = x + 0 = x$. 4. $x + (x \\times y) = (x \\times 1) + (x \\times y) = x \\times (1 + y) = x \\times 1 = x$. Are there quicker ways to do any of these? 1. $\\hspace{10px}$False: $a$ can never equal $a' x + x' = 1$, but $x + x = x.$ So $x = 1$. However $1 \\times 1 = 1$ and $1 \\times 1' = 0$. 2. $\\hspace{5px}$True: $1 = 1 \\times 1 = (x' + x) \\times (x' + x'') = x' + (x \\times x'')$ so $x = x \\times x''$ and $x'' = x$. 3. False by assuming that $y$ is also an inverse of $x: x' = x' \\times 1 = x' \\times (x + y) = (x' \\times y) + (x' \\times y) = x' \\times y$. But we can replace $x'$ by $y$ and $y$ by $x'$ in this series of equations to get $x' = x' \\times y = y \\times x'= y$. 1. To do this we need to prove that $(x + y) + (x' \\times y') = 1$ and", "# How do you translate to an algebraic expression twice the sum of x and y? $2 \\times \\left(x + y\\right)$ Twice = two times $2 \\times$", "$2 \\times 0$.", "is $$\\sqrt {{3 \\over x}}$$. The value of x will be _______________.", "# If 2cos^2x - cosx = 1 , what is the value of x? Apr 2, 2016 $x = 0$ #### Explanation: As we know: $\\cos 0 = 1$ Put $x = 0$ in equation: $2 {\\cos}^{2} 0 - \\cos 0 = 1$ $2 \\cdot 1 - 1 = 1$ $1 = 1$ $L . H . S = R . H . S$", "Q # Find the value of the following expressions, when x = -1: (iii) x^ 2 + 2x + 1 Q. 3. Find the value of the following expressions, when $x=-1$: (iii) $x^{2}+2x+1$ $\\\\x^2 + 2x + 1 \\\\= ( -1 )^2 + 2 \\times ( -1 ) + 1 \\\\= 1 - 2 + 2 \\\\= 0$" ]

[ "Y)+(X\\times Z).}$", "Factorise the following expressions $2 x^3 + 2xy^2 + 2 xz ^2$ We have, $2x^3 = 2 \\times x \\times x \\times x$ $2xy^2 = 2 \\times x \\times y \\times y$ $2xz^2 = 2 \\times x \\times z \\times z$ Therefore, $2 x^3 + 2xy^2 + 2 xz ^2$ $= (2 \\times x \\times x \\times x) + ( 2 \\times x \\times y \\times y) + ( 2 \\times x \\times z \\times z)$ $= (2 \\times x) [(x \\times x) + (y \\times y ) + (z \\times z)]$ $= 2x(x^2+y^2+z^2)$ Question:3(iv) Factorise the following expressions $am^2 + bm ^2 + bn ^2 + an^2$ We have, $am^2 + bm ^2 + bn ^2 + an^2$ $= m^2(a + b) + n^2(a + b)$ $= (a + b)$$(m^{2 }+n^{2})$ Question:3(v) Factorise the following expressions $( lm + l ) + m + 1$ We have, $( lm + l ) + m + 1$ $= lm + l + m + 1$ $= l(m + 1) +1(m + 1)$ $= (m + 1)(l + 1)$ Question:3(vi) Factorise the following expressions $y ( y + z ) + 9 ( y + z )$ We have, $y ( y + z ) + 9 ( y + z )$ Take ( y+z) common from this Therefore,", "Now, $$x!$$ is >$$1\\times 2\\times 3\\times$$ … $$(x-1)\\times x$$, so $$(x-1)$$ and 2 are included in $$x!$$ (that is, provided that $$x$$ is greater than 2). Hence, this is the correct answer. Incorrect. [[snippet]] Yes, statements II and III are true. But what about statement I? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I and III are true. But what about statement II? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I is true. But what about statements II and III? Can you find an example eliminating these statements? Incorrect. [[snippet]] Yes, statement II is true. But what about statements I and III? Can you find an example eliminating these statements? Correct. [[snippet]] __Plug In__ $$x = 3$$. In this case, the product could be $$3 \\times 4 \\times 5 = 60$$. >I. 60 is divisible by $$x - 1 = 3 - 1 = 2$$. >II. 60 is divisible by $$2x = 2 (3) = 6$$. >III. 60 is divisible by $$x! = 3! = 3 \\times 2 \\times 1 = 6$$. __Plug In__ another value of $$x$$ to be sure because the question says *must be*. This time try an even integer (such as $$x= 4$$). In this case, the product could be $$2 \\times 3 \\times", "means $$(n \\times n)-1!$$ according to the order of operations. $$n! = n \\times (n-1)!$$$OK, we have • $$n! = n \\times (n-1)!$$, and • $$(n-1)! = (n-1) \\times (n-2)!$$, and • $$(n-2)! = (n-2) \\times (n-3)!$$, and so on. So $$n! = n \\times (n-1)! = n \\times (n-1) \\times (n-2)! = n \\times (n-1) \\times (n-2) \\times (n-3)!$$$ and so on. If we just keep going, we’ll eventually reach $$n \\times (n-1) \\times (n-2) \\times \\dots \\times 5 \\times 4 \\times 3!$$$and we already know that $$3! = 6$$, which I’ll write as $$3 \\times 2 \\times 1$$ for reasons which are about to become obvious. So we have the following formula, which is how we define the factorial: $$n! = n \\times (n-1) \\times \\dots \\times 4 \\times 3 \\times 2 \\times 1$$ “$n!$” is read out loud as “$n\\$ factorial”, and it means “the number of ways to arrange $$n$$ objects into any order”. # Edge cases We’ve seen $$3!$$, but never $$2!$$ or $$1!$$. • It’s easy to see that there are two ways to arrange two objects into an order: $$1,2$$ or $$2,1$$. So $$2! = 2$$. • It’s also easy (if a bit weird) to see that there is just one way of arranging one object into an order: $$1$$ is", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "- e^{x_i}) = e - e^0 = \\boxed{e-1}.$ Done!", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "= {y - 2 \\over 0} = {z - 2 \\over -2} = t$$ or equivalently $$\\boxed{x = t + 1, \\ \\ y = 2, \\ \\ z = - 2 t + 2}$$", "$\\frac{13 - b}{7 - b} = 2$ Now we simplify the fraction : $13 - b = 2(7 - b)$ $13 - b = 14 - 2b$ $- b + 2b = 14 - 13$ $\\boxed{b = 1}$ 2. Find $a$ Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ : $13 = 4a + b$ Since $b = 1$, we get : $13 = 4a + 1$ $12 = 4a$ $3 = a$ $\\boxed{a = 3}$ 3. Conclusion You are done : you have found the linear relationship between $x$ and $f(x)$ : $\\boxed{f(x) = 3x + 1}$ Let's check those results : $f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ --> $f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ --> $f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ --> $f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ --> Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ : $f(1) = 1 \\times 3", "care­ful to put $$n-1$$ in brack­ets, be­cause oth­er­wise it looks like $$n \\times n - 1!$$, which means $$(n \\times n)-1!$$ ac­cord­ing to the or­der of op­er­a­tions. $$n! = n \\times (n-1)!$$$OK, we have • $$n! = n \\times (n-1)!$$, and • $$(n-1)! = (n-1) \\times (n-2)!$$, and • $$(n-2)! = (n-2) \\times (n-3)!$$, and so on. So $$n! = n \\times (n-1)! = n \\times (n-1) \\times (n-2)! = n \\times (n-1) \\times (n-2) \\times (n-3)!$$$ and so on. If we just keep go­ing, we’ll even­tu­ally reach $$n \\times (n-1) \\times (n-2) \\times \\dots \\times 5 \\times 4 \\times 3!$$$and we already know that $$3! = 6$$, which I’ll write as $$3 \\times 2 \\times 1$$ for rea­sons which are about to be­come ob­vi­ous. So we have the fol­low­ing for­mula, which is how we define the fac­to­rial: $$n! = n \\times (n-1) \\times \\dots \\times 4 \\times 3 \\times 2 \\times 1$$ “$n!$” is read out loud as “$n\\$ fac­to­rial”, and it means “the num­ber of ways to ar­range $$n$$ ob­jects into any or­der”. # Edge cases We’ve seen $$3!$$, but never $$2!$$ or $$1!$$. • It’s easy to see that there are two ways to ar­range two ob­jects into an or­der: $$1,2$$ or $$2,1$$. So $$2! = 2$$. • It’s also easy (if a bit weird)", "## Prealgebra (7th Edition) $x=20$ \"Twice\" means multiplied by 2. The term preceding \"subtracted from\" is the subtrahend. The comma tells us the \"twice\" applies to \"a number\" rather than to the difference. $60-2x=20$ $\\longrightarrow$addition property $60-2x-60=20-60$ $\\longrightarrow$combine like terms $-2x=-40$ $\\longrightarrow$multiplication property $-2x\\div-2=-40\\div-2$ $\\longrightarrow$divide $x=20$", "Q # Q. 5. Give expressions in the following cases. (a) 11 added to 2m Q. 5. Give expressions in the following cases. (a) $11$ added to $2m$ (b) $11$ subtracted from $2m$ c) $5$ times y to which $3$ is added (d) $5$ times y from which $3$ is subtracted (e) $y$ is multiplied by $-8$ Views (a) $11$ added to $2m$ $=2m+11$ (b) $11$ subtracted from $2m$ $=2m-11$ c) $5$ times y to which $3$ is added$=5y+3$ (d) $5$ times y from which $3$ is subtracted$=5y-3$ (e) $y$ is multiplied by $-8$$=-8y$ Exams Articles Questions", "two things times the third. You’ve seen that, though. 1 x (2 x 3) is the same as (1 x 2) x 3. And it has to distribute: something times the sum of two other things has to be the same as the sum of the something times the first thing and the something times the second. That is, 2 x (3 + 4) is the same as 2 x 3 plus 2 x 4. For example, the group we had before, 0 times anything will be 0. 1 times anything will be what we started with: 1 times 0 is 0, 1 times 1 is 1, 1 times 2 is 2, and 1 times 3 is 3. 2 times 0 is 0, 2 times 1 is 2, 2 times 2 will be 0 again, and 2 times 3 will be 2 again. 3 times 0 is 0, 3 times 1 is 3, 3 times 2 is 2, and 3 times 3 is 1. Believe it or not, this all works out. And “times” doesn’t get to look nearly so weird as “plus” does. And that’s all you need: a collection of things, an operation that looks a bit like addition, and an operation that looks even more vaguely like multiplication. Now the controversy. How much does something", "= $=3\\times mn+5=3mn+5$ Product of numbers $y$ and $z$ subtracted from $10$. Product of numbers y and z $=y\\times z=yz$ Product of numbers y and z subtracted from 10 $=10-yz$ Sum of numbers $a$ and $b$ subtracted from their product. Sum of numbers a and b = a + b Product of the numbers a and b $=a\\times b=ab$ Sum of numbers $a$ and $b$ subtracted from their product $= ab - (a+b) = ab - a - b$. Expression : x - 3 Terms in the above expression: x and -3 Tree diagram for the given expression Expression: $1 + x + x^2$ Tems in the above expression: $1, x \\and\\ x^2$ Factors of x2: x and x Tree diagram for the given expression Expression: y - y3 Tems in the above expression: y and -y3 Factors of -y3: -1, y, y and y Tree diagram for the given expression Expression: 5xy2 + 7x2y Terms in the above expression: 5xy2 and 7x2y Factors of 5xy2: 5, x, y and y Factors of 7x2y: 7, x, x and y Tree diagram for the given expression Expression: -ab + 2ab2 - 3a2 Terms in the above expression: -ab, 2ab2 and -3a2 Factors of -ab: -1, a and b Factors of 2ab2: 2, a, b and b Factors of -3a2:", "before addition, so $2\\times 3^k+3^k$ actually means $(2\\times 3^k)+3^k$. This is $(2\\times 3^k)+(1\\times 3^k)$, which is clearly just $3\\times 3^k$, or $3^{k+1}$. - It looks like you are confusing $2\\cdot 3^k+3^k$ with $2 \\cdot (3^k+3^k)$. The first gives $3 \\cdot 3^k=3^{k+1}$, while the second gives $2 \\cdot (2 \\cdot 3^k)=4\\cdot 3^k$. Parentheses are important. - $2\\times 3^k+3^k=(2+1)\\times3^k=3\\times3^k$ -", "16 at 16:09 Supposing that $\\times~(mult)$ is primitive recursive function. Then you could write: $exp(x,y)={ x }^{ y }$ 1) $exp(x,0)=x^0=1$ 2) $exp(x,y+1)=x^{y+1}=(x^y)\\times x=mult(exp(x,y),x)$ for $mult$ you could show that: $mult(x,y)=x\\times y$ 1)$mult(x,0)=x \\times 0=0$ 2)$\\operatorname{mult} \\left({x, y + 1}\\right) = x \\times \\left({y + 1}\\right) = \\left({x \\times y}\\right) + x = \\operatorname{add} \\left({\\operatorname{mult} \\left({x, y}\\right), x}\\right)$ and for addition $add$ the proof is straightforward. Hope these are useful! -", "to avoid confusion between the symbols $x$ and $\\times$. To indicate multiplication between things, we just place the two together. Consider this: Expand $5\\times(a+b)$ . $5\\times(a+b) = 5(a+b) = 5a + 5b$ All you really need to remember is too multiply everything in the brackets by what is outside of the brackets. A few more examples: Expand $6(x+2)$ . $6(x+2) = 6x+12$ Expand $10(3y+2)$ . $10(3y+2) =30y+20$ Expand $-2(9+z)$ . $-2(9+z) = -18-2z$ To be continued…", "$40\\times2^6$ = 2560 20) Applying the order of operations on expressions with powers involving negative exponents and variable bases. $3^{-3}\\times y^3\\div y^{-2}\\times y^3$ 1. $3^{-3} = \\frac{1}{9}$ 2. $y^{3-([-2]+3)} = y^2$ 3. $\\frac{y^2}{9}$ When answering these questions, you have to use BEDMAS, in the correct order. In this case, I start with Answering the exponents. with these exponents, there is still a way to simplify. I subtracted the exponents on the alike bases. Then I followed the rest in Order, remembering the negative exponents rule. Anything else that I know about exponents… Anything raised to the power of one, is it itself, $6^1 = 6$ Positive 1 raised to the power of any whole number will always be 1. $1^4 = 1\\times1\\times1\\times1 = 1$", "Sub these into $x^2+y^2+z^2 = 1$" ]

[ "Now, $$x!$$ is >$$1\\times 2\\times 3\\times$$ … $$(x-1)\\times x$$, so $$(x-1)$$ and 2 are included in $$x!$$ (that is, provided that $$x$$ is greater than 2). Hence, this is the correct answer. Incorrect. [[snippet]] Yes, statements II and III are true. But what about statement I? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I and III are true. But what about statement II? Can you find an example eliminating this statement? Incorrect. [[snippet]] Yes, statement I is true. But what about statements II and III? Can you find an example eliminating these statements? Incorrect. [[snippet]] Yes, statement II is true. But what about statements I and III? Can you find an example eliminating these statements? Correct. [[snippet]] __Plug In__ $$x = 3$$. In this case, the product could be $$3 \\times 4 \\times 5 = 60$$. >I. 60 is divisible by $$x - 1 = 3 - 1 = 2$$. >II. 60 is divisible by $$2x = 2 (3) = 6$$. >III. 60 is divisible by $$x! = 3! = 3 \\times 2 \\times 1 = 6$$. __Plug In__ another value of $$x$$ to be sure because the question says *must be*. This time try an even integer (such as $$x= 4$$). In this case, the product could be $$2 \\times 3 \\times", "Y)+(X\\times Z).}$", "# Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "# If x = 4 and y = 2, what is the value of the following expression: 3xy - 12y + 5x 20 Explanation To find the value of this expression, substitute the given values for x and y into the expression: 3(4)(2) - 12(2) + 5(4) Then, calculate the value of the expression: $$3 \\times 8-12 \\times 2+5 \\times 4=24-24+20=20$$", "\\times x) \\times x' = x \\times x' = 0$. 3. $x \\times (x + y) = (x + 0) \\times (x + y) = x + (0 \\times y) = x + 0 = x$. 4. $x + (x \\times y) = (x \\times 1) + (x \\times y) = x \\times (1 + y) = x \\times 1 = x$. Are there quicker ways to do any of these? 1. $\\hspace{10px}$False: $a$ can never equal $a' x + x' = 1$, but $x + x = x.$ So $x = 1$. However $1 \\times 1 = 1$ and $1 \\times 1' = 0$. 2. $\\hspace{5px}$True: $1 = 1 \\times 1 = (x' + x) \\times (x' + x'') = x' + (x \\times x'')$ so $x = x \\times x''$ and $x'' = x$. 3. False by assuming that $y$ is also an inverse of $x: x' = x' \\times 1 = x' \\times (x + y) = (x' \\times y) + (x' \\times y) = x' \\times y$. But we can replace $x'$ by $y$ and $y$ by $x'$ in this series of equations to get $x' = x' \\times y = y \\times x'= y$. 1. To do this we need to prove that $(x + y) + (x' \\times y') = 1$ and", "$\\frac{13 - b}{7 - b} = 2$ Now we simplify the fraction : $13 - b = 2(7 - b)$ $13 - b = 14 - 2b$ $- b + 2b = 14 - 13$ $\\boxed{b = 1}$ 2. Find $a$ Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ : $13 = 4a + b$ Since $b = 1$, we get : $13 = 4a + 1$ $12 = 4a$ $3 = a$ $\\boxed{a = 3}$ 3. Conclusion You are done : you have found the linear relationship between $x$ and $f(x)$ : $\\boxed{f(x) = 3x + 1}$ Let's check those results : $f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ --> $f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ --> $f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ --> $f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ --> Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ : $f(1) = 1 \\times 3", "\\times m} \\right)\\left[ {\\left( {2 \\times l} \\right) + \\left( {3 \\times a} \\right)} \\right]$ $= 10lm\\left( {2l + 3a} \\right)$ Therefore, $20{l^2}m + 30alm$ can be factored as $10lm\\left( {2l + 3a} \\right)$. (vi) Factorise the expression $5{x^2}y - 15x{y^2}$. Ans: To factorise the expression $5{x^2}y - 15x{y^2}$ write $5{x^2}y$ and $15x{y^2}$ as products of different numbers. $5{x^2}y$ can be written as $5 \\times x \\times x \\times y$. $15x{y^2}$ can be written as $3 \\times 5 \\times x \\times y \\times y$. Substitute $5 \\times x \\times x \\times y$ for $5{x^2}y$ and $3 \\times 5 \\times x \\times y \\times y$ for $15x{y^2}$ in expression $5{x^2}y - 15x{y^2}$. $5{x^2}y - 15x{y^2} = \\left( {5 \\times x \\times x \\times y} \\right) - \\left( {3 \\times 5 \\times x \\times y \\times y} \\right)$ $\\left( {5 \\times x \\times y} \\right)$ is the common factor of $5 \\times x \\times x \\times y$ and $3 \\times 5 \\times x \\times y \\times y$ Thus, take $\\left( {5 \\times x \\times y} \\right)$ as a common factor from right hand side of expression $5{x^2}y - 15x{y^2} = \\left( {5 \\times x \\times x \\times y} \\right) - \\left( {3 \\times 5 \\times x \\times y \\times y} \\right)$. $5{x^2}y - 15x{y^2} = \\left( {5 \\times x \\times", "$2^3$, since $2^2$ is not divisible by $2^3$. So, if we are trying to divide $N$, that has prime factorization $N=p_1^{a_1} \\times p_2^{a_2} \\times ...p_x^{a_x}$, then the divisors of $N$, $D$, will have prime factorization of form $D=p_1^{b_1} \\times p_2^{b_2} \\times ...p_x^{b_x}$, where $b_i <= a_i$. For example, divisor of $12=2^2 \\times 3$ will have prime factorization, $D=2^x \\times 3^y$, where $x <= 2, y <= 1$. When $x = 1$ and $y=1$, $D=2^1 \\times 3^1=6$ which is a divisor of $12$. If we use a different combination of $[x,y]$, then we get a different divisor. So how many different combination can we use here? We know that, $0<=x<=2$ and $0<=y<=1$. So for $x$ we can use ${0,1,2}$ and for y we can use ${0,1}$. So we can use $3 \\times 2=6$ combinations of ${x,y}$. And that is our NOD. So, if $N=p_1^{a_1} \\times p_2^{a_2} \\times ...p_x^{a_x}$, then $D=p_1^{b_1} \\times p_2^{b_2} \\times ...p_x^{b_x}$. We get new divisors for using different combination for ${b_1,b_2...b_x}$. How many different combinations can we make? $(a_1+1) \\times (a_2+1) \\times...(a_x+1)$. Therefore, $\\sigma_0(N) = (a_1+1) \\times (a_2+1) \\times...(a_x+1)$. So basically, in order to find NOD, all we need to do is factorize $N$, then take each of its prime factors power, increase them by $1$ and finally multiply all of them. Easy right? ## Examples", "\\times a \\times b} \\right) - \\left( {2 \\times 2 \\times c \\times a} \\right)$ $\\left( {2 \\times 2 \\times a} \\right)$ is the common factor of $2 \\times 2 \\times a \\times a$, $2 \\times 2 \\times a \\times b$ and $2 \\times 2 \\times c \\times a$ Thus, take $\\left( {2 \\times 2 \\times a} \\right)$ as a common factor from right hand side of expression $- 4{a^2} + 4ab - 4ca = - \\left( {2 \\times 2 \\times a \\times a} \\right) + \\left( {2 \\times 2 \\times a \\times b} \\right) - \\left( {2 \\times 2 \\times c \\times a} \\right)$. $- 4{a^2} + 4ab - 4ca = \\left( {2 \\times 2 \\times a} \\right)\\left[ { - \\left( a \\right) + \\left( b \\right) - \\left( c \\right)} \\right]$ $= 4a\\left( { - a + b - c} \\right)$ Therefore, $- 4{a^2} + 4ab - 4ca$ can be factorized as $4a\\left( { - a + b - c} \\right)$. (ix) Factorise the expression ${x^2}yz + x{y^2}z + xy{z^2}$. Ans: To factorise the expression ${x^2}yz + x{y^2}z + xy{z^2}$ write ${x^2}yz$, $x{y^2}z$ and $xy{z^2}$ as product of different numbers. ${x^2}yz$ can be written as $x \\times x \\times y \\times z$. $x{y^2}z$ can be written as $x \\times y \\times y \\times z$. $xy{z^2}$can be", "$SU(2)_L \\times SU(2)_R$ to $SU(2)_L$.", "## College Algebra (10th Edition) $4$ $(xy)^2$ if x=2, and y=-1. Substitute the values: $(2\\times-1)^2=$ $(-2)^2.$ Therefore, the answer is: $4.$", "note that there are two special solutions: one with $x=y=1$, and one with $y=z=1$. We counted each of them twice, hence we have to subtract two from the total. Therefore $m-n = 334 + 501 + 167 - 2 = 1000$, and the answer is $1000\\bmod 1000 = \\boxed{000}$.", "$x^2 + y^2 = 2019$. (2365) Solve the following equation in positive integers: $3\\times (5x + 1)=y^2$", "means $$(n \\times n)-1!$$ according to the order of operations. $$n! = n \\times (n-1)!$$$OK, we have • $$n! = n \\times (n-1)!$$, and • $$(n-1)! = (n-1) \\times (n-2)!$$, and • $$(n-2)! = (n-2) \\times (n-3)!$$, and so on. So $$n! = n \\times (n-1)! = n \\times (n-1) \\times (n-2)! = n \\times (n-1) \\times (n-2) \\times (n-3)!$$$ and so on. If we just keep going, we’ll eventually reach $$n \\times (n-1) \\times (n-2) \\times \\dots \\times 5 \\times 4 \\times 3!$$$and we already know that $$3! = 6$$, which I’ll write as $$3 \\times 2 \\times 1$$ for reasons which are about to become obvious. So we have the following formula, which is how we define the factorial: $$n! = n \\times (n-1) \\times \\dots \\times 4 \\times 3 \\times 2 \\times 1$$ “$n!$” is read out loud as “$n\\$ factorial”, and it means “the number of ways to arrange $$n$$ objects into any order”. # Edge cases We’ve seen $$3!$$, but never $$2!$$ or $$1!$$. • It’s easy to see that there are two ways to arrange two objects into an order: $$1,2$$ or $$2,1$$. So $$2! = 2$$. • It’s also easy (if a bit weird) to see that there is just one way of arranging one object into an order: $$1$$ is", "# How do you write \"twice y\" as an algebraic expression? $2 y$ \"Twice of $y$\" means $y$ times two. $2 \\times y = 2 y$", "$$X(2)=(X\\times X)/\\mathbb{Z}_2$$ does not.", "Q # Find the value of the following expressions when x = - 1, (i) 2x - 7 Q. 3. Find the value of the following expressions, when $x=-1$: (i) $2x-7$ $\\\\2x - 7 \\\\= 2 \\times ( -1 ) - 7 \\\\= -2 - 7 \\\\= -9$", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "# How do you simplify 2a * -a? Since a $\\times$ a = ${a}^{2}$, so $2 a \\times - a$ = $- 2 {a}^{2}$.", "\\times 2 = 2^3$ $2 \\times 2 \\times 2 \\times 2 = 2^4$ $2$ $4$ $8$ $16$ Each time we increment the exponent we multiply by the base. What happens if we decrement the exponent? $2 \\times 2 \\times 2 = 2^3$ $2 \\times 2 = 2^2$ $2 = 2^1$ $2^0 = 1$ $8$ $4$ $2$ $1$ You can see we are dividing by the base each time we decrement the exponent so it comes as no surprise that $2^0 = 1$. You can probably see that anything raised to the power of $0$ is equal to $1$ except for $0^0$. • $0^0$ is undefined. What happens if we keep decrementing the exponent? $2^1$ $2^0$ $2^{-1}$ $2^{-2}$ $2$ $1$ $1/2$ $1/4$ Which means $2^{-2} = 1/2^2$ or, more generally: • $x^{-a} = 1/x^a$ What is $3^2 \\times 3^3$ ? Well, $3^2 = 9$ and $3^3 = 27$. Now $9 \\times 27 = 243$ which is equal to $3^5$ which means $3^2 \\times 3^3 = 3^5$. More generally we can write • $x^a \\times x^b = x^{a+b}$ Psst - This is how logarithms and slide rules work. We know that $x^a \\times x^b = x^{a+b}$. What happens in the case of $(x^a)^b$ ? First, let’s try this with some easy numbers. Try $(2^3)^2 = (8)^2 = 64 = 2^6$" ]

[ "Sherlock Holmes, and A Christmas Carol along with tens of thousands of other titles are available to read for free. In 2010, they released a DVD that contains over 29,500 titles, but the project has grown still since then. 29,500 sounds like a big number, but it’s bigger still when you realize that it’s not 29,500 books, but 29,500 unique books. Each of these took time to write. Some of them took years. Authors slaved away on each of these titles, and many of them will scarcely be read or even remembered by their decedents centuries later. In a world where a cheap MP3 player can store 10,000 songs, we often forget exactly what this means. I have 4861 songs in my music library, but 2182 of them haven’t been played in the past two years. If I wanted to listen to all of them, it would take me over two weeks of continuous playback. Fourteen days of unique songs that each took weeks or months to write and produce. And my library is pretty small. Some people’s libraries would take months. In an effort to provide context to these kinds of numbers, I created a device that reads books. Day and night, it works its way through the 2010 English Project Gutenberg library paragraph by paragraph displaying", "and compact Compact-disk! 6 long years I took I don't really know this one I can remember a lot You have a pretty big storage capacity A very basic procedure Insert disk in the reader! White is very important to me Might be virgin disks? Beautiful for sure The rainbow behind CD! I'm pretty sure this isn't the answer Are you Music Music sheets can be read, Music takes no physical space, I guess there must be a song named 6 long years or maybe some music took 6 years to write, it can definitely be tied to memories (remember a lot) or symbolically represent something. Music is actually conceptually very very very simple (basic procedure), whole notes are empty(white) in music sheets, Music is obviously beautiful, there's only the slither that doesn't fit. • Does music slither, though? – Joe Z. Apr 10 '15 at 19:15 • As I said, it doesn't fit. Just couldn't think of anything else... You could always go with snake charmers that play music to charm 'em though. – Spacemonkey Apr 10 '15 at 19:17", "I appreciate your answer though. My library is only about 2000 tracks, so maybe taking an evening to sort them by hand would be the way to go. – Robert Lemiesz Mar 23 '18 at 18:14", "at the earliest. However, the other night Mark Pendleton boasted on Twitter that, \"I actually have 2.5 days' worth of Christmas music! [on my iPod]\". This sounded impressive but I knew I had a few Christmas CDs and was quietly confident I could match or beat him. But, as they say, pride comes before a fall. With only a few more tracks to add, it looks like I'll barely make 24 hours. :-( ...And this includes some tracks that appear on multiple albums but I ask you, is it possible to have too many copies of Slade's Merry Xmas Everybody - I think not! However, a desire to get over the 24 hour mark led me to the topic for today's Fun on Friday. I wondered if there was any legal, free source of Christmas music on the Internet. The answer is, of course, yes! The first free music I found was O Holiday Hits from (of all people) Oprah Winfrey. However these tracks are only available for 48 hours - so you'll have to be quick. Next, I found a blog dedicated to free Christmas music called (surprisingly enough) Free Christmas Music. So far, there is only one entry for this year (the Oprah downloads) but most/all of last year's downloads are still available. For example, I", "## Friday, November 28, 2008 ### Fun on Friday #11: It's the most wonderful time of the year (allegedly) I make it a rule (well, more of a guideline really) not to play any Christmas music until the 1st of December at the earliest. However, the other night Mark Pendleton boasted on Twitter that, \"I actually have 2.5 days' worth of Christmas music! [on my iPod]\". This sounded impressive but I knew I had a few Christmas CDs and was quietly confident I could match or beat him. But, as they say, pride comes before a fall. With only a few more tracks to add, it looks like I'll barely make 24 hours. :-( ...And this includes some tracks that appear on multiple albums but I ask you, is it possible to have too many copies of Slade's Merry Xmas Everybody - I think not! However, a desire to get over the 24 hour mark led me to the topic for today's Fun on Friday. I wondered if there was any legal, free source of Christmas music on the Internet. The answer is, of course, yes! The first free music I found was O Holiday Hits from (of all people) Oprah Winfrey. However these tracks are only available for 48 hours - so you'll have to be quick.", "1 month ago Get a good feel for the types of music you like. Listen to lots of music, even stuff you wouldn't normally like. Pay close attention to elements of music in everyday life. Experiment :3 - 6 years, 1 month ago Jake, what is your soundcloud account? I will follow you on there. - 6 years, 1 month ago", "conclude that, above a given throughput threshold it is unlikely that more throughput would have a significant effect on the consumers browsing experience. More work needs to be done in order to better understand the experience threshold after which more connection bandwidth has diminishing returns on the customer’s browsing experience. However, it would appear that 1-Gbps 5G connection speed would be far above that threshold. An average web page in 2016 was 2.2 MB which from an LTE speed perspective would take 568 ms to load fully provided connection speed was the only limitation (which is not the case). For 5G the same page would download within 18 ms assuming that connection speed was the only limitation. Now we surely are talking. If I wanted to download the whole Library of the US Congress (I like digital books!), I am surely in need for speed!? The US Congress have estimated that the whole print collection (i.e., 26 million books) adds up to 208 terabytes.Thus assuming I have 208+ TB of storage, I could within 20+ (at 1 Gbps) to 2+ (at 20 Gbps) days download the complete library of the US Congress. In fact, at 1 Gbps would allow me to download 15+ books per second (assuming 1 book is on average 3oo pages and formatted at 600", "7 years, I have managed to amass a mere 100000 views total. If I somehow got 20000 views of my songs every single month, the total cost of streaming 140 GB from Amazon S3 at$0.05 per GB would be $7 per month. That's how much SoundCloud is charging just to double my storage space! This makes even less sense when you calculate that 6 hours of FLAC would be 4.7 GB, or about 5 GB including the 128 kbps streaming MP3s. 5 GB of storage space costs a pathetic$0.12 cents a month to store on Amazon S3! All of the costs come down to bandwidth, which is relatively fixed by how many people are listening to songs, not how many songs there are. This means, if I'm paying any music service for the right to store music on their servers, I should get near unlimited storage space (maybe put in a sanity check of 10 GB max per month to limit abuse). I will point out that Clyp.it actually does this properly, giving you 6 hours of storage space for free and unlimited storage space if you pay them $6 a month. Unfortunately, Clyp.it does not try to be SoundCloud as it has no comments, no reshares, and well, not much of anything, really. It's like a giant", "Path hard rock roots, minor key tonality, melodic string accompaniment, mixed acoustic and electric instrumentation and mallet percussion Metallica - Jump In The Fire hard rock roots, a subtle use of vocal harmony, mild rhythmic syncopation, repetive melodic phrasing and minor key tonality As far as genres goes it clearly tends to play similar songs in a row, but still plays very different stuff. Generally it has played 4 or 5 songs that fit into one category then jumps to a new one. Again this is exactly what I would want it to do. If you only wanted to listen to one type of music at a time you could make different radio stations for the different types. The next two limitations I've not experienced first hand yet, but read them on Wikipedia. First off you only get 12 skips per 24 hours. So you can't be super picky about what it plays. If you want to hear one specific band or song you should just be listening to your media play anyway. The next limit though is a big downfall. Wikipedia claims you only get 40 hours per month for free. 40 hours is ridiculous short for a month. I'll listen to 40 hours of music in 3 days easily. I suppose for someone with less free", "First despite what I was told it does have audio ads. They are quite reasonable though. There is only ever one ad played at a time, and they are about 10-30 seconds long. It seems to go about 6 songs between ads. I'm sure this will increase as it become more popular, but even one 30 second ad per song would be fine with me. My biggest problem with the ads now is that I've only heard about 5 different ads, and they are just repeated. I don't see any visual ads on the page itself, although I'm using Adblock+. The next two limitations I've not experienced first hand yet, but read them on Wikipedia. First off you only get 12 skips per 24 hours. So you can't be super picky about what it plays. If you want to hear one specific band or song you should just be listening to your media play anyway. The next limit though is a big downfall. Wikipedia claims you only get 40 hours per month for free. 40 hours is ridiculous short for a month. I'll listen to 40 hours of music in 3 days easily. I suppose for someone with less free time than me (read: everyone) this would last longer, still I'm sure most people will reach this limit", "months. You automated the downloading process? xouper wrote: Thu Aug 16, 2018 12:07 amAt that rate it will take me maybe another 3.5 years to get to a million. 1 h / book, 8 h / day, it would take 300+ years to read that million. But I understand it's not for reading… xouper Posts: 10181 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Witness wrote: Thu Aug 16, 2018 12:19 am xouper wrote: Thu Aug 16, 2018 12:07 amThat's about 550 books per day, every day for six months. You automated the downloading process? The kindle app for windows automatically downloads my purchases to my hard drive. And it crashes regularly while trying to do that. I suspect the programmers did not anticipated this level of usage. I'm watching the memory usage and I predict that by 500,000 books, because it is a 32-bit app, it will hit the 2 gigabyte limit and crash permanently, even though I have 8 gb ram installed. I am hoping by then they will release an update that might (perhaps inadvertently) solve that problem. xouper Posts: 10181 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Of the 627 free", "60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means about 6.5% of songs were being played for the first time in 60 days. Honestly, that's not bad. However, the 50%", "much music as you want for a year. All the music is yours to keep forever and downloads can take place at any time during a whole year after the product is registered – and best of all the music is downloaded in MP3 format and DRM free to your PC which means you can sync them with any digital music player In order to run Datz Music Lounge software you must have the secure USB dongle that is supplied in the retail pack. Unfortunately the Mac version is not currently available. Please check back here after December 1st.", "much data does streaming a movie takes? A 2-hour-long movie consumes about 16 GB in 4K, 4 GB in HD, and 1 GB in SD resolutions, respectively. ### How to calculate monthly data usage? To calculate monthly data usage: 1. Calculate the data consumption for every activity. 2. Add the numbers to find the sum. 3. Multiply the sum with the number of days in a month. ### How much data per hour does multiplayer gaming takes? Multiplayer gaming platforms require high-speed connections and consume about 200 MB per hour on average. ### How to calculate my internet bill? To calculate your internet bill: 1. Multiply the data consumed by the cost per GB of data. 2. Add the taxes to the product. Rahul Dhari Movies (Netflix, Hulu, Amazon Prime, YouTube, etc.) Movie streaming in 4K hrs Movie streaming in HD hrs Movie streaming in SD hrs Movies data usage GB Music (Spotify, Apple music, Amazon music, etc.) Music streaming hrs Music data usage GB Gaming (Steam, Origin, Twitch, etc.) Multiplayer gaming/downloads hrs Gaming data usage GB Social Media (TikTok, Twitter, Instagram, etc.) Social media browsing hrs Social media data usage GB Video Calling SD Video calling hrs HD Video calling hrs Video call data usage GB Emails No. of emails per day Emails data usage GB Smart", "only listed the last few songs, or embedded it in flash. I began scraping on June 26th and August 24th makes 60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means", "some rough calculations, That's over 5,000 hours of standard quality MP3 audio. over 200 days! • #### Help! I need context! (Score:2, Funny) 500GB - how many Libraries of Congress is that? • #### Re:Help! I need context! (Score:2) I'd like to keep a personal copy of the Library of Congress. Maybe two, so that I can have one version to make notes in. Can I do that? • #### Re:Help! I need context! (Score:3, Interesting) Last I saw, the LOC was 13 TeraBYtes. You'd need 26 of these drives. Alternatively, Each drive holds ~4% of a LOC. Yes, I understand your question was kind of a joke, but I thought I'd do the math real quick just to consider the implications. I wonder if anyone has a good estimate of how long it will be until the typical consumer PC has sufficient storage space to hold the LOC? • #### Re:Help! I need context! (Score:3, Interesting) It's not \"how long until the typical PC can store the LOC?\", it's \"how long ago's LOC can the typical PC store?\" How big was the LoC 5 years ago? Under a few hundred gig, I'll bet. Today's LoC is (or so I read on Slashdot so it must be true) about 13TB. So how long will it take for desktops", "GB would be $7 per month. That’s how much SoundCloud is charging just to double my storage space! This makes even less sense when you calculate that 6 hours of FLAC would be 4.7 GB, or about 5 GB including the 128 kbps streaming MP3s. 5 GB of storage space costs a pathetic$0.12 cents a month to store on Amazon S3! All of the costs come down to bandwidth, which is relatively fixed by how many people are listening to songs, not how many songs there are. This means, if I’m paying any music service for the right to store music on their servers, I should get near unlimited storage space (maybe put in a sanity check of 10 GB max per month to limit abuse). I will point out that Clyp.it actually does this properly, giving you 6 hours of storage space for free and unlimited storage space if you pay them $6 a month. Unfortunately, Clyp.it does not try to be SoundCloud as it has no comments, no reshares, and well, not much of anything, really. It’s like a giant pastebin for sounds where you can follow people or favorite things. It’s also probably screwed. Even though I had a name and a website design, I never launched it because even if I could find a", "4 Exabytes (4 Billion gigabytes, GB), more than 5.5× the level of 2016. It is more than 600 million times the average mobile data consumption of 6.5 GB per month per customer (in 2021). Compare this with the Western European population of ca. 200 million. While big numbers, the 6.5 GB per month per customer is insignificant. Assuming that most of this volume comes from video streaming at an optimum speed of 3 – 5 Mbps (good enough for HD video stream), the 6.5 GB translates into approx. 3 – 5 hours of video streaming over a month. The above Figure 29 Illustrates a 24-hour workday total data demand on the mobile network infrastructure. A weekend profile would be more flattish. We spend at least 12 hours in our home, ca. 7 hours at work (including school), and a maximum of 5 hours (~20%) commuting, shopping, and otherwise being away from our home or workplace. Previous studies of mobile traffic load have shown that 80% of a consumer’s mobile demand falls in 3 main radio node sites around the home and workplace. The remaining 20% tends to be much more mobile-like in the sense of being spread out over many different radio-node sites. Daily we have an average of ca. 215 Megabytes per day (if spread equally over", "should you leave your music player playing to answer question number 1? I should probably make the drastic assumption that all the songs in the playlist have the same length, e.g. 3 minutes or so. - If the algorithm is pseudorandom, then shouldn't the expected answer to number 1 be 1/N, where N is the total number of songs? Of course this doesn't apply to things like iTunes which add weights to more frequently manually chosen songs. (Also, the music playing program I currently run literally \"shuffles\" the play list when you hit that button: between button-presses, the player deterministically loops through all entries using the current order given by shuffling. =p ) – Willie Wong Nov 22 '10 at 11:47 It sounds reasonable, but in the one I'm using (mplayer on Linux), I've found that sometimes it takes a while in between the instances a favorite song of mine in my playlist has played, and sometimes it takes a few songs (e.g. my playlist has a length of 100, my favorite song was the 96th one played in one cycle, and then the third one played in the next cycle). I definitely haven't seen the last song of one cycle being the first song of the following cycle, though. – Ｊ. Ｍ. Nov 22 '10 at 11:55", "You automated the downloading process? xouper wrote: Thu Aug 16, 2018 12:07 amAt that rate it will take me maybe another 3.5 years to get to a million. 1 h / book, 8 h / day, it would take 300+ years to read that million. But I understand it's not for reading… xouper Posts: 9751 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Witness wrote: Thu Aug 16, 2018 12:19 am xouper wrote: Thu Aug 16, 2018 12:07 amThat's about 550 books per day, every day for six months. You automated the downloading process? The kindle app for windows automatically downloads my purchases to my hard drive. And it crashes regularly while trying to do that. I suspect the programmers did not anticipated this level of usage. I'm watching the memory usage and I predict that by 500,000 books, because it is a 32-bit app, it will hit the 2 gigabyte limit and crash permanently, even though I have 8 gb ram installed. I am hoping by then they will release an update that might (perhaps inadvertently) solve that problem. xouper Posts: 9751 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Of the 627 free kindle" ]

[ "C. 12 D. 13 2- A floppy disk shows 937,036 bytes free and 739,352 bytes used. If you delete a file of size 652,159 bytes and create a new file of size 599,986 bytes, how many free bytes will the floppy disk have? A. 687,179 B. 791,525 C. 884,867 D. 989,209 3- 5 days 19 hours 35 minutes $$– 3$$ days 12 hours 22 minutes $$=$$? A. 3 days 10 hours 13 minutes B. 2 days 7 hours 13 minutes C. 3 days 10 hours 13 minutes D. 2 days 7 hours 23 minutes 4- The base of a right triangle is 2 feet, and the interior angles are $$45-45-90$$. What is its area? A. 2 foot squared B. 4 foot squared C. 3.5 foot squared D. 5.5 foot squared 5- Increased by $$50\\%$$, the numbers 84 becomes: A. 42 B. 100 C. 126 D. 130 6- Which equation represents the statement twice the difference between 6 times h and 3 gives 30. A. $$\\frac{6h + 3}{2}= 30$$ B. $$6(2h + 3) = 30$$ C. $$2(6h – 3) = 30$$ D. $$3\\frac{6h}{2}= 30$$ 7- A circle is inscribed in a square, as shown below. The area of the circle is $$16π$$cm$$^2$$. What is the area of the square? A. $$8 \\space$$cm$$^2$$ B. $$16 \\space$$cm$$^2$$ C. $$48 \\space$$cm$$^2$$ D.", "If it takes 57 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total=15, then Basic=10 and Deluxe=5. Now, if time needed to produce one Deluxe stereo is 1 unit than time needed to produce one Basic stereo would be 7/5 units. Total time for Basic would be 10*1=10 and total time", "a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total = 15, then Basic = 10 and Deluxe = 5. Now, if time needed to produce one Basic stereo is 1 unit than time needed to produce one Deluxe stereo would be 7/5 units. Total time for Basic would be 10*1 = 10 and total time for", "# work problems (ii) It is going to show the correct time again when it has lost 12 hours. It is going to take $60 \\cdot 12 = 720 hours$", "Path hard rock roots, minor key tonality, melodic string accompaniment, mixed acoustic and electric instrumentation and mallet percussion Metallica - Jump In The Fire hard rock roots, a subtle use of vocal harmony, mild rhythmic syncopation, repetive melodic phrasing and minor key tonality As far as genres goes it clearly tends to play similar songs in a row, but still plays very different stuff. Generally it has played 4 or 5 songs that fit into one category then jumps to a new one. Again this is exactly what I would want it to do. If you only wanted to listen to one type of music at a time you could make different radio stations for the different types. The next two limitations I've not experienced first hand yet, but read them on Wikipedia. First off you only get 12 skips per 24 hours. So you can't be super picky about what it plays. If you want to hear one specific band or song you should just be listening to your media play anyway. The next limit though is a big downfall. Wikipedia claims you only get 40 hours per month for free. 40 hours is ridiculous short for a month. I'll listen to 40 hours of music in 3 days easily. I suppose for someone with less free", "Tags 16 Jun 2018, 09:56 udaymathapati wrote: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 We can let the total number of stereos = 12; thus, 8 are basic and 4 are deluxe. We can let the number of hours to produce a basic stereo = 10, and thus it takes 7/5 x 10 = 14 hours to produce a deluxe stereo. The number of hours to produce the deluxe stereos last month was 14 x 4 = 56 hours, and the number of hours to produce the basic stereos was 8 x 10 = 80 hours. Thus, the time to make the deluxe stereos was 56/(56 + 80) = 56/136 = 7/17 of the total time taken to produce all stereos. _________________ # Jeffrey Miller [email protected] 181 Reviews 5-star rated online GMAT quant self study course See why", "which means that the movie lasts for $2$ hours and $57$ minutes. Math Vault Would you like to solve exercises about seconds, minutes and hours? Try Math Vault!", "the overall time into 4 sessions. Let us divide 100 minutes by 4. We get: So, the duration of each session is 25 minutes. Example 3: You watch a movie for 150 minutes. There is a 15-minute intermission, and there are commercials that run for 15 minutes. If the overall time spent at the theater is supposed to be divided into three equal parts, how long would one part be? Solution: Step 1: Find the overall time spent at the theater: The overall time spent at the movie theater: = Movie time + Time taken for the intermission + Time taken to run the commercials. = 150 + 15 + 15 = 180 Step 2: Divide the overall time into 3 equal parts: 180 $$\\div$$ 3 = = 60 So, we can say that the overall movie time can be divided into three equal parts, each 60 minutes long. Example 4: A group of students attends a concert from 5.30 pm to 8.00 pm. The band in the concert performs 10 songs, and each song is performed for an equal amount of time. For how many minutes was each song played at the concert? Solution: Step 1: Find the overall time spent at the concert: From 5:30 pm to 8:00 pm, the total minutes can be found as", "60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means about 6.5% of songs were being played for the first time in 60 days. Honestly, that's not bad. However, the 50%", "week, and _____ days in two weeks. Since _____ × _____ = _____ , there are about _____ hours in 2 weeks on Earth. Type below: _________ On Earth, there are about 23.93 hours in a day, 7 days in 1 week, and 14 days in two weeks. Since 23.93 × 14 = 335.02, there are about 335.02 hours in 2 weeks on Earth. Question 12. Michael’s favorite song is 3.19 minutes long. If he listens to the song 15 times on repeat, how long will he have listened to the same song? _____ minutes 47.85 minutes Explanation: Michael’s favorite song is 3.19 minutes long. If he listens to the song 15 times, 15 x 3.19 = 47.85 minutes Question 13. Test Prep A car travels 56.7 miles in an hour. If it continues at the same speed, how far will the car travel in 12 hours? Options: a. 68.004 miles b. 680.04 miles c. 680.4 miles d. 6,804 miles c. 680.4 miles Explanation: A car travels 56.7 miles in an hour. In 12 hours, 12 x 56.7 = 680.4 hours ### Share and Show – Page No. 179 Question 1. Manuel collects $45.18 for a fundraiser. Gerome collects$18.07 more than Manuel. Cindy collects 2 times as much as Gerome. How much money does Cindy collect for the", "minutes in one hour. $1\\,h=60\\,min=1000\\cdot60\\cdot60\\,ms$ Also, a day has $24$ hours, so the number of milliseconds in one day is calculated by multiplying the number of milliseconds in one hour by the number of hours in one day. $1\\,day=24\\,h=1000\\cdot60\\cdot60\\cdot24\\,ms$ Finally, we make the assumption that a year has 365 days. After that, the number of milliseconds in one year is calculated by multiplying the number of milliseconds in one day by the number of days in one year. $1\\,year=365\\,days=1000\\cdot60\\cdot60\\cdot24\\cdot365\\,ms$ So, from the given options, it can be seen that: $1000\\cdot 60\\cdot 60\\cdot 24\\cdot 365$ is the correct option. Example $1$ Convert $6$ days and $7$ hours into hours. Since $1$ day is equal to $24$ hours, this means $6$ days $7$ hours will be equal to: $(6\\times 24)\\,h+7\\, h$ $=151\\,h$ Example $2$ Convert $2$ years into seconds. $2$ years is equal to $2(365)=730\\, days$ $1$ day is equal to $24\\,h$ $1\\,h$ is equal to $60$ minutes, and $1$ minute is equal to $60$ seconds Hence, $2$ years $= 730 \\cdot 24 \\cdot 60 \\cdot 60 = 63,072,000\\,s$", "stereo as it does to produce a basic stereo, It takes 7/5 as many hours.. So if it takes 5 hrs to produce a basic one, a deluxe one takes 7/5 * 5 = 7 hrs. The question says 7/5 as many hours so you just multiply the hours it takes to produce a basic one with 7/5. Think of it this way - if the question said, \"it takes twice as many hours as it takes to produce a basic one...\" would you just multiply by 2 or would you follow your method? then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? Hours taken for 2 basic stereos = 2*5 Hours taken for 1 deluxe = 1*7 Total hours = 17 Required fraction = 7/17 _________________ Karishma Veritas Prep GMAT Instructor Director Joined: 01 Feb 2011 Posts: 528 ### Show Tags 05 Aug 2011, 11:29 B/D = 2/1 lets say time required to produce 1 Basic stereo = 5 hours => time required to produced 1 deluxe stereo = 7 hours total time required to produce B basic stereo's = 5B hours total time required to produce D deluxe stereo's = 7D hours ratio of total", "for $9$ hours and $35$ minutes. The shortest day in the winter lasts for $9$ hours and $55$ minutes. The night lasts for $14$ hours and $5$ minutes.", "## anonymous one year ago Simple question for a fan and medal! Check below !!! Yesterday, a local radio station played 92 songs prior to 10 a.m. The station also played 10 songs per hour for the remainder of the day. If a person listened to the radio station yesterday afternoon for x hours, which of the following equations can be used to find the number of songs the person listened to? A. y = x + 10 B. y = 10x + 92 C. y = 92x + 10 D. y = 10x 1. anonymous If the person listened in the afternoon, would they have heard any of the 92 songs played before 10 a.m.? 2. anonymous I suppose not, considering that is crucial informationt that should be added the the question 3. anonymous It's there in the question, but you have to read it very carefully. Now, during the afternoon the station played 10 songs every hour. If the listener listened for x hours, how many songs would they hear? 4. anonymous 140? 5. anonymous Remember, we don't know that the listener listened for 14 hours. We're told the listener listened for $$x$$ hours. What do you think? 6. anonymous but it says prior to ten am they played songs for the rest of the day", "20/2 = 10 pages for each day Mai reads 35 pages in 3 days. 35/3 = 11.66 pages for each day Mai reads at a faster rate Question 4. An online music store offers 5 downloads for$6.25. Another online music store offers 12 downloads for $17.40. Which store offers the better deal? ___________ Answer: An online music store offers 5 downloads for$6.25 offers the better deal Explanation: An online music store offers 5 downloads for $6.25.$6.25/5 = $1.25 Another online music store offers 12 downloads for$17.40. $17.40/12 =$1.45 An online music store offers 5 downloads for $6.25 offers the better deal On Your Own Write the rate as a fraction. Then find the unit rate. Question 5. A company packed 108 items in 12 boxes. Type below: ___________ Answer: 9 Explanation: A company packed 108 items in 12 boxes. 108/12 Divide 108/12 with 12/12 108/12 ÷ 12/12 = 9 Question 6. There are 112 students for 14 teachers. Type below: ___________ Answer: 8 Explanation: There are 112 students for 14 teachers. 112/14 Divide 112/14 with 14/14 112/14 ÷ 14/14 = 8 Question 7. Geoff charges$27 for 3 hours of swimming lessons. Anne charges $31 for 4 hours. How much more does Geoff charge per hour than Anne?$ _____ $1.25 Explanation: Geoff charges$27 for 3 hours of swimming lessons.", "network has twice as many layers so, as Dan Huang pointed out in his article, it’s reasonable to infer that twice the amount of time was used per move. So ~ 0.8 seconds rather than ~ 0.4 seconds. 1. Over 40 days, ~ 29 million games were played. 2. Each move of self-play used ~ 0.8 seconds of computer time where each self-play machine consisted of 4 TPUs. 3. How many self-play machines $N_{SP}$ were used? 4. If the average game has 200 moves we have: $$\\frac{40 \\cdot 24 \\cdot 3600 \\cdot N_{SP}}{200 \\cdot 29 \\cdot 10^6} \\approx 0.8 \\implies N_{SP} \\approx 1300$$ 5. Given that each self-play machine contained 4 TPUs we have: $$N_{TPU} = 4 \\cdot N_{SP} \\approx 5000$$ If we use the Google’s TPU pricing as of March 2019, the cost for an organisation outside of Google to replicate this experiment is therefore: $$\\text{Cost} > 5000 \\cdot 960 \\cdot 4.5 \\approx 2 \\cdot 10^7 \\quad \\text{US dollars}$$ It goes without saying that this is well outside the budget of any AI lab in academia. ## The true cost of Google’s 40 day experiment: Climate agreements since 1990: progress or the illusion thereof This in my opinion is the more important calculation. While it’s not at all clear that AI research will ‘save the world’ in", "3 hours at the concert and more than 4 hours at the park. $You have at most 8 hours to spend at a concert and at park. You want to spend at least 3 hours at t$...", "harder. Consumers will therefore choose like this: Over 10 years, buy 1 house, 10,000 meals, and as many performances as you can afford after that. Further suppose that each person had$2,100 per year to spend in 1940-1950, and $50,000 per year to spend in 1990-2000. (This is approximately true for actual nominal US GDP per capita.) 1940-1950: Total funds:$21,000 1 house = $10,000 10,000 meals =$10,000 Remaining funds: $1,000 Performances purchased: 10 1990-2000: Total funds:$500,000 1 house = $200,000 10,000 meals =$50,000 Remaining funds: $250,000 Performances purchased: 250,000 (Do you really listen to this much music? 250,000 performances over 10 years is about 70 songs per day. If each song is 3 minutes, that’s only about 3.5 hours per day. If you listen to music while you work or watch a couple of movies with musical scores, yes, you really do listen to this much music! The unrealistic part is assuming that people in 1950 listen to so little, given that radio was already widespread. But if you think of music as standing in for all media, the general trend of being able to consume vastly more media in the digital age is clearly correct.) Now consider how we would compute a price index for each time period. We would construct a basket of goods and determine the", "play live music, and we’re always fighting that same battle. “What do you mean, you charge more than 100 per person for an evening performance? That’s 400 for the whole band!!!” Yep. It is. Let’s break that down. 100 per person for the three hours of performance: 33 per hour. Not bad! But I need to get there 30 minutes early to meet the promoters and do sound check; I need an additional hour before that to load in the sound gear and set it up, and I need an hour of hard work at the end of the gig to pack everything up and put it in the car. That’s now 100 for 5.5 hours, or 18 an hour. But I also have to get to the gig. If I’m lucky enough to play gigs in my area, we’ll call that an hour of travel each way, bringing my hourly wage to 13. However, I’m more frequently traveling 3-6 hours each way, which puts my hourly wage between 7.40 and 5.12 per hour–assuming that you paid me that original 100. Then there’s gas. My car will go about 400 miles on a tank of gas, which costs 50. So if I’m originally paid 100 and I just count gas prices, my pay has now dropped to between", "only listed the last few songs, or embedded it in flash. I began scraping on June 26th and August 24th makes 60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means" ]

[ "Path hard rock roots, minor key tonality, melodic string accompaniment, mixed acoustic and electric instrumentation and mallet percussion Metallica - Jump In The Fire hard rock roots, a subtle use of vocal harmony, mild rhythmic syncopation, repetive melodic phrasing and minor key tonality As far as genres goes it clearly tends to play similar songs in a row, but still plays very different stuff. Generally it has played 4 or 5 songs that fit into one category then jumps to a new one. Again this is exactly what I would want it to do. If you only wanted to listen to one type of music at a time you could make different radio stations for the different types. The next two limitations I've not experienced first hand yet, but read them on Wikipedia. First off you only get 12 skips per 24 hours. So you can't be super picky about what it plays. If you want to hear one specific band or song you should just be listening to your media play anyway. The next limit though is a big downfall. Wikipedia claims you only get 40 hours per month for free. 40 hours is ridiculous short for a month. I'll listen to 40 hours of music in 3 days easily. I suppose for someone with less free", "If it takes 57 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total=15, then Basic=10 and Deluxe=5. Now, if time needed to produce one Deluxe stereo is 1 unit than time needed to produce one Basic stereo would be 7/5 units. Total time for Basic would be 10*1=10 and total time", "a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total = 15, then Basic = 10 and Deluxe = 5. Now, if time needed to produce one Basic stereo is 1 unit than time needed to produce one Deluxe stereo would be 7/5 units. Total time for Basic would be 10*1 = 10 and total time for", "Tags 16 Jun 2018, 09:56 udaymathapati wrote: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 We can let the total number of stereos = 12; thus, 8 are basic and 4 are deluxe. We can let the number of hours to produce a basic stereo = 10, and thus it takes 7/5 x 10 = 14 hours to produce a deluxe stereo. The number of hours to produce the deluxe stereos last month was 14 x 4 = 56 hours, and the number of hours to produce the basic stereos was 8 x 10 = 80 hours. Thus, the time to make the deluxe stereos was 56/(56 + 80) = 56/136 = 7/17 of the total time taken to produce all stereos. _________________ # Jeffrey Miller [email protected] 181 Reviews 5-star rated online GMAT quant self study course See why", "harder. Consumers will therefore choose like this: Over 10 years, buy 1 house, 10,000 meals, and as many performances as you can afford after that. Further suppose that each person had$2,100 per year to spend in 1940-1950, and $50,000 per year to spend in 1990-2000. (This is approximately true for actual nominal US GDP per capita.) 1940-1950: Total funds:$21,000 1 house = $10,000 10,000 meals =$10,000 Remaining funds: $1,000 Performances purchased: 10 1990-2000: Total funds:$500,000 1 house = $200,000 10,000 meals =$50,000 Remaining funds: $250,000 Performances purchased: 250,000 (Do you really listen to this much music? 250,000 performances over 10 years is about 70 songs per day. If each song is 3 minutes, that’s only about 3.5 hours per day. If you listen to music while you work or watch a couple of movies with musical scores, yes, you really do listen to this much music! The unrealistic part is assuming that people in 1950 listen to so little, given that radio was already widespread. But if you think of music as standing in for all media, the general trend of being able to consume vastly more media in the digital age is clearly correct.) Now consider how we would compute a price index for each time period. We would construct a basket of goods and determine the", "60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means about 6.5% of songs were being played for the first time in 60 days. Honestly, that's not bad. However, the 50%", "the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 37183 Followers: 7289 Kudos [?]: 96982 [9] , given: 10822 ### Show Tags 08 Sep 2010, 11:25 9 KUDOS Expert's post 4 This post was BOOKMARKED udaymathapati wrote: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 32 were basic and the rest were deluxe. If it takes 57 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two", "to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total = 15, then Basic = 10 and Deluxe = 5. Now, if time needed to produce one Basic stereo is 1 unit than time needed to produce one Deluxe stereo would be 7/5 units. Total time for Basic would be 10*1 = 10 and total", "the repository of one of the largest commercially available sound effects libraries in the world. https://www.sound-ideas.com/ on the flight home from CGDC '96 I finally broke down and bought The General 6000 Series (50 CDs!) for $1000. collections start at about$100.", "3 hours at the concert and more than 4 hours at the park. $You have at most 8 hours to spend at a concert and at park. You want to spend at least 3 hours at t$...", "only listed the last few songs, or embedded it in flash. I began scraping on June 26th and August 24th makes 60 days worth of data. Thanks to my recent post, we know classic rock songs average 4:48 in length. There are 86,400 minutes in 60 days. That would be enough time for 18,000 songs. If we assume only 75% of the time is actually music that's 13,500 songs. During these 60 days WMGK played 14,286 songs. I won't speculate about what this means about the actual percentage of music on the air. Slight changes in average song length have a big effect on numbers. So, now it is time for the interesting questions. How many of those were unique songs? How many songs would be needed to represent 25% or 75% coverage? In case it isn't clear what I mean by coverage, I mean how many songs represent 10% (or whatever) of the total songs played. For example, if a station played 1 song 10 times, and then 10 other songs 1 time each, for a total of 20 plays, that 1 top song would give 50% coverage. So, without further ado here are the stats: Coverage Songs 10.04% 28 25.27% 77 50.23% 172 75.13% 279 90.03% 373 100.00% 924 924 unique plays out of 14,286 means", "7 years, I have managed to amass a mere 100000 views total. If I somehow got 20000 views of my songs every single month, the total cost of streaming 140 GB from Amazon S3 at$0.05 per GB would be $7 per month. That's how much SoundCloud is charging just to double my storage space! This makes even less sense when you calculate that 6 hours of FLAC would be 4.7 GB, or about 5 GB including the 128 kbps streaming MP3s. 5 GB of storage space costs a pathetic$0.12 cents a month to store on Amazon S3! All of the costs come down to bandwidth, which is relatively fixed by how many people are listening to songs, not how many songs there are. This means, if I'm paying any music service for the right to store music on their servers, I should get near unlimited storage space (maybe put in a sanity check of 10 GB max per month to limit abuse). I will point out that Clyp.it actually does this properly, giving you 6 hours of storage space for free and unlimited storage space if you pay them $6 a month. Unfortunately, Clyp.it does not try to be SoundCloud as it has no comments, no reshares, and well, not much of anything, really. It's like a giant", "I appreciate your answer though. My library is only about 2000 tracks, so maybe taking an evening to sort them by hand would be the way to go. – Robert Lemiesz Mar 23 '18 at 18:14", "the overall time into 4 sessions. Let us divide 100 minutes by 4. We get: So, the duration of each session is 25 minutes. Example 3: You watch a movie for 150 minutes. There is a 15-minute intermission, and there are commercials that run for 15 minutes. If the overall time spent at the theater is supposed to be divided into three equal parts, how long would one part be? Solution: Step 1: Find the overall time spent at the theater: The overall time spent at the movie theater: = Movie time + Time taken for the intermission + Time taken to run the commercials. = 150 + 15 + 15 = 180 Step 2: Divide the overall time into 3 equal parts: 180 $$\\div$$ 3 = = 60 So, we can say that the overall movie time can be divided into three equal parts, each 60 minutes long. Example 4: A group of students attends a concert from 5.30 pm to 8.00 pm. The band in the concert performs 10 songs, and each song is performed for an equal amount of time. For how many minutes was each song played at the concert? Solution: Step 1: Find the overall time spent at the concert: From 5:30 pm to 8:00 pm, the total minutes can be found as", "much music as you want for a year. All the music is yours to keep forever and downloads can take place at any time during a whole year after the product is registered – and best of all the music is downloaded in MP3 format and DRM free to your PC which means you can sync them with any digital music player In order to run Datz Music Lounge software you must have the secure USB dongle that is supplied in the retail pack. Unfortunately the Mac version is not currently available. Please check back here after December 1st.", "preferences, but socially isolating. Music affects everyone, but listening habits vary with age. Children and teenagers are the highest consumers of music. A 2010 survey by Victoria J. Rideout, Ulla G. Foehr, and Donald F. Roberts, “Generation M2: Media in the Lives of 8- to 18-Year-Olds,” indicated that 8- to 18-year-olds listen to roughly 2.5 hours of music each day—a dramatic increase from 1.48 hours in 1999. Children aged 8 to 10 reported listening to music for just over an hour each day, whereas teenagers aged 15 to 18 reported listening to music for well over three hours every single day. The youth in society are vigorous consumers of music, and in 2009 they had an estimated market value in the United States of $140 billion. More than ever before, listening to music is deeply enmeshed into our social-economy. Throughout the 20th century and motivated by social, economic, and technological forces, music gradually shifted from a collective to an individual activity. The invention of the phonograph allowed public performances to be heard at home, transistor radios meant that music could be experienced in the backyard or bedroom, and today's smartphones deliver music as a fully individualized experience that can be summoned any time, anywhere. Such experiences are themselves a commodity, but they can also add meaning and value", "months. You automated the downloading process? xouper wrote: Thu Aug 16, 2018 12:07 amAt that rate it will take me maybe another 3.5 years to get to a million. 1 h / book, 8 h / day, it would take 300+ years to read that million. But I understand it's not for reading… xouper Posts: 10181 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Witness wrote: Thu Aug 16, 2018 12:19 am xouper wrote: Thu Aug 16, 2018 12:07 amThat's about 550 books per day, every day for six months. You automated the downloading process? The kindle app for windows automatically downloads my purchases to my hard drive. And it crashes regularly while trying to do that. I suspect the programmers did not anticipated this level of usage. I'm watching the memory usage and I predict that by 500,000 books, because it is a 32-bit app, it will hit the 2 gigabyte limit and crash permanently, even though I have 8 gb ram installed. I am hoping by then they will release an update that might (perhaps inadvertently) solve that problem. xouper Posts: 10181 Joined: Fri Jun 11, 2004 4:52 am Title: mere ghost of his former self ### Re: Free kindle ebooks Of the 627 free", "Tags 10 Apr 2018, 09:59 Bunuel wrote: udaymathapati wrote: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 32 were basic and the rest were deluxe. If it takes 57 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 udaymathapati please do check the questions when posting. Correct question is as follows: Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos? A. 7/17 B. 14/31 C. 7/15 D. 17/35 E. 1/2 # of basic stereos was 2/3 of total and # of deluxe stereos was 1/3 of total, let's assume total=15, then Basic=10", "play live music, and we’re always fighting that same battle. “What do you mean, you charge more than 100 per person for an evening performance? That’s 400 for the whole band!!!” Yep. It is. Let’s break that down. 100 per person for the three hours of performance: 33 per hour. Not bad! But I need to get there 30 minutes early to meet the promoters and do sound check; I need an additional hour before that to load in the sound gear and set it up, and I need an hour of hard work at the end of the gig to pack everything up and put it in the car. That’s now 100 for 5.5 hours, or 18 an hour. But I also have to get to the gig. If I’m lucky enough to play gigs in my area, we’ll call that an hour of travel each way, bringing my hourly wage to 13. However, I’m more frequently traveling 3-6 hours each way, which puts my hourly wage between 7.40 and 5.12 per hour–assuming that you paid me that original 100. Then there’s gas. My car will go about 400 miles on a tank of gas, which costs 50. So if I’m originally paid 100 and I just count gas prices, my pay has now dropped to between", "GB would be $7 per month. That’s how much SoundCloud is charging just to double my storage space! This makes even less sense when you calculate that 6 hours of FLAC would be 4.7 GB, or about 5 GB including the 128 kbps streaming MP3s. 5 GB of storage space costs a pathetic$0.12 cents a month to store on Amazon S3! All of the costs come down to bandwidth, which is relatively fixed by how many people are listening to songs, not how many songs there are. This means, if I’m paying any music service for the right to store music on their servers, I should get near unlimited storage space (maybe put in a sanity check of 10 GB max per month to limit abuse). I will point out that Clyp.it actually does this properly, giving you 6 hours of storage space for free and unlimited storage space if you pay them $6 a month. Unfortunately, Clyp.it does not try to be SoundCloud as it has no comments, no reshares, and well, not much of anything, really. It’s like a giant pastebin for sounds where you can follow people or favorite things. It’s also probably screwed. Even though I had a name and a website design, I never launched it because even if I could find a" ]

[ "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]", "area of $CMN$ in square inches is $[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label(\"A\", (0,0), S); label(\"B\", (1,0), S); label(\"C\", (1,1), N); label(\"D\", (0,1), N); label(\"M\", (0,.76), W); label(\"N\", (.82,0), S);[/asy]$ $\\mathrm{(A)\\ } 2\\sqrt{3}-3 \\qquad \\mathrm{(B) \\ }1-\\frac{\\sqrt{3}}{3} \\qquad \\mathrm{(C) \\ } \\frac{\\sqrt{3}}{4} \\qquad \\mathrm{(D) \\ } \\frac{\\sqrt{2}}{3} \\qquad \\mathrm{(E) \\ }4-2\\sqrt{3}$ ## Problem 20 Let $$T=\\frac{1}{3-\\sqrt{8}}-\\frac{1}{\\sqrt{8}-\\sqrt{7}}+\\frac{1}{\\sqrt{7}-\\sqrt{6}}-\\frac{1}{\\sqrt{6}-\\sqrt{5}}+\\frac{1}{\\sqrt{5}-2}.$$ Then $\\mathrm{(A)\\ } T<1 \\qquad \\mathrm{(B) \\ }T=1 \\qquad \\mathrm{(C) \\ } 12 \\qquad$ $\\mathrm{(E) \\ }T=\\frac{1}{(3-\\sqrt{8})(\\sqrt{8}-\\sqrt{7})(\\sqrt{7}-\\sqrt{6})(\\sqrt{6}-\\sqrt{5})(\\sqrt{5}-2)}$ ## Problem 21 In a geometric series of positive terms the difference between the fifth and fourth terms is $576$, and the difference between the second and first terms is $9$. What is the sum of the first five terms of this series? $\\mathrm{(A)\\ } 1061 \\qquad \\mathrm{(B) \\ }1023 \\qquad \\mathrm{(C) \\ } 1024 \\qquad \\mathrm{(D) \\ } 768 \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 22 The minimum of $\\sin\\frac{A}{2}-\\sqrt{3}\\cos\\frac{A}{2}$ is attained when $A$ is $\\mathrm{(A)\\ } -180^\\circ \\qquad \\mathrm{(B) \\ }60^\\circ \\qquad \\mathrm{(C) \\ } 120^\\circ \\qquad \\mathrm{(D) \\ } 0^\\circ \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 23 In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is", "triangle. $$5 \\boldcdot 2 = 10$$, $$4 \\boldcdot 2 = 8$$, $$(6.4) \\boldcdot 2 = 12.8$$ 2. Copy 2 is a scaled copy of the original triangle. The scale factor is $$\\frac12$$ or 0.5, because each side in Copy 2 is half as long as the corresponding side in the original triangle. $$5 \\boldcdot (0.5) = 2.5$$, $$4 \\boldcdot (0.5) = 2$$, $$(6.4) \\boldcdot (0.5) = 3.2$$ 3. Copy 3 is not a scaled copy of the original triangle. The shape has been distorted. The angles are different sizes and there is not one number we can multiply by each side length of the original triangle to get the corresponding side length in Copy 3. 1. Answers vary. Sample response: A right triangle with side lengths of 12, 15, and 19.2 units would be a scaled copy of the original triangle using a scale factor of 3. ### Scale Drawings This week your student will be learning about scale drawings. A scale drawing is a two-dimensional representation of an actual object or place. Maps and floor plans are some examples of scale drawings. The scale tells us what some length on the scale drawing represents in actual length. For example, a scale of “1 inch to 5 miles” means that 1 inch on the drawing represents 5 actual", "find the area of the entire paper, we square our side length and multiply by $4$. So, the answer is $4\\cdot\\left(\\frac{w}{\\sqrt{2}} + \\frac{h}{\\sqrt{2}}\\right)^2 = \\boxed{\\textbf{(A) } 2(w+h)^2}$. ~IronicNinja ## Solution 3 The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below. $[asy] /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); fill((2,0)--(3,1)--(3,-1)--cycle,yellow); fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow); fill((0,2)--(1,3)--(-1,3)--cycle,yellow); fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow); draw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),linewidth(.5)); dot((-3,3)); label(\"A\",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype(\"2.5 2.5\")+linewidth(.5)); draw((3,0)--(-3,0),linetype(\"2.5 2.5\")+linewidth(.5)); label(\"w\",(-1,-1),SW); label(\"w\",(1,-1),SE); label(\"w\",(1,1),NE); label(\"w\",(-1,1),NW); label(\"h\",(2.5,0.5),NW); label(\"h\",(2.5,-0.5),SW); label(\"h\",(-2.5,0.5),NE); label(\"h\",(-2.5,-0.5),SE); label(\"h\",(0.5,2.5),SE); label(\"h\",(-0.5,2.5),SW); label(\"h\",(0.5,-2.5),NE); label(\"h\",(-0.5,-2.5),NW); [/asy]$ The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\\frac{h^2}{2}$. There are four of them, so multiplied by four is $2h^2$. Together, paper's area is $2w^2 + 4wh + 2h^2$. This can be factored and written as $\\boxed{\\textbf{(A) } 2(w+h)^2} \\qquad$. ~Yee2121 (Solution) ~MRENTHUSIASM (Diagram)", "curve? $[asy] size(5cm); defaultpen(fontsize(6pt)); dotfactor=4; label(\"\\circ\",(0,1)); label(\"\\circ\",(0.865,0.5)); label(\"\\circ\",(-0.865,0.5)); label(\"\\circ\",(0.865,-0.5)); label(\"\\circ\",(-0.865,-0.5)); label(\"\\circ\",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]$ $\\textbf{(A)}\\ 2\\pi+6\\qquad\\textbf{(B)}\\ 2\\pi+4\\sqrt{3}\\qquad\\textbf{(C)}\\ 3\\pi+4\\qquad\\textbf{(D)}\\ 2\\pi+3\\sqrt{3}+2\\qquad\\textbf{(E)}\\ \\pi+6\\sqrt{3}$ ## Problem 19 Paula the painter and her two helpers each paint at constant, but different, rates. They always start at $8:00$ AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at $4:00$ PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at $2:12$ PM. On Wednesday Paula worked by herself and finished the house by working until $7:12$ P.M. How long, in minutes, was each day's lunch break? $\\textbf{(A)}\\ 30\\qquad\\textbf{(B)}\\ 36\\qquad\\textbf{(C)}\\ 42\\qquad\\textbf{(D)}\\ 48\\qquad\\textbf{(E)}\\ 60$ ## Problem 20 A $3 \\times 3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated $90\\,^{\\circ}$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black? $\\textbf{(A)}\\ \\frac{49}{512}\\qquad\\textbf{(B)}\\ \\frac{7}{64}\\qquad\\textbf{(C)}\\ \\frac{121}{1024}\\qquad\\textbf{(D)}\\", "of length $\\frac{2\\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve? $[asy] size(5cm); defaultpen(fontsize(6pt)); dotfactor=4; label(\"\\circ\",(0,1)); label(\"\\circ\",(0.865,0.5)); label(\"\\circ\",(-0.865,0.5)); label(\"\\circ\",(0.865,-0.5)); label(\"\\circ\",(-0.865,-0.5)); label(\"\\circ\",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]$ $\\textbf{(A)}\\ 2\\pi+6\\qquad\\textbf{(B)}\\ 2\\pi+4\\sqrt{3}\\qquad\\textbf{(C)}\\ 3\\pi+4\\qquad\\textbf{(D)}\\ 2\\pi+3\\sqrt{3}+2\\qquad\\textbf{(E)}\\ \\pi+6\\sqrt{3}$ ## Problem 19 Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break? $\\textbf{(A)}\\ 30\\qquad\\textbf{(B)}\\ 36\\qquad\\textbf{(C)}\\ 42\\qquad\\textbf{(D)}\\ 48\\qquad\\textbf{(E)}\\ 60$ ## Problem 20 A $3$ x $3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated $90\\,^{\\circ}$ clockwise about its center, and every white square in a position formerly occupied", "the equation in this form. Testing $s = \\sqrt{7}$, We obtain $7\\sqrt{3}$ on both sides, revealing that our answer is in fact $\\boxed{\\textbf{(B) } \\sqrt{7}}$ ~ siluweston ~ edits by aopspandy ## Solution 4 (Area) Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length. Begin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$. Connect these points to the vertices of the equilateral triangle, as well as to each other. $[asy] size(300); draw((0,3.5)--(4,9.15)--(8,3.5)--cycle); label(\"A\", (4,9.15), N, p = fontsize(10pt)); label(\"C\", (8,3.5), SE, p = fontsize(10pt)); label(\"B\", (0,3.5), SW, p = fontsize(10pt)); label(\"P\", (3.5,7), NW, p = fontsize(10pt)); draw((0,3.5)--(3.5,7)); draw((8,3.5)--(3.5,7)); draw((4,9.15)--(3.5,7)); label(\"P'\", (3.5, 0), S, p = fontsize(10pt)); draw((8,3.5)--(3.5,0)); draw((0,3.5)--(3.5,0)); label(\"P''\",(6,8.5), NE, p = fontsize(10pt)); draw((4,9.15)--(6,8.5)); draw((8,3.5)--(6,8.5)); label(\"P'''\",(2.25,8), NW, p = fontsize(10pt)); draw((0,3.5)--(2.25,8)); draw((4,9.15)--(2.25,8)); draw((3.5,0)--(6,8.5)--(2.25,8)--cycle); [/asy]$ Observe that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$. Note that $AP=AP''=AP'''$. The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \\sqrt{3}$. Similarly (pun intended), $P'P'''=3$ and $P'P''=2\\sqrt{3}$. Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are", "2 = 8$$, $$(6.4) \\boldcdot 2 = 12.8$$ 2. Copy 2 is a scaled copy of the original triangle. The scale factor is $$\\frac12$$ or 0.5, because each side in Copy 2 is half as long as the corresponding side in the original triangle. $$5 \\boldcdot (0.5) = 2.5$$, $$4 \\boldcdot (0.5) = 2$$, $$(6.4) \\boldcdot (0.5) = 3.2$$ 3. Copy 3 is not a scaled copy of the original triangle. The shape has been distorted. The angles are different sizes and there is not one number we can multiply by each side length of the original triangle to get the corresponding side length in Copy 3. 1. Answers vary. Sample response: A right triangle with side lengths of 12, 15, and 19.2 units would be a scaled copy of the original triangle using a scale factor of 3. Scale Drawings This week your student will be learning about scale drawings. A scale drawing is a two-dimensional representation of an actual object or place. Maps and floor plans are some examples of scale drawings. The scale tells us what some length on the scale drawing represents in actual length. For example, a scale of “1 inch to 5 miles” means that 1 inch on the drawing represents 5 actual miles. If the drawing shows a road that is", "- x$$ and $$12$$ $$(18-x)^2 + 144 = x^2$$ $$324 - 36*x + x^2 + 144 = x^2$$ $$X = 13$$ which means that $$BC = 5$$ and that means that area of triangle ABC = $$12*5*0.5 = 30$$ which means that the resulting area is $$108 - 30 = 78$$ B Director Joined: 07 Aug 2011 Posts: 578 GMAT 1: 630 Q49 V27 A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure [#permalink] Show Tags 04 May 2015, 09:15 4 KUDOS 3 This post was BOOKMARKED Bunuel wrote: A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure 1. The sheet is then folded along the segment CF so that points A and D coincide after the paper is folded, as shown in Figure 2 (The shaded area represents a portion of the back side of the paper, not visible in Figure 1). What is the area, in square inches, of the shaded triangle shown? A) 72 B) 78 C) 84 D) 96 E) 108 [Reveal] Spoiler: Attachment: The attachment Screen-Shot-2013-10-25-at-12.13.36-PM.png is no longer available Attachment: The attachment Screen-Shot-2013-10-25-at-12.13.45-PM.png is no longer available Kudos for a correct solution. From fig. 1 in attached image , $$BC=FE$$ as triangle ABC and DEF are congruent (as shown in fig. ). From fig", "triangle. 0.002778. In the previous example, the calculation yields the scale factor: 200.0000 ÷ 273.5781 = 0.73105; Enter the SCALE (Command). 12:3 ratio simplified is 4:1 Method 1 Adjusting Image Size by Hand When the drawings are printed for production, they’re represented much smaller than they actually are. For example, if you would like to apply a scale factor of 1:6 and the length of the item is 60 cm, you simply divide 60 / 6 = 10 cm to get the new dimension. Enter the obtained scale factor to adjust all objects in the drawing model to their correct size. The dimensions of our scale drawing are 6 by 8 which gives us an area of 48 square units. 1\"=30' 1:360. Or wondered how close a model ship or airplane looks to the real thing? The simplest way to calculate the scale factor is by using these simple formulas. 24 × 4 = 96 it helps us calculate the length quickly. 1/8\" = 1'-0\". Likewise, if you have a drawing at 1\" = 30'-0\" and you want to change it to 1/2\" = 1'-0\": 1\" = 30'-0\" has a Scale Factor of 360. To scale an object to a smaller size, you simply divide each dimension by the required scale factor. For example: 1/4” = 1” Invert", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "words, the “claim” with ID one is for a spot that starts at 483 inches from the left of the fabric and at 830 inches from the top. This piece is 24 inches wide and 18 inches tall. How many square inches of fabric are within two or more claims? Our challenge is to determine how many squares of fabric are claimed more than once. In over words, if we take all the squares from all the elves, how many square inches are claimed by more than one elf? Let’s start by reading the file: library(tidyverse) df <- read.delim(\"input3.txt\", header = FALSE) %>% as.tibble() %>% print() %>% # Clean the column mutate(V1 = str_replace_all(V1, \"[#@,:x]\", \" \"), V1 = str_replace_all(V1, \" {2,}\", \" \"), V1 = str_replace_all(V1, \"^ \", \"\")) %>% # Separate the column separate(V1, into = c(\"id\", \"left\", \"top\", \"wide\", \"tall\")) %>% # Make sure everything is numeric mutate_all(as.numeric) %>% # Create the right and down element mutate( right = left + wide, down = top + tall ) ## # A tibble: 1,375 x 1 ## V1 ## <fct> ## 1 #1 @ 483,830: 24x18 ## 2 #2 @ 370,498: 21x17 ## 3 #3 @ 403,823: 25x21 ## 4 #4 @ 619,976: 20x15 ## 5 #5 @ 123,385: 15x26 ## 6 #6 @ 484,592: 11x19", "# Problem #2129 2129 A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet? $[asy] size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label(\"3\", (-1.375,9.05), dir(260), fontsize(7)); label(\"A\", (0,15), N); label(\"B\", (0,-1), NE); label(\"30\", (0, -3), S); label(\"80\", (-6, 8), W);[/asy]$ $\\mathrm{(A) \\ } 120 \\qquad \\mathrm{(B) \\ } 180 \\qquad \\mathrm{(C) \\ } 240 \\qquad \\mathrm{(D) \\ } 360 \\qquad \\mathrm{(E) \\ } 480$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "(triangle[2][1] - triangle[0][1]) \\ + triangle[2][0] * (triangle[0][1] - triangle[1][1]) area /= 2 area = abs(area) print(area) # prints 20.0", "# Difference between revisions of \"1987 AHSME Problems/Problem 21\" ## Problem There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure 1 below, then one finds that the area of the square is $441 \\text{cm}^2$. What is the area (in $\\text{cm}^2$) of the square inscribed in the same $\\triangle ABC$ as shown in Figure 2 below? $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label(\"A\", (-25,10), W); label(\"B\", (-25,0), W); label(\"C\", (-15,0), E); label(\"Figure 1\", (-20, -5)); label(\"Figure 2\", (5, -5)); label(\"A\", (0,10), W); label(\"B\", (0,0), W); label(\"C\", (10,0), E); [/asy]$ $\\textbf{(A)}\\ 378 \\qquad \\textbf{(B)}\\ 392 \\qquad \\textbf{(C)}\\ 400 \\qquad \\textbf{(D)}\\ 441 \\qquad \\textbf{(E)}\\ 484$ ## Solution We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$. $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label(\"A\", (0,10), W); label(\"B\", (0,0), W); label(\"C\", (10,0), E); label(\"S\", (25/3,11/6), E); label(\"S\", (11/6,25/3), E); label(\"S\", (5,5), NE); [/asy]$ We now have that $3S=42\\sqrt{2}$, so $S=14\\sqrt{2}$. But we want the area of the square which is $S^2=(14\\sqrt{2})^2= \\boxed{\\mathrm{(B)}\\ 392}$", "minus the area of the circle. RonL 4. 2) area of circle 3.141 * 25 squared = 1963.125cm2 area of rectangles = Length * Width or Base * Height 220mm * 30mm = 6600mm2 or 66cm2 130mm * 30mm = 3900mm2 or 39cm2 area of triangle = 1/2 * base * height 1/2*130mm*125mm=8125mm2 how do i find out the volume in cubic millimetres if the template is 3mm thick? 5. Hello, turbod15b! Your dimensions are a bit off . . . Code: 30 *-------* | | | |45 | | | * 170 | : * | (2) : * | : * | :125 * | : * | : 130 * 60 * - - - * - - - - - - *---------------* | | 30| (1) |30 | | *-------------------------------------* 220 Rectangle (1) : . $220 \\times 30\\:= 6600\\text{ mm}^2$ Rectangle (2): . $170 \\times 30 \\:=\\:5100\\text{ mm}^2$ Triangle: . $\\frac{1}{2}(130)(125) \\:=\\:8125\\text{ mm}^2$ Circle: . $\\pi(25^2)\\:\\approx\\:1963.5\\text{ mm}^2$ Total area: . $6600 + 5100 + 8125 - 1963.5\\:=\\:17,861.5\\text{ mm}^2$ Total volume: . $17,861.5\\text{ mm}^2 \\times 3\\text{ mm} \\;=\\;53,584.5\\text{ mm}^3$ 6. ty", "+0 # HELP ASAP 0 32 1 The area of a square poster is 92 in2 Find the length of one side of the poster to the nearest tenth of an inch. Mar 11, 2021 #1 +311 0 Really, man? sigh Alright... $$\\sqrt92$$$$\\approx$$9.59166 Yay! Mar 11, 2021", "+0 i need help -2 188 2 In the diagram below, the area of $\\triangle BCD$ is 60. What is the length of side $\\overline{BC}$? [asy] size(5cm); pair a=(9,8); pair b=(0,8); pair c=(0,0); pair d=(15,0); dot(a); dot(b); dot(c); dot(d); draw(b--c--d--a--b--d); label(\"$A$\",a,NE); label(\"$B$\",b,NW); label(\"$C$\",c,SW); label(\"$D$\",d,SE); label(\"9\",(a+b)/2,N); label(\"15\",(c+d)/2,S); draw(b/8--b/8+d/15--d/15); draw(b+(a-b)/9--b+(a-b)/9+(c-b)/8--b+(c-b)/8); [/asy] Apr 8, 2020 #1 +924 0 Can you write this question in laTex so it is readable and you need to attach the diagram aforemention in the question. Thanks! Apr 8, 2020 #2 +924 +1 Your question describes the diagram here but asks for BC instead of the Area. If that is the diagram described, then BC= 8 Hope it helps! Apr 8, 2020", "width is 6 inches (fig.width = 6) and fig.asp = 0.7, the figure height will be automatically calculated from fig.width * fig.asp = 6 * 0.7 = 4.2. Figure 2.1 is an example using the chunk options fig.asp = 0.7, fig.width = 6, and fig.align = 'center', generated from the code below: par(mar = c(4, 4, 0.1, 0.1)) plot(pressure, pch = 19, type = \"b\") The actual size of a plot is determined by the chunk options fig.width and fig.height (the size of the plot generated from a graphical device), and we can specify the output size of plots via the chunk options out.width and out.height. The possible value of these two options depends on the output format of the document. For example, out.width = '30%' is a valid value for HTML output, but not for LaTeX/PDF output. However, knitr will automatically convert a percentage value for out.width of the form x% to (x / 100) \\linewidth, e.g., out.width = '70%' will be treated as .7\\linewidth when the output format is LaTeX. This makes it possible to specify a relative width of a plot in a consistent manner. Figure 2.2 is an example of out.width = 70%. par(mar = c(4, 4, 0.1, 0.1)) plot(cars, pch = 19) If you want to put multiple plots in one figure environment," ]

[ "\\:=\\:x^2 \\quad\\Rightarrow\\quad x^2 - 4x + 2 \\:=\\:0$ Quadratic Formula: . $x \\;=\\;\\frac{\\text{-}(\\text{-}4) \\pm \\sqrt{(\\text{-}4)^2 - 4(1)(2)}}{2(1)} \\;=\\;\\frac{4 \\pm\\sqrt{8}}{2} \\:=\\:\\frac{4 \\pm2\\sqrt{2}}{2} \\:=\\:2 \\pm \\sqrt{2}$ The side of the triangle is: . $2 + \\sqrt{2}$ cm. The side of the square is: . $1 + \\sqrt{2}$ cm. The area of the square is: . $(1 + \\sqrt{2})^2 \\;=\\;1 + 2\\sqrt{2} + 2 \\;=\\;\\boxed{3 + 2\\sqrt{2}\\,\\text{ cm}^2}$ 4. Ok, thanks guys. I understand better now", "area of $625$ square units occurs when $k = 0$, that is, when the rectangle is a square. Since $x+y= 100$ and $xy \\geq 500$, we have $x(50-x) \\geq 500$. Thus $x^2 -50x + 500 \\leq 0$. Consequently, $x$ must lie between the roots of the quadratic $x^2-50x+500 = 0$. The roots are $25 \\pm 5\\sqrt{5}$. Thus $25-5\\sqrt{5} \\leq x \\leq 25+5\\sqrt{5}$.", "$4T + 4P = S + 3T + 3P$ We discover that $T + P = S$ This lets us substitute $T + P$ in our formula for the area of our original square. $S + 4T + 4P = 400$ $S + 4(T + P) = 400$ $S + 4S + 400$ $5S = 400$ And so we can just solve for s $S = 80$ So the area of the shaded area is 80. #### Using Visual Similarity Between Components If we look at the multi-square image and see that the black squares and the small triangle + trapezoid little squares are the same size, then we can just say that $T + P = S$ directly, without having to figure that out in terms of algebra on the bigger squares. Then we can use that information in the equation for our original square as before and solve the problem: I. Equality of Black square and other small squares: $T + P = S$. II. Original square: $S + 4P + 4T = 400$ Substitution of I into II: $S + 4S = 400$ $5S = 400$ $S = 80$ The multi-square image is very helpful to see this solution because there are no \"whole\" squares made out of a trapezoid and small triangle in the", "Difference between revisions of \"2013 AIME I Problems/Problem 7\" Problem 7 A rectangular box has width $12$ inches, length $16$ inches, and height $\\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$. Solution After using the pythagorean formula three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{(x/2)^2 + 64}$, and $\\sqrt{(x/2)^2 + 36}$. Therefore, we can use Heron's formula to set up an equation for the area of the triangle. The semi perimeter is (10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 900 = $\\frac{1}{2}$((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - 10)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - $\\sqrt{(x/2)^2 + 64}$)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - $\\sqrt{(x/2)^2 + 36}$). Solving, we get $\\boxed{041}$.", "If the diagonals of two squares are in the ratio of 2 : 5, there area will be in the ratio of [A] 4 : 25 [B] 4 : 5 [C] 4 : 2 [D] $\\sqrt{2} : \\sqrt{5}$", "to the area of the smaller square? First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. We can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is 4 inches×2=8 inches\\begin{align*}4 \\ inches \\times 2 = 8 \\ inches\\end{align*}. Next find the area of both squares and compare. Area of smaller square: AAA=lw=(4 inches)(4 inches)=16 inches2 Area of larger square: AAA=lw=(8 inches)(8 inches)=64 inches2 Now compare the two areas. You want to know how the area of the larger square compares to the area of the smaller square. Write a ratio comparing the two areas. 64 inches216 inches2=4\\begin{align*}\\frac{64 \\ inches^2}{16 \\ inches^2} = 4\\end{align*} The area of the larger square is 4 times larger than the area of the smaller square. In the previous example, the scale factor that changed the smaller square to the larger square was 2, but the area was 4 times as large. This leads to a rule when comparing areas of similar figures. Write this rule down in your notebook. Let’s look at another example. Example Sandy has a garden with a length of 6 feet and a width of 10 feet. He wants", "+0 Help 0 135 2 Work out the base of a square with area, A = 225cm2. Oct 3, 2018 #1 +4833 +1 really.... $$A = s^2 \\\\ 225 = s^2 \\\\ s = \\pm \\sqrt{225} \\\\ s = \\pm 15 \\\\ \\text{negative lengths don't make sense}\\\\ s = 15$$ . Oct 3, 2018 #1 +4833 +1 really.... $$A = s^2 \\\\ 225 = s^2 \\\\ s = \\pm \\sqrt{225} \\\\ s = \\pm 15 \\\\ \\text{negative lengths don't make sense}\\\\ s = 15$$ Rom Oct 3, 2018 #2 0 Work out the area of a rectangle with base, b = 18mm and perimeter, P = 50mm. Oct 3, 2018", "## Algebra 1 $13$ inches A square with area $160$ square inches will have side lengths that are the square root of the area. This is because, for a square, Area = $s^2$, so $s=\\sqrt A$ With an area of $160$ inches squared, $s=\\sqrt {160}$ We will use method 2 in the book to estimate the square root. According to a calculator's square root function, $\\sqrt {160}\\approx12.65$ inches Since $12.65$ is closer to $13$ than to $12$, we round to the nearest integer which is $13$ inches.", "# The diagonals of a square each measure 7 feet. How do you find the area of the square? Oct 19, 2015 4.95 feet #### Explanation: You must use trigonometry. Since it is a square, and each angle in a square is 90 degrees, the two angles of the triangle the diagonal makes is half of that, 45. $\\sin \\theta =$(opposite)/hypoteneus $\\sin 45 = \\frac{o}{7}$ $7 \\sin 45 = o$ $o = 4.94974746831$ You can use the Pythagorean Theorem $A = 24.5$ square feet #### Explanation: The Pythagorean Theorem says that in a square triangle the square of the hypotenuse (a) is equal to the square of one leg (b) plus the square of the other lag (c): ${a}^{2} = {b}^{2} + {c}^{2}$ The diagonal of a square form a square triangle with two of the sides (see figure below) and (b) is equal to (c). Therefore, the equation can be written as below: ${a}^{2} = {b}^{2} + {b}^{2}$ ${7}^{2} = {b}^{2} + {b}^{2}$ $49 = 2 {b}^{2}$ ${b}^{2} = \\frac{49}{2}$ $b = \\sqrt{\\frac{49}{2}}$ feet (side of the square) The area is the product of two sides $A = \\sqrt{\\frac{49}{2}} \\times \\sqrt{\\frac{49}{2}} = \\frac{49}{2} = 24.5$ square feet", "cm × 5 cm) + (4 cm × 5 cm) + (4 cm × 5 cm) Ella’s Expression: 2(3 cm × 4 cm) + 2(3 cm × 5 cm) + 2(4 cm × 5 cm) Sofia’s and Ella’s expressions are the same, but Ella used the distributive property to make her expression more compact than Sofia’s. b. What fact about the surface area of a rectangular prism does Ella’s expression show more clearly than Sofia’s? A rectangular prism is composed of three pairs of sides with identical areas. ### Eureka Math Grade 6 Module 5 Lesson 17 Exit Ticket Answer Key Question 1. Name the shape, and then calculate the surface area of the figure. Assume each box on the grid paper represents a 1 in. × 1 in. square. Name of Shape: Rectangular Pyramid Area of Base: 5 in. × 4 in. = 20 in2 Area of Triangles: $$\\frac{1}{2}$$ × 4 in. × 4 in. = 8 in2, $$\\frac{1}{2}$$ × 5 in. × 4 in. = 10 in2 SurfaceArea: 20 in2 +8 in2 + 8in2 + 10 in2 + 10 in2 = 56 in2 Addition and Subtraction Equations – Round 1: Directions: Find the value of m in each equation. Question 1. m + 4 = 11 m = 7 Question 2. m + 2 = 5", "What is the correct answer? 4 # The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas. $4:15$ $4:25$ $3:15$ $3:25$ ### Correct Answer : B. $4:25$ Let the diagonals of the squares be 2x and 5x. Then ratio of their areas will be", "unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of $n$ if the the area of the small square is exactly $\\frac1{1985}$. ## Problem 5 A sequence of integers $a_1, a_2, a_3, \\ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \\ge 3$. What is the sum of the first $2001$ terms of this sequence if the sum of the first $1492$ terms is $1985$, and the sum of the first $1985$ terms is $1492$? ## Problem 6 As shown in the figure, $\\triangle ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of $\\triangle ABC$. ## Problem 7 Assume that $a$, $b$, $c$ and $d$ are positive integers such that $a^5 = b^4$, $c^3 = d^2$ and $c - a = 19$. Determine $d - b$. ## Problem 8 The sum of the following seven numbers is exactly 19: $a_1 = 2.56$, $a_2 = 2.61$, $a_3 = 2.65$, $a_4 = 2.71$, $a_5 = 2.79$, $a_6 = 2.82$, $a_7 = 2.86$. It is desired to replace each $a_i$ by", "= \\sqrt{(2a^2)}$$ . Now, diagonal of the square $$= 2\\sqrt{(x)} = \\sqrt{(2a^2)}$$. • Finding the side of the square (a) in terms of ‘x’: $$2a^2 = 4x$$. o $$a^2 = 2x$$. Therefore $$a = \\sqrt{(2x)}$$. Area of a square $$= a^2 = (\\sqrt{(2x)})^2 = 2x$$. _________________ Re: If the length of a diagonal of a square is 2x^(1/2), what is the area [#permalink] 31 May 2020, 07:08", "Problem Solving forum “The important thing to recognize here is that there are two SIMILAR TRIANGLES hiding in this diagram. http://s24.postimg.cc/50bvp89dt/sim_tri.jpg Notice that these two triangles share both a 90-degree angle AND the angle I''ve denoted with a purple dot. Since the 3 angles in each triangle must ...” May 24, 2019 Max@Math Revolution posted a new topic called A rectangular board measuring 9 inches by 11 inches is to be in the Problem Solving forum “[GMAT math practice question] A rectangular board measuring 9 inches by 11 inches is to be covered with 2 inch by 3 inch rectangular tiles and 1-inch square tiles. What is the smallest number of tiles that can be used to cover the board? A. 19 B. 20 C. 22 ...” May 24, 2019 Max@Math Revolution posted a reply to How many roots does the equation (x^2-1)/(x+1)=x have? in the Problem Solving forum “=> (x^2-1)/(x+1)=x => (x+1)(x-1) / (x+1) = x => x-1 = x by canceling (x+1). But x – 1 = x has no solution. Therefore, the answer is A. Answer: A” May 24, 2019 Max@Math Revolution posted a new topic called If the average (arithmetic mean) of 5 numbers is 21, what is in the Data Sufficiency forum “[GMAT math practice question] If the average (arithmetic mean) of 5", "by 7$$\\frac{3}{4}$$ inches. If 2$$\\frac{1}{4}$$ inches are added to each dimension as the rectangles get larger, what is the total area of the entire piece? The area of the smallest rectangle = 4 1/2 x 7 3/4 = 24 + 3 + 3 1/2 + 3/8 = 34 7/8 square inches b. Now, 4 1/2 + 2 1/4 = 4 2/4 + 2 1/4 = 6 3/4 7 3/4 + 2 1/4 = 10 36 3/4 x 10 = 60 + 7 2/4 = 67 1/2 square inches c. 6 3/4 + 2 1/4 =9 10 + 2 1/4 = 12 1/4 9 x 12 1/4 = = 108 + 2 1/4 = 110 1/4 square inches D. d. 9 + 2 1/4 = 11 1/4 12 1/4 + 2 1/4 = 14 2/4 11 1/4 x 14 2/4 = 154 + 3 2/4 + 5 1/2 + 1/8 = 163 1/8 square inches. So, the total area of the figure = 110 + 34 + 67 + 163 = 374 + 1 3/4 = 375 3/4 square inches Scroll to Top", "$${-5 + \\sqrt{623} \\over 2}+5\\,\\approx\\,14.98$$ feet Aug 31, 2017 #1 +8866 +2 The wall and the ground form a right angle...so we can use the Pythagorean theorem to find x . x2 + (x + 5)2 = 182 Multiply out (x + 5)2 . x2 + (x + 5)(x + 5) = 324 x2 + x2 + 10x + 25 = 324 Combine x2 and x2 , and subtract 324 from both sides. 2x2 + 10x - 299 = 0 Now we can use the quadratic formula to solve for x . $$x = {-10 \\pm \\sqrt{10^2-4(2)(-299)} \\over 2(2)} \\\\~\\\\ x={-10 \\pm \\sqrt{100+2392} \\over 4} \\\\~\\\\ x={-10 \\pm \\sqrt{2492} \\over 4} \\\\~\\\\ x={-10 \\pm 2\\sqrt{623} \\over 4} \\\\~\\\\ x={-5 + \\sqrt{623} \\over 2}\\,\\approx\\,9.98 \\qquad \\text{ or }\\qquad x={-5 - \\sqrt{623} \\over 2}\\, \\approx\\, -14.98$$ Since -14.98 causes a negative length for the side on the ground..... x must be ≈ 9.98 So, the length of the side on the wall = $${-5 + \\sqrt{623} \\over 2}+5\\,\\approx\\,14.98$$ feet hectictar Aug 31, 2017", "square is $$(\\text{side})^2$$. Upper bound of area is $$8.55^2 = 73.10\\;\\text{m}^2$$.", "+0 # three squares +1 388 2 +303 Three squares are drawn inside a circle as shown. The area of the circle is $60\\pi$ square inches. How many square inches are in the area of one square? Feb 20, 2021 #1 +36420 +2 https://web2.0calc.com/questions/help-would-be-appreciated_12#r1 Feb 20, 2021 #2 +303 0 oh ok lol notsmart2.0 Feb 20, 2021", "## Geometry: Common Core (15th Edition) The area of this rectangle is $25 ft^2$. The perimeter of a square is given by the formula: $P = 4s$, where $s$ is the measure of one side of the square. The area of a square is given by the formula: $A = s^2$, where $s$ is the measure of one side of the square. In order to find the area of the square, we need to find the measure of one of its sides. Let us use what we know. We actually know that the square has a perimeter of $20$ ft., so we plug in this information into the formula for the perimeter of a square and solve for $s$: $20 = 4s$ Divide both sides of the equation by $4$ to solve for $s$: $s = 5$ Now, we can plug in $5$ for $s$ in the formula for the area of a square: $A = 5^2$ Evaluate the exponent to solve for $A$: $A = 25$ The area of this rectangle is $25 ft^2$.", "My thinking was the sides of the squares were $\\frac{s}{\\sqrt{2}}$, so their areas were 1/2, and the areas of pairs of shaded triangles were 1/4. So I put down this series... $\\frac{1}{4}\\sum(\\frac{s}{(\\sqrt{2})^{n-1}})^2$ Which I see now, can be simplified to $\\sum\\frac{s^2}{2^{n+1}}$, the same thing Spimon shows above. -Scott" ]

[ "area of $CMN$ in square inches is $[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((.82,0)--(1,1)--(0,.76)--cycle); label(\"A\", (0,0), S); label(\"B\", (1,0), S); label(\"C\", (1,1), N); label(\"D\", (0,1), N); label(\"M\", (0,.76), W); label(\"N\", (.82,0), S);[/asy]$ $\\mathrm{(A)\\ } 2\\sqrt{3}-3 \\qquad \\mathrm{(B) \\ }1-\\frac{\\sqrt{3}}{3} \\qquad \\mathrm{(C) \\ } \\frac{\\sqrt{3}}{4} \\qquad \\mathrm{(D) \\ } \\frac{\\sqrt{2}}{3} \\qquad \\mathrm{(E) \\ }4-2\\sqrt{3}$ ## Problem 20 Let $$T=\\frac{1}{3-\\sqrt{8}}-\\frac{1}{\\sqrt{8}-\\sqrt{7}}+\\frac{1}{\\sqrt{7}-\\sqrt{6}}-\\frac{1}{\\sqrt{6}-\\sqrt{5}}+\\frac{1}{\\sqrt{5}-2}.$$ Then $\\mathrm{(A)\\ } T<1 \\qquad \\mathrm{(B) \\ }T=1 \\qquad \\mathrm{(C) \\ } 12 \\qquad$ $\\mathrm{(E) \\ }T=\\frac{1}{(3-\\sqrt{8})(\\sqrt{8}-\\sqrt{7})(\\sqrt{7}-\\sqrt{6})(\\sqrt{6}-\\sqrt{5})(\\sqrt{5}-2)}$ ## Problem 21 In a geometric series of positive terms the difference between the fifth and fourth terms is $576$, and the difference between the second and first terms is $9$. What is the sum of the first five terms of this series? $\\mathrm{(A)\\ } 1061 \\qquad \\mathrm{(B) \\ }1023 \\qquad \\mathrm{(C) \\ } 1024 \\qquad \\mathrm{(D) \\ } 768 \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 22 The minimum of $\\sin\\frac{A}{2}-\\sqrt{3}\\cos\\frac{A}{2}$ is attained when $A$ is $\\mathrm{(A)\\ } -180^\\circ \\qquad \\mathrm{(B) \\ }60^\\circ \\qquad \\mathrm{(C) \\ } 120^\\circ \\qquad \\mathrm{(D) \\ } 0^\\circ \\qquad \\mathrm{(E) \\ }\\text{none of these}$ ## Problem 23 In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is", "box indicates a 90º (right) angle. $A=\\pi{r}^2\\\\$ If your triangle is as pictured at left, then the height is drawn and measured outside the triangle. The area formula is the same. ### Example 10 Find the area for each described situation. Create a drawing of the shape with the included information. Show all work. As in the examples, if units are included then units should be present in your final result. Use 3.14 for π and round answers to tenths as needed. 1. Find the area of a rectangle whose length is 12.9 meters and height is one-third that amount. 2. Find the area of a triangle with base $\\displaystyle{24}\\frac{1}{2}\\\\$ inches and height 7 inches. 3. Find the area of a circle with radius $\\displaystyle{2}\\frac{1}{3}\\\\$ inches. Present answer in exact form and also compute rounded form using 3.14 for π. Present rounded form to the nearest tenth. #### Solutions 1. 55.5 m2 or 55.5 square meters (rounded) 2. 85.8 in2 or 85.8 square inches (rounded) 3. Exact 49/9 π in2, Rounded 17.1 in2 ### Example 11 Find the area given each described situation. Include a drawing of the shape with the included information. Show all work. As in the examples, if units are included then units should be present in your final result. Round answers to tenths unless", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "the equation in this form. Testing $s = \\sqrt{7}$, We obtain $7\\sqrt{3}$ on both sides, revealing that our answer is in fact $\\boxed{\\textbf{(B) } \\sqrt{7}}$ ~ siluweston ~ edits by aopspandy ## Solution 4 (Area) Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length. Begin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$. Connect these points to the vertices of the equilateral triangle, as well as to each other. $[asy] size(300); draw((0,3.5)--(4,9.15)--(8,3.5)--cycle); label(\"A\", (4,9.15), N, p = fontsize(10pt)); label(\"C\", (8,3.5), SE, p = fontsize(10pt)); label(\"B\", (0,3.5), SW, p = fontsize(10pt)); label(\"P\", (3.5,7), NW, p = fontsize(10pt)); draw((0,3.5)--(3.5,7)); draw((8,3.5)--(3.5,7)); draw((4,9.15)--(3.5,7)); label(\"P'\", (3.5, 0), S, p = fontsize(10pt)); draw((8,3.5)--(3.5,0)); draw((0,3.5)--(3.5,0)); label(\"P''\",(6,8.5), NE, p = fontsize(10pt)); draw((4,9.15)--(6,8.5)); draw((8,3.5)--(6,8.5)); label(\"P'''\",(2.25,8), NW, p = fontsize(10pt)); draw((0,3.5)--(2.25,8)); draw((4,9.15)--(2.25,8)); draw((3.5,0)--(6,8.5)--(2.25,8)--cycle); [/asy]$ Observe that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$. Note that $AP=AP''=AP'''$. The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \\sqrt{3}$. Similarly (pun intended), $P'P'''=3$ and $P'P''=2\\sqrt{3}$. Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are", "# Difference between revisions of \"1987 AHSME Problems/Problem 21\" ## Problem There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure 1 below, then one finds that the area of the square is $441 \\text{cm}^2$. What is the area (in $\\text{cm}^2$) of the square inscribed in the same $\\triangle ABC$ as shown in Figure 2 below? $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label(\"A\", (-25,10), W); label(\"B\", (-25,0), W); label(\"C\", (-15,0), E); label(\"Figure 1\", (-20, -5)); label(\"Figure 2\", (5, -5)); label(\"A\", (0,10), W); label(\"B\", (0,0), W); label(\"C\", (10,0), E); [/asy]$ $\\textbf{(A)}\\ 378 \\qquad \\textbf{(B)}\\ 392 \\qquad \\textbf{(C)}\\ 400 \\qquad \\textbf{(D)}\\ 441 \\qquad \\textbf{(E)}\\ 484$ ## Solution We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$. $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label(\"A\", (0,10), W); label(\"B\", (0,0), W); label(\"C\", (10,0), E); label(\"S\", (25/3,11/6), E); label(\"S\", (11/6,25/3), E); label(\"S\", (5,5), NE); [/asy]$ We now have that $3S=42\\sqrt{2}$, so $S=14\\sqrt{2}$. But we want the area of the square which is $S^2=(14\\sqrt{2})^2= \\boxed{\\mathrm{(B)}\\ 392}$", "of this triangle is 60 sq ft, which is often written as 60 ft 2 . Find the area of the rectangle. Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction $\\frac{\\text{12 in}\\text{.}}{\\text{1 ft}}$ since it has inches in the numerator. Then, $\\begin{array}{ccc}\\hfill 4\\text{ft}& =& \\frac{\\text{4 ft}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{\\text{1 ft}}\\hfill \\\\ & =& \\frac{4\\overline{)\\text{ft}}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{1\\overline{)\\text{ft}}}\\hfill \\\\ & =& \\text{48}\\text{in}\\text{.}\\hfill \\end{array}$ Thus, $\\text{4 ft 2 in}\\text{.}=\\text{48 in}\\text{.}+\\text{2 in}\\text{.}=\\text{50 in}\\text{.}$ The area of this rectangle is 400 sq in. Find the area of the parallelogram. The area of this parallelogram is 63.86 sq cm. Find the area of the trapezoid. $\\begin{array}{ccc}\\hfill {A}_{\\mathit{\\text{Trap}}}& =& \\frac{1}{2}\\cdot \\left({b}_{1}+{b}_{2}\\right)\\cdot h\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{14.5 mm},+,\\text{20.4 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{34.9 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{143.09 sq mm}\\right)\\hfill \\\\ & =& \\text{71.545 sq mm}\\hfill \\end{array}$ The area of this trapezoid is 71.545 sq mm. Find the approximate area of the circle. $\\begin{array}{ccc}\\hfill {A}_{c}& =& \\pi \\cdot {r}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot {\\left(\\text{16.8 ft}\\right)}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot \\left(\\text{282.24 sq ft}\\right)\\hfill \\\\ & \\approx & \\text{888.23 sq ft}\\hfill \\end{array}$ The area of this circle is approximately 886.23 sq ft. ## Practice set a Find the area of each of the following geometric figures. 36 sq cm", "of this triangle is 60 sq ft, which is often written as 60 ft 2 . Find the area of the rectangle. Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction $\\frac{\\text{12 in}\\text{.}}{\\text{1 ft}}$ since it has inches in the numerator. Then, $\\begin{array}{ccc}\\hfill 4\\text{ft}& =& \\frac{\\text{4 ft}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{\\text{1 ft}}\\hfill \\\\ & =& \\frac{4\\overline{)\\text{ft}}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{1\\overline{)\\text{ft}}}\\hfill \\\\ & =& \\text{48}\\text{in}\\text{.}\\hfill \\end{array}$ Thus, $\\text{4 ft 2 in}\\text{.}=\\text{48 in}\\text{.}+\\text{2 in}\\text{.}=\\text{50 in}\\text{.}$ The area of this rectangle is 400 sq in. Find the area of the parallelogram. The area of this parallelogram is 63.86 sq cm. Find the area of the trapezoid. $\\begin{array}{ccc}\\hfill {A}_{\\mathit{\\text{Trap}}}& =& \\frac{1}{2}\\cdot \\left({b}_{1}+{b}_{2}\\right)\\cdot h\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{14.5 mm},+,\\text{20.4 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{34.9 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{143.09 sq mm}\\right)\\hfill \\\\ & =& \\text{71.545 sq mm}\\hfill \\end{array}$ The area of this trapezoid is 71.545 sq mm. Find the approximate area of the circle. $\\begin{array}{ccc}\\hfill {A}_{c}& =& \\pi \\cdot {r}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot {\\left(\\text{16.8 ft}\\right)}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot \\left(\\text{282.24 sq ft}\\right)\\hfill \\\\ & \\approx & \\text{888.23 sq ft}\\hfill \\end{array}$ The area of this circle is approximately 886.23 sq ft. ## Practice set a Find the area of each of the following geometric figures. 36 sq cm", "of this triangle is 60 sq ft, which is often written as 60 ft 2 . Find the area of the rectangle. Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction $\\frac{\\text{12 in}\\text{.}}{\\text{1 ft}}$ since it has inches in the numerator. Then, $\\begin{array}{ccc}\\hfill 4\\text{ft}& =& \\frac{\\text{4 ft}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{\\text{1 ft}}\\hfill \\\\ & =& \\frac{4\\overline{)\\text{ft}}}{1}\\cdot \\frac{\\text{12}\\text{in}\\text{.}}{1\\overline{)\\text{ft}}}\\hfill \\\\ & =& \\text{48}\\text{in}\\text{.}\\hfill \\end{array}$ Thus, $\\text{4 ft 2 in}\\text{.}=\\text{48 in}\\text{.}+\\text{2 in}\\text{.}=\\text{50 in}\\text{.}$ The area of this rectangle is 400 sq in. Find the area of the parallelogram. The area of this parallelogram is 63.86 sq cm. Find the area of the trapezoid. $\\begin{array}{ccc}\\hfill {A}_{\\mathit{\\text{Trap}}}& =& \\frac{1}{2}\\cdot \\left({b}_{1}+{b}_{2}\\right)\\cdot h\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{14.5 mm},+,\\text{20.4 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{34.9 mm}\\right)\\cdot \\left(\\text{4.1 mm}\\right)\\hfill \\\\ & =& \\frac{1}{2}\\cdot \\left(\\text{143.09 sq mm}\\right)\\hfill \\\\ & =& \\text{71.545 sq mm}\\hfill \\end{array}$ The area of this trapezoid is 71.545 sq mm. Find the approximate area of the circle. $\\begin{array}{ccc}\\hfill {A}_{c}& =& \\pi \\cdot {r}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot {\\left(\\text{16.8 ft}\\right)}^{2}\\hfill \\\\ & \\approx & \\left(3.14\\right)\\cdot \\left(\\text{282.24 sq ft}\\right)\\hfill \\\\ & \\approx & \\text{888.23 sq ft}\\hfill \\end{array}$ The area of this circle is approximately 886.23 sq ft. ## Practice set a Find the area of each of the following geometric figures. 36 sq cm", "# Difference between revisions of \"1993 AJHSME Problems/Problem 17\" ## Problem Square corners, 5 units on a side, are removed from a $20$ unit by $30$ unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is $[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,0)--(20,0)--(20,5)--(0,5)--cycle); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)); draw((20+5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5*sqrt(2))--(20,0)); draw((5*sqrt(2),5+5*sqrt(2))--(5*sqrt(2),5*sqrt(2))--(5,5),dashed); draw((5*sqrt(2),5*sqrt(2))--(15+5*sqrt(2),5*sqrt(2)),dashed); [/asy]$ $\\text{(A)}\\ 300 \\qquad \\text{(B)}\\ 500 \\qquad \\text{(C)}\\ 550 \\qquad \\text{(D)}\\ 600 \\qquad \\text{(E)}\\ 1000$ ## Solution If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $((20)(30)-4(5)(5)) = 600-100 = \\boxed{\\text{(B)}\\ 500}$.", "Development (Energy and Utilities) Re: The top surface area of a square tabletop was changed so that one of t [#permalink] Show Tags 20 Nov 2019, 04:33 Official solution: S1:Let initial side of square=x After changes,area=$$(x-1)*(x+2)=70$$ solving we get, $$x^2+x-72 = 0$$ or, $$(x-8)(x+9)= 0$$ or, $$x$$= 8 or -9 since, the dimension cannot be negative, x = 8 SUFFICIENT S2: there was 25% change in one of the dimension if there is 25 % increase, that means 25% of x = 2 or x = 8 if there is 25% decrease, that means 25% of x = 1 or x = 4. Hence INSUFFICENT gmatbusters wrote: The top surface area of a square tabletop was changed so that one of the dimensions was reduced by 1 inch and the other dimension was increased by 2 inches. What was the surface area before these changes were made? (1) After the changes were made, the surface area was 70 square inches. (2) There was a 25 percent change in one of the dimensions. GMATbuster's Collection _________________ Re: The top surface area of a square tabletop was changed so that one of t [#permalink] 20 Nov 2019, 04:33 Display posts from previous: Sort by", "81 \\;\\text{in}^2$. This is because when the sides of two squares are in the ratio $9:1$, their areas are in the ratio $81:1$, and we have defined the units $\\text{in}^2$ so that $1\\;\\text{in}^2$ is the area of a square of side $1\\;\\text{in}$. From this you might get the idea that units of measurement are just another algebraic quantity that is multiplied by the number on the left of the units. And indeed they do seem to work that way, because we have defined them so that they do. So if we have a quantity $ab$, with numeric value $a$ and units $b$, then when it make sense to square this quantity (such as when taking the area of a square of side $ab$), the result will be $(ab)^2 = a^2 b^2$. This serves well when different units of the same dimension occur. For example, $3\\;\\text{inch} = 0.25\\;\\text{foot}$, so a square of side $3\\;\\text{inch}$ is also a square of side $0.25\\;\\text{foot}$, and its area is $9\\;\\text{inch}^2 = 0.0625\\;\\text{foot}^2$. It appears that we can reverse this process by taking a square root, that is, $9\\;\\text{inch}^2 = 0.0625\\;\\text{foot}^2$ is the area of a square of side $3\\;\\text{inch} = 0.25\\;\\text{foot}$, so we might think that $(9\\;\\text{inch}^2)^{1/2} = 3\\;\\text{inch}$ and $(0.0625\\;\\text{foot}^2)^{1/2} = 0.25\\;\\text{foot}$. If you suppose that it is possible for something to", "+0 Help 0 135 2 Work out the base of a square with area, A = 225cm2. Oct 3, 2018 #1 +4833 +1 really.... $$A = s^2 \\\\ 225 = s^2 \\\\ s = \\pm \\sqrt{225} \\\\ s = \\pm 15 \\\\ \\text{negative lengths don't make sense}\\\\ s = 15$$ . Oct 3, 2018 #1 +4833 +1 really.... $$A = s^2 \\\\ 225 = s^2 \\\\ s = \\pm \\sqrt{225} \\\\ s = \\pm 15 \\\\ \\text{negative lengths don't make sense}\\\\ s = 15$$ Rom Oct 3, 2018 #2 0 Work out the area of a rectangle with base, b = 18mm and perimeter, P = 50mm. Oct 3, 2018", "minus the area of the circle. RonL 4. 2) area of circle 3.141 * 25 squared = 1963.125cm2 area of rectangles = Length * Width or Base * Height 220mm * 30mm = 6600mm2 or 66cm2 130mm * 30mm = 3900mm2 or 39cm2 area of triangle = 1/2 * base * height 1/2*130mm*125mm=8125mm2 how do i find out the volume in cubic millimetres if the template is 3mm thick? 5. Hello, turbod15b! Your dimensions are a bit off . . . Code: 30 *-------* | | | |45 | | | * 170 | : * | (2) : * | : * | :125 * | : * | : 130 * 60 * - - - * - - - - - - *---------------* | | 30| (1) |30 | | *-------------------------------------* 220 Rectangle (1) : . $220 \\times 30\\:= 6600\\text{ mm}^2$ Rectangle (2): . $170 \\times 30 \\:=\\:5100\\text{ mm}^2$ Triangle: . $\\frac{1}{2}(130)(125) \\:=\\:8125\\text{ mm}^2$ Circle: . $\\pi(25^2)\\:\\approx\\:1963.5\\text{ mm}^2$ Total area: . $6600 + 5100 + 8125 - 1963.5\\:=\\:17,861.5\\text{ mm}^2$ Total volume: . $17,861.5\\text{ mm}^2 \\times 3\\text{ mm} \\;=\\;53,584.5\\text{ mm}^3$ 6. ty", "- x$$ and $$12$$ $$(18-x)^2 + 144 = x^2$$ $$324 - 36*x + x^2 + 144 = x^2$$ $$X = 13$$ which means that $$BC = 5$$ and that means that area of triangle ABC = $$12*5*0.5 = 30$$ which means that the resulting area is $$108 - 30 = 78$$ B Director Joined: 07 Aug 2011 Posts: 578 GMAT 1: 630 Q49 V27 A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure [#permalink] Show Tags 04 May 2015, 09:15 4 KUDOS 3 This post was BOOKMARKED Bunuel wrote: A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure 1. The sheet is then folded along the segment CF so that points A and D coincide after the paper is folded, as shown in Figure 2 (The shaded area represents a portion of the back side of the paper, not visible in Figure 1). What is the area, in square inches, of the shaded triangle shown? A) 72 B) 78 C) 84 D) 96 E) 108 [Reveal] Spoiler: Attachment: The attachment Screen-Shot-2013-10-25-at-12.13.36-PM.png is no longer available Attachment: The attachment Screen-Shot-2013-10-25-at-12.13.45-PM.png is no longer available Kudos for a correct solution. From fig. 1 in attached image , $$BC=FE$$ as triangle ABC and DEF are congruent (as shown in fig. ). From fig", "# Difference between revisions of \"1960 AHSME Problems/Problem 4\" ## Problem Each of two angles of a triangle is $60^{\\circ}$ and the included side is $4$ inches. The area of the triangle, in square inches, is: $\\textbf{(A)} 8\\sqrt{3}\\qquad \\textbf{(B)} 8\\qquad \\textbf{(C)} 4\\sqrt{3}\\qquad \\textbf{(D)} 4\\qquad \\textbf{(E)} 2\\sqrt{3}$ ## Solution If two of the angles are $60^{\\circ}$, then the other angle is $60^{\\circ}$ because angles in triangle add up to $180^{\\circ}$. That makes the triangle an equilateral triangle, so all sides are $4$ inches long. $[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); label(\"4\",(10,25)); label(\"2\",(12.5,-5)); label(\"2\",(37.5,-5)); label(\"4\",(40,25)); draw((25,43.301)--(25,0)); label(\"2\\sqrt{3}\",(20,15)); draw((25,3)--(28,3)--(28,0)); [/asy]$ Using the area formula $A = \\frac{s^2\\sqrt{3}}{4}$, the area of the triangle is $\\frac{4^2\\sqrt{3}}{4} = 4\\sqrt{3}$ square inches, which is answer choice $\\boxed{\\textbf{(C)}}$.", "of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each side of this square is $\\frac{h}{\\sqrt2}$. Thus, the area is the side length squared which is $\\frac{h^2}{2}$. Similarly, the combined figure of the two triangles with base $w$ is a square with area $\\frac{w^2}{2}$. Adding all of these together, we get $\\frac{w^2}{2} + \\frac{h^2}{2} + wh$. Since we have four of these areas in the entire wrapping paper, we multiply this by $4$, getting $4\\left(\\frac{w^2}{2} + \\frac{h^2}{2} + wh\\right) = 2\\left(w^2 + h^2 + 2wh\\right) = \\boxed{\\textbf{(A) } 2(w+h)^2}$. The diagram for this solution is shown below: $[asy] /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey); dot((-3,3)); label(\"A\",(-3,3),NW); draw((-3,0)--(-3,3)--(0,3),linewidth(.5)); draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5)); draw((0,2)--(-2,0),linewidth(.5)); draw((0,3)--(0,0),linetype(\"2.5 2.5\")+linewidth(.5)); draw((0,0)--(-3,0),linetype(\"2.5 2.5\")+linewidth(.5)); label(\"w\",(-1,1),NW); label(\"w\",(-2,2),SE); label(\"h\",(-2.5,0.5),NE); label(\"h\",(-0.5,2.5),SW); label(\"\\frac{w}{\\sqrt{2}}\",(-2,3),N); label(\"\\frac{h}{\\sqrt{2}}\",(-0.5,3),N); label(\"\\frac{h}{\\sqrt{2}}\",(0,2.5),E); label(\"\\frac{w}{\\sqrt{2}}\",(0,1),E); label(\"\\frac{w}{\\sqrt{2}}\",(-1,0),S); label(\"\\frac{h}{\\sqrt{2}}\",(-2.5,0),S); label(\"\\frac{h}{\\sqrt{2}}\",(-3,0.5),W); label(\"\\frac{w}{\\sqrt{2}}\",(-3,2),W); label(\"wh\",(-1.5,1.5),red); label(\"\\frac{w^2}{4}\",centroid((-3,3),(-1,3),(-3,1)),red); label(\"\\frac{w^2}{4}\",centroid((0,0),(-2,0),(0,2)),red); label(\"\\frac{h^2}{4}\",centroid((-3,0),(-2,0),(-3,1)),red); label(\"\\frac{h^2}{4}\",centroid((0,3),(-1,3),(0,2)),red); [/asy]$ ~Hydroquantum (Solution) ~MRENTHUSIASM (Diagram) ## Solution 2 The sheet of paper is made out of $4$ squares, as shown in the diagram in Solution 1. Each square has a side length of $\\frac{w}{\\sqrt{2}} + \\frac{h}{\\sqrt{2}}$, which we get from the Pythagorean Theorem (a $45^\\circ\\text{-}45^\\circ\\text{-}90^\\circ$ triangle's legs is the hypotenuse divided by $\\sqrt2$). Thus, to", "y if and only if it belongs to the shaded triangle bounded by the lines x=y, y=1, and x=0, the area of which is 1/2. The ratio of the area of the triangle to the area of the rectangle is \\frac{1/2}{4} = \\boxed{\\frac{1}{8}}. [asy] draw((-1,0)--(5,0),Arrow); draw((0,-1)--(0,2),Arrow); for (int i=1; i<5; ++i) { draw((i,-0.3)--(i,0.3)); } fill((0,0)--(0,1)--(1,1)--cycle,gray(0.7)); draw((-0.3,1)--(0.3,1)); draw((4,0)--(4,1)--(0,1),linewidth(0.7)); draw((-0.5,-0.5)--(1.8,1.8),dashed); [/asy]$$ Mellie Apr 27, 2015 #3 +17711 +10 Best Answer You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... ) If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values. Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2. geno3141 Apr 27, 2015 #4 +80983 +5 I get something a little different here than geno.....see the following pic........ All points inside triangle ADF will have x values less than their associated y values. And the area of this triangle = 1/2 sq units And the area of the whole rectangle = (4)(1) = 4 sq units So....the probability", "# Problem #2185 2185 A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral? $[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.25,0.2)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); [/asy]$ $\\mathrm{(A) \\ } 15\\qquad \\mathrm{(B) \\ } 17\\qquad \\mathrm{(C) \\ } \\frac{35}{2}\\qquad \\mathrm{(D) \\ } 18\\qquad \\mathrm{(E) \\ } \\frac{55}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "the figure. We know that its top edge is 22 inches. We also know that the length of the rectangle in the middle is 9 inches. We need to subtract that so we don’t include it as part of the bases of the triangles. That means there are $22 - 9 = 13$ inches left of the bottom edge. If we divide this equally, we find that each triangle has a base of 6.5 inches. Now we have the height and base of each triangle (they have the same height and base), so we can calculate the area. $A & = \\frac{1}{2} bh\\\\A & = \\frac{1}{2} 6.5(5)\\\\A & = \\frac{1}{2} (32.5)\\\\A & = 16.25 \\ in.^2$ Great! Now we have the area of each shape within the figure. All we have to do is add these together to find the area of the whole figure. $& \\text{triangle} \\ 1 \\qquad \\qquad \\quad \\text{rectangle} \\ \\qquad \\qquad \\ \\text{triangle}\\ 2 \\qquad \\quad \\qquad \\text{whole figure}\\\\& 16.25 \\ in.^2 \\qquad + \\qquad 45 \\ in.^2 \\qquad + \\qquad 16.25 \\ in.^2 \\qquad = \\qquad 77.5 \\ in.^2$ The area of the whole figure is 77.5 square inches. Now it's time for you to try a few on your own. #### Example A A figure is made up of three triangles.", "two-dimensional region, without any gaps or overlaps. For example, the area of region A is 8 square units. The area of the shaded region of B is $$\\frac12$$ square unit. • corresponding When part of an original figure matches up with part of a copy, we call them corresponding parts. These could be points, segments, angles, or distances. For example, point $$B$$ in the first triangle corresponds to point $$E$$ in the second triangle. Segment $$AC$$ corresponds to segment $$DF$$. • reciprocal Dividing 1 by a number gives the reciprocal of that number. For example, the reciprocal of 12 is $$\\frac{1}{12}$$, and the reciprocal of $$\\frac25$$ is $$\\frac52$$. • scale factor To create a scaled copy, we multiply all the lengths in the original figure by the same number. This number is called the scale factor. In this example, the scale factor is 1.5, because $$4 \\boldcdot (1.5) = 6$$, $$5 \\boldcdot (1.5)=7.5$$, and $$6 \\boldcdot (1.5)=9$$. • scaled copy A scaled copy is a copy of a figure where every length in the original figure is multiplied by the same number. For example, triangle $$DEF$$ is a scaled copy of triangle $$ABC$$. Each side length on triangle $$ABC$$ was multiplied by 1.5 to get the corresponding side length on triangle $$DEF$$." ]

[ "45$ And simplify.", "+ 4$ 4. Simplify And that's it", "simp; 99 99", "# How do you simplify 34 - (13 + x) ? This gives us $34 - 13 - x$. Combine like terms and we get $21 - x$.", "90.000", "and actual cost was$750,000. Using the data provided above, and assuming a... ##### How do you simplify ( 8x^2 -9x - 1) - ( -6x^2 + 4x - 8 )? How do you simplify ( 8x^2 -9x - 1) - ( -6x^2 + 4x - 8 )?...", "# How do you simplify 950 (700 + 250)? ##### 2 Answers Apr 27, 2018 $902500$ #### Explanation: $700 + 250$ $= 950$ So: $950 \\left(950\\right) = 902500$ Apr 27, 2018 $902500$ #### Explanation: To simplify this, we use the distributive property: Based on this image, this problem would be expanded to: $950 \\cdot 700 + 950 \\cdot 250$ Now let's simplify: $665000 + 237500$ So the answer is: $902500$ Hope this helps!", "InfiniteRepitition Basic Problem - 3233 Simplify $$\\sqrt{10+4\\sqrt{3-2\\sqrt{2}}}$$ The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "is : 90", "# How do you simplify 2.82/1.41? Jan 26, 2017 To simplify divide 2.82 by 1.41 ($2.82 \\div 1.41$) giving $2$", "90$)", "# How do you simplify 7m - 2m? Just arrange it and get $5 m$ $7 m - 2 m = 5 m$ Your answer is $5 m$", "required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required New \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 S M L XL 2XL This option is required \\$ 45.90 \\$ 45.90 S M L XL This option is required", "$(6)$: integrate $(7)$: simplify", "# How do you simplify and write (6.5times10^3)(5.2times10^-8) in standard form? I found: $0.000338$", "\\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL 3XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 45.90 \\$ 39.90 \\$ 45.90 \\$ 39.90 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL 3XL This option is required \\$ 23.99 \\$ 23.99 This option is required S M L XL 2XL 3XL This option is required", "simplify to 3 : 2 : 7.", "# How do you simplify 2 * 8 -4? Apr 18, 2018 Using P arentheses E xponents M ultiplication D ivision A ddition S ubtraction we need to first simplify $2 \\times 8$ to $16$ Now we have $16 - 4$ which is $12$", "# How do you simplify (-11i)(-10i)? $- 110$ $- 11 i \\left(- 10 i\\right) = 110 {i}^{2} = 110 \\left(- 1\\right) = - 110$", "# How do you simplify 4^3 + 5 ^4? Sep 27, 2015 The answer is $689$. ${4}^{3} + {5}^{4}$ $64 + 625 = 689$" ]

[ "Hence the answer is $\\boxed{34}$", "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "so $e = \\boxed{58}$.", "answer is $\\boxed{3}$.", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "that $332.5$ is closer to $385$ than any other answer choices, so our answer is $385$ which is $\\boxed{\\textbf{(A)}}$.", "answer is \\boxed{021}$", "result $\\boxed{492}$.", "answer of $\\boxed{A}$.", "a final answer of $\\boxed{400}$. ~AopsUser101", "and the answer is $\\boxed{D}$.", "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "-36.$$ Summing, the answer is $\\boxed{\\textbf{(A)} -100}.$ ~Leonard_my_dude~", "answer is $\\boxed{D}$.", "= \\boxed{108}$.", "$4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\\boxed{071}$. -Stormersyle", "- 400 \\ m = \\boxed {76.1 \\ m} {/eq}", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "$\\boxed{\\text{(A)}18\\text{m}}.$" ]

[ "while $871^2$ ends in $41$. Thus, the answer is $\\boxed{869}$.", "45$ And simplify.", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "that $332.5$ is closer to $385$ than any other answer choices, so our answer is $385$ which is $\\boxed{\\textbf{(A)}}$.", "-36.$$ Summing, the answer is $\\boxed{\\textbf{(A)} -100}.$ ~Leonard_my_dude~", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "sequence.) digits. Our answer is $40-14=\\boxed{26}$. ~IceMatrix", "= \\boxed{108}$.", "= \\boxed{\\textbf{(B)}\\ 24}.$", "$4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\\boxed{071}$. -Stormersyle", "a final answer of $\\boxed{400}$. ~AopsUser101", "and $$\\boxed{÷10}$$ are pressed after them. Similarily, $$\\boxed{-10}$$ may be worth $$-10$$, $$-100$$, $$-1000$$... or $$-1$$, $$-0.1$$, $$-0.01$$... The required number needs to be constructed from these final worths of $$\\boxed{+10}$$ and $$\\boxed{-10}$$. The easiest way to write $$2019$$ in this way is $$2019 = 1000 + 1000 +10 + 10 -1$$ That means you need: • Two instances of pressing $$\\boxed{+10}$$ that have their value increased 2 times • Two instances of pressing $$\\boxed{+10}$$ that have their value neither increased or decreased • One instances of pressing $$\\boxed{-10}$$ that have their value decreased 1 time Therefore in total you'll need to press $$\\boxed{+10}$$ four times, and $$\\boxed{-10}$$ one time. Somewhere between them you'll need to press $$\\boxed{\\times 10}$$ and $$\\boxed{÷10}$$ appropriate number of times, so that two of the $$\\boxed{+10}$$ had their value increased twice, two $$\\boxed{+10}$$ had their value unchanged, and $$\\boxed{-10}$$ had their value decreased once. There are actually three ways to do this in a minimal number of operations. One is the one you've found: $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}$$ The other two are $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{+10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}$$ $$\\boxed{+10}\\boxed{+10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{\\times 10}\\boxed{-10}\\boxed{÷10}\\boxed{+10}\\boxed{+10}$$", "$m+n=2+89=\\boxed{091}$.", "$m+n=\\boxed{831}$.", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "# How do you simplify x-46x? $- 45 x$ Given: $x - 46 x$ Since $x = 1 x$ and $46 x$ are \"alike\", they can be added or subtracted. Simplify: $1 x - 46 x = - 45 x$", "$$0$$ , $$-18 a - 36$$ , $$60 a + 100\\bigr]$$", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910}$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys." ]

[ "= intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite and", "T = intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite", "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant.", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "+0 # Some old geometry problem for you to solve 0 70 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)$$ $$PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "+0 # Some old geometry problem for you to solve 0 133 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "+0 # geometry +1 87 2 +2764 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 ### Best Answer #1 +866 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +866 +5 Best Answer Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +2764 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and", "rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label(\"A\", A, lsf * dir(110)); label(\"B\", B, lsf * unit(B-M)); label(\"C\", C, lsf * unit(C-M)); label(\"P\", P, lsf * unit(P-M) * 1.8); label(\"Q\", Q, lsf * dir(90) * 1.6); label(\"R\", R, lsf * unit(R-M) * 2); label(\"X\", X, lsf * dir(-60) * 2); label(\"Y\", Y, lsf * dir(45)); label(\"Z\", Z, lsf * dir(5)); label(\"M\", M, lsf * dir(M-P)*2); label(\"D\", D, lsf * dir(150)); label(\"E\", E, lsf * dir(5));[/asy]$ In this solution, all lengths and angles are directed. Firstly, it is easy to see by that $\\omega_A, \\omega_B, \\omega_C$ concur at a point $M$. Let $XM$ meet $\\omega_B, \\omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have $$XM \\cdot XE = XZ \\cdot XP \\quad\\text{and}\\quad XM \\cdot XD = XY \\cdot XP$$ Thusly $$\\frac{XY}{XZ} = \\frac{XD}{XE}$$ But we claim that $\\triangle XDP \\sim \\triangle PBM$. Indeed, $$\\measuredangle XDP = \\measuredangle MDP = \\measuredangle MBP = - \\measuredangle PBM$$ and $$\\measuredangle DXP = \\measuredangle MXY = \\measuredangle MXA = \\measuredangle MRA = \\measuredangle MRB = \\measuredangle MPB = -\\measuredangle BPM$$ Therefore, $\\frac{XD}{XP} = \\frac{PB}{PM}$. Analogously we find that $\\frac{XE}{XP} = \\frac{PC}{PM}$ and", "We reflect $\\overline{CD}$ about the $x$-axis to get $\\overline{C'D'}$ with $D'=(7,-5),$ then reflect $\\overline{C'D'}$ about the $y$-axis to get $\\overline{C'D''}$ with $D''=(-7,-5).$ We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(225); int xMin = -9; int xMax = 9; int yMin = -7; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (7,-5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(C--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(\"A(3,5)\",A,(0,2),linewidth(3.5)); dot(\"B\",B,(-2,0),linewidth(3.5)); dot(\"C\",C,(0,-2),linewidth(3.5)); dot(\"D(7,5)\",D,(0,2),linewidth(3.5)); dot(\"C'\",E,(0,-2),linewidth(3.5)); dot(\"D'(7,-5)\",F,(0,-2),linewidth(3.5)); dot(\"D''(-7,-5)\",G,(0,-2),linewidth(3.5)); [/asy]$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC+CD' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Parallelogram) Define points $A,B,C,$ and $D$ as Solution 1 does. Moreover, let $E$ be a point on $\\overline{CD}$ such that $\\overline{BE}$ is perpendicular to the $y$-axis, and $F$ be a point on $\\overline{BE}$ such that $\\overline{CF}$ is perpendicular to the $x$-axis, as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (4,2); pair F = (2,2); draw(A--B--C--D,red);", "= intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$, we have $$\\dfrac{BF}{FC} \\cdot \\dfrac{CA}{DA} \\cdot \\dfrac{DE}{BE} = 3\\dfrac{BF}{FC} = 1 \\implies \\dfrac{BF}{FC} = \\dfrac13 \\implies \\dfrac{BF}{BC} = \\dfrac14.$$ Therefore, $$[EBF] = \\dfrac{BE}{BD}\\cdot\\dfrac{BF}{BC}\\cdot [BCD] = \\dfrac12 \\cdot \\dfrac 14 \\cdot \\left( \\dfrac23 \\cdot [ABC]\\right) = \\boxed{\\textbf{(B) }30}.$$", "+0 # geometry +1 148 2 +3039 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +3039 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018", "anglemark(M,R,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label(\"A\", A, lsf * dir(110)); label(\"B\", B, lsf * unit(B-M)); label(\"C\", C, lsf * unit(C-M)); label(\"P\", P, lsf * unit(P-M) * 1.8); label(\"Q\", Q, lsf * dir(90) * 1.6); label(\"R\", R, lsf * unit(R-M) * 2); label(\"X\", X, lsf * dir(-60) * 2); label(\"Y\", Y, lsf * dir(45)); label(\"Z\", Z, lsf * dir(5)); label(\"M\", M, lsf * dir(M-P)*2); label(\"D\", D, lsf * dir(150)); label(\"E\", E, lsf * dir(5));[/asy]$ In this solution, all lengths and angles are directed. Firstly, it is easy to see by that $\\omega_A, \\omega_B, \\omega_C$ concur at a point $M$. Let $XM$ meet $\\omega_B, \\omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have $$XM \\cdot XE = XZ \\cdot XP \\quad\\text{and}\\quad XM \\cdot XD = XY \\cdot XP$$ Thusly $$\\frac{XY}{XZ} = \\frac{XD}{XE}$$ But we claim that $\\triangle XDP \\sim \\triangle PBM$. Indeed, $$\\measuredangle XDP = \\measuredangle MDP = \\measuredangle MBP = - \\measuredangle PBM$$ and $$\\measuredangle DXP = \\measuredangle MXY = \\measuredangle MXA = \\measuredangle MRA = 180^\\circ - \\measuredangle MRB = \\measuredangle MPB = -\\measuredangle BPM$$ Therefore," ]

[ "+0 # geometry +1 87 2 +2764 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 ### Best Answer #1 +866 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +866 +5 Best Answer Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +2764 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and", "+0 # geometry +1 148 2 +3039 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +3039 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018", "+0 # help geometry 0 38 4 In the diagram below, AB is parallel to CD and angle AXE is 98 degrees less than 3 times angle CYX . Find angle BXY. Apr 18, 2022 #1 +122 -3 ummm hang tight Apr 18, 2022 #2 +122390 +1 This is impossible If AB and CD are parallel, then angles AXE and CYX are equal since they are corresponding angles Apr 18, 2022 #3 +745 +1 You are missing angle measures. There is no way to solve it without more. Following my knowledge of proofs and just proving angles in a nutshell there is no way to do this.... If you can give us more angle measures I'll be more than happy to help! ILP Apr 18, 2022 #4 +1384 +1 I think it is possible... Let \\(\\angle AXE = \\angle CYX = x\\) We have the equation: \\(x = 3x-98\\) Solving for x, we find that \\(x = 49\\). Now we have: \\( \\angle AXE = \\angle BXY = \\color{brown}\\boxed{49}\\) What am I missing? Apr 18, 2022 edited by BuilderBoi Apr 18, 2022", "is given. Question 30. MODELING REAL LIFE A tributary joins a river at an angle x. Find the value of x. Explain. Answer: x = 21 Explanation: (2x + 21 ) = x 2x – x = 21 x = 21 Question 31. MODELING REAL LIFE The iron cross is a skiing trick in which the tips of the skis are crossed while the skier is airborne. Find the value of x in the iron cross shown. Answer: The value of x in the iron cross = 43 Explanation: (2x + 41) = 127 2x = 127 – 41 2x = 86 x = 43 FINDING ANGLE MEASURES Find all angle measures in the diagram. Question 32. Answer: x = 90˚ Question 33. Answer: 23.33 Explanation: (3x + 5) = 75 3x = 75 – 5 3x = 70 x = (70/3) x = 23.33 Question 34. Answer: x = 68 x = 67 Explanation: (2x + 4) = 140 2x = (140 – 4) 2x = 136 x = (136/2) x = 68 (2x + 6) = 140 2x = (140 – 6) 2x = 134 x = (134/2) x = 67 OPEN-ENDED Draw a pair of adjacent angles with the given description. Question 35. Both angles are acute. Answer: Question 36. One angle is acute, and", "axe^(-bx)(-b) + e^(-bx)(a) dy/dx = -abxe^(-bx) + ae^(-bx) dy/dx = 0 = ae^(-bx) (-bx + 1) $\\textcolor{red}{x = \\frac{1}{b}}$ ... 3. Originally Posted by skeeter ... i was just doing something wrong in the second derivative (to prove max) due to which it was turning out be a minimum. thanks anyways for spending your time on it.", "Y be points where the triangle intersects the square and [AXE]=[YIM]=$\\frac{100-80}{2}$=10 then AX=YI=2 units then XY=10-4=6 units triangle GXY is similar to triangle GEM where h=height of triangle GXY then by similarity $\\frac{h+10}{10}=\\frac{h}{6}$ then h=15 and h+10=25.", "# The transformed equation of $3x^2+3y^2,2xy=2$ when the coordinate axes are rotated through an angle of $45^{\\circ}$ is $\\begin {array} {1 1} (1)\\;x^2+2y^2=1 & \\quad (2)\\;2x^2+y^2=1 \\\\ (3)\\;x^2+y^2=1 & \\quad (4)\\;x^2+3y^2=1 \\end {array}$ $(2)\\;2x^2+y^2=1$", "|| CD then the value of x is (a) 87 (b) 93 (c) 147 (d) 141 Construction: Draw a line PQ parallel to AB which is also parallel to CD ∠FCD + Reflex∠FCD = 360 (Complete angle) ⇒ ∠FCD + 273 = 360 ⇒ ∠FCD = 87 Since, PQ || CD ∴∠QFC + ∠FCD = 180 (Angles on the same side of a transversal line are supplementary) ⇒ ∠QFC + 87 = 180 ⇒ ∠QFC = 93 Now, ∠ABF = ∠BFQ (Corresponding angles) = ∠BFC + ∠QFC = 54 + 93 = 147 x = 147 x = 147 Hence, the correct answer is option (c). #### Question 31: In Fig. 109, if AB || CD then the value of x is (a) 34 (b) 124 (c) 24 (d) 158 Construction: Draw a line PQ parallel to AB which is also parallel to CD ∠QEC + ∠ECD = 180 [Angles on the same side of a transversal line are supplementary] ⇒ ∠QEC + 56 = 180 ⇒ ∠QEC = 124 Now, ∠BEQ + ∠QEC = ∠BEC ⇒ ∠BEQ + 124 = 158 ⇒ ∠BEQ = 34 Now, ∠ABE = ∠BEQ = 34 [Corresponding angles] x = 34 x = 34 Hence, the correct answer is option (a). #### Question 32: In Fig. 110, if AB || CD. The", "If the altitudes intersect at a $123^\\circ$ angle, and $\\angle YXH = 26^\\circ$, then what is $\\angle HZX$ in degrees? Geometry Altitudes $\\overline{XD}$ and $\\overline{YE}$ of acute triangle $\\triangle XYZ$ intersect at point $H$. If the altitudes intersect at a $123^\\circ$ angle, and $\\angle YXH = 26^\\circ$, then what is $\\angle HZX$ in degrees? Geometry In $\\triangle ABC$, we have $AB = AC = 13$ and $BC = 10$. Let $M$ be the midpoint of $\\overline{AB}$ and $N$ be on $\\overline{BC}$ such that $\\overline{AN}$ is an altitude of $\\triangle ABC$. If $\\overline{AN}$ and $\\overline{CM}$ intersect at $X$, then what is $AX$? Geometry If you could fold a football field in half, the fifty yard line would the ? Line of symmetry Geometry Imagine a pie cut into four equal pieces. The angles formed by two adjacent pieces are A) supplementary B) Complementary I choose B is this correct? Imagine a pie cut into eight equal pieces. The angles formed by two adjacent pieces are? Supplementary (A) Am I correct?? Geometry If you could fold a football field in half the fifty yard line would be the ? Find geometry term Geometry Casey sights the top of an 84 foot tall lighthouse at an angle of elevation of 58 degrees. If Casey is 6 feet tall, how far is", "Feeds: Posts ## Isosceles Problems One of my students gave me a problem last fall that was very interesting. The problem is posed with the following diagram: What is the measure of $\\angle CDA$? That is, what is the value of $x$? I’ve given a similar problem in my trigonometry class for the past few years, except that version of the problem has a side length included and the triangle is not isosceles. A pdf of this problem is linked below: jun_7_mth_112_river_problem Working from my experience with the other version of this problem, I began to write in values for the various unlabeled angles in the diagram – if we label the intersection of $\\overline{AD}$ and $\\overline{BC}$ as $K$, then $\\angle CKD$ and $\\angle AKB$ are both $70^{\\circ}$, $\\angle CKA$ and $\\angle DKB$ are both $110^{\\circ}$, which makes $\\angle KCA$ $50^{\\circ}$ and $\\angle ADB$ is $40^{\\circ}$. I added in new variables and created a system of four equations with four unknowns, but it was a dependent system. The solution for this problem that was devised by the student who gave it to me is after the jump… His solution was primarily synthetic and not algebraic! His first step was to draw line segment $\\overline{AE}$ such that $\\overline{AE}$ and $\\overline{AB}$ are congruent, creating a small isosceles triangle in the", "+ 6Be^x = 36xe^x $ Rewritting as (adding a $0e^{x}$) may help you see it So then $(5A + 6B)e^x + 6Axe^x = 36xe^x + 0e^{x} $ We get a \"system\" of equations for A and B $5A+6B=0$ $6A=36$ So we can see A=6 and B=-5. I hope this helps. 5. Originally Posted by Jim Newt Thanks for the input EmptySet. So from this point: 1. yp = (Ax + B)e^x y'p= Ae^x + (Ax + B )e^x y\"p=2Ae^x + Axe^x + Be^x 2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x how do I make the jump to yp = (6x-5)e^x? Compare the coefficients: \\begin{aligned} 5A+6B&=0 \\\\ 6A&=36 \\end{aligned} This implies that A=6 and B=-5 Therefore, $y_p=(6x-5)e^x$. Hope this clarified things", "and $\\overline{PB}$. If $x$ represents the length of $\\overline{PB}$, $\\overline{AP}$ is equal to $(26 – x)$ ft. Using the angle bisector theorem, the ratio of $\\overline{AC}$ and $\\overline{AP}$ is equal to $\\overline{CB}$ and $\\overline{PB}$. \\begin{aligned}\\dfrac{AC}{AP} &= \\dfrac{CB}{PB}\\\\\\dfrac{36}{26- x} &= \\dfrac{42}{x}\\end{aligned} Apply cross-multiplication to simplify and solve the resulting equation. Find the length of $\\overline{PB}$ by finding the value of $x$. \\begin{aligned}36x &= 42(26- x)\\\\36x &= 1092- 42x\\\\36x + 42x &= 1092\\\\78x &= 1092\\\\x&= 14\\end{aligned} Hence, the length of $\\overline{PB}$ is equal to $14$ ft. ### Practice Question 1. In the triangle $\\Delta LMN$ the line $\\overline{MO}$ bisects $\\angle LMO$. Suppose that $\\overline{LM} = 20$ cm, $\\overline{MN} = 81$ cm, and $\\overline{LO} = 64$ cm, what is the length of line segment $\\overline{ON}$? A. $\\overline{ON} = 45$ cm B. $\\overline{ON} = 64$ cm C. $\\overline{ON} = 72$ cm D. $\\overline{ON} = 81$ cm 2. In the triangle $\\Delta ACB$, the line $\\overline{CP}$ bisects $\\angle ACB$. Suppose that $\\overline{AC} = 38$ ft, $\\overline{CB} = 57$ ft, and $\\overline{AB} = 75$ ft, what is the length of line segment $\\overline{PB}$? A. $\\overline{PB} = 38$ ft B. $\\overline{PB} = 45$ ft C. $\\overline{PB} = 51$ ft D. $\\overline{PB} = 57$ ft 3. The angle bisector $\\overline{AD}$ divides the line segment $AC$ that forms the triangle $\\Delta ACB$. Suppose that $\\overline{AC} = 12$", "Browse Questions # The transformed equation of $x^2+6xy+8y^2=10$ when the axes are rotated through an angle $\\large\\frac{\\pi}{4}$ is : $\\begin {array} {1 1} (a)\\;15x^2-14xy+3y^2=20 & \\quad (b)\\;15x^2+14xy-3y^2=20 \\\\ (c)\\;15^2+14xy+3y^2=20 & \\quad (d)\\;15x^2-14xy-3y^2=20 \\end {array}$ $(c)\\;15^2+14xy+3y^2=20$", "of a tetrahedron. | ax ay az 1 | 1 | bx by bz 1 | - | cx cy cz 1 | 6 | dx dy dz 1 | In general, the d-by-d determinant is 1/d! times time volume of the corresponding d-dimensional simplex. 27. In-circle test. Determine if the point d lies inside or outside the circle defined by the three points a, b, and c in the plane. Application: primitive operation in Delaunay triangulation algorithms. Assuming that a, b, c are labeled in counter-clockwise order around the circle, compute the determinant: | ax ay ax^2 + ay^2 1 | | bx by bx^2 + by^2 1 | | cx cy cx^2 + cy^2 1 | | dx dy dx^2 + dy^2 1 | adx = ax - dx; ady = ay - dy; bdx = bx - dx; bdy = by - dy; cdx = cx - dx; cdy = cy - dy; abdet = adx * bdy - bdx * ady; bcdet = bdx * cdy - cdx * bdy; cadet = cdx * ady - adx * cdy; alift = adx * adx + ady * ady; blift = bdx * bdx + bdy * bdy; clift = cdx * cdx + cdy * cdy; return alift * bcdet + blift * cadet", "# Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Solution I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows; here. $$ax^2+2hxy+by^2=0$$ Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides : $$ax+hy=\\pm y\\sqrt{h^2-ab}.$$ What should I do next? ## 6 Answers If $ax^2+2hxy+by^2=0$ bisects the coordinate axes then the points $(x,x)$ and $(x,-x)$ belong to these lines. If we consider $(x,x)$ we get $$ax^2+2hx^2+bx^2=0$$ and if we consider $(x,-x)$ we get $$ax^2-2hx^2+bx^2=0.$$ If $x\\ne 0$ then it is $$a+b=\\pm 2h\\implies (a+b)^2=4h^2.$$ Conversely, if $4h^2=(a+b)^2$ then, if $2h=a+b$ we have $ax^2+2hxy+by^2=(x+y)(ax+by)$ and if $2h=-(a+b)$ we have $ax^2+2hxy+by^2=(x-y)(ax+by).$ In any case, $ax^2+2hxy+by^2=0$ contains one of the lines $x+y=0$ or $x-y=0,$ which bisect the coordinate axis. • how did you get $$a+b=\\pm2h$$ – Iaamuser user Feb 9 '16 at 17:32 • If $(a+b)^2=4h^2$ then $|a+b|=\\sqrt{(a+b)^2}=\\sqrt{4h^2}=2|h|,$ from where it follows. – mfl Feb 9 '16 at 17:34 • how did you write.directly $a+b=\\pm2h$ after you consider $x\\ne 0$? – Iaamuser user Feb 9 '16 at 17:39 • If $x\\ne 0$ in $ax^2+2hx^2+bx^2=0$ we can divide both terms of the equality by $x^2$ to get $a+2h+b=0.$ That is, $a+b=-2h.$", "+0 # holp -1 178 1 +56 In the diagram below, we have $\\overline{AB}\\parallel\\overline{DE}$, $AC = 9$, $CE = 6$, and $BD = 24$. Find $CD$. Jul 8, 2021", "# The line $$\\overline{CD}$$ meet the line $$\\overline{AB}$$ at C. If $$\\overline{CE}$$ and $$\\overline{CF}$$ are the angle bisector of the angles ∠ACD and ∠BCD respectively then find the value of ∠ECF: 1. 80° 2. 100° 3. 70° 4. 90° Option 4 : 90° ## Detailed Solution Given: The line $$\\overline{CD}$$ meets the line $$\\overline{AB}$$ at C. The bisector of the angle ∠ACD is $$\\overline{CE}$$. The bisector of the angle ∠BCD is $$\\overline{CF}$$. Concept used: In the above figure, the sum of a and b is always 180°. Calculation: CE is the bisector of ∠ACD, So, ∠ACD = ∠ECD = x (Let) CF is the bisector of ∠DCB, So, ∠DCF = ∠FCB = y (Let) According to the question, According to the concept, ∠ACD + ∠BCD = 180° ⇒ 2x + 2y = 180° ⇒ x + y = 90° ⇒ ∠ECD + ∠DCF = 90° ∴ The value of the angle ∠ECF is 90°.", "RY in the diagram). On doing so, the diagram will look as follows: Solution: Since BX is a median in triangle ABC, we can use the Special Property to conclude that, Area of ABX = Area of BXC It is given that the area of triangle ABX = 32 square units. Hence, Area of BXC = 32 square units. Since XY is a median in triangle BXC, we can conclude that • Area of BXY = Area of XYC = $$32/2$$ = 16 Since RY is a median in triangle XYC, we can conclude that • Area of RYX = Area of RYC = $$16/2$$= 8 And since RS is a median in triangle YRC we can conclude that • Area of YRS = Area of RCS =$$8/2$$ = 4 So, is this question really tough? Definitely, not as tough as it looked at the beginning. We used one simple and special property to elegantly solve a complicated looking question. 2. DISCOVER THE PROPERTY OF SUM OF EXTERNAL ANGLES OF A TRIANGLE We know that, an external angle = sum of two opposite interior angles. In the above diagram, ABC is a triangle with sides extended as shown. Therefore, using the above property we can write, Ext. Angle ACD = int. angle CAB + int. angle CBA But", "+0 # holp -1 38 1 +28 In the diagram below, we have $\\overline{AB}\\parallel\\overline{DE}$, $AC = 9$, $CE = 6$, and $BD = 24$. Find $CD$. Jul 8, 2021", "This is an interesting problem which may still be looking for an answer I get the following y=-.3955x +1.486 Method used employs trig to convert slopes to elevation angles.The angle for the bisector is the angle of lower line plus half the angle between the lines 21.58 degrees slope -.3955. line equation from point slope formula using the point (2/3,11/9) bj • July 4th 2009, 10:42 AM earboth Quote: Originally Posted by bjhopper posted by mj.alawami and earboth This is an interesting problem which may still be looking for an answer ..... No I get the following y=-.3955x +1.486 ... The result from my previous post was: $\\left(\\dfrac{2\\cdot \\sqrt{13}}{13} + \\dfrac{\\sqrt{37}}{37}\\right) \\cdot x + \\left(\\dfrac{3 \\cdot \\sqrt{13}}{13} + \\dfrac{6 \\cdot \\sqrt{37}}{37}\\right) \\cdot y - \\left(\\dfrac{5\\cdot \\sqrt{13}}{13} + \\dfrac{8 \\cdot \\sqrt{37}}{37}\\right) = 0$ If you solve this equation for y and use approximate values you'll get: $y = - 0.3954474799\\cdot x + 1.485853875$" ]

[ "+0 # geometry +1 87 2 +2764 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 ### Best Answer #1 +866 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +866 +5 Best Answer Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +2764 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and", "+0 # geometry +1 148 2 +3039 In the diagram below, $$\\overline{AB}\\parallel \\overline{CD}$$ and $$\\angle AXE$$ is $$108^\\circ$$ less than 3 times $$\\angle CYX$$. Find $$\\angle BXY$$. tertre Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #1 +942 +5 Hi tertre, Since line AB is parallel to line CD, angle AXE is equal to CYX. This means we can set angle AXE = angle CYX = x Since AXE is 108º less than 3 times CYX, we have this: $$x+108=3x$$ Solving for x, we get: $$x=54$$ We know that AXE and CYX = 54º Since AXE and BXY are congruent angles, we know that $$BXY=54º$$ I hope this helps, Gavin GYanggg Apr 21, 2018 #2 +3039 +2 Thanks for the detailed explanation, Gavin! tertre Apr 21, 2018", "T = intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite", "= intersectionpoints(p1,p2)[0]; //Time to start drawing dot(A); dot(B); dot(C); dot(W); dot(H); dot(M); dot(L); dot(X); dot(N); dot(Y); dot(T); dot(circumcenter(N,B,W)); dot(circumcenter(C,M,W)); draw(p1); draw(p2); draw(A--B--C--Y--W--X--B); draw(C--A); draw(A--L); draw(B--M); draw(C--N); label(\"A\", A, E); label(\"B\", B, S); label(\"C\", C, E); label(\"L\", L, S); label(\"M\", M, S); label(\"N\", N, S); label(\"H\", H, E); label(\"T\", T, E); label(\"W\", W, S); label(\"X\", X, E); label(\"Y\", Y, E); [/asy]$ Let $T$ be the intersection of $\\omega_1$ and $\\omega_2$ other than $W$. Lemma 1: $T$ is on $XY$. Proof: We have $\\angle{XTW} = \\angle{YTW} = 90^\\circ$ because they intercept semicircles. Hence, $\\angle{XTY} = \\angle{XTW} + \\angle{YTW} = 180^\\circ$, so $XTY$ is a straight line. Lemma 2: $T$ is on $AW$. Proof: Let the circumcircles of $NBW$ and $MWC$ be $\\omega_1$ and $\\omega_2$, respectively, and, as $BNMC$ is cyclic (from congruent $\\angle{BNC} = \\angle{BMC} = 90^\\circ$), let its circumcircle be $\\omega_3$. Then each pair of circles' radical axises, $BN, TW,$ and $MC$, must concur at the intersection of $BN$ and $MC$, which is $A$. Lemma 3: $YT$ is perpendicular to $AW$. Proof: This is immediate from $\\angle{YTW} = 90^\\circ$. Let $AH$ meet $BC$ at $L$, which is also the foot of the altitude to that side. Hence, $\\angle{ALB} = 90^\\circ.$ Lemma 4: Quadrilateral $THLW$ is cyclic. Proof: We know that $NHLB$ is cyclic because $\\angle{BNH}$ and $\\angle{BLH}$, opposite and", "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)$$ $$PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "We reflect $\\overline{CD}$ about the $x$-axis to get $\\overline{C'D'}$ with $D'=(7,-5),$ then reflect $\\overline{C'D'}$ about the $y$-axis to get $\\overline{C'D''}$ with $D''=(-7,-5).$ We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(225); int xMin = -9; int xMax = 9; int yMin = -7; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (7,-5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(C--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(\"A(3,5)\",A,(0,2),linewidth(3.5)); dot(\"B\",B,(-2,0),linewidth(3.5)); dot(\"C\",C,(0,-2),linewidth(3.5)); dot(\"D(7,5)\",D,(0,2),linewidth(3.5)); dot(\"C'\",E,(0,-2),linewidth(3.5)); dot(\"D'(7,-5)\",F,(0,-2),linewidth(3.5)); dot(\"D''(-7,-5)\",G,(0,-2),linewidth(3.5)); [/asy]$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC+CD' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Parallelogram) Define points $A,B,C,$ and $D$ as Solution 1 does. Moreover, let $E$ be a point on $\\overline{CD}$ such that $\\overline{BE}$ is perpendicular to the $y$-axis, and $F$ be a point on $\\overline{BE}$ such that $\\overline{CF}$ is perpendicular to the $x$-axis, as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (4,2); pair F = (2,2); draw(A--B--C--D,red);", "# Control distance of label from point in asymptote I'm making a diagram for a solution on the AoPS wiki and some of the labels are a little too close for comfort. Is there a way I can move a label farther from what I'm labeling? I have the following code so far: import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label(\"$$A$$\",A,W); label(\"$$B$$\",B,E); label(\"$$C$$\",C,W); label(\"$$D$$\",D,NE); label(\"$$O_a$$\",O,W); label(\"$$O_b$$\",P,E); label(\"$$3$$\",(A+C)/2,W); label(\"$$4$$\",(B+C)/2,S); label(\"$$\\frac{15}{7}$$\",(A+D)/2,NE); label(\"$$\\frac{20}{7}$$\",(B+D)/2,NE); label(\"$$M$$\", inter1, W); label(\"$$N$$\", inter2, E); and the labels for M and N are a little too close to the lines I've drawn. Thanks! On a side note, is having so many pictures defined good form? I just started asymptote and I feel like it's not very elegant. From the docs (asymptote.pdf, 4.4 label): ... If align is NoAlign, the label will be centered at user coordinate position; otherwise it will be aligned in the", "rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label(\"A\", A, lsf * dir(110)); label(\"B\", B, lsf * unit(B-M)); label(\"C\", C, lsf * unit(C-M)); label(\"P\", P, lsf * unit(P-M) * 1.8); label(\"Q\", Q, lsf * dir(90) * 1.6); label(\"R\", R, lsf * unit(R-M) * 2); label(\"X\", X, lsf * dir(-60) * 2); label(\"Y\", Y, lsf * dir(45)); label(\"Z\", Z, lsf * dir(5)); label(\"M\", M, lsf * dir(M-P)*2); label(\"D\", D, lsf * dir(150)); label(\"E\", E, lsf * dir(5));[/asy]$ In this solution, all lengths and angles are directed. Firstly, it is easy to see by that $\\omega_A, \\omega_B, \\omega_C$ concur at a point $M$. Let $XM$ meet $\\omega_B, \\omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have $$XM \\cdot XE = XZ \\cdot XP \\quad\\text{and}\\quad XM \\cdot XD = XY \\cdot XP$$ Thusly $$\\frac{XY}{XZ} = \\frac{XD}{XE}$$ But we claim that $\\triangle XDP \\sim \\triangle PBM$. Indeed, $$\\measuredangle XDP = \\measuredangle MDP = \\measuredangle MBP = - \\measuredangle PBM$$ and $$\\measuredangle DXP = \\measuredangle MXY = \\measuredangle MXA = \\measuredangle MRA = \\measuredangle MRB = \\measuredangle MPB = -\\measuredangle BPM$$ Therefore, $\\frac{XD}{XP} = \\frac{PB}{PM}$. Analogously we find that $\\frac{XE}{XP} = \\frac{PC}{PM}$ and", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$. Case 3: $g$ passes through $CD$ or $BC$ $[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label(\"A\",A,SW); dot(B); label(\"B\",B,SE); dot(C); label(\"C\",C,NE); dot(D); label(\"D\",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label(\"E\",e,N); dot(XA); label(\"X_1\",XA,N); dot(XB); label(\"X_2\",XB,E); dot(XC); label(\"X_3\",XC,S); [/asy]$ Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \\perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$. By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\\angle EAB = \\angle AED = \\angle CEX_1$. Thus, by AA Similarity, $\\triangle X_2BA \\sim \\triangle X_3DE \\sim \\triangle X_1CE$. Using similar triangle ratios, we have $DX_3 = \\tfrac{bx}{a}$ and $X_1 = \\tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \\frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that", "dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### Solution 3 Again, draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\\angle ABE = 65^\\circ$ and $m\\angle DCE = 115^\\circ$. Applying the Law of Sines to $\\triangle AEB$ and $\\triangle EDC$, it follows that $DE = \\frac{2\\sin 115^\\circ}{\\sin \\angle DEC} = \\frac{2\\sin 65^\\circ}{\\sin \\angle AEB} = EA$, so $\\triangle AED$ is isosceles. We finish as we did in solution 2. $[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"E\",P,W); [/asy]$ ## Solution 4 Start off with the same diagram as solution 1. Now draw $\\overline{CA}$ which creates isosceles $\\triangle EAB$. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that its is $\\boxed{85°}$ (Error compiling LaTeX. ! Package inputenc Error: Unicode char \\u8:° not set up for use with LaTeX.)", "# 2001 IMO Problems/Problem 5 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem $ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$. ## Solution $[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)*A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (0,0.1); dot(\"B\", B, NW); dot(\"Y\", Y, NE); dot(\"D\", D, W); dot(\"E\", E, E); dot(\"A\",A,N); dot(\"X\",X,S); label(\"C\",E+(0,-0.1),E); draw(A--D--E--cycle); draw(B--Y); draw(B--E); // draw(B--Xx--E,dashed); // draw(Y--Xx, dashed); draw(A--X--D, dashed); [/asy]$ Let $D$ be on extension of $AB$ and $BD=BX$. Let $E$ be on $YC$ and $YE=YB$, then $\\[AD=AB+BD=AB+BX=AY+YB=AE\\]$ Since $A=60$, $\\triangle{ADE}$ is equilateral. Let $\\angle{ABY}=x$, then, $\\[\\angle{YBX}=\\angle{BDX}=\\angle{BXD}=\\angle{YEX}=x\\]$ We claim that $X$ must be on $BE$, i.e., $C=E$. If $X$ is not on $BE$, then $\\angle{EBX}=\\angle{YBX}-\\angle{YBE}=\\angle{YEX}-\\angle{YEB}=\\angle{BEX}$, which leads to $BX=EX=DX$, and $\\triangle{BDX}$ is equilateral, which is not possible. With that, we have, in $\\triangle{ABE}$, $60+2x+x=180$, $x=40$, and $\\angle{ABE}=80$. Solution by $Mathdummy$. 2001 IMO (Problems) • Resources Preceded byProblem 4 1 • 2 •", "# User talk:Bobthesmartypants/Sandbox ## Solution 1 $[asy] path Q; Q=(0,0)--(1,2)--(5,2)--(4,0)--cycle; draw(Q); draw((0,0)--(1.5,1)); label(\"D\",(0,0),S); draw((1,2)--(1.5,1)); label(\"A\",(1,2),N); draw((5,2)--(1.5,1)); label(\"B\",(5,2),N); draw((4,0)--(1.5,1)); label(\"C\",(4,0),S); draw((2,0)--(1.5,1),linetype(\"8 8\")); label(\"E\",(2,0),S); draw((2/3,4/3)--(1.5,1),linetype(\"8 8\")); label(\"F\",(2/3,4/3),W); label(\"P\",(1.5,1),NNE); [/asy]$ First, continue $\\overline{AP}$ to hit $\\overline{CD}$ at $E$. Also continue $\\overline{CP}$ to hit $\\overline{AD}$ at $F$. We have that $\\angle PAB=\\angle PCB$. Because $\\overline{AB}\\parallel\\overline{CD}$, we have $\\angle PAB=\\angle PED$. Similarly, because $\\overline{AD}\\parallel\\overline{BC}$, we have $\\angle PCB=\\angle PFD$. Therefore, $\\angle PAB=\\angle PED=\\angle PCB=\\angle PFD$. We also have that $\\angle ADC=\\angle ABC$ because $ABCD$ is a parallelogram, and $\\angle APC=\\angle FPE$. Therefore, $ABCP\\sim FDEP$. This means that $\\dfrac{FD}{AB}=\\dfrac{FP}{AP}=\\dfrac{DP}{BP}$, so $\\Delta ABP\\sim\\Delta FDP$. Therefore, $\\angle PBA=\\angle PDA$. $\\Box$ ## Solution 2 Note that $\\dfrac{1}{n}$ is rational and $n$ is not divisible by $2$ nor $5$ because $n>11$. This means the decimal representation of $\\dfrac{1}{n}$ is a repeating decimal. Let us set $a_1a_2\\cdots a_x$ as the block that repeats in the repeating decimal: $\\dfrac{1}{n}=0.\\overline{a_1a_2\\cdots a_x}$. ($a_1a_2\\cdots a_x$ written without the overline used to signify one number so won't confuse with notation for repeating decimal) The fractional representation of this repeating decimal would be $\\dfrac{1}{n}=\\dfrac{a_1a_2\\cdots a_x}{10^x-1}$. Taking the reciprocal of both sides you get $n=\\dfrac{10^x-1}{a_1a_2\\cdots a_x}$. Multiplying both sides by $a_1a_2\\cdots a_n$ gives $n(a_1a_2\\cdots a_x)=10^x-1$. Since $10^x-1=9\\times \\underbrace{111\\cdots 111}_{x\\text{ times}}$ we divide $9$ on both sides of the equation to get $\\dfrac{n(a_1a_2\\cdots a_x)}{9}=\\underbrace{111\\cdots 111}_{x\\text{ times}}$. Because", "= Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"\\angle C obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"\\angle B obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: 1. The region in which $\\angle B$ is obtuse is determined by construction. 2. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"A\", A, 1.5*S, linewidth(4.5)); dot(\"B\", B, 1.5*S, linewidth(4.5)); dot(\"D\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"C_2\", C2, 1.5*N, linewidth(4.5)); dot(\"C_1\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: • If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$," ]

[ "teach the course two times per school year, so over the years I've seen about a dozen different groups of students. Every single year, my classes are filled mostly with boys. Out of a class of 15 students, there are usually 2-3 girls, sometimes none. I always wondered why young women did not sign up for the class; I assumed that the course title \"Web Page Design\" simply was not interesting to them. Or, maybe their mothers were just slower to register them -- there is, unfortunately, a limited number of seats in the class, it's first-come, first-serve registration. Or, maybe it's a cultural thing -- girls are pushed away from the computers by their brothers and male classmates, and if they don't get a chance to use the computer, how can they know how engaging and wonderful it can be? But, I'm a girl, I have 3 brothers, and somehow I managed to wriggle my way into technology. Who knows. Regardless of the reason, the statistics are obvious: -something- has been keeping girls away from technology. Until now. My eyes boggled, my brow furrowed, and I scratched my head in wonder at the class list in my hands. This term, there are MORE GIRLS in my class than boys! I can't believe it! What is happening?!?! Is", "In Miss Quaffley's class, the girls make up more than $45$% of the pupils, but less than $50$%. What is the smallest possible number of girls in her class?", "choices. 3 appetizers, 4 main courses, and 3 desserts. How many different combinations to choose from? • Boys and girls There are 20 boys and 10 girls in the class. How many different dance pairs can we make of them? • Desks A class has 20 students. The classroom consists of 20 desks, with 4 desks in each of 5 different rows. Amy, Bob, Chloe, and David are all friends, and would like to sit in the same row. How many possible seating arrangements are there such that Amy, Bob, • Playing cards How many possible ways are to shuffle 7 playing cards?", "to In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?", "there are 7 boys and 14 girls. Right on!", "girls and one boy c) there will be at least two boys • Competition 15 boys and 10 girls are in the class. On school competition of them is selected 6-member team composed of 4 boys and 2 girls. How many ways can we select students? • Math logic There are 20 children in the group, each two children have a different name. Alena and John are among them. How many ways can we choose 8 children to be among the selected A) was John B) was John and Alena C) at least one was Alena, John D) maximum one wa • 2nd class combinations From how many elements you can create 4560 combinations of the second class? • School parliament There are 18 boys and 14 girls in the class. In how many ways can 3 representatives be elected to the school parliament, if these are to be: a) the boys themselves b) one boy and two girls • Honored students Of the 25 students in the class, 10 are honored. How many ways can we choose 5 students from them, if there are to be exactly two honors between them? • Commitee A class consists of 6 males and 7 females. How many committees of 7 are possible if the committee must consist of 2 males", "that left a) only boys b) just two boys • Classroom There are eighty more girls in the class than boys. Boys are 40 percent and girls are 60 percent. How many are boys and how many girls?", "more daffodils. How many spring flowers grown together? 12. The school The school has 268 boys. Girls are 60 more than boys. How many children are there at school?", "half the guppies in the bowl, plus had a guppy. 2. Heather said - I'll take half of what you have, plus half a guppy. The third customer, Nancy • Today in school There are 9 girls and 11 boys in the class today. What is the probability that Suzan will go to the board today?", "a note of the number of story books and story cards read by boys and girls in the school. This is what she found: Students Story books Story cards Girls 145 254 Boys 66 120 2Vijaya's age is 22 years.Laxmi's age is 25 years more than Vijaya's age.Whats is Laxmi's age?", "of female students in class MCA-III ? (A) 40 (B) 45 (C) 50 (D) 55 5. What is the percentage of vegetarian students in class MCA-I? (A) 40 (B) 45 (C) 50 (D) 55 6. How many total non-vegetarian students are there in class MCA-I and class MCA-II? (A) 72 (B) 88 (C) 78 (D) 92", "a class are 32 pupils. Of these are 8 boys. What percentage of girls are in the class? • Classroom Of the 26 pupils in the classroom, 12 boys and 14 girls, four representatives are picked to the odds of being: a) all the girls b) three girls and one boy c) there will be at least two boys • Girls The boys and girls in the class formed groups without the rest of the fives, two girls and three boys. There are six girls missing to create mixed pairs (1 boy and one girl). How many girls are in the classroom? • Classroom In a classroom are 30 students, boys are four times more than girls. How many boys are in the class? • Pupils There are 350 girls in the school, and the other 30% of the total number of pupils are boys. How many pupils does the school have? • Ski class Class attend 30 boys and some number of girls. Ski training was attended by 28 boys and all the girls, which was 95% of all students. How many girls attend this class? How many percent is that? • Cross-country competition Cross-country competition was attended by 76% of pupils of class 9.B. Class consis of 16 boys, two of whom did not run and", "is by 1/5 less than in the third. How many stamps are in the first album? • Ski class Class attend 30 boys and some number of girls. Ski training was attended by 28 boys and all the girls, which was 95% of all students. How many girls attend this class? How many percent is that? • Columns of two and three When students in one class stand in columns of two there is none left. When he stands in columns of three, there is one student left. There are 5 more double columns than three columns. How many students are in the class? • Three siblings Three siblings have birthday in one day-today. Together they have 35 years today. The youngest is three years younger than middle and the oldest is 5 years older than middle. How old is each? • Pair They have together 39 years - he has 2 times as she was when he was equal as she has today ....", "The student sample consisted of 33 males and 15 females, all attended schools in a midwest, rural school system.", "will get the required number of students in the class.", "Class pairs In a class of 34 students, including 14 boys and 20 girls. How many couples (heterosexual, boy-girl) we can create? By what formula?", "these are 8 boys. What percentage of girls are in the class?", "There are 14 girls and 11 boys in the class. How many ways can a four-member team be chosen so that there are exactly two boys in it? • Five girls Five girls traveling by bus. Each holds in each hand two baskets. In each basket are four cats. Each cat has three kittens. Two girls exits bus. How many legs are in the bus? • Classroom In a class are 32 pupils. Boys are 8 less than girls. How many boys and girls are in the classroom? • A library A library has 12,500 fiction books and 19,000 non fiction books. Currently 2/5 of the fiction books are checked out. Currently 2/5 of the non fiction books are checked out. Of the books checked out, only 1/10 are due back this week. How many books are due • Marbles Dave had 40 marbles. Junjun has 2 1/5 more than Dave’s marbles. How many marbles do they have altogether? • Equivalent fractions 2 Write the equivalent multiplication expression. 2 1/6÷3/4", "and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is", "what percent of the class is composed of girls" ]

[ "In Miss Quaffley's class, the girls make up more than $45$% of the pupils, but less than $50$%. What is the smallest possible number of girls in her class?", "Difference between revisions of \"2009 AMC 8 Problems/Problem 23\" Problem On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class? $\\textbf{(A)}\\ 26\\qquad\\textbf{(B)}\\ 28\\qquad\\textbf{(C)}\\ 30\\qquad\\textbf{(D)}\\ 32\\qquad\\textbf{(E)}\\ 34$ Solution If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans. \\begin{align*} x^2+(x+2)^2 &= 394\\\\ x^2+x^2+4x+4 &= 394\\\\ 2x^2 + 4x &= 390\\\\ x^2 + 2x &= 195\\\\ \\end{align*} From here, we can see that $x = 13$ as $13^2 + 26 = 195$, so there are $13$ girls, $13+2=15$ boys, and $13+15=\\boxed{\\textbf{(B)}\\ 28}$ students.", "first: $21 = 3 + k$ $18 = k$ $k = 18$ Now let's find $a$: $a = 4 \\left(21\\right)$ $a = 84$ In order to check our answers, let's plug the number of dolls for $c$, $k$, and $a$ all back into the last equation: $21 + 18 + 84 = 123$ $123 = 123$ So... Krista has 18 dolls, Carlos has 21 dolls, and Amanda has 84 dolls.", "Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \\boxed{\\textbf{(C) } 3}$ students that are taking all $3$ classes. ## Solution 3 (Algebra) Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\\boxed{\\textbf{(C) 3}}$.", "be $4!*4!$. However, now we have 5 women. We need to multiply $^5C_4=5$ to choose 4 women. Then the last women can be paired with any other women, which has 4 ways. So we have $4!*4!*4*5=11520$ ways.", "Therefore, when there is one extra student, the number is divisible by $3$, $4$, $5$ and $6$. This means it is a common multiple of these numbers. The lowest common multiple of $3$, $4$, $5$ and $6$ is $60$, so there must be $60$ students with the extra. Therefore the number of students (without the imaginary student) is $59$.", "- I'll take half the guppies in the bowl, plus had a guppy. 2. Heather said - I'll take half of what you have, plus half a guppy. The third customer, Nancy • Elimination method Solve system of linear equations by elimination method: 5/2x + 3/5y= 4/15 1/2x + 2/5y= 2/15 • Today in school There are 9 girls and 11 boys in the class today. What is the probability that Suzan will go to the board today?", "groups + 2nd, 3rd, & 4th groups = $2^4 + 2^3 + 2^3 + 2^3$ = 40 The total number of sets with all of the groups. = $2^2$ = 4 $(512-(384-160+40-4)) \\implies \\boxed{252}$.", "assumption that they all lie implies that each boy is sitting between two girls. Hence they divide the $$28 - k$$ girls into $$k$$ segments, with each segment containing at least one girl. Let $$a_1, \\dotsc, a_k$$ be the length of these segments. We then have $$\\sum_{i = 1}^k a_k = 28 - k.$$ For each segment $$a_i$$, if $$a_i = 1$$, then the unique girl in that segment lies; otherwise, exactly $$a_i - 2$$ girls in that segment lies. Let $$u$$ be the number of segments with $$a_i = 1$$. It is clear that $$u \\leq 3$$. Moreover, the number of lying girls is equal to $$2u + \\sum_{i = 1}^k(a_i - 2) = 2u - 2k + \\sum_{i = 1}^ka_i = 2u - 2k + (28 - k) = 28 + 2u - 3k.$$ Therefore we obtain $$28 + 2u - 3k = 3$$, or $$3k - 2u = 25$$. Together with $$u \\leq 3$$, an argument mod $$3$$ shows that $$u = 1$$ and $$k = 9$$ is the only possibility. Hence there are $$9$$ boys and $$19$$ girls. Let's say there are all girls or all boys: They are all lying since their neighbors have the same gender So all boys are lying, what does that mean? If we have a B, their neighbors", "Mark Bennet Aug 17 '14 at 18:20 Correct me if I am wrong. Does it mean? $$5! \\times {6 \\choose 6-5} = 86400$$ – TommyLeeJones Aug 17 '14 at 18:38 @TommyLeeJones Close.. You are undercounting by a factor of $5!$ which arises from the fact that once you choose $5$ of the $6$ spots where the boys will go (your factor of ${6 \\choose 6-5}$) you still need to account for the fact that different permutations of the boys result in a different arrangement. – Tom Aug 18 '14 at 1:54 Imagine that the positions of the boys are like this: $$\\text{X}\\qquad \\text{X}\\qquad \\text{X}\\qquad \\text{X}\\qquad \\text{X}$$ Where can the $5$ girls go? There are $3$ kinds of possible pattern: $1$.) A girl at the left end, and $1$ girl between each boy; $2$.) A girl at the right end, and $1$ girl between each boy; $3$.) No girl at either end. Then one of the $4$ gaps between boys must have $2$ girls, and the rest $1$ girl each. That gives a total of $6$ patterns. Multiply by $5!5!$, to take account of the fact that the girls/boys are all distinct. Added: Note that the solution by Tom is better. For $5$ and $5$, it is somewhat more efficient. But it is far better for the problem of", "correct, and defend their choice to their partner. Select students to share their reasoning with the class who have different ways of knowing that “at least \\$3,000” means “more than \\$3,000.”", "# Boys dancing with girls In a party there are $r$ boys and $m$ girls. The first boy dances with $5$ girls, the second boy dances with $6$ girls, and the last boy dances with all girls. What is the relation between $r$ and $m$? I managed to find that $m=r+4,$ but it was not using combinatorics. Let me explain how I found that answer: first boy dances with $a_{1}$ girls, second boy dances with $a_{2}$girls, last boy dances with $a_{r}$ girls. Then $5=a_{1}=1+4, 6=a_{2}=2+4,\\cdots, m=a_{r}=4+r.$ My question: How does one solve this question by using combinatorics? - What you've done is fine, if a bit informal. You can make it a bit nicer as follows. Clearly the intended pattern is that each succeeding boy dances with one more girl. If $a_k$ is the number of girls whom the $k$-th boy dances with, then you have $a_1=5,a_2=6,\\dots,a_r=m$. In other words, $\\langle 5,6,\\dots m\\rangle$ is supposed to be a sequence of $r$ consecutive integers. This sequence has $m-4$ terms, so we must have $r=m-4$. Bring in $4$ more boys from the other class, the first to dance with $1$ girl, the second with $2$ girls, the third with $3$, the fourth with $4$, and let the rest of the action be as described. Then the number of boys would", "of $x^5y^5$ in $((1-x)^{-1}(1-y)^{-1}-1)^5$. – Gerry Myerson Jun 24 '16 at 12:50 So one car to each girl and one doll to each boy is clear, and we have $5$ cars and $5$ dolls left. Choose $k$ girls in $\\binom5k$ ways and give at least one extra car to each of them and none to the others, in $\\binom{5-1}{k-1}$ ways. Now give an extra doll to each boy whose sister didn't get an extra car, and distribute the other $k$ dolls arbitrarily among the boys, in $\\binom{k+5-1}{5-1}$ ways, for a total of $$\\sum_{k=1}^5\\binom5k\\binom4{k-1}\\binom{k+4}4=4251\\;.$$", "= 7.5$. 3) Solve for x: $\\frac{x + 1}{4} = \\frac{3}{8}$ Cross multiply: $8 \\cdot (x+1) = 3 \\cdot 4$ $8x + 8 = 12$ Now, solve for x: Subtract 8 from both sides: $8x = 4$ Divide both sides by 8: $x = \\frac{1}{2}$ 4) A 9th-grade algebra classroom has a ratio of boys to girls of \\frac{1} {2}. If there are 14 girls, how many boys are in the class? We can create a proportion that compares the ratios of boys to girls, where x is the number of boys in the class: $\\frac{1}{2} = \\frac{x}{14}$ Solve the proportion by cross-multiplying: $2x = 14$ $x = 7$ There are 7 boys in the class. Practice Games A list of games/resources about RATIOS and PROPORTIONS: Ratios and Proportions from eThemes Practice Problems Reduce each ratio. 1) $\\frac{2}{10}$ 2) $\\frac{12}{48}$ 3) $\\frac{3x}{9x}$ Solve each proportion. 4) $\\frac{x}{4} = \\frac{8}{12}$ 5) $\\frac{10}{17} = \\frac{14}{x}$ 6) $\\frac{2}{x-2} = \\frac{3}{7}$ Set up a proportion to answer the following questions. 7) If the ratio of boys to girls in a particular classroom was 2 to 3, and there are 12 boys, how many girls are in the class? 8) A recipe for pizza dough calls for 10 cups of flour and 2 cups of water. If Susan wants to only use 3", "# Solve question no. 3 of page no. 78 from the chapter \"Average\" of textbook \"Shaping Maths AB 5B Part I - Solutions\". 1.50 + 1.80 + 2.50 + 2.20 = 8 Press $\\overline{)\\mathrm{C}}$ 1.50 $\\overline{)+}$ 1.80 $\\overline{)+}$ 2.50 $\\overline{)+}$ 2.20 $\\overline{)=}$ Display: 8 The total pocket money of the four girls is \\$8. \\$8 $÷$ 4 = \\$2 The average amount of pocket money the girls get is \\$2. • 0 What are you looking for?", "girls and $$6!$$ ways to place the boys. So you have $$7 \\cdot 6! \\cdot 6! = 7! \\cdot 6!$$ possible ways to violate the rule. Thus there are $$12! - 7! \\cdot 6!$$ ways that respect the rule.", "are 346 girls and 254 boys. About how many students are there in the school? A. 500 B. 600 C. 700 Step: 1 Number of girls in the school = 346. Step: 2 346 rounded to the nearest ten is 350. Step: 3 Number of boys in the school = 254. Step: 4 254 rounded to the nearest ten is 250. Step: 5 Total number of students = 350 + 250 [Substitute the rounded values.] Step: 6 = 600 [Add.] Step: 7 Therefore, about 600 students are there in the school. Correct Answer is : 600 Q13Jim's father gifted him$205 on his birthday. He spent $181 for shopping. Estimate the amount left with him. [Round to the nearest ten.] A.$50 B. $30 C.$20 Step: 1 Money with Jim = $205. Step: 2 205 rounded to the nearest ten is 210. Step: 3 Money he spent =$181. Step: 4 181 rounded to the nearest ten is 180. Step: 5 Money left with Jim = 210 - 180 = 30. [Substitute the rounded values.] Step: 6 So, the amount left with Jim is about $30. Correct Answer is :$30 Q14Lucas has 192 chocolates. He distributed 132 of them. Estimate the number of chocolates he has now. A. 300 B. 200 C. 100 Step: 1 Number of chocolates that Lucas has", "## College Algebra (6th Edition) The girl has $1000$, the mother has $2000$, the boy has $4000$. Let $x$ denote the amount of money the girl has. Then $2x$ denotes the amount of money the mother has and $4x$ denotes the amount of money the boy has. We know that in total they have $14000$, thus $x+2x+4x=14000\\\\7x=14000\\\\7x/7=14000/7\\\\x=2000.$ Thus the girl has $1000$, the mother has $2000$, the boy has $4000$.", "is assumed by the other partner.$5,000 - $1,500 =$3,500.", "Maths Revision Cards (252 Reviews) £8.99 ### Example Questions a) In order to work out the fraction of students that have blonde hair, we need to add up the ratio parts. The sum of the ratio is $4 + 5 = 9$. This means that we are dealing with $9$ths. Since the ratio share for blond students is $4$, this means that the fraction of blonde students is $\\dfrac{4}{9}$ b) We know from the previous question that $\\dfrac{4}{9}$ of the students have blonde hair. Therefore, the fraction of students with brown hair is $\\dfrac{5}{9}$. If there is a total of $450$ students in the school, we need to work out what $\\dfrac{5}{9}$ of $450$ is: $\\dfrac{5}{9} \\times 450 = 250$ students. In total, there are $7$ parts to this ratio $(2 + 5 = 7)$. If $7$ shares have a value of $35$, then $1$ share has a value of $5$ $(35 \\div 7 = 5)$ Since $1$ share has a value of $5$, then $2$ shares will have a value of $10$ $(2 \\times 5 = 10)$. Since $1$ share has a value of $5$, then $5$ shares will have a value of $25$ $(5 \\times 5 = 25)$. To work out the total cost of the tiles Lucy buys, we need to work out how many" ]

[ "In Miss Quaffley's class, the girls make up more than $45$% of the pupils, but less than $50$%. What is the smallest possible number of girls in her class?", "bracket in which their entire income falls. Mr. Jackson earned $34,287 last year; Mrs. Jackson earned$25,879. How much will the couple pay in income tax for that year (nearest hundred dollars)? Explanation: The Jackson's income totaled . This puts them in the 1.7% tax bracket, so they will pay in taxes. The correct response is \\$1,000. ### Example Question #3 : How To Find The Answer From A Table Jefferson Elementary School has seven eighth-grade teachers; each teacher has the number of boys and girls listed above. How many teachers have classes in which at least 40% of the students are boys? Six Four Seven Five Six Explanation: Five teachers each have 30 students, and 40% of 30 is . Two teachers each have 29 students, and 40% of 29 is . Therefore, to have a class that is at least 40% boys, a teacher must have 12 boys at minimum. Only Mr. Georghiou does not have at least 12 boys, so the correct response is six. ### Example Question #6 : How To Find The Answer From A Table Adams Elementary School has seven fifth-grade teachers; each teacher has been assigned the number of boys and girls listed above. However, just before the start of the school year, Mrs. Henry has been hired as an additional fifth-grade", "matches boys will play $6 \\times 5 = 30$ For Girls – No of girls in team $1 = 4$ No of girls in team $2 = 3$ As we did for boys same goes for girls i.e. every girl of team $1$ will compete with every girl of team $2$ Therefore, total no of matches girls will play$= 4 \\times 3 = 12$ Total no of matches $=$ matches of boys $+$ matches of girls $= 30 + 12 = 42$ Note:In this question you just need to take care of the point that every boy of team $1$ will compete with every boy of team $2$ and every girl of team $1$ will compete with every girl of team $2$ . The chances of mistakes anyone could make is while counting the number of matches boys or girls will play . So you need to do that carefully.", "a class are girls, then what percent of the students in the class are boys? (ii) We have a basket full of apples, oranges, and mangoes if 50% are apples, 30% are oranges, then what per cent are mangoes? Solution:", "together is $$20*6*120=14,400$$. _________________ Kudos [?]: 135519 [4], given: 12697 Manager Joined: 16 Feb 2011 Posts: 195 Kudos [?]: 247 [0], given: 78 Schools: ABCD Re: In how many ways can 5 boys and 3 girls [#permalink] Show Tags 24 Sep 2012, 12:37 Bunuel wrote: voodoochild wrote: In how many ways can 5 boys and 3 girls be seated on 8 chairs so that no two girls are together? A 5760 B 14400 C 480 D 56 E 40320 HEre's what I did : 5! * (4C3) *2* 3! = 120*6*4*2= 5760. Am I correct? Consider the following arrangement: *B*B*B*B*B* Now, if girls occupy the places of 6 stars no girls will be together. # of ways 3 girls can occupy the places of these 6 stars is $$C^3_6=20$$; # of ways 3 girls can be arranged on these places is $$3!=6$$; # of ways 5 boys can be arranged is $$5!=120$$. So total # of ways to arrange 3 Girls and 5 boys so that no girls are together is $$20*6*120=14,000$$. Bunuel, Here's what I thought: _ O _ O _ O _ O 3 G can occupy any of the 4 \"_\" positions in 4C3 ways. Similar Girls could also occupy any of the 4 \"O\" positions in 4C3 ways. Boys can be permuted in 5!", "are 346 girls and 254 boys. About how many students are there in the school? A. 500 B. 600 C. 700 Step: 1 Number of girls in the school = 346. Step: 2 346 rounded to the nearest ten is 350. Step: 3 Number of boys in the school = 254. Step: 4 254 rounded to the nearest ten is 250. Step: 5 Total number of students = 350 + 250 [Substitute the rounded values.] Step: 6 = 600 [Add.] Step: 7 Therefore, about 600 students are there in the school. Correct Answer is : 600 Q13Jim's father gifted him$205 on his birthday. He spent $181 for shopping. Estimate the amount left with him. [Round to the nearest ten.] A.$50 B. $30 C.$20 Step: 1 Money with Jim = $205. Step: 2 205 rounded to the nearest ten is 210. Step: 3 Money he spent =$181. Step: 4 181 rounded to the nearest ten is 180. Step: 5 Money left with Jim = 210 - 180 = 30. [Substitute the rounded values.] Step: 6 So, the amount left with Jim is about $30. Correct Answer is :$30 Q14Lucas has 192 chocolates. He distributed 132 of them. Estimate the number of chocolates he has now. A. 300 B. 200 C. 100 Step: 1 Number of chocolates that Lucas has", "is the ratio of the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school? Solution a) ratio of girls to boys 260:240 or 13:12 b) ratio of boys to the total number of pupils 240:(240+260) or 240:500 or 12:25 5. If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend? Solution The tip is 20% of what he paid for lunch. Hence tip = 20% of 50.50 = (20/100)*50.50 = 101/100 =$10.10 Total spent 50.50 + 10.10 = $60.60 6. Find k if 64 ÷ k = 4. Solution Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then k = 16 7. Little John had$8.50. He spent $1.25 on sweets and gave to his two friends$1.20 each. How much money was left? Solution John spent and gave to his two friends a total of 1.25 + 1.20 + 1.20 = $3.65 Money left 8.50 - 3.65 =$4.85 8. What is x if x + 2y = 10 and y = 3? Solution Substitute y by 3 in x + 2y = 10 x + 2(3) = 10 x + 6 = 10 If we substitute x by 4 in x", "Questions & Answers Question Answers # A team consists of $6$ boys and $4$ girls and the other has $5$ boys and $3$ girls. How many single matches can be arranged between the two teams where a boy plays against a boy and girl plays against a girl? Answer Verified Hint:Take $1$ boy from team $1$and then start counting the no of matches he will play against the boys of team $2$. now take every boy from team $1$ and count the no of matches they will play with the boys of team $2$. Similarly, you can get the total no of matches that girls of team $1$ will play with girls of team $2$. Complete step-by-step answer: Let us find the total number of matches that can be arranged between two teams where a boy plays against a boy and girl plays against a girl. For Boys - No of boys in team $1$$= 6$ No of boys in team $2$ $= 5$ We have to arrange single matches. This means that every boy from team $1$ will go against every boy of team $2$ . Hence every boy from team $1$ will play $5$ matches i.e. $1$ match with every boy from team $2$ Since there are $6$ boys in team $1$ Therefore, total no of", "the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school? Solution a) ratio of girls to boys 260:240 or 13:12 b) ratio of boys to the total number of pupils 240:(240+260) or 240:500 or 12:25 If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend? Solution The tip is 20% of what he paid for lunch. Hence tip = 20% of 50.50 = (20/100)*50.50 = 101/100 =$10.10 Total spent 50.50 + 10.10 = $60.60 Find k if 64 ÷ k = 4. Solution Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then k = 16 Little John had$8.50. He spent $1.25 on sweets and gave to his two friends$1.20 each. How much money was left? Solution John spent and gave to his two friends a total of 1.25 + 1.20 + 1.20 = $3.65 Money left 8.50 - 3.65 =$4.85 What is x if x + 2y = 10 and y = 3? Solution Substitute y by 3 in x + 2y = 10 x + 2(3) = 10 x + 6 = 10 If we substitute x by 4 in x + 6 = 10, we have 4 +", "the number of girls to the number of boys? b) What is the ratio of the number of boys to the total number of pupils in the school? Solution a) ratio of girls to boys 260:240 or 13:12 b) ratio of boys to the total number of pupils 240:(240+260) or 240:500 or 12:25 If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend? Solution The tip is 20% of what he paid for lunch. Hence tip = 20% of 50.50 = (20/100)*50.50 = 101/100 =$10.10 Total spent 50.50 + 10.10 = $60.60 Find k if 64 ÷ k = 4. Solution Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then k = 16 Little John had$8.50. He spent $1.25 on sweets and gave to his two friends$1.20 each. How much money was left? Solution John spent and gave to his two friends a total of 1.25 + 1.20 + 1.20 = $3.65 Money left 8.50 - 3.65 =$4.85 What is x if x + 2y = 10 and y = 3? Solution Substitute y by 3 in x + 2y = 10 x + 2(3) = 10 x + 6 = 10 If we substitute x by 4 in x + 6 = 10, we have 4 +", "year there were 612 boys in a school. The number decreased by 25 percent this year. How many girls are there in the school if the number of girls is $$166\\frac{2}{3}\\;$$percent of the total number of boys in the school this year? 1. 765 2. 762 3. 798 4. 782 5. None of these Que. 9 In a company, there are 75% skilled workers and remaining ones are unskilled. 80% of skilled workers and 20% of unskilled workers are permanent. If number of temporary workers is 126, then find the number of total workers. 1. 124 2. 388 3. 324 4. 360 5. None of these Que. 10 The average age of 80 boys in a class is 15. The average age of a group of 15 boys in the class is 16 and the average age of another 25 boys in the class is 14. What is the average age of the remaining boys in the class? 1. 11.5 years 2. 14 years 3. 15.25 years 4. 18.75 years 5. None of these Did you like this Average and Percentage Quiz 3 for Banking & Insurance Exams? Let us know! You may also like – • Save 11 hours ago 1 day ago 2 days ago 3 days ago", "choices. 3 appetizers, 4 main courses, and 3 desserts. How many different combinations to choose from? • Boys and girls There are 20 boys and 10 girls in the class. How many different dance pairs can we make of them? • Desks A class has 20 students. The classroom consists of 20 desks, with 4 desks in each of 5 different rows. Amy, Bob, Chloe, and David are all friends, and would like to sit in the same row. How many possible seating arrangements are there such that Amy, Bob, • Playing cards How many possible ways are to shuffle 7 playing cards?", "1,335 3. 1,365 4. 1,325 783 x 9 = 1. 7,047 2. 7,057 3. 6,357 4. 6,447 Micah's school has 800 students. Laura's school has 10 times as many students. How many students go to Laura's school? 1. 80 2. 8,000 3. 80,000 4. 8 45 x 10 = 1. 45 2. 405 3. 450 4. 4,500 49 x 34 = 1. 1,686 2. 2,666 3. 1,666 4. 1,696 321 x 22 = 1. 7,162 2. 7,052 3. 7,062 4. 6,962 $52 xx 93 =$", "number of girls. How many girls are in the class? Solution: Let ‘x’ be the number of girls in the class. Therefore, $4x + 3 = 43$ $4x = 43 – 3$ $x = \\frac{43 – 3}{4}$ $x = 10$ Hence, there are 10 girls present in the class. Example 2: Nazneen’s father is 5 times older than Nazneen and Nazneen is twice as old as her sister Jasmine. In two years time, the sum of their ages will be 58. How old is Nazneen now? Let the present age of Nazneen is ‘x’ Then the Present Age of – Nazneen = x Jasmine = $\\frac{1}{2}x$ Nazneen’s Father = 5x In 2 years, their respective age will be Nazneen = x + 2 Jasmine = $\\frac{1}{2}x + 2$ Nazneen’s Father = 5x + 2 As, the sum of their age after 2 years is 58 Therefore, $5x + 2 + x + 2 +\\frac{1}{2}x + 2 = 58$ $5x + x +\\frac{1}{2}x = 58 – (2 + 2 + 2)$ $6\\frac{1}{2}x = 52$ $\\Rightarrow x = 8$ Therefore the present age of Nazneen is 8 Years.", "Marbles] Red Marbles = 125 – (5 + 75) = 125 – 80 = 45 Question 6. There are 32 students in Mr. Moreno’s class and 62.5% of the students are girls. How many boys are in the class? _______ students Answer: 12 students Explanation: There are 32 students in Mr. Moreno’s class 62.5% of the students are girls = 32 × 62.5/100 = 20 boys = 32 – 20 = 12 ### On Your Own – Page No. 291 Find the percent of the quantity. Question 7. 60% of 90 _______ Answer: 54 Explanation: Write the percent as a rate per 100 60% = 60/100 60/100 × 90 = 54 Question 8. 25% of 32.4 _______ Answer: 8.1 Explanation: Write the percent as a rate per 100 25% = 25/100 25/100 × 32.4 = 8.1 Question 9. 110% of 300 _______ Answer: 330 Explanation: Write the percent as a rate per 100 110% = 110/100 110/100 × 300 = 330 Question 10. 0.2% of 6500 _______ Answer: 13 Explanation: Write the percent as a rate per 100 0.2% = 0.2/100 0.2/100 × 6500 = 13 Question 11. A baker made 60 muffins for a cafe. By noon, 45% of the muffins were sold. How many muffins were sold by noon? _______ muffins Answer: 27 muffins Explanation: A", "is : a. $\\sqrt \\pi :1$ b. $1:\\sqrt \\pi$ c. $1:\\pi$ d. $\\pi :1$ 23. In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls is : a. 11:4 b. 14:11 c. 25:28 d. 28:25 24. A, B and C do a work in 20, 25 and 30 days respectively. They undertook to finish the work together for Rs.2220, then the share of A exceeds that of B by : a. Rs.120 b. Rs.180 c. Rs.300 d. Rs.600", "# There are 32 students in a class. Nine of those students are women. What percent are men? ##### 1 Answer Jul 16, 2016 There is a total of 32 students. 9 of the students are female. If we assume the only gender options are man and woman we can conclude that the number of men is $32 - 9 = 23$. %Men = 23/32 *100 = 71.875%", "Question # A class has three teachers, Mr. $$P$$, Ms. $$Q$$ and $$Mrs. R$$ and six students $$A, B, C, D, E, F$$. Number of ways in which they can be seated in a line of $$9$$ chairs, if between any two teachers there are exactly two students, is $$k!(18)$$, then the value of $$k$$ is. Solution ## Let $$T$$ and $$S$$ denotes teacher and student respectively.Then we have following possible patterns according to question(i) $$T\\ S\\ S\\ T\\ S\\ S\\ T\\ S\\ S$$(ii) $$S\\ T\\ S\\ S\\ T\\ S\\ S\\ T\\ S$$(iii) $$S\\ S\\ T\\ S\\ S\\ T\\ S\\ S\\ T$$Hence total number of arrangements are $$3 \\cdot (3!) 6! = 18\\times 6! \\Rightarrow k = 6$$.Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "then $(n+2)/4=5+$, so take $k=6$. Any class with 108 attendants or more can be opened, now you just have to look at the smaller classes. - With $n=20$, I think any class with $90$ attendants or more can be opened. – Henry Jun 29 '11 at 7:16 @Henry, yes, 89 is the largest class that can't be opened with $n=20$. You can open 18 to 22, 36 to 44, 54 to 66, 72 to 88, and everything from 90 up. – Gerry Myerson Jun 29 '11 at 13:06", "# Choosing representatives in class Probability Level 1 Suppose of all possible ways to choose two representatives among 10 students in a class, there are exactly $30$ different ways to choose one or more female as a representative. How many males are there in the class? ×" ]

[ "drawn. Then the area of the quadrilateral PQOR is (a) 60 cm² (b) 65 cm² (c) 30 cm² (d) 32.5 cm² = $$\\frac{1}{2}$$ × 12 × 5 + $$\\frac{1}{2}$$ × 12 × 5 = 60 cm²", "the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are congruent. In particular $\\angle PQR$ is a right angle, so that $PQ$ is tangent to the semicircle and $QR$ is tangent to the quarter-circle. Also, $PR$ bisects the angle $\\theta =\\angle QPS$, so that $\\angle RPS = \\frac\\theta2$ and $\\tan\\frac\\theta2 = \\frac{RS}{PS} = \\frac12.$ Therefore $$\\tan\\theta = \\frac{2\\tan\\frac\\theta2}{1 - \\tan^2\\frac\\theta2} = \\frac1{1-\\frac14} = \\frac43.$$ It follows that $\\sin\\theta = \\frac45$", "it. Given that the side of the square is 10 cm, find the combined areas of the regions labelled A and C. Any hints, perhaps, on how to solve this seemingly simple geometry problem? This is a primary math problem, so one should refrain from using geometry or trigonometric method to solve it...any help is much appreciated! The area of the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are", "quadrilateral PQRS are P(0, 0), Q [#permalink] ### Show Tags 09 Feb 2018, 01:02 Bunuel wrote: The coordinates of the vertices of quadrilateral PQRS are P(0, 0), Q(9, 0), R(10, 3) and S(1, 3), respectively. The area of PQRS is (A) $$9\\sqrt{10}$$ (B) $$\\frac{9}{2}*\\sqrt{10}$$ (C) 27/2 (D) 27 (E) $$27\\sqrt{10}$$ On plotting the points on the x-y coordinate, we see that it forms a parallelogram. Area of parallelogram = Base x height = 9 x 3 =27 _________________ Please give kudos, if you like my post When the going gets tough, the tough gets going... Manager Joined: 23 Sep 2016 Posts: 217 Re: The coordinates of the vertices of quadrilateral PQRS are P(0, 0), Q [#permalink] ### Show Tags 09 Feb 2018, 23:24 Bunuel wrote: The coordinates of the vertices of quadrilateral PQRS are P(0, 0), Q(9, 0), R(10, 3) and S(1, 3), respectively. The area of PQRS is (A) $$9\\sqrt{10}$$ (B) $$\\frac{9}{2}*\\sqrt{10}$$ (C) 27/2 (D) 27 (E) $$27\\sqrt{10}$$ Answer is D as base is 9 and height is 3 if we plot on graph so area of quadrilateral is B*H= 9*3= 27. Oa plz Re: The coordinates of the vertices of quadrilateral PQRS are P(0, 0), Q &nbs [#permalink] 09 Feb 2018, 23:24 Display posts from previous: Sort by", "ABCD into four right triangles and one square, as shown. Find the coordinates of the vertices for the five smaller polygons. The coordinates of ΔAQD are: (1, 4), Q (1, 0), and D (4, 0), The coordinates of ΔBPA are B (-3, 1), P (1, 1), and A (1, 4), The coordinates of ΔDRC are D (4, 0), R (0, 0), and C (0, -3), The coordinates of ΔCSB are C (0, -3), S (0, 1), and B (-3, 1), The coordinates of square PQRS are P (1, 1), Q (1, 0), R (0, 0), and S (0, 1), Explanation: From the figure The five smaller polygons are ΔAQD, ΔBPA, ΔDRC, ΔCSB, and square PQRS Now the coordinates for the given vertices polygons are : The coordinates of ΔAQD are A (1, 4), Q (1, 0), and D (4, 0), The coordinates of ΔBPA are B (-3, 1), P (1, 1), and A (1, 4), The coordinates of ΔDRC are D (4, 0), R (0, 0), and C (0, -3), The coordinates of ΔCSB are C (0, -3), S (0, 1), and B (-3, 1), The coordinates of square PQRS are P (1, 1), Q (1, 0), R (0, 0), and S (0, 1). b. Find the areas of the five smaller polygons. Area of ΔBPA is 6,", "### Set Square A triangle PQR, right angled at P, slides on a horizontal floor with Q and R in contact with perpendicular walls. What is the locus of P? ### Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. ### Strange Rectangle ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles. Consider any convex quadrilateral $Q$ made from four rigid rods with flexible joints at the vertices so that the shape of $Q$ can be changed while keeping the lengths of the sides constant. If the diagonals of the quadrilateral cross at an angle $\\theta$ in the range $(0 \\leq \\theta < \\pi/2)$, as we deform $Q$, the angle $\\theta$ and the lengths of the diagonals will change. Using the results of the two problems on quadrilaterals Diagonals for Area and Flexi Quads prove that the area of $Q$ is proportional to $\\tan\\theta$.", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "figure, PQRS is a square with side x + 4. Each of the [#permalink] ### Show Tags 20 Sep 2017, 05:13 Bunuel wrote: In the above figure, PQRS is a square with side x + 4. Each of the four smaller squares has side of length 2. If the area of the shaded region is 48, what is the value of x? (A) 1 (B) 4 (C) 4√2 (D) 8 (E) 12 [Reveal] Spoiler: Attachment: 2017-09-20_1020.png Find numeric value for area of large square. Use (x + 4) to derive equation for the same area. Set them equal to solve for $$x$$. Numeric area (Area of small squares) + (Area of shaded region) = Area of large square Small square side = 2. Each small square's area = 4. Their total area = 4*4 = 16 Area of shaded region = 48 Area of large square = 16 + 48 = 64 Equation for same area One side of large square = $$(x + 4)$$ Equation for area of large square is $$(x + 4)^{2}$$ = $$x^2 + 8x + 16$$ From above, numeric area of large square = 64. Set equation equal to numeric value: $$x^2 + 8x + 16 = 64$$ $$x^2 + 8x - 48 = 0$$ $$(x + 12)(x - 4) = 8$$ $$x$$", "+0 # HELP ASAP PLZ -2 70 1 In cyclic quadrilatera $$PQRS,$$ $$\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$$ Find the largest angle of quadrilateral PQRS, in degrees. Apr 9, 2020 #1 +4569 0 The sum of the opposite angles in a cyclic quadrilateral sum to 180 degrees. Thus, p=2/3q, q=q, r=4/3q 2/3q+4/3q=180 6/3q=180 q=90 p=60 r=120 s=90 And, the largest angle is angle r, which measures 120 degrees. Apr 9, 2020", "angle PQR is (A) (√3/2)x + y = 0 (B) √3y + x = 0 (C) √3x + y = 0 (D) (√3/2)y + x = 0 Solution: Since the angle PQR = 120ᵒ ⇒ The slope of the required angle bisector is = Tan(120ᵒ), and the bisector passes through the origin ⇒ The equation of the bisector of the angle PQR is √3x + y = 0 Option (C) is correct Q.8₂₀₀₂ The straight line through the origin O meet the parallel straight lines 4x + 2y = 9 and 2x + y + 6 = 0 at the points P and Q respectively, then the point O divides the segment PQ in the ratio (A) 1 : 2 (B) 3 : 2 (C) 2 : 1 (D) 4 : 3 Solution: OP/OQ = 9/(4×3) = 3/4 Option (D) is correct Q.9₂₀₀₁ The area of the parallelogram formed by the straight lines y = mx, y = mx + 1, y = nx, and y = nx + 1 is (A) |m + n|/(m – n) (B) 2/|m + n| (C) 1/|m + n| (D) 1/|m – n| Solution: The vertices of the parallelogram are O(0, 0), A(0, 1), B(1/(m – n), a), and C(1/(m – n), b), Here a, and b are any arbitrary real numbers", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "is geometry. Answers: A, C, C, D, C, B, C Dec 25: Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded L-shaped region is $[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label($ $\\text{(A)}\\ 7 \\qquad \\text{(B)}\\ 10 \\qquad \\text{(C)}\\ 12.5 \\qquad \\text{(D)}\\ 14 \\qquad \\text{(E)}\\ 15$ Dec 26: What is the degree measure of the smaller angle formed by the hands of a clock at 10 o’clock? $[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label($ $\\text{(A)}\\ 30 \\qquad \\text{(B)}\\ 45 \\qquad \\text{(C)}\\ 60 \\qquad \\text{(D)}\\ 75 \\qquad \\text{(E)}\\ 90$ Dec 27: In triangle $CAT$, we have $\\angle ACT = \\angle ATC$ and $\\angle CAT = 36^\\circ$. If $\\overline{TR}$ bisects $\\angle ATC$, then $\\angle CRT =$ $[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label($ $\\text{(A)}\\ 36^\\circ \\qquad \\text{(B)}\\ 54^\\circ \\qquad \\text{(C)}\\ 72^\\circ \\qquad \\text{(D)}\\ 90^\\circ \\qquad \\text{(E)}\\ 108^\\circ$ Dec 28: In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is $[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label($ $\\text{(A)}\\ 27 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 34 \\qquad \\text{(E)}\\", "4.2.1 Polygons, PT3 Practice 4.2.1 Polygons, PT3 Practice Question 1: Diagram below shows a pentagon PQRST. TPU and RSV are straight lines. Find the value of x. Solution: Question 2: In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\angle QPU={180}^{o}-{160}^{o}={20}^{o}\\\\ \\text{Reflex}\\angle PUT={360}^{o}-{80}^{o}={280}^{o}\\\\ \\angle UTS={180}^{o}-{120}^{o}={60}^{o}\\\\ \\angle TSR={180}^{o}-{35}^{o}={145}^{o}\\\\ \\\\ \\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\\\ {x}^{o}+{y}^{o}+{145}^{o}+{60}^{o}+{280}^{o}+{20}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{505}^{o}\\\\ \\text{}={215}^{o}\\\\ \\text{}x+y=215\\end{array}$ Question 3: Diagram below shows a regular hexagon PQRSTU. PUV is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular hexagon}\\\\ =\\frac{\\left(6-2\\right)×{180}^{o}}{6}\\\\ ={120}^{o}\\\\ {x}^{o}=\\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\\\ {y}^{o}={180}^{o}-{120}^{o}\\\\ \\text{}={60}^{o}\\\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\\\ \\text{}={90}^{o}\\\\ \\text{}x+y=90\\end{array}$ 3 thoughts on “4.2.1 Polygons, PT3 Practice” 1. answer in Question 1 is error. The answer is 115. • Thanks for pointing out our mistake. We had done the correction accordingly. 2. I like these questions .", "pI=extension(D,M,R,Q); label(\"O\",pI+(-0.2,0.166),f); draw(anglemark2(Sp,P,R,17)); label(\"p\",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label(\"q\",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label(\"n\",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label(\"m\",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label(\"d\",D+(-0.75,0.095)); label(\"R\",R,N,f); label(\"M\",M+(-.07,.07),f); label(\"N\",Np+(-.08,.15),f); label(\"P\",P,S,f); label(\"S\",Sp,S,f); label(\"Q\",Q,S,f); label(\"D\",D,S,f);[/asy]$ Looking at triangle PRQ, we have $\\angle RPD+\\angle RQS+\\angle MRN=180$ and from the given statement $\\angle NMR=\\frac{1}{2}\\angle MRN$, so looking at triangle MOR $\\angle NMR=90-\\frac{\\angle RPD+\\angle RQS}{2}$, which rules out 1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 37 Followed byProblem 39 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 15 ## Problem In the diagram, the area of rectangle $PQRS$ is $24$. If $TQ = TR$, what is the area of quadrilateral $PTRS$? $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 16 \\qquad \\text{(D)}\\ 6 \\qquad \\text{(E)}\\ 15$ $[asy] size(100); draw((0,0)--(6,0)--(6,4)--(0,4)--cycle); draw((0,4)--(6,2)); draw((5.8,1.1)--(6.2,1.1)); draw((5.8,.9)--(6.2,.9)); draw((5.8,3.1)--(6.2,3.1)); draw((5.8,2.9)--(6.2,2.9)); label(\"P\",(0,4),NW); label(\"S\",(0,0),SW); label(\"R\",(6,0),SE); label(\"T\",(6,2),E); label(\"Q\",(6,4),NE); [/asy]$ ## Solution 1 Since the area of rectangle $PQRS$ is $24$, let us assume that $PQ = 6$ and $QR = 4$. Since $QT = TR$, then $QR = 2QT$ so $QT = 2$. Therefore, triangle $PQT$ has base $PQ$ of length $6$ and height $QT$ of length 2, so its area is $\\frac{1}{2}(6)(2) = \\frac{1}{2}(12) = 6$. So the area of quadrilateral $PTRS$ is equal to the area of rectangle $PQRS$ (which is $24$) minus the area of triangle $PQT$ (which is $6$), or $18$. Therefore, the answer is $A$. ## Solution 2 Draw a line through $T$ parallel to $PQ$ across the rectangle parallel so that it cuts $PS$ at point $V$. Since $T$ is halfway between $Q$ and $R$, then $V$ is halfway between $P$ and $S$. Therefore, $SVTR$ is a rectangle which has an area equal to half the area of rectangle $PQRS$, or $12$. Similarly, $VPQT$ is a rectangle of area $12$, and $VPQT$", "# 2.2.1 Polygons II, PT3 Practice 2.2.1 Polygons II, PT3 Practice Question 1: Diagram below shows a pentagon PQRST. TPU and RSV are straight lines. Find the value of x. Solution: Question 2: In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\angle QPU={180}^{o}-{160}^{o}={20}^{o}\\\\ \\text{Reflex}\\angle PUT={360}^{o}-{80}^{o}={280}^{o}\\\\ \\angle UTS={180}^{o}-{120}^{o}={60}^{o}\\\\ \\angle TSR={180}^{o}-{35}^{o}={145}^{o}\\\\ \\\\ \\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\\\ {x}^{o}+{y}^{o}+{145}^{o}+{60}^{o}+{280}^{o}+{20}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{505}^{o}\\\\ \\text{}={215}^{o}\\\\ \\text{}x+y=215\\end{array}$ Question 3: Diagram below shows a regular hexagon PQRSTU. PUV is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular hexagon}\\\\ =\\frac{\\left(6-2\\right)×{180}^{o}}{6}\\\\ ={120}^{o}\\\\ {x}^{o}=\\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\\\ {y}^{o}={180}^{o}-{120}^{o}\\\\ \\text{}={60}^{o}\\\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\\\ \\text{}={90}^{o}\\\\ \\text{}x+y=90\\end{array}$ Question 4: In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines. Find the value of x + y. Solution: $\\begin{array}{l}\\text{Size of each interior angle of a regular pentagon}\\\\ =\\frac{\\left(5-2\\right)×{180}^{o}}{5}\\\\ ={108}^{o}\\\\ \\angle PKS=\\angle PNQ={180}^{o}-{108}^{o}={72}^{o}\\\\ \\text{Reflex angle}\\angle KPN={360}^{o}-{108}^{o}={252}^{o}\\end{array}$ $\\begin{array}{l}\\text{Sum of interior angles of a hexagon}\\\\ =\\left(6-2\\right)×{180}^{o}\\\\ ={720}^{o}\\\\ \\therefore {x}^{o}+{y}^{o}+{72}^{o}+{252}^{o}+{72}^{o}+{100}^{o}={720}^{o}\\\\ {x}^{o}+{y}^{o}={720}^{o}-{496}^{o}\\\\ \\text{}={224}^{o}\\\\ \\text{}x+y=224\\end{array}$ Question 5: In Diagram below, PQR is an isosceles triangle and PRU is a straight line. Find the value of x + y. Solution: $\\begin{array}{l}{x}^{o}={180}^{o}-{20}^{o}-{20}^{o}={140}^{o}\\\\ \\angle PRS={180}^{o}-{110}^{o}={70}^{o}\\\\ {y}^{o}+{85}^{o}+{75}^{o}+{70}^{o}={360}^{o}\\\\ {y}^{o}+{230}^{o}={360}^{o}\\\\ \\text{}{y}^{o}={130}^{o}\\\\ {x}^{o}+{y}^{o}={140}^{o}+{130}^{o}\\\\ \\text{}={270}^{o}\\\\ \\text{}x+y=270\\end{array}$", "= \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH$$ so $H$ lies on this line as well and we may conclude $\\square$ --mathguy623 03:10, 12 August 2016 (EDT) ## Solution 3 (Complex Bash) $[asy] unitsize(0.8cm); draw((0,0)--(14,0)--cycle); draw((0,0)--(5,12)--cycle); draw((5,12)--(14,0)--cycle); draw((2.071,4.97)--(14,0)--cycle); draw((8.96,6.72)--(0,0)--cycle); draw((2.071,4.97)--(9,0)--cycle); draw((8.96,6.72)--(9,0)--cycle); draw((0,0)--(0,2.083)--cycle); draw((14,0)--(14,6.75)--cycle); draw((0,2.083)--(14,6.75)--cycle); draw((0,2.083)--(2.071,4.97)--cycle); draw((0,2.083)--(9,0)--cycle); draw((8.96,6.72)--(14,6.75)--cycle); draw((14,6.75)--(9,0)--cycle); draw(circle((4.5,1.0415),4.61895)); draw(circle((11.5,3.375),4.20007)); label((0,0),\"C\",SW); label((14,0),\"B\",SE); label((5,12),\"A\",N); label((8.96,7),\"N\",N); label((2.071,4.97),\"M\",NW); label((8.25,0),\"W\",S); label((5,4),\"H\",N); label((0,2.083),\"Y\",NW); label((14,6.75),\"X\",NE); [/asy]$ For any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$, we must have $\\angle{YMW}=\\angle{YCH}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{MYW}=\\angle{MCW}=\\angle{C}.$ This means that $\\frac{|w-m|}{|y-m|}=\\tan{\\angle{C}}.$ Combining this with the fact that $\\angle{YMW}$ is right, we find that $$i(y-m)\\tan{\\angle{C}}=(w-m).$$ Solving, we find that $$y=\\frac{w+(i\\tan{\\angle{C}}-1)m}{i\\tan{\\angle{C}}}.$$ We wish to simplify $(i\\tan{\\angle{C}}-1)m$ first. Note that $$m=\\frac{|CM|}{|CA|}\\cdot(a)=\\frac{(|BC|)\\cos{\\angle{C}}}{|CA|}\\cdot(|CA|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})=(|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}).$$ This means that \\begin{align*} (i\\tan{\\angle{C}}-1)m&=(i\\tan{\\angle{C}}-1)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=\\left(\\frac{i\\sin{\\angle{C}}}{\\cos{\\angle{C}}}-1\\right)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=(i\\sin{\\angle{C}}-\\cos{\\angle{C}})(|BC|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})\\\\ &=-|BC| \\end{align*} This means that $$y=\\frac{w-|BC|}{i\\tan{\\angle{C}}}=\\frac{|BC|-w}{\\tan{\\angle{C}}}i.$$ Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\\angle{WNX}=\\angle{WBX}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{NXW}=\\angle{NBW}=\\angle{B}.$ This means that $\\frac{|x-n|}{|w-n|}=\\cot{\\angle{B}}.$ Combining this with the fact that $\\angle{WNX}$ is right, we find that $$(x-n)=i\\cot{\\angle{B}}(w-n).$$ Solving, we find that $$x=wi\\cot{\\angle{B}}+n(1-i\\cot{\\angle{B}}).$$ We wish to simplify $n(1-i\\cot{\\angle{B}})$ first. Note that $$n=e^{(90-\\angle{B})i}|CN|=(\\cos{(90-\\angle{B})}+i\\sin{(90-\\angle{B})})(|BC|)\\sin{\\angle{B}}=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}})).$$ This means that \\begin{align*} n(1-i\\cot{\\angle{B}})&=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))(1-i\\cot{\\angle{B}})\\\\ &=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))\\left(1-\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}i\\right)\\\\ &=(|BC|)(\\sin{\\angle{B}}+i\\cos{\\angle{B}})(\\sin{\\angle{B}}-i\\cos{\\angle{B}})\\\\ &=(|BC|)(1)", "and LY = 7cm. Given PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm Steps: • Drawn a line segment PL = 7 cm • With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A. • Joined PA and LA. • With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y. • Joined LY, PY and AY. • PLAY is the required quadrilateral. Calculation of Area: Area of the quadrilateral PLAY = $$\\frac { 1 }{ 2 }$$ × d × (h1 + h2) sq. units = $$\\frac { 1 }{ 2 }$$ × 8 × (5.1 + 1.4) cm2 $$\\frac { 1 }{ 2 }$$ × 8 × 6.5 cm2 = 26 cm2 Area of the quadrilateral = 26 cm2 Question 3. PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°. Given PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120° Steps: • Draw a line segment PQ = 3.5 cm • Made ∠Q = 120°. Drawn the ray QX. • With Q as centre drawn an", "# In the following figure, Question: In the following figure, PQRS is a square of side 4 cm. Find the area of the shaded square. Solution: Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle. We know that area of the four sectors with radius is equal to area of one circle. $\\therefore$ Area of the shaded region $=4^{2}-\\pi \\times 1^{2}-\\pi \\times 1^{2}$ $\\therefore$ Area of the shaded region $=4^{2}-2 \\times \\pi \\times 1$ $\\therefore$ Area of the shaded region $=16-2 \\pi$ Therefore, area of the shaded region is $(16-2 \\pi) \\mathrm{cm}^{2}$", "in a society. Community smaller than society. • It is a network of social relationships which cannot see or touched. • common interests and common objectives are not necessary for society. ##### Constructions Construct: Quadrilateral PQRS and $\\triangle$PSZ Results: Area of quadrilteral PQRS = Areaof $\\triangle$PSZ where; PQ = 4.2 cm, QR = 6.5 cm, RS = 8 cm, SP = 5.3 cm, SQ = 7.4 cm. To construct: Quadrilateral PQRS and $\\\\triangle$PST Result: Area of quadrilateral PQRS = Area of $\\triangle$PST where, PQ = QR = 5.1 cm, PS = RS = 6.2 cm and QS= 5.6 cm. To construct: parallelogram ABCD and $\\triangle$AEF Result: Area of pallelogram ABCD = Area of $\\triangle$AEF where, AB = 6 cm, BC = 4 cm, $\\angle$BAD = 60° and one side of $\\triangle$AEF = 7.5 cm. To construct: quadrilateral ABCD and $\\triangle$ABE Result: Area of quadrilateral ABCD = Area of $\\triangle$ABE where,AB = BC = 5.6 cm, CD = AD = 4.9 cm and $\\angle$BAD = 60°. Construct: quadrilateral ABCD and $\\triangle$ADE Results: Area of quadrilateral ABCD = Area of $\\triangle$ADE where, Ab = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8 cm. To construct: Result: area of quadrilateral ABCD = Area of rectangle ASTU where, AC = 6.6 cm, BD = 8 cm, AB" ]

[ "Pythagorean theorem: $h^2+5^2 = e^2~\\implies~h^2 = (5 \\cdot \\sqrt{3})^2-5^2 = 50$ Thus $h = 5 \\cdot \\sqrt{2}$ 3. The area of the projection is $A = \\frac 12 \\cdot 10 \\cdot 5 \\cdot \\sqrt{2} \\approx 35.355339...$", "# The side of a square is 4 centimeters shorter than the side of a second square. If the sum of their areas is 40 square centimeters, how do you find the length of one side of the larger square? Let 'a' be the side of the shorter square. Then by condition, 'a+4' is the side of larger square. We know the area of a square is equal to the square of it's side. So ${a}^{2} + {\\left(a + 4\\right)}^{2} = 40$ (given) or $2 {a}^{2} + 8 \\cdot a - 24 = 0$ or ${a}^{2} + 4 \\cdot a - 12 = 0$ or $\\left(a + 6\\right) \\cdot \\left(a - 2\\right) = 0$ So either $a = 2 \\mathmr{and} a = - 6$ Side length canot be negative . $\\therefore a = 2$. Hence the length of the side of larger square is $a + 4 = 6$ [Answer]", "given by the formula of the square of its sides. Area of square =${{(side)}^{2}}$. Area of square =${{a}^{2}}$ . We got the area of the square as 12$c{{m}^{2}}$. We need to find the side of a square PQRS. ${{a}^{2}}$ = 12. a = $\\sqrt{12}$ =$\\sqrt{3\\times 4}$. a =$2\\sqrt{3}cm$. Therefore the side of the square =$2\\sqrt{3}cm$. I.e. PQ = QR = RS = PS =$2\\sqrt{3}cm$. Now let us draw the square PQRS with side$2\\sqrt{3}cm$. Note: The side of the square can be calculated directly by equating the area of the rectangle to the area of the square. As it is mentioned in the question that both the areas are the same: Area of rectangle = Area of square. Length $\\times$ breadth =${{(side)}^{2}}$. 4 $\\times$ 3 =${{a}^{2}}$. ${{a}^{2}}$ = 12 cm. a =$2\\sqrt{3}cm$.", "of $$KD$$ is the desired square. Its area is $$KD^2$$ and by the Pythagorean Theorem, ${\\rm{Area}}(ACDB)+{\\rm{Area}}(EGHF)=KL^2 + LD^2=KD^2.$ In the applet, notice as we slide point $$I$$ from $$B$$ to $$K,$$ the area of the constructed square increases, finally meeting the sum of the given areas when $$I$$ meets $$K.$$ Notice that when the areas of the two given squares are equal, the side of the desired square is the diagonal of a given square. A similar technique is used to find the difference of two squares, also from Āpastamba's Śulba-sūtra, Section 2.5 [Plofker2, p. 21]: Removing a [square] quadrilateral from a [square] quadrilateral: Cut off a part of the larger, as much as the side of the one to be removed. Bring the [long] side of the larger [part] diagonally against the other [long] side. Cut off that [other side] where it falls. With the cut-off [side is made a square equal to] the difference. Figure 7. This applet demonstrates the method in the Śulba-sūtra of Āpastamba for constructing a square with area equal to the difference of the areas of two given squares. Click \"Go\" to advance to the next step. The areas of the two given squares can be adjusted by sliding points $$D$$ and $$H.$$ The area of square $$PNDJ$$ can also be adjusted", "the area of the square is 8 cm$^2$. Using Pythagoras' Theorem on the whole triangle The hypotenuse of the triangle is equal to $3k$, and can also be found using Pythagoras' Theorem: $$6^2+6^2=(3k)^2\\\\\\Rightarrow 72=9k^2\\\\ \\Rightarrow 8=k^2$$ But $k^2$ is the area of the square. So the area of the square is $8$ cm$^2$. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "\\\\ A_{front} &= 12 \\qquad \\qquad A_{side} = 6 \\qquad \\qquad \\; \\; A_{top} = 8 \\end{split}$$ Notice for each of the three faces you see, there is an identical opposite face that does not show. $$\\begin{split} S &= (front + back)+(left\\; side + right\\; side) + (top + bottom) \\\\ S &= (2 \\cdot front) + (2 \\cdot left\\; side) + (2 \\cdot top) \\\\ S &= 2 \\cdot 12 + 2 \\cdot 6 + 2 \\cdot 8 \\\\ S &= 24 + 12 + 16 \\\\ S &= 52\\; sq.\\; units \\end{split}$$ The surface area S of the rectangular solid shown in Figure 9.30 is 52 square units. In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its length and its width (see Figure 9.31). Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side. $$S = 2LH + 2LW + 2WH$$ Figure 9.31 - For each face of the rectangular solid facing you, there is another face on the opposite side. There are 6 faces in all. Definition: Volume and Surface Area of a Rectangular Solid For a rectangular solid with length L, width", "of the rhombus, Area of rhombus = Product of both diagonals/ 2, ⇒ 840 = P × Q /2, ⇒ P × Q = 1680, Using Pythagorean Theorem we get, ⇒ (P/2)2 + (Q/2)2 = 372 ⇒ P2 + Q2 = 1369 × 4 ⇒ P2 + Q2 = 5476 Using perfect square formula we get, ⇒ (P + Q)2 = P2 + 2PQ + Q2 ⇒ (P + Q)2 = 5476 + 2 × 1680 ⇒ P + Q = 94 Hence option 4 is correct.", "$$X^{\\prime}$$ reaches $$X$$ relies on the Pythagorean Theorem. Indeed, assume that $$X^{\\prime}$$ and $$X$$ coincide. We can divide rectangle $$EFHG$$ into three smaller sub-rectangles, thus obtaining ${\\rm{Area}}(EFHG)={\\rm{Area}}(EFP_2P_1)+{\\rm{Area}}(P_1P_2P_5P_4)+ {\\rm{Area}}( P_4P_5HG).$ We can divide square $$P_1P_3TG$$ into two sub-rectangles and two sub-squares with ${\\rm{Area}}( P_1P_3TG)={\\rm{Area}}(P_1P_2P_5P_4)+{\\rm{Area}}( P_5P_6TH)+ {\\rm{Area}}( P_4P_5HG)+{\\rm{Area}}(P_2P_3P_6P_5).$ Further, by symmetry, $${\\rm{Area}}(P_1P_2P_5P_4)={\\rm{Area}}(EFP_2P_1),$$ which allows us to infer ${\\rm{Area}}( P_1P_3TG)={\\rm{Area}}(EFHG)+{\\rm{Area}}(P_2P_3P_6P_5).$ Thus, this becomes a problem of “subtracting squares.” We seek a square with area equal to ${\\rm{Area}}( P_1P_3TG)-{\\rm{Area}}(P_2P_3P_6P_5).$ Observe that ${\\rm{Area}}(P_2P_3P_6P_5)=HT^2$ and ${\\rm{Area}}( P_1P_3TG)=P_3T^2=XT^2.$ The last equalities follow since $$P_3T$$ and $$XT$$ are radii of the same circular arc. By the Pythagorean Theorem applied to right triangle $$XTH,$$ $$XH^2+HT^2=XT^2,$$ so that $XH^2=XT^2-HT^2={\\rm{Area}}( P_1P_3TG)-{\\rm{Area}}(P_2P_3P_6P_5)= {\\rm{Area}}(EFHG).$ Finally observe that $$XH$$ is the side length of $$QRTS.$$ (For more details consult \"Ancient Indian Rope Geometry in the Classroom\" [Huffman and Thuong].) So we have ${\\rm{Area}}(QRTS)= {\\rm{Area}}(EFHG)=2\\cdot{\\rm{Area}}(ABDC).$ As a matter of fact, this was the most difficult part of the construction. Now take any point $$Y$$ on side $$QS.$$ The area of triangle $$RTY$$ is half the area of $$QRTS.$$ To see this, note that its base and height are both equal to the side length of $$QRTS,$$ so that ${\\rm{Area}}(RTY) =\\frac{1}{2}{\\rm{(Base)(Height)}} =\\frac{1}{2}{\\rm{Area}}(QRTS)={\\rm{Area}}(ABDC).$ To make triangle $$RTY$$ isosceles, simply slide point $$Y$$ to $$Y^{\\prime},$$ the midpoint of segment $$QS.$$ Figure 4. A soma cart with", "using vectors. In this unit we will do a geometric proof. ## Geometric Proof of the Pythagorean Theorem We start from the generic right triangle introduced above to state the theorem. Then, we construct a square with a side measure that equals the sum of the legs, i.e., a square on the side $(b+c)$. We will agree that the area of ​​this square is $(b+c)^2$. We have arranged the measures $b$ and $c$ in a way that, if we draw the hypotenuses, we construct four right triangles like the generic one, leaving an inner square on the side $a$. Now we can write the area of ​​the large square, which we have previously calculated as $(b+c)^2$, by using the sum of the areas of the four triangles plus the inner square. We have four right triangles with an area $A_t=\\dfrac{b\\cdot c}{2}$ and a square with the area $a^2$. Thus, we get to the following equality: $$4\\cdot\\dfrac{b\\cdot c}{2}+a^2 = (b+c)^2$$ Let's develop both sides: $$2\\cdot b\\cdot c + a^2 = b^2+2\\cdot b \\cdot c + c^2$$ $$2\\cdot b\\cdot c + a^2 - 2\\cdot b \\cdot c= b^2 + c^2$$ $$a^2= b^2 + c^2$$ Like this we get the relationship stated by the Pythagorean theorem. We can also find statements of the Pythagorean theorem using other letters/variables to express equality.", "SEARCH HOME Math Central Quandaries & Queries Question from shubham, a student: Why is the area of square not conserved when it changes to rhombus, both have equal sides still rhombus have less area than square.?? Hi Shubham, First print a copy of one centimeter graph paper. Here is a printable copy. Step 1. Starting at one corner of the graph paper draw a square which is 5 cm by 5 cm. If you count the square you will see that the 5 by 5 square covers 25 of the 1 centimeter squares. Thus the area of the 5 by 5 square is 25 square centimeters. Step 2. Starting at $P$ an intersection of the grid line in a different place on the graph paper draw along the horizontal grid line to a point $Q,$ 5 centimeters from $P.$ Draw the line $L_{1},$ 3 centimeters above the line segment $PQ.$ Using a ruler find the point $R$ on $L_1$ which is 5 centimeters from $P.$ Again with a ruler find the point $S$ on $L_1$ which is 5 centimeters from $R.$ If you did this carefully you will find that the line segment $QS$ has length 5 centimeters. How many 1 centimeter squares are covered by the rhombus $PQSR?$ You can count the number of complete 1 centimeter", "Pythagorean Theorem we have that $A D=\\sqrt{3^{2}-1^{2}}=\\sqrt{8}$ Again from $\\triangle A D O_{1} \\sim \\triangle A F C$ $\\frac{A D}{A F}=\\frac{D O_{1}}{C F} \\Longrightarrow \\frac{2 \\sqrt{2}}{8}=\\frac{1}{C F} \\Rightarrow C F=2 \\sqrt{2}$ can you finish the problem…….. The area of the triangle is $\\frac{1}{2} \\cdot A F \\cdot B C=\\frac{1}{2} \\cdot A F \\cdot(2 \\cdot C F)=A F \\cdot C F=8(2 \\sqrt{2})$=$16\\sqrt2$", "49 cm2 + 19.25 cm2 = 68.25 cm2. 2. In the adjoining figure, PQRS is a square of side 14 cm and O is the centre of the circle touching all sides of the square. Find the area of the shaded region. Solution: Step I: First we divide the combined figure into its simple geometrical shapes. The given combined shape is combination of a square and a circle. Step II: Then calculate the area of these simple geometrical shapes separately. Area of the square PQRS = 142 cm2 = 196 cm2 Area of the circle with centre O = π ∙ 72 cm2, [Since, diameter SR = 14 cm] = $$\\frac{22}{7}$$ ∙ 49 cm2 = 22 × 7 cm2 = 154 cm2 Step III: Finally, to find the required area of the combined figure we need to subtract the area of the circle from the area of the square. Therefore, the required area = 196 cm2 - 154 cm2 = 42 cm2 3. In the adjoining figure alongside, there are four equal quadrants of circles each of radius 3.5 cm, their centres being P, Q, R and S. Find the area of the shaded region. Solution: Step I: First we divide the combined figure into its simple geometrical shapes. The given combined shape is combination of a square", "49 cm2 + 19.25 cm2 = 68.25 cm2. 2. In the adjoining figure, PQRS is a square of side 14 cm and O is the centre of the circle touching all sides of the square. Find the area of the shaded region. Solution: Step I: First we divide the combined figure into its simple geometrical shapes. The given combined shape is combination of a square and a circle. Step II: Then calculate the area of these simple geometrical shapes separately. Area of the square PQRS = 142 cm2 = 196 cm2 Area of the circle with centre O = π ∙ 72 cm2, [Since, diameter SR = 14 cm] = $$\\frac{22}{7}$$ ∙ 49 cm2 = 22 × 7 cm2 = 154 cm2 Step III: Finally, to find the required area of the combined figure we need to subtract the area of the circle from the area of the square. Therefore, the required area = 196 cm2 - 154 cm2 = 42 cm2 3. In the adjoining figure alongside, there are four equal quadrants of circles each of radius 3.5 cm, their centres being P, Q, R and S. Find the area of the shaded region. Solution: Step I: First we divide the combined figure into its simple geometrical shapes. The given combined shape is combination of a square", "figure, PQRS is a square with side x + 4. Each of the [#permalink] ### Show Tags 20 Sep 2017, 05:13 Bunuel wrote: In the above figure, PQRS is a square with side x + 4. Each of the four smaller squares has side of length 2. If the area of the shaded region is 48, what is the value of x? (A) 1 (B) 4 (C) 4√2 (D) 8 (E) 12 [Reveal] Spoiler: Attachment: 2017-09-20_1020.png Find numeric value for area of large square. Use (x + 4) to derive equation for the same area. Set them equal to solve for $$x$$. Numeric area (Area of small squares) + (Area of shaded region) = Area of large square Small square side = 2. Each small square's area = 4. Their total area = 4*4 = 16 Area of shaded region = 48 Area of large square = 16 + 48 = 64 Equation for same area One side of large square = $$(x + 4)$$ Equation for area of large square is $$(x + 4)^{2}$$ = $$x^2 + 8x + 16$$ From above, numeric area of large square = 64. Set equation equal to numeric value: $$x^2 + 8x + 16 = 64$$ $$x^2 + 8x - 48 = 0$$ $$(x + 12)(x - 4) = 8$$ $$x$$", "# Area of a rectangle Calculate the area of a rectangle with a diagonal of u = 12.5cm and a width of b = 3.5cm. Use the Pythagorean theorem. Result S = 42 cm2 #### Solution: $u = 12.5 \\ cm \\ \\\\ b = 3.5 \\ cm \\ \\\\ \\ \\\\ a^2 + b^2 = u^2 \\ \\\\ \\ \\\\ a = \\sqrt{ u^2-b^2 } = \\sqrt{ 12.5^2-3.5^2 } = 12 \\ cm \\ \\\\ \\ \\\\ S = a \\cdot \\ b = 12 \\cdot \\ 3.5 = 42 = 42 \\ cm^2$ Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! #### Following knowledge from mathematics are needed to solve this word math problem: Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator. ## Next similar math problems: 1. Billboard Rectangular billboard is 2.5 m long with a diagonal 2.8 m long. Calculate the perimeter and the content area of the billboard. 2. Rectangle diagonal The rectangle, one side of which is 5 cm long, is divided by a 13 cm diagonal into two triangles. Calculate the area of one of these triangles in cm2. 3. The field The player crossed the field diagonally", "square $PQRS$ we need to calculate the length of one of its sides, say $PQ$. Using the Pythagorean theorem we find \\begin{eqnarray} |PQ|^2 &=& |OP|^2 + |OQ|^2 \\\\ &=& r^2 + r^2 \\\\ &=& 2r^2. \\end{eqnarray} Thus $|PQ| = \\sqrt{2} r$. So the area of square $PQRS$ is $2 r^2$ versus $\\pi r^2$, the area of the circle. The square contains slightly more than $63$ percent of the area of its circumscribing circle.", "| * - - - - - * s A square with side $s$ has area $s^2.$ Code: * - - - - - * | _ * | | √2s * | | * | s | * | | * | * - - - - - * s The diagonal $d$ of the square can be found with Pythagorus: . . $d^2 \\:=\\:s^2+s^2 \\:=\\:2s^2 \\quad\\Rightarrow\\quad d \\:=\\:\\sqrt{2}\\,s$ The area of the square of side $d$ is: . $\\left(\\sqrt{2}\\,s\\right)^2 \\;=\\;2s^2$ Therefore, the ratio of the two areas is . . . 5. ## hello let the side of the square be (a),so the area of the first square is a^2, then by pythegoras , it's diagonal is a√2,and the area of the second square ia (a√2)^2=2(a^2). therefore the ratio is:a^2 / 2(a^2)=1/2 D is correct.", "### Home > CCA > Chapter 5 > Lesson 5.2.1 > Problem5-47 5-47. The area of a square is $225$ square centimeters. 1. Make a diagram and determine the length of each side. Find the length of one of the sides of the square and use the sides to find the diagonal. $\\sqrt{225}=15\\text{ cm}$ 2. Use the Pythagorean Theorem to find the length of the square's diagonal. • The Pythagorean Theorem is the following: $a^2+b^2=c^2$ • The length of the diagonal is • $15\\sqrt{2}$", "2$ area of the square=$\\pi (\\sqrt2)^2$=$2\\pi$ Area of the shaded region= 28-2$\\pi$", "found using the Pythagoras Theorem $D = \\sqrt{L^{2}+W^{2}}$ Example- Find the Area and Perimeter of a rectangle where length and width are given as 12 and 8 cm respectively. Also, find the length of the Diagonal. Solution- We know that the area of a rectangle is given by $A = Length \\times Width$. $\\Rightarrow A = 12 \\times 8$ $\\Rightarrow A = 96 cm^{2}$ Now Perimeter is given by $P = 2 (Length + Width)$ $\\Rightarrow P = 2 (12 + 8)$ $\\Rightarrow P = 40$ Diagonal Length, $D = \\sqrt{L^{2}+W^{2}}$ $\\Rightarrow D = \\sqrt{12^{2}+8^{2}}$ $\\Rightarrow D = \\sqrt{144 + 64}$ $\\Rightarrow D = \\sqrt{208}$ $\\Rightarrow D = 4\\sqrt{13}$ Register at BYJU’S to learn more properties of different shapes and figures in a fun and creative way. Learn more on Rectangles Surface Area Of Rectangle Special parallelograms :Rhombus, Square, Rectangle #### Practise This Question The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are:" ]

[ "the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are congruent. In particular $\\angle PQR$ is a right angle, so that $PQ$ is tangent to the semicircle and $QR$ is tangent to the quarter-circle. Also, $PR$ bisects the angle $\\theta =\\angle QPS$, so that $\\angle RPS = \\frac\\theta2$ and $\\tan\\frac\\theta2 = \\frac{RS}{PS} = \\frac12.$ Therefore $$\\tan\\theta = \\frac{2\\tan\\frac\\theta2}{1 - \\tan^2\\frac\\theta2} = \\frac1{1-\\frac14} = \\frac43.$$ It follows that $\\sin\\theta = \\frac45$", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The", "it. Given that the side of the square is 10 cm, find the combined areas of the regions labelled A and C. Any hints, perhaps, on how to solve this seemingly simple geometry problem? This is a primary math problem, so one should refrain from using geometry or trigonometric method to solve it...any help is much appreciated! The area of the regions A+D is $25\\pi$. The area of the regions C+D is $12.5\\pi$. So to find the area of A+C we need to know the area of D. [TIKZ] \\draw (0,-3) rectangle (6,3); \\begin{scope} \\draw (6,-3) arc (-90:-270:3cm); \\end{scope} \\begin{scope} \\draw (6,-3) arc (0:90:6cm); \\end{scope} \\coordinate[label=above: A] (A) at (2,0); \\coordinate[label=above: B] (B) at (3.6,2.4); \\coordinate[label=above: C] (C) at (5.4,0.8); \\coordinate[label=above: D] (D) at (4,-1); \\coordinate[label=left: P] (P) at (0,-3); \\coordinate[label=above: Q] (Q) at (3.6,1.8); \\coordinate[label=right: R] (R) at (6,0); \\coordinate[label=right: S] (S) at (6,-3); \\draw (P) -- (Q) -- (R) -- (P) ; \\draw (Q) -- (S) ; \\draw (0.4,-2.8) node {$\\theta$} ; \\draw (3,-3.2) node {$10$} ; \\draw (6.2,-1.5) node {$5$} ;[/TIKZ] Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are", "# Find the volume of the parallelepiped with adjacent edges PQ, Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS where P(3,0,1), Q(-1,2,5), R(5,1,-1) and S(0,4,2). You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Ben Owens As we have three vectors, there is a product, called scalar triple product, that gives the volume of the parallelepiped that has the three vectors as dimensions. $P{Q}^{\\to }=\\left(3+1,0-2,1-5\\right)=\\left(4,-2,-4\\right)$ $P{R}^{\\to }=\\left(3-5,0-1,1+1\\right)=\\left(-2,-1,2\\right)$ $P{S}^{\\to }=\\left(3-0,0-4,1-2\\right)=\\left(3,-4,-1\\right)$ The derminant is given for example with the Laplace rule: $4×\\left[\\left(-1\\right)\\left(-1\\right)-\\left(2\\right)\\left(-4\\right)\\right]-\\left(-2\\right)\\left[\\left(-2\\right)\\left(-1\\right)-\\left(2\\right)×\\left(3\\right)+\\left(-4\\right)\\left[\\left(-2\\right)\\left(-4\\right)-\\left(-1\\right)\\left(3\\right)\\right]=4\\left(1+8\\right)+2\\left(2-6\\right)-4\\left(8+3\\right)=36-8-444=-16$ Thus, $V=16$ rodclassique4r $PQ=\\left(3+1,0-2,1-5\\right)=\\left(4,-2,-4\\right)$ $PR=\\left(3-5,0-1,1+1\\right)=\\left(-2,-1,2\\right)$ $PS=\\left(3-0,0-4,1-2\\right)=\\left(3,-4,-1\\right)$ Thus, $V=16$ is the volume", "C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"P\", P, NW); label(\"O\", O, 1.5*S); label(\"\\theta\", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy]$ Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get $$PA = 2r\\sin\\left(\\dfrac{\\theta}2\\right)$$ $$PC = 2r\\sin\\left(\\dfrac{180-\\theta}2\\right) = 2r\\sin\\left(90 - \\dfrac{\\theta}2\\right) = 2r\\cos\\left(\\dfrac{\\theta}2\\right)$$ Multiplying and simplifying these 2 equations gives $$PA \\cdot PC = 4r^2 \\sin \\left(\\dfrac{\\theta}2 \\right) \\cos \\left(\\dfrac{\\theta}2 \\right) = 2r^2 \\sin\\left(\\theta \\right) = 56$$ Similarly $PB = 2r\\sin\\left(\\dfrac{90 +\\theta}2\\right)$ and $PD =2r\\sin\\left(\\dfrac{90 -\\theta}2\\right)$. Again, multiplying gives $$PB \\cdot PD = 4r^2 \\sin\\left(\\dfrac{90 +\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right) = 4r^2 \\sin\\left(90 -\\dfrac{90 -\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right)$$ $$=4r^2 \\sin\\left(\\dfrac{90 -\\theta}2\\right) \\cos\\left(\\dfrac{90 -\\theta}2\\right) = 2r^2 \\sin\\left(90 - \\theta \\right) = 2r^2 \\cos\\left(\\theta \\right) = 90$$ Dividing $2r^2 \\sin \\left(\\theta \\right)$ by $2r^2 \\cos \\left( \\theta \\right)$ gives $\\tan \\left(\\theta \\right) = \\dfrac{28}{45}$, so $\\theta = \\tan^{-1} \\left(\\dfrac{28}{45} \\right)$. Pluging this back into one of the equations, gives $$2r^2 = \\dfrac{90}{\\cos\\left(\\tan^{-1}\\left(\\dfrac{28}{45}\\right)\\right)}$$ If we imagine a $28$-$45$-$53$ right triangle, we see that if", "of the area of square PQRS ? A. $$\\frac{1}{6}$$ B. $$\\frac{1}{8}$$ C. $$\\frac{1}{5}$$ D. $$\\frac{1}{4}$$ E. $$\\frac{1}{3}$$ $$area:TSR = PQU = (x/2)x/2=x^2/4$$ $$area:Square-(TSR+PQU)=PTRU…x^2-2x^2/4=2x^2/4=x^2/2$$ $$area:PTRU/2=shaded…(x^2/2)/2=shaded…shaded=x^2/4$$ $$area:shaded/square=(x^2/4)/x^2=1/4$$ Ans (D) Manager Joined: 10 May 2018 Posts: 59 Re: In the figure shown, PQRS is a square, T is the midpoint of side PS [#permalink] ### Show Tags 22 Nov 2019, 10:20 PTUR is a parallelogram so the area is x/2*x = x^2/2 the shaded portion is half of this so it is x^2/4 Area of square x^2 (x^2/4)/x^2 =1/4 Re: In the figure shown, PQRS is a square, T is the midpoint of side PS [#permalink] 22 Nov 2019, 10:20 Display posts from previous: Sort by", "figure, PQRS is a square with side x + 4. Each of the [#permalink] ### Show Tags 20 Sep 2017, 05:13 Bunuel wrote: In the above figure, PQRS is a square with side x + 4. Each of the four smaller squares has side of length 2. If the area of the shaded region is 48, what is the value of x? (A) 1 (B) 4 (C) 4√2 (D) 8 (E) 12 [Reveal] Spoiler: Attachment: 2017-09-20_1020.png Find numeric value for area of large square. Use (x + 4) to derive equation for the same area. Set them equal to solve for $$x$$. Numeric area (Area of small squares) + (Area of shaded region) = Area of large square Small square side = 2. Each small square's area = 4. Their total area = 4*4 = 16 Area of shaded region = 48 Area of large square = 16 + 48 = 64 Equation for same area One side of large square = $$(x + 4)$$ Equation for area of large square is $$(x + 4)^{2}$$ = $$x^2 + 8x + 16$$ From above, numeric area of large square = 64. Set equation equal to numeric value: $$x^2 + 8x + 16 = 64$$ $$x^2 + 8x - 48 = 0$$ $$(x + 12)(x - 4) = 8$$ $$x$$", "PM. Since $\\angle PQR + \\angle RQM =$ 180º (Linear pair) $\\Rightarrow$ 110º + $\\angle RQM$ = 180º $\\Rightarrow$ $\\angle RQM =$ 70º Since $\\angle QRS$ = 180º – (70º + 50º) = 60º [Since sum of the angles of a triangle is 180º] Q.5 In figure , if AB || CD, $\\angle APQ =$ 50º and $\\angle PRD =$ 127º , find x and y. Sol. Therefore AB || CD and transversal PQ intersects them at P and Q respectively. Therefore $\\angle PQR = \\angle APQ$ [Alternate angles] $\\Rightarrow$ x = 50º [Since $\\angle APQ =$ 50º (given)] Since AB || CD and transversal PR intersects them at P and R respectively. Therefore $\\angle APR = \\angle PRD$ [Alternate angles] $\\Rightarrow$ $\\angle APQ + \\angle QPR$ = 127º [Since $\\angle PRD =$ 127º] $\\Rightarrow$ 50º + y = 127º [Since $\\angle APQ =$ 50º] $\\Rightarrow$ y = 127º – 50º = 77º Hence , x = 50º and y = 77º Q.6 In figure , PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB|| CD. Sol. Two plane mirrors PQ and", "given by the formula of the square of its sides. Area of square =${{(side)}^{2}}$. Area of square =${{a}^{2}}$ . We got the area of the square as 12$c{{m}^{2}}$. We need to find the side of a square PQRS. ${{a}^{2}}$ = 12. a = $\\sqrt{12}$ =$\\sqrt{3\\times 4}$. a =$2\\sqrt{3}cm$. Therefore the side of the square =$2\\sqrt{3}cm$. I.e. PQ = QR = RS = PS =$2\\sqrt{3}cm$. Now let us draw the square PQRS with side$2\\sqrt{3}cm$. Note: The side of the square can be calculated directly by equating the area of the rectangle to the area of the square. As it is mentioned in the question that both the areas are the same: Area of rectangle = Area of square. Length $\\times$ breadth =${{(side)}^{2}}$. 4 $\\times$ 3 =${{a}^{2}}$. ${{a}^{2}}$ = 12 cm. a =$2\\sqrt{3}cm$.", "+0 # Help Geometry 0 154 1 In right triangle $ABC,$ $AC = BC$ and $\\angle C = 90^\\circ.$ and Let $P$ and $Q$ be points on hypotenus $\\overline{AB},$ as shown below, such that $\\angle PCQ = 45^\\circ.$ Show that $AP^2 + BQ^2 = PQ^2.$ $$[asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label(\"A\", A, SE); label(\"B\", B, NW); label(\"C\", C, SW); label(\"P\", P, NE); label(\"Q\", Q, NE); [/asy]$$ off-topic Feb 16, 2020 edited by Guest Feb 16, 2020", "drawn. Then the area of the quadrilateral PQOR is (a) 60 cm² (b) 65 cm² (c) 30 cm² (d) 32.5 cm² = $$\\frac{1}{2}$$ × 12 × 5 + $$\\frac{1}{2}$$ × 12 × 5 = 60 cm²", "### Set Square A triangle PQR, right angled at P, slides on a horizontal floor with Q and R in contact with perpendicular walls. What is the locus of P? ### Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. ### Strange Rectangle ABCD is a rectangle and P, Q, R and S are moveable points on the edges dividing the edges in certain ratios. Strangely PQRS is always a cyclic quadrilateral and you can find the angles. Consider any convex quadrilateral $Q$ made from four rigid rods with flexible joints at the vertices so that the shape of $Q$ can be changed while keeping the lengths of the sides constant. If the diagonals of the quadrilateral cross at an angle $\\theta$ in the range $(0 \\leq \\theta < \\pi/2)$, as we deform $Q$, the angle $\\theta$ and the lengths of the diagonals will change. Using the results of the two problems on quadrilaterals Diagonals for Area and Flexi Quads prove that the area of $Q$ is proportional to $\\tan\\theta$.", "RS = PQ (by CPCT) ………… (i) In QCRandSAP$\\bigtriangleup QCR \\;\\; and \\;\\; \\bigtriangleup SAP$, RC = PA (Halves of the opposite sides of the rhombus) RCQ=PAS$\\angle RCQ=\\angle PAS$ (Opposite angles of the rhombus) CQ = AS (Halves of the opposite sides of the rhombus) Thus, QCRSAP$\\bigtriangleup QCR \\cong \\bigtriangleup SAP$ (by SAS congruence condition). RQ = SP (by CPCT) ……(ii) Now, In CBD$\\bigtriangleup CBD$, R and Q are the mid points of CD and BC respectively. QRBD$\\Rightarrow QR \\parallel BD$ also, P and S are the mid points of AD and AB respectively. PSBD$\\Rightarrow PS \\parallel BD$ QRPS$\\Rightarrow QR \\parallel PS$ Thus, PQRS is a parallelogram. also, PQR=90$\\angle PQR=90^{\\circ}$ Now, In PQRS, RS = PQ and RQ = SP from (i) and (ii) Q=90$\\angle Q=90^{\\circ}$ Thus, PQRS is a rectangle. Q3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. Solution: Given, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, AC and BD are joined. To Prove, PQRS is a rhombus. Proof, In ABC$\\bigtriangleup ABC$ P and Q are the mid-points of AB and BC respectively Thus, PQAC$PQ\\parallel AC$ and PQ=12AC$PQ= \\frac{1}{2}AC$ (Mid", "Category : JEE Main & Advanced A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S. Let a cyclic quadrilateral be such that $PQ=a,\\,QR=b,\\,RS=c$ and $SP=d$. Then$\\angle Q+\\angle S={{180}^{o}}$,$\\angle A+\\angle C={{180}^{o}}$ Let $2s=a+b+c+d$ Area of cyclic quadrilateral = $\\frac{1}{2}(ab+cd)\\sin Q$ Also, area of cyclic quadrilateral = $\\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $2s=a+b+c+d$ and $\\cos Q=\\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{d}^{2}}}{2(ab+cd)}$. (1) Circumradius of cyclic quadrilateral : Circum circle of quadrilateral PQRS is also the circumcircle of $\\Delta PQR$. $R=\\frac{1}{4\\Delta }\\sqrt{(ac+bd)(ad+bc)(ab+cd)}=\\frac{1}{4}\\sqrt{\\frac{(ac+bd)(ad+bc)(ab+cd)}{(s-a)(s-b)(s-c)(s-d)}}$. (2) Ptolemy's theorem : In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS $PR.QS=PQ.RS+RQ.PS.$ LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL 31st March 2018 ONLY! You need to login to perform this action. You will be redirected in 3 sec", "Similarly, ∠PQR = 90° and ∠SRQ = 90° So, PQRS is a quadrilateral in which all angles are right angles. We have shown that ∠PSR = ∠PQR = 90° and ∠SPQ = ∠SRQ = 90°. So both pairs of opposite angles are equal. Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and hence, PQRS is a rectangle. Example 6: ABCD is a parallelogram in which P and Q are midpoints of opposite sides AB and CD (see the given figure). If AQ intersects DP at S and BQ intersects CP at R, show that: (i) APCQ is a parallelagrair. (ii) DPBQ is a parallelogram. (iii) PSQR is a parallelogram AP ∥ QC (v AB ∥ CD) ……..(1) AP = $$\\frac{1}{2}$$AB, CQ = $$\\frac{1}{2}$$CD (Given) Also, AB = CD (Opposite sides of a parallelogram) So, AP = QC ……………(2) Therefore, APCQ is a parallelogram. [From (1) and (2) and theorem 8.8] ∴AQ ∥CP (ii) Similarly, quadrilateral DPBQ is a parallelogram, because DQ ∥ PB and DQ = PB ∴ DP ∥ QB SP ∥ QR (∵ DP ∥ QB) Similarly, SQ ∥ PR (∵ AQ ∥ CP) So, PSQR is a parallelogram. Following properties are proved in the sums of Exercise: • The diagonals of a parallelogram bisect each other. • The", "# 2005 CEMC Gauss (Grade 7) Problems/Problem 15 ## Problem In the diagram, the area of rectangle $PQRS$ is $24$. If $TQ = TR$, what is the area of quadrilateral $PTRS$? $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 16 \\qquad \\text{(D)}\\ 6 \\qquad \\text{(E)}\\ 15$ $[asy] size(100); draw((0,0)--(6,0)--(6,4)--(0,4)--cycle); draw((0,4)--(6,2)); draw((5.8,1.1)--(6.2,1.1)); draw((5.8,.9)--(6.2,.9)); draw((5.8,3.1)--(6.2,3.1)); draw((5.8,2.9)--(6.2,2.9)); label(\"P\",(0,4),NW); label(\"S\",(0,0),SW); label(\"R\",(6,0),SE); label(\"T\",(6,2),E); label(\"Q\",(6,4),NE); [/asy]$ ## Solution 1 Since the area of rectangle $PQRS$ is $24$, let us assume that $PQ = 6$ and $QR = 4$. Since $QT = TR$, then $QR = 2QT$ so $QT = 2$. Therefore, triangle $PQT$ has base $PQ$ of length $6$ and height $QT$ of length 2, so its area is $\\frac{1}{2}(6)(2) = \\frac{1}{2}(12) = 6$. So the area of quadrilateral $PTRS$ is equal to the area of rectangle $PQRS$ (which is $24$) minus the area of triangle $PQT$ (which is $6$), or $18$. Therefore, the answer is $A$. ## Solution 2 Draw a line through $T$ parallel to $PQ$ across the rectangle parallel so that it cuts $PS$ at point $V$. Since $T$ is halfway between $Q$ and $R$, then $V$ is halfway between $P$ and $S$. Therefore, $SVTR$ is a rectangle which has an area equal to half the area of rectangle $PQRS$, or $12$. Similarly, $VPQT$ is a rectangle of area $12$, and $VPQT$", "Clearly, we can see that PR+ QR= PQ2. Therefore the triangle with vertices P, Q and R is right angled (at R). Here is a figure to illustrate On to some more examples ! Example 3. Show that the points (i) A(0, 0), B(3, 0), C(4, 1) and D(1, 1) form a parallelogram (ii) P(0, 0), Q(4, 0), R(4, 1) and S(0, 1) form a rectangle (iii) K(-1, 0), L(0, -2), M(1, 0) and N(0, 2) form a rhombus (iv) E (-2, 0), F(0, -2), G(2, 0) and H(0, 2) form a square Solution. To solve the problem, we need to recall the properties of the above quadrilaterals, in terms of the relation between their sides, and diagonals. I’ll skip the calculations now. Here are figures to illustrate Fig. 4: The other two quadrilaterals (i) For a paralellogram, both pairs of opposite sides should be equal AB=3, BC= $$\\sqrt{2}$$ , CD = 3 and DA= $$\\sqrt{2}$$ We can see that, AB = CD and AD = BC .. and we’re done. (ii) Same conditions as above. Additionally, we need to show that the diagonals are of equal length Here, PQ = 4, QR = 1, RS = 4 and SP = 1. This makes PQRS a parallelogram. Now, PR= $$\\sqrt{17}$$ and also QS= $$\\sqrt{17}$$ , i.e. the diagonals", "in a society. Community smaller than society. • It is a network of social relationships which cannot see or touched. • common interests and common objectives are not necessary for society. ##### Constructions Construct: Quadrilateral PQRS and $\\triangle$PSZ Results: Area of quadrilteral PQRS = Areaof $\\triangle$PSZ where; PQ = 4.2 cm, QR = 6.5 cm, RS = 8 cm, SP = 5.3 cm, SQ = 7.4 cm. To construct: Quadrilateral PQRS and $\\\\triangle$PST Result: Area of quadrilateral PQRS = Area of $\\triangle$PST where, PQ = QR = 5.1 cm, PS = RS = 6.2 cm and QS= 5.6 cm. To construct: parallelogram ABCD and $\\triangle$AEF Result: Area of pallelogram ABCD = Area of $\\triangle$AEF where, AB = 6 cm, BC = 4 cm, $\\angle$BAD = 60° and one side of $\\triangle$AEF = 7.5 cm. To construct: quadrilateral ABCD and $\\triangle$ABE Result: Area of quadrilateral ABCD = Area of $\\triangle$ABE where,AB = BC = 5.6 cm, CD = AD = 4.9 cm and $\\angle$BAD = 60°. Construct: quadrilateral ABCD and $\\triangle$ADE Results: Area of quadrilateral ABCD = Area of $\\triangle$ADE where, Ab = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8 cm. To construct: Result: area of quadrilateral ABCD = Area of rectangle ASTU where, AC = 6.6 cm, BD = 8 cm, AB", "$MN$ down so that $N$ coincides with $B$, and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$, the longer leg is the same as $MN=x$, and the hypotenuse is $AB=12$. We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$, and solve), you find that $x=\\sqrt{130}$, so $x^2 = \\boxed{130}$. Solution by Kinglogic ### Full Proof that R, A, B are collinear $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); label(\"T\", t , NW); [/asy]$ Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$, since the problem told us $QP = PR$. We will show that $R$ lies on $AB$. Let $T$ be the intersection of circle centered at $B$", "# 2019 AIME I Problems/Problem 3 ## Problem In $\\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\\overline{PQ}$, points $C$ and $D$ lie on $\\overline{QR}$, and points $E$ and $F$ lie on $\\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$. ## Diagram $[asy] dot((0,0)); dot((15,0)); dot((15,20)); draw((0,0)--(15,0)--(15,20)--cycle); dot((5,0)); dot((10,0)); dot((15,5)); dot((15,15)); dot((3,4)); dot((12,16)); draw((5,0)--(3,4)); draw((10,0)--(15,5)); draw((12,16)--(15,15)); [/asy]$ ## Solution 1 We know the area of the hexagon $ABCDEF$ to be $\\triangle PQR- \\triangle PAF- \\triangle BCQ- \\triangle RED$. Since $PR^2+RQ^2=PQ^2$, we know that $\\triangle PRQ$ is a right triangle. Thus the area of $\\triangle PQR$ is $150$. Another way to compute the area is $$\\frac12 \\cdot PQ\\cdot RQ \\sin \\angle PQR = \\frac12 \\cdot 500 \\cdot \\sin \\angle PQR=150 \\implies \\sin \\angle PQR = \\frac35.$$ Then the area of $\\triangle BCQ = \\frac12 \\cdot BQ \\cdot CQ \\cdot \\sin \\angle PQR= \\frac{25}{2}\\cdot \\frac{3}{5}=\\frac{15}{2}$. Preceding in a similar fashion for $\\triangle PAF$, the area of $\\triangle PAF$ is $10$. Since $\\angle ERD = 90^{\\circ}$, the area of $\\triangle RED=\\frac{25}{2}$. Thus our desired answer is $150-\\frac{15}{2}-10-\\frac{25}{2}=\\boxed{120}$ ## Solution 2 Let $R$ be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$, and $F=(0,10)$. Using the shoelace theorem, the area is $\\boxed{120}$. Shoelace theorem:Suppose the" ]

[ "By observations, we circle all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student", "all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57", "you missing the first four years of data? The series in Figure 7.1 in the book seems close to this series of personal saving as a percentage of disposable personal income for the period 1955-1979. – javlacalle Jun 5 '15 at 21:11 • Thanks, javlacalle. I modified the question after using your more complete data set. – StatSmartWannaB Jun 5 '15 at 22:33 This is what I get fitting the following model in R: $$y_t - \\mu = \\phi (y_{t-1} - \\mu) + \\epsilon_t + \\theta \\epsilon_{t-2} \\,,$$ x <- structure(c(9.3, 9.5, 10, 9.9, 10.6, 11.1, 11.3, 11.5, 11.1, 11.6, 11.4, 10.8, 11.3, 11.2, 11.6, 11.5, 10.7, 10.7, 9.6, 10.1, 10.3, 9.7, 10.2, 10, 10.9, 10.9, 11.7, 11.6, 11.5, 11.3, 11, 10.6, 10.7, 10.6, 10.3, 11, 11, 11.8, 11.3, 12, 11.1, 11.1, 11.9, 11.3, 10.9, 10.9, 10.9, 11.6, 12.3, 11.8, 12.2, 12.4, 11.8, 11.9, 10.5, 10.6, 9.8, 10.1, 11.6, 11.5, 11.8, 12.5, 13.1, 13.1, 13.2, 13.6, 13.4, 12.9, 12.2, 11.4, 11.7, 13.2, 12.2, 13, 13, 14.1, 13.6, 12.5, 12.2, 13.3, 12.4, 15, 12.5, 12.3, 11.7, 11.3, 11.1, 10.4, 9.4, 10, 10.5, 10.8, 10.9, 9.9, 10.1, 10.1, 10.6, 9.8, 9.4, 9.5), .Dim = c(100L, 1L), .Dimnames = list(NULL, \"V1\"), .Tsp = c(1959, 1983.75, 4), class = \"ts\") arima(x, order=c(1,0,2), include.mean=TRUE, fixed=c(NA,0,NA,NA)) # Coefficients: # ar1 ma1 ma2 intercept #", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Because Steve only wants to see if the monthly payment is above or below400, he can estimate the answer by dividing the leading digits. First, he should identify the two leading digits of each number. • The leading digits of 21,045.68 are 21,000 (remember that you have to maintain the placement of the decimal point!) • The leading digits of 48 are 48 Now, note that 21,000 is your dividend and 48 is your divisor. Because your divisor is already a whole number, you do not need to move any decimal points. Now, divide as usual using long division. 48)21000.0¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 437.5 192 180 144 360 336 240 240 0\\begin{align*}\\begin{array}{rcl} && \\overset{ \\ \\ \\ \\ \\ \\ \\ \\ 437.5}{48 \\overline{ ) {21000.0 \\;}}}\\\\ && \\underline{\\ - 192 \\;\\;\\;\\;\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ 180 \\\\ && \\underline{\\ \\ \\ - 144 \\;\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ 360 \\\\ && \\underline{\\ \\ \\ \\ \\ - 336\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ \\ \\ 240 \\\\ && \\underline{\\ \\ \\ \\ \\ \\ \\ \\ - 240\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0 \\\\ \\end{array}\\end{align*} The answer is that the monthly payments will be approximately $437.50. Steve’s dad shouldn’t buy", "the original datapoints and the calculated ones figure hold on plot(t, x); plot(t, y); plot(t, xv, 'k--', 'linewidth', 3); plot(t, yv, 'g--', 'linewidth', 3); grid on xlabel('Time (s)'); ylabel('Position'); legend('X-position from Q5', 'Y-position from Q5', 'X-position from Q6', 'Y-position from Q6'); %% Question 8 % Plot the trajectory figure plot(xv, yv), grid on xlabel('X-Position (m)'); ylabel('Y-Position (m)'); %% Question 9 % Write a new file file = fopen('output.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); for k = 1:length(t) fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t(k),xv(k), yv(k)); end fclose(file); %% Question 10 % Write a new file file = fopen('output2.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t',xv', yv'); fclose(file); clc, clear all, close all P = 0; % initial money in the account deposit_per_year = [288, 345, 355, 382, 392]; % amounts deposited each month, per year (year 1, year2, ... year5) withdraw = 2331; % amount to be withdrawn each year r = 0.04; % interes of the saving account rcd = 0.08; % interest of the CD n_CD = 0; % number of CDs bought year = 1; deposited = 0; for i = 1:5*12 % 5 years, 12 months per year: total 60 months % if i < 60 % deposited = deposit_per_year(year); % else % deposited = deposit_per_year(5); % end deposited = deposit_per_year(year); P = P*(1+r); P = P", "### Re: Correct answer marked incorrect with Currency context by Jason Aubrey - Here's what webwork tells me$5253.12 $5253.12 incorrect$5798.47 incorrect Yes, you're right it marks $5253.12 as incorrect, but it's also telling me this is the correct answer (which I believed). Is this what you see when you try this, or am I missing something? ## http://wwrk.maa.org/moodle/mod/forum/discuss.php?d=459 ## seed 3515 ######################################################################## DOCUMENT(); loadMacros( \"PGstandard.pl\", # Standard macros for PG language \"MathObjects.pl\", \"contextCurrency.pl\", #\"source.pl\", # allows code to be displayed on certain sites. #\"PGcourse.pl\", # Customization file for the course ); # Show which answers are correct and which ones are incorrect$showPartialCorrectAnswers = 1; ############################################################## # # Setup # # TEXT(&beginproblem); Context(\"Currency\"); $showPartialCorrectAnswers = 1; @periods = (\"annually\",\"semiannually\",\"quarterly\",\"monthly\",\"weekly\",\"daily\"); @ms = (1,2,4,12,52,365);$mm = random(1,4); $period =$periods[$mm];$m = $ms[$mm]; $rate = random(5,10);$r = $rate/100;$t1 = random(1,6); $t2 =$t1 + random(1,6); $p = 5000*random(1,10);$P = Currency($p);$a1 = $p*(1+$r/$m)**($m*$t1);$a2 = $p*(1+$r/$m)**($m*$t2);$A1 = Currency($a1);$A2 = Currency($a2); BEGIN_TEXT; If$P is invested at $rate$PERCENT compounded $period, what is the amount after$HR (A) $t1 years?$PAR $PAR (B)$t2 years? $PAR Answer: \\{ans_rule(8)\\}$PAR (Use dollar signs for monetary values) END_TEXT ANS($A1->cmp); ANS($A2->cmp); ENDDOCUMENT(); ### Re: Correct answer marked incorrect with Currency context by Davide Cervone - Jason: Yes, that is what I'm seeing as well. The reason is that the correct answer is shown with two decimals, and", "• Payment Amount PMT = 100.00 • Payments per Period q = 12 (monthly) Using equation (8) we have \\begin{align} FV&=PV(1+\\frac{r}{m})^{mt} \\\\[0.5em] &+\\dfrac{PMT}{\\frac{r}{m}}((1+\\frac{r}{m})^{mt}-1) \\\\[0.5em] &\\times(1+(\\frac{r}{m})T) \\end{align} \\begin{align} FV&=15,000(1+0.015/12)^{12*10} \\\\[0.5em] &+\\dfrac{100}{0.015/12}((1+0.015/12)^{12*10}-1) \\\\[0.5em] &\\times(1+(0.015/12)*0) \\end{align} \\begin{align} FV&=15,000(1.00125)^{120} \\\\[0.5em] &+\\dfrac{100}{0.00125}((1.00125)^{120}-1) \\\\[0.5em] &\\times 1 \\end{align} \\begin{align} FV&=15,000(1.16173) \\\\[0.5em] &+80,000(1.16173-1) \\\\[0.5em] &\\times 1 \\end{align} \\begin{align} FV&=17,425.88 \\\\[0.5em] &+92,938.03-80,000 \\\\[0.5em] &=30,361.91 \\end{align} At the end of 10 years your savings account will be worth30,363.91 Suppose you find a bank that offers you daily compounding (365 times per year). $F V = P V ( 1 + i ) n$ Cite this content, page or calculator as: Furey, Edward \"Future Value Calculator\" at https://www.calculatorsoup.com/calculators/financial/future-value-calculator.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators", "amounts spent on the first 6 purchases. We use the dollar sign to pull out a specific column of the data and focus (only) on that column. head(myDF$SPEND) ## [1] -1.50 -1.50 2.19 0.99 2.50 4.50 These first 6 values in the SPEND column add up to a total sum of 7.18 (you can check by hand if you like!) sum(head(myDF$SPEND)) ## [1] 7.18 The average of the first 6 values in the SPEND column is 1.196667 mean(head(myDF$SPEND)) ## [1] 1.196667 The first 100 values in the SPEND column are: head(myDF$SPEND, n=100) ## [1] -1.50 -1.50 2.19 0.99 2.50 4.50 3.49 2.79 1.00 9.98 1.29 1.79 ## [13] 3.99 1.00 2.00 10.80 3.49 1.00 3.99 1.88 0.49 2.49 1.99 2.50 ## [25] 1.67 1.99 5.50 7.89 6.49 1.00 2.78 3.69 1.19 0.69 3.00 5.99 ## [37] 8.19 3.49 4.29 5.66 0.99 5.99 0.99 8.11 12.82 7.99 4.19 1.49 ## [49] 4.96 3.49 4.49 2.79 2.99 5.49 3.99 12.00 3.79 0.89 4.99 2.29 ## [61] 1.69 5.78 6.99 2.00 3.89 6.77 2.69 4.99 3.20 14.40 6.93 2.50 ## [73] 1.00 5.98 1.75 1.19 4.25 3.00 1.11 0.98 8.17 13.10 17.98 4.38 ## [85] 5.79 3.59 4.99 11.56 3.42 2.99 17.99 1.50 -0.38 3.14 2.49 3.99 ## [97] 3.39 1.49 0.53 1.25 Note that, in the line above, we have", "fine for going to lines and h-j-k-l’ing around. -John ## Line Graph of Financial Data in Google Sheets Month Notes January Notes February Net Worth \\$500 \\$600 Liabilities \\$50 \\$40 Credit Cards \\$10 \\$20 Credit Card 1 \\$5 \\$10 Credit Cards 2 \\$5 \\$10 Savings & Checking \\$400 \\$500 Investments \\$100 \\$200 I have a google spreadsheet I use for tracking monthly financials that looks something like this. I use a new tab or sheet per year. I was looking to add a Dashboard to the first tab that would display an individual line for each row. Essentially I want the row and column labels to be the same I have here but translate the numerical values into a line graph. I have attempted to select all of the cells with the respective data manually. I have tried selecting the ranges – but with my notes columns and there are detailed rows underneath each of the rows displayed here i.e. credit cards have a row each for each card, etc. So the rows and columns in the sheet are not exactly contiguous. I’m pretty sure I’ll have to manually click each item. I have also tried selecting a row with the column intended become that row’s label and the numeric values in that row as the plotted lines", "motivate the number$e, which plays a significant role in the (F-LE) domain of tasks. Note that banks report account balances to the nearest cent. However, in this task we have kept several decimal places of accuracy in the account balance to clearly show changes in the account balance beyond the second decimal place. ## Solution 1. At the end of March, the man will earn one quarter of the interest: \\begin{align} \\1000 + \\frac{5 \\%}{4} \\cdot \\1000 &= \\1000 + \\frac{0.05}{4} \\cdot \\1000 \\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right) \\\\ &= \\1012.5. \\end{align} At the end of June, the man will earn the next quarter of the interest: \\begin{align} \\1012.5 + \\frac{5 \\%}{4} \\cdot \\1012.5 &= \\1012.5 + \\frac{0.05}{4} \\cdot \\1012.5\\\\ &= \\1012.5\\left(1+\\frac{0.05}{4}\\right)\\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right)\\left(1+\\frac{0.05}{4}\\right)\\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right)^2\\\\ &=\\1025.15625. \\end{align} At the end of September, the man will earn the next quarter of the interest: \\begin{align}\\1025.15625 + \\frac{5 \\%}{4} \\cdot \\1025.15625 &= \\1025.15625 + \\frac{0.05}{4} \\cdot \\1025.15625\\\\ &= \\1025.15625\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^2\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^3\\\\ &=\\1037.970703125. \\end{align} At the end of December, the man will earn the next quarter of the interest: \\begin{align}\\1037.970703125+ \\frac{5 \\%}{4} \\cdot \\1037.970703125 &= \\1037.970703125 + \\frac{0.05}{4} \\cdot \\1037.970703125\\\\ &= \\1037.970703125\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^3\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^4\\\\ &\\approx \\1050.95.\\end{align} 2. Organizing and generalizing our previous work, we have, putting the number of times the interest is compounded in the first column and the expression showing the", "(18.3/5) ## checking_balance in {< 0 DM,1 - 200 DM}: ## :...credit_history = perfect: ## :...percent_of_income <= 3: no (16.5/5.9) ## : percent_of_income > 3: yes (6.9) ## credit_history = poor: ## :...percent_of_income <= 1: no (7.2) ## : percent_of_income > 1: ## : :...savings_balance in {< 100 DM,> 1000 DM, ## : : 500 - 1000 DM}: yes (19.2/3.9) ## : savings_balance in {100 - 500 DM,unknown}: no (12.8/2.3) ## credit_history = very good: ## :...other_credit = none: yes (10.6) ## : other_credit in {bank,store}: ## : :...months_loan_duration <= 9: yes (4.7) ## : months_loan_duration > 9: no (20.6/8.1) ## credit_history = critical: ## :...years_at_residence <= 1: no (7.5) ## : years_at_residence > 1: ## : :...savings_balance in {> 1000 DM,100 - 500 DM, ## : : unknown}: no (16.3/2.4) ## : savings_balance = 500 - 1000 DM: yes (2.8/0.5) ## : savings_balance = < 100 DM: ## : :...dependents > 1: no (12.9/0.5) ## : dependents <= 1: ## : :...other_credit = bank: yes (6.7/0.5) ## : other_credit = store: no (1.7/0.5) ## : other_credit = none: ## : :...age > 61: no (6.4) ## : age <= 61: ## : :...existing_loans_count > 2: no (3.3) ## : existing_loans_count <= 2: ## : :...job in {management,unemployed, ## : : unskilled}: yes (14.6/2.7) ## : job =", "1 0 0 # [10,] 0 0 0 0 0 0 0 0 0 0 1 0 # [11,] 0 0 0 0 0 0 0 0 0 0 0 1 # [12,] 1 0 0 0 0 0 0 0 0 0 0 0 # [13,] 0 1 0 0 0 0 0 0 0 0 0 0 # [14,] 0 0 1 0 0 0 0 0 0 0 0 0 # [15,] 0 0 0 1 0 0 0 0 0 0 0 0 For the monthly trends we can reuse monthly.means': monthly.trends <- monthly.means * seq_along(x) / S round(monthly.trends[1:15,], 2) # Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec # [1,] 0 0.08 0.00 0.00 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [2,] 0 0.00 0.17 0.00 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [3,] 0 0.00 0.00 0.25 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [4,] 0 0.00 0.00 0.00 0.33 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [5,] 0 0.00 0.00 0.00 0.00 0.42 0.0 0.00 0.00 0.00 0.00 0.00 # [6,] 0 0.00 0.00 0.00 0.00 0.00 0.5 0.00 0.00 0.00 0.00 0.00 # [7,] 0 0.00 0.00 0.00 0.00 0.00 0.0 0.58 0.00 0.00 0.00 0.00 # [8,] 0", "1-Apr-15$6,640,341.00 6 1-May-15 $5,476,163.28 these are both strings. I wanted to change Month to a date. I do mydata$Month <-as.Date(mydata$Month,format= \"%d/%m/%y\") and it works str(mydata$Month) Date[1:65], format:.... Ok I got this half solved. But now I need to change Spend which is a character into a number. The data looks like this (Spend is the 2nd variable). 1 2020-01-12 5,790,000.00 2 2020-01-01 5,114,841.08 I need to get rid of the , and the periods and columns. All of them I tried a lot of different ways, but this works. mydata$Spend<-as.numeric(gsub(\"[['punct:]]\", \"\",mydata$Spend)) without the ' inside the bracket since that creates a smiley here In about a year I will be actually able to run data.... Last edited: #### noetsi ##### No cake for spunky I was not sure if this is wrong or simply how time series formats. But there are 16405 days between 12/1/2014 and 1/1/1970 when R calculates dates. Need to learn how to get the TS function to show months and dates in standard English not numbers My original data looks like this 1 12/1/2014 5,790,000.00 2 1/1/2015 5,114,841.08 3 2/1/2015 7,240,820.54 4 3/1/2015 7,482,837.26 5 4/1/2015 6,640,341.00 6 5/1/2015 5,476,163.28 I did this to make it a date mydata$Month <-as.Date(mydata$Month,format= \"%m/%d/%Y\") Not sure why it makes it year first then month. Or if there", "rise (%) over 21 years value after 21 years 0 20,790 43,111 1 23,007 46,734 2 25,526 50,791 Broadly speaking, this investment doubles the value of the contributions over two decades. Note: Rebalancing fees are not included in the simulation. Below is the code used to run the simulation. If you have Mathematica you can try with different funds. funds = {\"VTSMX\", \"VGTSX\", \"VBMFX\"}; (* Plotting the fund indices *) {tsm, ism, tbm} = FinancialData[#, {\"April 29, 1996\", DateList[], \"Month\"}] & /@ funds; DateListPlot[ Transpose[{First /@ #, 100 Last /@ #/#[[1, 2]]}] & /@ {tsm, ism, tbm}, PlotLegends -> funds, PlotLabel -> \"Indices from month-end April 1996 rebased to 100\"] (* Plotting the investment contributions *) salary = 550; investment = salary*0.15; inflation = 2; nmonths = Length[tsm] - 1; ny = Quotient[nmonths, 12]; iy = Array[investment/3 (1 + inflation/100)^(# - 1) &, ny]; d = Take[Flatten[ConstantArray[#, 12] & /@ iy], 12 ny]; DateListPlot[Transpose[{Take[First /@ tsm, 12 ny], 3 d}], PlotLabel -> Row[{\"Monthly contributions with \", inflation, \"% inflation - Total = \", Total[3 d]}], PlotRange -> {Automatic, {0, Automatic}}, PlotMarkers -> {Automatic, 6}, FrameLabel -> {\"Time\", Rotate[Style[\"$\", 12], Pi/2]}, ImageSize -> 380] (* Calculating & plotting the investment values *) {tsm2, ism2, tbm2} = Take[Ratios@# - 1, 12 ny] & /@ Map[Last, {tsm, ism, tbm}, {2}]; d2", "row in the other dataframe, that correponds to the balance in that account on that day. So the result in the toy example would be this: df.result <- data.frame(date=as.Date(paste0('2018-12-',c(11,15,16,18,22,25,27,29))), balance.salary=c(-500,-250,-250,0,250,-300,-300,500), balance.budget=c(1000,1000,1000,1000,700,700,250,250)) Notice how even though I don’t have information for the budget-account from the first date that the salay-account has a row, I’m using the information from the first time there is a row from the budget account. here I have changed the column names for the balance-variable, so that one row can have the balance for both, but this is not the essential part of the solution, only that the result can be computed like this: df.result$balance.total <- df.result$ balance.salary + df.result\\$ balance.budget I have tried using crossing() as per this answer, Copying row from one df into everyone row in another, but isn’t useful in this case, as far as I can tell. Thank you.", "number of dimes in the collection?... Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4); Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4);... Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!! Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!!...", "-0.364)$); %bridge middle right \\filldraw[bridge] ($(0, 1) + (0.658, -0.590)$) -- ($(0, 1) + (0.668, -0.546)$) -- ($(0, 1) + (0.778, -0.524)$) -- ($(0, 1) + (0.792, -0.554)$) -- ($(0, 1) + (0.796, -0.590)$) --cycle; \\draw[contour] ($(0, 1) + (0.668, -0.546)$) -- ($(0, 1) + (0.778, -0.524)$) ($(0, 1) + (0.658, -0.590)$) -- ($(0, 1) + (0.796, -0.590)$); %bridge bottom 1 \\filldraw[bridge] ($(0, 1) + (0.350, -0.826)$) -- ($(0, 1) + (0.410, -0.804)$) -- ($(0, 1) + (0.422, -0.794)$) -- ($(0, 1) + (0.412, -0.708)$) -- ($(0, 1) + (0.372, -0.702)$) -- ($(0, 1) + (0.336, -0.695)$) -- cycle; \\draw[contour] ($(0, 1) + (0.422, -0.794)$) -- ($(0, 1) + (0.412, -0.708)$) ($(0, 1) + (0.350, -0.826)$) -- ($(0, 1) + (0.336, -0.695)$); %bridge bottom 2 \\filldraw[bridge] ($(0, 1) + (0.596, -0.700)$) -- ($(0, 1) + (0.535, -0.708)$) -- ($(0, 1) + (0.523, -0.784)$) -- ($(0, 1) + (0.581, -0.777)$) -- cycle; \\draw[contour] ($(0, 1) + (0.535, -0.708)$) -- ($(0, 1) + (0.523, -0.784)$) ($(0, 1) + (0.596, -0.700)$) -- ($(0, 1) + (0.581, -0.777)$); %bridge bottom 3 \\filldraw[bridge] ($(0, 1) + (0.581, -0.777)$) -- ($(0, 1) + (0.660, -0.862)$) -- ($(0, 1) + (0.858, -0.717)$) -- ($(0, 1) + (0.837, -0.702)$) -- ($(0, 1) + (0.826, -0.688)$) -- ($(0, 1) + (0.785, -0.668)$) -- cycle; \\draw[contour] ($(0, 1)", ":...savings_balance in {> 1000 DM,100 - 500 DM,unknown}: no (23.6/5.9) ## savings_balance = 500 - 1000 DM: yes (6.2/1.6) ## savings_balance = < 100 DM: ## :...dependents > 1: no (12.2/1.1) ## dependents <= 1: ## :...age > 61: no (5.7) ## age <= 61: ## :...years_at_residence <= 1: no (6.1/1) ## years_at_residence > 1: ## :...other_credit in {bank,store}: yes (8.3/2.1) ## other_credit = none: ## :...amount > 5998: yes (7) ## amount <= 5998: ## :...existing_loans_count <= 1: no (11.6/3.9) ## existing_loans_count > 1: ## :...existing_loans_count > 2: no (2.7) ## existing_loans_count <= 2: ## :...age > 54: yes (3.9) ## age <= 54: ## :...age > 44: no (5) ## age <= 44: ## :...amount <= 5324: yes (30.9/10.8) ## amount > 5324: no (2.7) ## ## ----- Trial 5: ----- ## ## Decision tree: ## ## checking_balance = unknown: ## :...other_credit = store: no (17.3/6.3) ## : other_credit = bank: ## : :...purpose = business: yes (8.9/4.1) ## : : purpose in {car0,education,renovations}: no (7.2/1.8) ## : : purpose = car: ## : : :...credit_history in {critical,perfect,poor,very good}: yes (17.2/4.4) ## : : : credit_history = good: no (3.5) ## : : purpose = furniture/appliances: ## : : :...months_loan_duration <= 13: yes (6.5/1.6) ## : : months_loan_duration > 13: no (17.4/2.5) ## : other_credit =", "833,333.33 98,708,628.86 2 98,708,628.86 2,124,704.47 822,571.91 97,406,496.30 3 97,406,496.30 2,124,704.47 811,720.80 96,093,512.63 4 96,093,512.63 2,124,704.47 800,779.27 94,769,587.43 5 94,769,587.43 2,124,704.47 789,746.56 93,434,629.52 56 10,363,013.71 2,124,704.47 86,358.45 8,324,667.69 57 8,324,667.69 2,124,704.47 69,372.23 6,269,335.45 58 6,269,335.45 2,124,704.47 52,244.46 4,196,875.44 59 4,196,875.44 2,124,704.47 34,973.96 2,107,144.93 60 2,107,144.93 2,124,704.47 17,559.54 -0.00 New bond: Initial Balance Payment Interest Final Balance Month 1 100,000,000.00 2,075,835.52 750,000.00 98,674,164.48 2 98,674,164.48 2,075,835.52 740,056.23 97,338,385.19 3 97,338,385.19 2,075,835.52 730,037.89 95,992,587.55 4 95,992,587.55 2,075,835.52 719,944.41 94,636,696.44 5 94,636,696.44 2,075,835.52 709,775.22 93,270,636.14 56 10,149,672.43 2,075,835.52 76,122.54 8,149,959.45 57 8,149,959.45 2,075,835.52 61,124.70 6,135,248.63 58 6,135,248.63 2,075,835.52 46,014.36 4,105,427.47 59 4,105,427.47 2,075,835.52 30,790.71 2,060,382.65 60 2,060,382.65 2,075,835.52 15,452.87 0.00 Now we calculate the NPV of refi. We make it a function so that we can ask questions about parameters later. In [4]: def npv_refi(rnew, term, bnow, rnow, penalty): # rnew, term, bnow, rnow, penalty, here, are dummy names '''This function returns the npv of refi given rnew: new rate term: remaining term on the bond bnow: remaining principal rnow: current rate penalty: prepayment fee as a fraction of principal''' oldpayment=npf.pmt(rnow/12,term,-bnow) newpayment=npf.pmt(rnew/12,term,-bnow) return npf.pv(rnew/12,term,-(oldpayment-newpayment))-penalty*bnow print(\"The NPV of refi is: \" +put_comma(npv_refi(rnew, term, bnow, rnow, penalty))) The NPV of refi is: 354,182.11 Now we can ask all kinds of intersting questions. For instance, how low must the new rate be to justify refinancing?" ]

[ "## The table below shows the monthly balances a person had on their credit card over a five-month period. Month Balance Janua Question The table below shows the monthly balances a person had on their credit card over a five-month period. Month Balance January -$50 February$25 March -$70 April -$45 May $25 What was the average monthly balance over the five-month period? Enter your answers and your explanations or work in the space provided. in progress 0 2 months 2021-07-24T17:43:07+00:00 1 Answers 0 views 0 ## Answers ( ) 1. Answer: average monthly balance =$43 Step-by-step explanation: If we have a set of N numbers: {x₁, x₂, x₃, …, xₙ} The mean (or average) is calculated as: Mean = (x₁ + x₂ + … + xₙ)/N In this case, the values on the set are the balances for every month. Then our set is: {$50,$25, $70,$45, $25} This is a set of 5 elements. Then the average monthly balance is: average monthly balance = ($50 + $25 +$70 + $45 +$25)/5 = $43 average monthly balance =$43", "## Intermediate Algebra (6th Edition) First, we can find the number of people who have an average monthly balance of 0 to 100 dollars by finding the height associated with the bar that represents 0 to 100 dollars. The height of this bar is 5. To find the percentage of people polled who have an average monthly balance of 0 to 100 dollars, we must divide this number by 30 (which represents the total amount of people polled). $\\frac{5}{30}=\\frac{1}{6}\\approx$16.7%", "## Intermediate Algebra (6th Edition) First, we can find the number of people who have an average monthly balance of 201 to 300 dollars by finding the height associated with the bar that represents 201 to 300 dollars. The height of this bar is 5. To find the percentage of people polled who have an average monthly balance of 201 to 300 dollars, we must divide this number by 30 (which represents the total amount of people polled). $\\frac{5}{30}=\\frac{1}{6}\\approx$16.7%", "## Intermediate Algebra (6th Edition) We can find this number by finding the sum of the heights of the bars that represent the number of people who have an average monthly balance of 0 to 100 dollars and 101 to 200 dollars. The heights of these bars are 5 and 9, respectively, so we know that $5+9=14$ people have an average monthly balance of less than 200 dollars.", "average daily balance for the 30-day billing cycle is the average (arithmetic mea by deloitte247 » Tue Mar 23, 2021 2:36 pm ## Timer 00:00 ## Your Answer A B C D E ## Global Stats $$Average\\ balance\\ at\\ the\\ end\\ of\\ 30\\ days=\\frac{sum\\ of\\ daily\\ balance}{30\\ days}$$ Jane's credit card account had a balance of$600, and Jand made a payment of $300 during the cycle. So, we can say the balance on the credit card account is either$600 or $300 and once it decreases to$300, it stays at $300 for the rest of the month. For the sum of the daily balance, we have to figure out when Jane made the$300 payment to estimate the sum of daily balances. Statement 1: Jane's payment was credited on the 21st day of the billing cycle. The account had a $600 balance for 20 days starting from the 21 daily balance became$300 for the next 30 days. Sum of daily balance = (20 * 600) + (10 * 300) = 12000 + 3000 = 15,000 $$Average\\ balance\\ at\\ the\\ end\\ of\\ 30\\ days=\\frac{15000}{30}=500$$ Statement 1 is SUFFICIENT Statement 2: The average daily balance through the 25th day of the billing cycle was $540. Let the number of days for$600 balance = x no. of days for $300 balance = 25", "## Intermediate Algebra (6th Edition) We can find this number by finding the sum of the heights of the bars that represent the number of people who have an average monthly balance of 301 to 400 dollars, 401 to 500 dollars and over 500 dollars. The heights of these bars are 6, 2, and 3, respectively, so we know that $6+2+3=11$ people have an average monthly balance of 301 dollars or more.", "of the monthly proportions, as a percent, we take $\\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\\%$, which is wildly different from the true average of $20\\%$. For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth. - Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39", "card account, the average daily balance for 30-day bi [#permalink] ### Show Tags 02 Sep 2009, 11:30 Economist wrote: maliyeci wrote: 1)suff. balance of first 20 days are 600, last 10 days are 300. We can find average. 2)suff. balance of first 25 days are 450, last 5 days are 300. We can find average. D. for stmt2 we cannot say that last 5 days the balance is 300. Since the average balance is less than 600 for the first 25 days, the 300 credit has already been made on one of the days from 1-25. However, if we consider the average for the first 25 days to be 450 then we can find the exact day on which the credit of 300 was made using: (n-1)*600 + [ 25-(n-1)]*300 / 25 = 450, where n is the day on which 300 was credited no need to calculate, but once we get this exact day we can calculate the monthly average since for n days balance was 600 and rest 30-n days the balance was 300...just like in stmt 1 I agree with the answer D. However, I think you can make it easier by calculating the balance directly: $$\\frac{25\\times 450+5\\times 300}{30}=425$$ But this value (425) contradicts the balance of 500 from stmt 1. I think the balance", "the balance of 500 from stmt 1. I think the balance in statement 2 is wrong...it should be 540...that is the only value for the first 25 days that will give the average 30 day balance of 500. I think you have assumed that for the last 5 days the balance was 300. This is only possible if on the 25th day 300 was credited into the account. At the beginning of the month bal was 600, on 25th day the average balance reduced, which means that the 300 credit has taken place in the first 25 days (not necessarily on the 25th day)...if it was on the first day itself then average will be less, if it is on 24th day for example then average will be higher. So in any case we need to find the exact day on which 600 became 600-300=300. Anyhow, the answer is D. Manager Joined: 10 Aug 2009 Posts: 130 Re: On Jane's credit card account, the average daily balance for 30-day bi [#permalink] Show Tags 02 Sep 2009, 12:22 1 This post received KUDOS I did not assume that. I assumed that the charge ocurred during the first 25 days If you denote $$x_i$$ to be the balance on the day i, i={1,2,...30}, $$\\frac{\\sum_{i=1}^{25} x_i}{25}=450$$ $$\\sum_{i=1}^{25} x_i=450\\times 25$$ $$\\frac{\\sum_{i=1}^{30} x_i}{30}=\\frac{\\sum_{i=1}^{25}", "percent (%) difference between multi-year monthly average value, $<\\text{SRAD}>_{pj}$, and the maximum value of $<\\text{SRAD}>_{pj}$ over the time period. The following summarizes the procedure for calculating the multi-year monthly average value of $\\text{Max}/\\text{p-day}$: $\\normalsize{ <\\text{SRADmax}>^p_{j} = MAX(<\\text{SRAD}>^p_{j})}$", "solving gives: LRR = (4000-3500) / 4000* 100 = 12.5 (%/month)", "the first x days the balance will be$600 and then for (30-x) days the balance will be $600-$300=$300 (meaning that Jane made a payment of$300 on the (x+1)st day). So all we need to know is on which day did Jane make the payment of $300, the value of x+1. (1) Jane's payment was credited on the 21st day of the billing cycle --> directly gives the value of x+1 --> x+1=21 --> x=20: for 20 days the balance was$600 and for the rest 10 days (from 21st) it was $300 --> we can find the average daily balance. Sufficient. (2) The average daily balance through the 25th day of the billing cycle was$540 --> average_{25 \\ days}=540=\\frac{x*600+(25-x)*300}{25} --> we can find x --> we can find the average daily balance. Sufficient. Bunuel, what is the purpose of the first sentence in the question. Has it been mentioned to simply confuse or am I missing something? Re: On Jane's credit card account, the average daily balance for [#permalink] 28 Nov 2013, 04:55 Similar topics Replies Last post Similar Topics: On Jane's credit card account, the average daily balance for 1 15 Dec 2008, 21:03 On jane's credit card account, the average daily balance for 8 14 Jul 2008, 07:48 On Jane's credit card account, the average daily balance", "We can find average. D. for stmt2 we cannot say that last 5 days the balance is 300. Since the average balance is less than 600 for the first 25 days, the 300 credit has already been made on one of the days from 1-25. However, if we consider the average for the first 25 days to be 450 then we can find the exact day on which the credit of 300 was made using: (n-1)*600 + [ 25-(n-1)]*300 / 25 = 450, where n is the day on which 300 was credited no need to calculate, but once we get this exact day we can calculate the monthly average since for n days balance was 600 and rest 30-n days the balance was 300...just like in stmt 1 I agree with the answer D. However, I think you can make it easier by calculating the balance directly: \\frac{25\\times 450+5\\times 300}{30}=425 But this value (425) contradicts the balance of 500 from stmt 1. I think the balance in statement 2 is wrong...it should be 540...that is the only value for the first 25 days that will give the average 30 day balance of 500. I think you have assumed that for the last 5 days the balance was 300. This is only possible if on the 25th day", "we can calculate the monthly average since for n days balance was 600 and rest 30-n days the balance was 300...just like in stmt 1 I agree with the answer D. However, I think you can make it easier by calculating the balance directly: $$\\frac{25\\times 450+5\\times 300}{30}=425$$ But this value (425) contradicts the balance of 500 from stmt 1. I think the balance in statement 2 is wrong...it should be 540...that is the only value for the first 25 days that will give the average 30 day balance of 500. I think you have assumed that for the last 5 days the balance was 300. This is only possible if on the 25th day 300 was credited into the account. At the beginning of the month bal was 600, on 25th day the average balance reduced, which means that the 300 credit has taken place in the first 25 days (not necessarily on the 25th day)...if it was on the first day itself then average will be less, if it is on 24th day for example then average will be higher. So in any case we need to find the exact day on which 600 became 600-300=300. Anyhow, the answer is D. Kudos [?]: 843 [0], given: 18 Manager Joined: 10 Aug 2009 Posts: 129 Kudos [?]: 74", "average daily balance on Jane's account for the billing cycle? (1) Jane's payment was credited on the 21st day of the billing cycle. (2) The average daily balance through the 25th day of the billing cycle was$540. Question basically ask on which day did Jane make the payment of $300. Or algebraically: average \\ daily \\ balance=\\frac{x*600+(30-x)*300}{30} --> for the first x days the balance will be$600 and then for (30-x) days the balance will be $600-$300=$300 (meaning that Jane made a payment of$300 on the (x+1)st day). So all we need to know is on which day did Jane make the payment of $300, the value of x+1. (1) Jane's payment was credited on the 21st day of the billing cycle --> directly gives the value of x+1 --> x+1=21 --> x=20: for 20 days the balance was$600 and for the rest 10 days (from 21st) it was $300 --> we can find the average daily balance. Sufficient. (2) The average daily balance through the 25th day of the billing cycle was$540 --> average_{25 \\ days}=540=\\frac{x*600+(25-x)*300}{25} --> we can find x --> we can find the average daily balance. Sufficient. _________________ Manager Status: UC Berkeley 2012 Joined: 03 Jul 2010 Posts: 191 Location: United States Concentration: Economics, Finance GPA: 4 WE: Consulting (Investment Banking) Followers: 3 Kudos", "• SDB is the sum of daily balances ($) • D is the number of days in the period To calculate Average Daily Balance, divide the sum of all daily balances by the number of days in the period. ## How to Calculate Average Daily Balance? The following steps outline how to calculate the Average Daily Balance. 1. First, determine the sum of daily balances ($). 2. Next, determine the number of days in the period. 3. Next, gather the formula from above = ADB = SDB / #D. 4. Finally, calculate the Average Daily Balance. 5. After inserting the variables and calculating the result, check your answer with the calculator above. Example Problem : Use the following variables as an example problem to test your knowledge. sum of daily balances (\\$) = 500,000 number of days in the period = 30", "way to calculate: 410 for all 52 installments. So, average = $$\\frac{410*52}{52}$$ = 410 For the 32 installments, 65 is more for each. So, average= $$\\frac{32*65}{52}$$ = 40 Therefore, 410+40 = 450 Re: A certain debt will be paid in 52 installments from January 1 to Decem &nbs [#permalink] 24 Feb 2018, 07:32 Display posts from previous: Sort by", "## Algebra: A Combined Approach (4th Edition) People with an average checking balance between 0 and 100 is 5. People with an average checking balance between 101 and 200 is 9. People with an average checking balance between 201 and 300 is 5. People with an average checking balance between 301 and 400 is 6. People with an average checking balance between 401 and 500 is 2. People with an average checking balance of over 500= 3. Total number of people polled=30. % of people with an average checking balance between 201 and 300 is $\\frac{5}{30}\\times100=16.7$%.", "payment of$300, the value of x+1. (1) Jane's payment was credited on the 21st day of the billing cycle --> directly gives the value of $$x+1$$ --> $$x+1=21$$ --> $$x=20$$: for 20 days the balance was $600 and for the rest 10 days (from 21st) it was$300 --> we can find the average daily balance. Sufficient. (2) The average daily balance through the 25th day of the billing cycle was \\$540 --> $$average_{25 \\ days}=540=\\frac{x*600+(25-x)*300}{25}$$ --> we can find $$x$$ --> we can find the average daily balance. Sufficient. OPEN DISCUSSION OF THIS QUESTION IS HERE: on-jane-s-credit-card-account-the-average-daily-balance-for-106680.html _________________ Kudos [?]: 128723 [1], given: 12182 Re: On Jane's credit card account, the average daily balance for 30-day bi [#permalink] 27 Sep 2014, 05:52 Display posts from previous: Sort by", "# Should you average or add together daily bank balanace averages? I work in the finance field and I am looking at average balances on bank statements balances. When I have more than 1 bank statement I need to reflect the overall average balance between the accounts. I am not clear on whether the average balance from each account should be added together or averaged. Avg. Bal Bank Acct #1: \\$14,973 Avg. Bal Bank Acct #2: \\$8,652 Total: \\$23,625 - Should this be used for the average balances or \\$23,625 /2 = \\$11,813 ? • What does the average balance from one account mean? Is it the average over a month? – ShreevatsaR Jan 10 '14 at 19:33 ## 3 Answers You can multiply them, too (though what square dollars mean, I don't know). Probably you want to add them together, which will give the average of the total of the accounts over the month. This is the average amount of money you had in total. • Nice One!! The financial world is full of numbers like this , In the name of reconciliation from different systems of data, I go crazy every day. Thanks for your humor. – Satish Ramanathan Jan 10 '14 at 20:28 • +1: it's impressive when you can find a cartoon which hits" ]

[ "By observations, we circle all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student", "all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57", "you missing the first four years of data? The series in Figure 7.1 in the book seems close to this series of personal saving as a percentage of disposable personal income for the period 1955-1979. – javlacalle Jun 5 '15 at 21:11 • Thanks, javlacalle. I modified the question after using your more complete data set. – StatSmartWannaB Jun 5 '15 at 22:33 This is what I get fitting the following model in R: $$y_t - \\mu = \\phi (y_{t-1} - \\mu) + \\epsilon_t + \\theta \\epsilon_{t-2} \\,,$$ x <- structure(c(9.3, 9.5, 10, 9.9, 10.6, 11.1, 11.3, 11.5, 11.1, 11.6, 11.4, 10.8, 11.3, 11.2, 11.6, 11.5, 10.7, 10.7, 9.6, 10.1, 10.3, 9.7, 10.2, 10, 10.9, 10.9, 11.7, 11.6, 11.5, 11.3, 11, 10.6, 10.7, 10.6, 10.3, 11, 11, 11.8, 11.3, 12, 11.1, 11.1, 11.9, 11.3, 10.9, 10.9, 10.9, 11.6, 12.3, 11.8, 12.2, 12.4, 11.8, 11.9, 10.5, 10.6, 9.8, 10.1, 11.6, 11.5, 11.8, 12.5, 13.1, 13.1, 13.2, 13.6, 13.4, 12.9, 12.2, 11.4, 11.7, 13.2, 12.2, 13, 13, 14.1, 13.6, 12.5, 12.2, 13.3, 12.4, 15, 12.5, 12.3, 11.7, 11.3, 11.1, 10.4, 9.4, 10, 10.5, 10.8, 10.9, 9.9, 10.1, 10.1, 10.6, 9.8, 9.4, 9.5), .Dim = c(100L, 1L), .Dimnames = list(NULL, \"V1\"), .Tsp = c(1959, 1983.75, 4), class = \"ts\") arima(x, order=c(1,0,2), include.mean=TRUE, fixed=c(NA,0,NA,NA)) # Coefficients: # ar1 ma1 ma2 intercept #", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Because Steve only wants to see if the monthly payment is above or below400, he can estimate the answer by dividing the leading digits. First, he should identify the two leading digits of each number. • The leading digits of 21,045.68 are 21,000 (remember that you have to maintain the placement of the decimal point!) • The leading digits of 48 are 48 Now, note that 21,000 is your dividend and 48 is your divisor. Because your divisor is already a whole number, you do not need to move any decimal points. Now, divide as usual using long division. 48)21000.0¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 437.5 192 180 144 360 336 240 240 0\\begin{align*}\\begin{array}{rcl} && \\overset{ \\ \\ \\ \\ \\ \\ \\ \\ 437.5}{48 \\overline{ ) {21000.0 \\;}}}\\\\ && \\underline{\\ - 192 \\;\\;\\;\\;\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ 180 \\\\ && \\underline{\\ \\ \\ - 144 \\;\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ 360 \\\\ && \\underline{\\ \\ \\ \\ \\ - 336\\;\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ \\ \\ 240 \\\\ && \\underline{\\ \\ \\ \\ \\ \\ \\ \\ - 240\\;\\;}\\\\ && \\quad \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0 \\\\ \\end{array}\\end{align*} The answer is that the monthly payments will be approximately $437.50. Steve’s dad shouldn’t buy", "the original datapoints and the calculated ones figure hold on plot(t, x); plot(t, y); plot(t, xv, 'k--', 'linewidth', 3); plot(t, yv, 'g--', 'linewidth', 3); grid on xlabel('Time (s)'); ylabel('Position'); legend('X-position from Q5', 'Y-position from Q5', 'X-position from Q6', 'Y-position from Q6'); %% Question 8 % Plot the trajectory figure plot(xv, yv), grid on xlabel('X-Position (m)'); ylabel('Y-Position (m)'); %% Question 9 % Write a new file file = fopen('output.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); for k = 1:length(t) fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t(k),xv(k), yv(k)); end fclose(file); %% Question 10 % Write a new file file = fopen('output2.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t',xv', yv'); fclose(file); clc, clear all, close all P = 0; % initial money in the account deposit_per_year = [288, 345, 355, 382, 392]; % amounts deposited each month, per year (year 1, year2, ... year5) withdraw = 2331; % amount to be withdrawn each year r = 0.04; % interes of the saving account rcd = 0.08; % interest of the CD n_CD = 0; % number of CDs bought year = 1; deposited = 0; for i = 1:5*12 % 5 years, 12 months per year: total 60 months % if i < 60 % deposited = deposit_per_year(year); % else % deposited = deposit_per_year(5); % end deposited = deposit_per_year(year); P = P*(1+r); P = P", "### Re: Correct answer marked incorrect with Currency context by Jason Aubrey - Here's what webwork tells me$5253.12 $5253.12 incorrect$5798.47 incorrect Yes, you're right it marks $5253.12 as incorrect, but it's also telling me this is the correct answer (which I believed). Is this what you see when you try this, or am I missing something? ## http://wwrk.maa.org/moodle/mod/forum/discuss.php?d=459 ## seed 3515 ######################################################################## DOCUMENT(); loadMacros( \"PGstandard.pl\", # Standard macros for PG language \"MathObjects.pl\", \"contextCurrency.pl\", #\"source.pl\", # allows code to be displayed on certain sites. #\"PGcourse.pl\", # Customization file for the course ); # Show which answers are correct and which ones are incorrect$showPartialCorrectAnswers = 1; ############################################################## # # Setup # # TEXT(&beginproblem); Context(\"Currency\"); $showPartialCorrectAnswers = 1; @periods = (\"annually\",\"semiannually\",\"quarterly\",\"monthly\",\"weekly\",\"daily\"); @ms = (1,2,4,12,52,365);$mm = random(1,4); $period =$periods[$mm];$m = $ms[$mm]; $rate = random(5,10);$r = $rate/100;$t1 = random(1,6); $t2 =$t1 + random(1,6); $p = 5000*random(1,10);$P = Currency($p);$a1 = $p*(1+$r/$m)**($m*$t1);$a2 = $p*(1+$r/$m)**($m*$t2);$A1 = Currency($a1);$A2 = Currency($a2); BEGIN_TEXT; If$P is invested at $rate$PERCENT compounded $period, what is the amount after$HR (A) $t1 years?$PAR $PAR (B)$t2 years? $PAR Answer: \\{ans_rule(8)\\}$PAR (Use dollar signs for monetary values) END_TEXT ANS($A1->cmp); ANS($A2->cmp); ENDDOCUMENT(); ### Re: Correct answer marked incorrect with Currency context by Davide Cervone - Jason: Yes, that is what I'm seeing as well. The reason is that the correct answer is shown with two decimals, and", "• Payment Amount PMT = 100.00 • Payments per Period q = 12 (monthly) Using equation (8) we have \\begin{align} FV&=PV(1+\\frac{r}{m})^{mt} \\\\[0.5em] &+\\dfrac{PMT}{\\frac{r}{m}}((1+\\frac{r}{m})^{mt}-1) \\\\[0.5em] &\\times(1+(\\frac{r}{m})T) \\end{align} \\begin{align} FV&=15,000(1+0.015/12)^{12*10} \\\\[0.5em] &+\\dfrac{100}{0.015/12}((1+0.015/12)^{12*10}-1) \\\\[0.5em] &\\times(1+(0.015/12)*0) \\end{align} \\begin{align} FV&=15,000(1.00125)^{120} \\\\[0.5em] &+\\dfrac{100}{0.00125}((1.00125)^{120}-1) \\\\[0.5em] &\\times 1 \\end{align} \\begin{align} FV&=15,000(1.16173) \\\\[0.5em] &+80,000(1.16173-1) \\\\[0.5em] &\\times 1 \\end{align} \\begin{align} FV&=17,425.88 \\\\[0.5em] &+92,938.03-80,000 \\\\[0.5em] &=30,361.91 \\end{align} At the end of 10 years your savings account will be worth30,363.91 Suppose you find a bank that offers you daily compounding (365 times per year). $F V = P V ( 1 + i ) n$ Cite this content, page or calculator as: Furey, Edward \"Future Value Calculator\" at https://www.calculatorsoup.com/calculators/financial/future-value-calculator.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators", "amounts spent on the first 6 purchases. We use the dollar sign to pull out a specific column of the data and focus (only) on that column. head(myDF$SPEND) ## [1] -1.50 -1.50 2.19 0.99 2.50 4.50 These first 6 values in the SPEND column add up to a total sum of 7.18 (you can check by hand if you like!) sum(head(myDF$SPEND)) ## [1] 7.18 The average of the first 6 values in the SPEND column is 1.196667 mean(head(myDF$SPEND)) ## [1] 1.196667 The first 100 values in the SPEND column are: head(myDF$SPEND, n=100) ## [1] -1.50 -1.50 2.19 0.99 2.50 4.50 3.49 2.79 1.00 9.98 1.29 1.79 ## [13] 3.99 1.00 2.00 10.80 3.49 1.00 3.99 1.88 0.49 2.49 1.99 2.50 ## [25] 1.67 1.99 5.50 7.89 6.49 1.00 2.78 3.69 1.19 0.69 3.00 5.99 ## [37] 8.19 3.49 4.29 5.66 0.99 5.99 0.99 8.11 12.82 7.99 4.19 1.49 ## [49] 4.96 3.49 4.49 2.79 2.99 5.49 3.99 12.00 3.79 0.89 4.99 2.29 ## [61] 1.69 5.78 6.99 2.00 3.89 6.77 2.69 4.99 3.20 14.40 6.93 2.50 ## [73] 1.00 5.98 1.75 1.19 4.25 3.00 1.11 0.98 8.17 13.10 17.98 4.38 ## [85] 5.79 3.59 4.99 11.56 3.42 2.99 17.99 1.50 -0.38 3.14 2.49 3.99 ## [97] 3.39 1.49 0.53 1.25 Note that, in the line above, we have", "fine for going to lines and h-j-k-l’ing around. -John ## Line Graph of Financial Data in Google Sheets Month Notes January Notes February Net Worth \\$500 \\$600 Liabilities \\$50 \\$40 Credit Cards \\$10 \\$20 Credit Card 1 \\$5 \\$10 Credit Cards 2 \\$5 \\$10 Savings & Checking \\$400 \\$500 Investments \\$100 \\$200 I have a google spreadsheet I use for tracking monthly financials that looks something like this. I use a new tab or sheet per year. I was looking to add a Dashboard to the first tab that would display an individual line for each row. Essentially I want the row and column labels to be the same I have here but translate the numerical values into a line graph. I have attempted to select all of the cells with the respective data manually. I have tried selecting the ranges – but with my notes columns and there are detailed rows underneath each of the rows displayed here i.e. credit cards have a row each for each card, etc. So the rows and columns in the sheet are not exactly contiguous. I’m pretty sure I’ll have to manually click each item. I have also tried selecting a row with the column intended become that row’s label and the numeric values in that row as the plotted lines", "motivate the number$e, which plays a significant role in the (F-LE) domain of tasks. Note that banks report account balances to the nearest cent. However, in this task we have kept several decimal places of accuracy in the account balance to clearly show changes in the account balance beyond the second decimal place. ## Solution 1. At the end of March, the man will earn one quarter of the interest: \\begin{align} \\1000 + \\frac{5 \\%}{4} \\cdot \\1000 &= \\1000 + \\frac{0.05}{4} \\cdot \\1000 \\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right) \\\\ &= \\1012.5. \\end{align} At the end of June, the man will earn the next quarter of the interest: \\begin{align} \\1012.5 + \\frac{5 \\%}{4} \\cdot \\1012.5 &= \\1012.5 + \\frac{0.05}{4} \\cdot \\1012.5\\\\ &= \\1012.5\\left(1+\\frac{0.05}{4}\\right)\\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right)\\left(1+\\frac{0.05}{4}\\right)\\\\ &= \\1000\\left(1+\\frac{0.05}{4}\\right)^2\\\\ &=\\1025.15625. \\end{align} At the end of September, the man will earn the next quarter of the interest: \\begin{align}\\1025.15625 + \\frac{5 \\%}{4} \\cdot \\1025.15625 &= \\1025.15625 + \\frac{0.05}{4} \\cdot \\1025.15625\\\\ &= \\1025.15625\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^2\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^3\\\\ &=\\1037.970703125. \\end{align} At the end of December, the man will earn the next quarter of the interest: \\begin{align}\\1037.970703125+ \\frac{5 \\%}{4} \\cdot \\1037.970703125 &= \\1037.970703125 + \\frac{0.05}{4} \\cdot \\1037.970703125\\\\ &= \\1037.970703125\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^3\\left(1+\\frac{0.05}{4}\\right)\\\\ &=\\1000\\left(1+\\frac{0.05}{4}\\right)^4\\\\ &\\approx \\1050.95.\\end{align} 2. Organizing and generalizing our previous work, we have, putting the number of times the interest is compounded in the first column and the expression showing the", "(18.3/5) ## checking_balance in {< 0 DM,1 - 200 DM}: ## :...credit_history = perfect: ## :...percent_of_income <= 3: no (16.5/5.9) ## : percent_of_income > 3: yes (6.9) ## credit_history = poor: ## :...percent_of_income <= 1: no (7.2) ## : percent_of_income > 1: ## : :...savings_balance in {< 100 DM,> 1000 DM, ## : : 500 - 1000 DM}: yes (19.2/3.9) ## : savings_balance in {100 - 500 DM,unknown}: no (12.8/2.3) ## credit_history = very good: ## :...other_credit = none: yes (10.6) ## : other_credit in {bank,store}: ## : :...months_loan_duration <= 9: yes (4.7) ## : months_loan_duration > 9: no (20.6/8.1) ## credit_history = critical: ## :...years_at_residence <= 1: no (7.5) ## : years_at_residence > 1: ## : :...savings_balance in {> 1000 DM,100 - 500 DM, ## : : unknown}: no (16.3/2.4) ## : savings_balance = 500 - 1000 DM: yes (2.8/0.5) ## : savings_balance = < 100 DM: ## : :...dependents > 1: no (12.9/0.5) ## : dependents <= 1: ## : :...other_credit = bank: yes (6.7/0.5) ## : other_credit = store: no (1.7/0.5) ## : other_credit = none: ## : :...age > 61: no (6.4) ## : age <= 61: ## : :...existing_loans_count > 2: no (3.3) ## : existing_loans_count <= 2: ## : :...job in {management,unemployed, ## : : unskilled}: yes (14.6/2.7) ## : job =", "1 0 0 # [10,] 0 0 0 0 0 0 0 0 0 0 1 0 # [11,] 0 0 0 0 0 0 0 0 0 0 0 1 # [12,] 1 0 0 0 0 0 0 0 0 0 0 0 # [13,] 0 1 0 0 0 0 0 0 0 0 0 0 # [14,] 0 0 1 0 0 0 0 0 0 0 0 0 # [15,] 0 0 0 1 0 0 0 0 0 0 0 0 For the monthly trends we can reuse monthly.means': monthly.trends <- monthly.means * seq_along(x) / S round(monthly.trends[1:15,], 2) # Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec # [1,] 0 0.08 0.00 0.00 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [2,] 0 0.00 0.17 0.00 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [3,] 0 0.00 0.00 0.25 0.00 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [4,] 0 0.00 0.00 0.00 0.33 0.00 0.0 0.00 0.00 0.00 0.00 0.00 # [5,] 0 0.00 0.00 0.00 0.00 0.42 0.0 0.00 0.00 0.00 0.00 0.00 # [6,] 0 0.00 0.00 0.00 0.00 0.00 0.5 0.00 0.00 0.00 0.00 0.00 # [7,] 0 0.00 0.00 0.00 0.00 0.00 0.0 0.58 0.00 0.00 0.00 0.00 # [8,] 0", "1-Apr-15$6,640,341.00 6 1-May-15 $5,476,163.28 these are both strings. I wanted to change Month to a date. I do mydata$Month <-as.Date(mydata$Month,format= \"%d/%m/%y\") and it works str(mydata$Month) Date[1:65], format:.... Ok I got this half solved. But now I need to change Spend which is a character into a number. The data looks like this (Spend is the 2nd variable). 1 2020-01-12 5,790,000.00 2 2020-01-01 5,114,841.08 I need to get rid of the , and the periods and columns. All of them I tried a lot of different ways, but this works. mydata$Spend<-as.numeric(gsub(\"[['punct:]]\", \"\",mydata$Spend)) without the ' inside the bracket since that creates a smiley here In about a year I will be actually able to run data.... Last edited: #### noetsi ##### No cake for spunky I was not sure if this is wrong or simply how time series formats. But there are 16405 days between 12/1/2014 and 1/1/1970 when R calculates dates. Need to learn how to get the TS function to show months and dates in standard English not numbers My original data looks like this 1 12/1/2014 5,790,000.00 2 1/1/2015 5,114,841.08 3 2/1/2015 7,240,820.54 4 3/1/2015 7,482,837.26 5 4/1/2015 6,640,341.00 6 5/1/2015 5,476,163.28 I did this to make it a date mydata$Month <-as.Date(mydata$Month,format= \"%m/%d/%Y\") Not sure why it makes it year first then month. Or if there", "rise (%) over 21 years value after 21 years 0 20,790 43,111 1 23,007 46,734 2 25,526 50,791 Broadly speaking, this investment doubles the value of the contributions over two decades. Note: Rebalancing fees are not included in the simulation. Below is the code used to run the simulation. If you have Mathematica you can try with different funds. funds = {\"VTSMX\", \"VGTSX\", \"VBMFX\"}; (* Plotting the fund indices *) {tsm, ism, tbm} = FinancialData[#, {\"April 29, 1996\", DateList[], \"Month\"}] & /@ funds; DateListPlot[ Transpose[{First /@ #, 100 Last /@ #/#[[1, 2]]}] & /@ {tsm, ism, tbm}, PlotLegends -> funds, PlotLabel -> \"Indices from month-end April 1996 rebased to 100\"] (* Plotting the investment contributions *) salary = 550; investment = salary*0.15; inflation = 2; nmonths = Length[tsm] - 1; ny = Quotient[nmonths, 12]; iy = Array[investment/3 (1 + inflation/100)^(# - 1) &, ny]; d = Take[Flatten[ConstantArray[#, 12] & /@ iy], 12 ny]; DateListPlot[Transpose[{Take[First /@ tsm, 12 ny], 3 d}], PlotLabel -> Row[{\"Monthly contributions with \", inflation, \"% inflation - Total = \", Total[3 d]}], PlotRange -> {Automatic, {0, Automatic}}, PlotMarkers -> {Automatic, 6}, FrameLabel -> {\"Time\", Rotate[Style[\"$\", 12], Pi/2]}, ImageSize -> 380] (* Calculating & plotting the investment values *) {tsm2, ism2, tbm2} = Take[Ratios@# - 1, 12 ny] & /@ Map[Last, {tsm, ism, tbm}, {2}]; d2", "row in the other dataframe, that correponds to the balance in that account on that day. So the result in the toy example would be this: df.result <- data.frame(date=as.Date(paste0('2018-12-',c(11,15,16,18,22,25,27,29))), balance.salary=c(-500,-250,-250,0,250,-300,-300,500), balance.budget=c(1000,1000,1000,1000,700,700,250,250)) Notice how even though I don’t have information for the budget-account from the first date that the salay-account has a row, I’m using the information from the first time there is a row from the budget account. here I have changed the column names for the balance-variable, so that one row can have the balance for both, but this is not the essential part of the solution, only that the result can be computed like this: df.result$balance.total <- df.result$ balance.salary + df.result\\$ balance.budget I have tried using crossing() as per this answer, Copying row from one df into everyone row in another, but isn’t useful in this case, as far as I can tell. Thank you.", "number of dimes in the collection?... Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4); Which set of ordered pairs represents a function? {(0, 1), (1,3), (1,5), (2, 6)} {(0, 0), (1, 2), (2, 4), (3, 4); {(0, 1), (1, 2), (2, 3), (2, 4); {(0,0), (0, 2), (2, 2), (2, 4);... Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!! Melody paid off a loan of $120,000 in 10 years. If she made a payment of$1,225 each month, then what was the simple interest rate for her loan? PLEASE HELP!!!...", "-0.364)$); %bridge middle right \\filldraw[bridge] ($(0, 1) + (0.658, -0.590)$) -- ($(0, 1) + (0.668, -0.546)$) -- ($(0, 1) + (0.778, -0.524)$) -- ($(0, 1) + (0.792, -0.554)$) -- ($(0, 1) + (0.796, -0.590)$) --cycle; \\draw[contour] ($(0, 1) + (0.668, -0.546)$) -- ($(0, 1) + (0.778, -0.524)$) ($(0, 1) + (0.658, -0.590)$) -- ($(0, 1) + (0.796, -0.590)$); %bridge bottom 1 \\filldraw[bridge] ($(0, 1) + (0.350, -0.826)$) -- ($(0, 1) + (0.410, -0.804)$) -- ($(0, 1) + (0.422, -0.794)$) -- ($(0, 1) + (0.412, -0.708)$) -- ($(0, 1) + (0.372, -0.702)$) -- ($(0, 1) + (0.336, -0.695)$) -- cycle; \\draw[contour] ($(0, 1) + (0.422, -0.794)$) -- ($(0, 1) + (0.412, -0.708)$) ($(0, 1) + (0.350, -0.826)$) -- ($(0, 1) + (0.336, -0.695)$); %bridge bottom 2 \\filldraw[bridge] ($(0, 1) + (0.596, -0.700)$) -- ($(0, 1) + (0.535, -0.708)$) -- ($(0, 1) + (0.523, -0.784)$) -- ($(0, 1) + (0.581, -0.777)$) -- cycle; \\draw[contour] ($(0, 1) + (0.535, -0.708)$) -- ($(0, 1) + (0.523, -0.784)$) ($(0, 1) + (0.596, -0.700)$) -- ($(0, 1) + (0.581, -0.777)$); %bridge bottom 3 \\filldraw[bridge] ($(0, 1) + (0.581, -0.777)$) -- ($(0, 1) + (0.660, -0.862)$) -- ($(0, 1) + (0.858, -0.717)$) -- ($(0, 1) + (0.837, -0.702)$) -- ($(0, 1) + (0.826, -0.688)$) -- ($(0, 1) + (0.785, -0.668)$) -- cycle; \\draw[contour] ($(0, 1)", ":...savings_balance in {> 1000 DM,100 - 500 DM,unknown}: no (23.6/5.9) ## savings_balance = 500 - 1000 DM: yes (6.2/1.6) ## savings_balance = < 100 DM: ## :...dependents > 1: no (12.2/1.1) ## dependents <= 1: ## :...age > 61: no (5.7) ## age <= 61: ## :...years_at_residence <= 1: no (6.1/1) ## years_at_residence > 1: ## :...other_credit in {bank,store}: yes (8.3/2.1) ## other_credit = none: ## :...amount > 5998: yes (7) ## amount <= 5998: ## :...existing_loans_count <= 1: no (11.6/3.9) ## existing_loans_count > 1: ## :...existing_loans_count > 2: no (2.7) ## existing_loans_count <= 2: ## :...age > 54: yes (3.9) ## age <= 54: ## :...age > 44: no (5) ## age <= 44: ## :...amount <= 5324: yes (30.9/10.8) ## amount > 5324: no (2.7) ## ## ----- Trial 5: ----- ## ## Decision tree: ## ## checking_balance = unknown: ## :...other_credit = store: no (17.3/6.3) ## : other_credit = bank: ## : :...purpose = business: yes (8.9/4.1) ## : : purpose in {car0,education,renovations}: no (7.2/1.8) ## : : purpose = car: ## : : :...credit_history in {critical,perfect,poor,very good}: yes (17.2/4.4) ## : : : credit_history = good: no (3.5) ## : : purpose = furniture/appliances: ## : : :...months_loan_duration <= 13: yes (6.5/1.6) ## : : months_loan_duration > 13: no (17.4/2.5) ## : other_credit =", "833,333.33 98,708,628.86 2 98,708,628.86 2,124,704.47 822,571.91 97,406,496.30 3 97,406,496.30 2,124,704.47 811,720.80 96,093,512.63 4 96,093,512.63 2,124,704.47 800,779.27 94,769,587.43 5 94,769,587.43 2,124,704.47 789,746.56 93,434,629.52 56 10,363,013.71 2,124,704.47 86,358.45 8,324,667.69 57 8,324,667.69 2,124,704.47 69,372.23 6,269,335.45 58 6,269,335.45 2,124,704.47 52,244.46 4,196,875.44 59 4,196,875.44 2,124,704.47 34,973.96 2,107,144.93 60 2,107,144.93 2,124,704.47 17,559.54 -0.00 New bond: Initial Balance Payment Interest Final Balance Month 1 100,000,000.00 2,075,835.52 750,000.00 98,674,164.48 2 98,674,164.48 2,075,835.52 740,056.23 97,338,385.19 3 97,338,385.19 2,075,835.52 730,037.89 95,992,587.55 4 95,992,587.55 2,075,835.52 719,944.41 94,636,696.44 5 94,636,696.44 2,075,835.52 709,775.22 93,270,636.14 56 10,149,672.43 2,075,835.52 76,122.54 8,149,959.45 57 8,149,959.45 2,075,835.52 61,124.70 6,135,248.63 58 6,135,248.63 2,075,835.52 46,014.36 4,105,427.47 59 4,105,427.47 2,075,835.52 30,790.71 2,060,382.65 60 2,060,382.65 2,075,835.52 15,452.87 0.00 Now we calculate the NPV of refi. We make it a function so that we can ask questions about parameters later. In [4]: def npv_refi(rnew, term, bnow, rnow, penalty): # rnew, term, bnow, rnow, penalty, here, are dummy names '''This function returns the npv of refi given rnew: new rate term: remaining term on the bond bnow: remaining principal rnow: current rate penalty: prepayment fee as a fraction of principal''' oldpayment=npf.pmt(rnow/12,term,-bnow) newpayment=npf.pmt(rnew/12,term,-bnow) return npf.pv(rnew/12,term,-(oldpayment-newpayment))-penalty*bnow print(\"The NPV of refi is: \" +put_comma(npv_refi(rnew, term, bnow, rnow, penalty))) The NPV of refi is: 354,182.11 Now we can ask all kinds of intersting questions. For instance, how low must the new rate be to justify refinancing?" ]

[ "# Proportional Thinking A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ of one cup of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? 3. How many different methods can you use to solve this problem? Check solutions here.", "How many quarts of milk containing 4% butter fat and how many quarts of cream containing 29% butter fat must be mixed to make 40 quarts of cream containing 20% butter fat? Aug 18, 2015 Derive two linear equations from the description and solve by substitution to find that $14.4$ quarts of milk must be mixed with $25.6$ quarts of cream. Explanation: Suppose we combine $m$ quarts of milk and $c$ quarts of cream. We want: $m + c = 40$ $0.04 m + 0.29 c = 0.20 \\times 40 = 8$ From the first of these we get $c = 40 - m$ Substitute this into the second equation to get: $8 = 0.04 m + 0.29 \\cdot \\left(40 - m\\right) = 0.04 m + 11.6 - 0.29 m = 11.6 - 0.25 m$ Add $0.25 m - 8$ to both ends to get: $0.25 m = 3.6$ Multiply both sides by $4$ to get $m = 14.4$ Then $c = 40 - m = 25.6$ Check: $0.04 m + 0.29 c = 0.04 \\times 14.4 + 0.29 \\times 25.6$ $= 0.576 + 7.424 = 8$", "# RATL 1 | Lesson 3 | Explore (Proportional Reasoning) Solutions A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ cups of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? Methods vary. Two methods are described below: If $$\\dfrac{2}{3}$$ cups of milk makes 12 cookies, then • 2 times $$\\dfrac{2}{3}= \\dfrac{4}{3}=1\\dfrac{1}{3}$$ cups of milk makes $$2(12)=24$$ cookies • 3 times $$\\dfrac{2}{3}=\\dfrac{6}{3}=2$$ cups of milk makes $$3(12)=36$$ cookies • 6 times $$\\dfrac{2}{3}=\\dfrac{12}{3}=4$$ cups of milk makes $$6(12)=72$$ cookies • 13.5 times $$\\dfrac{2}{3}= \\dfrac{27}{3}=9$$ cups of milk makes $$13.5(12)=162$$ cookies Method 2: Division and multiplication $$\\dfrac{9}{\\dfrac{2}{3}}$$$$=9(\\dfrac{3}{2})=\\dfrac{27}{2}=13.5$$ dozens of cookies. $$13.5(12)=162$$ cookies. 3. Which of the two methods above did you use? Did you use another method?", "Emily bought 1.06 quarts of milk. a. $\\frac{1}{2}$ in. b. 1 in. c. 1$\\frac{1}{2}$ in. d. 2$\\frac{1}{2}$ in.", "Suppose you have a budget of $8 and cookies and pretzels cost$1 each. What is the utility maximizing...", "Main content Proportion word problem: cookies CCSS Math: 6.RP.A.3b Video transcript A recipe for oatmeal cookies calls for 2 cups of flour for every 3 cups of oatmeal. How much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? So let's think about what they're saying. They're saying 2 cups of flour. So 2 cups of flour for every 3 cups of oatmeal. And so they're saying, how much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? Now we're going to go to a situation where we are using 9 cups of oatmeal. Let me write it this way-- 9 cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So one way to think about it, so we're wondering. We're going to say, look, we know if we have 3 cups of oatmeal, we should use 2 cups of flour. But what we don't know is if we have 9 cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from 3 cups of oatmeal to 9 cups of oatmeal, how much more oatmeal are we using? Well, we're using three", "Help. a recipe for eggnog calls for 2 eggs for every 3 cups of milk. If there are cups in a quart, how many eggs will be needed to mix with 6 quarts of milk? If there are 4 cups in every quart, and there are 6 quarts, that mean there are 24 cups of milk. If you take 24 and divide it by 3 cups each, you will get 8 sets with 3 cups in each set. For every set, you need to add 2 eggs, so that would be 16 eggs for the total of 6 quarts.", "cups of milk on hand and makes 2 batches of cookies, [#permalink] ### Show Tags 29 Jan 2020, 02:20 9/4 - 2*2/3 = 11/12 = 0.9 Re: Stephanie has 9/4 cups of milk on hand and makes 2 batches of cookies, [#permalink] 29 Jan 2020, 02:20", "## Elementary Algebra A gallon of milk costs 4.16 dollars. First, we need to convert this price from dollars per gallon to dollars per pint.To do this, we multiply $\\frac{4.16\\ dollars}{1\\ gallon}$ by the ratio that converts gallons to pints. The ratio that converts gallons to pints is $\\frac{1\\ gallon}{8\\ pints}$. So, $\\frac{4.16\\ dollars}{1\\ gallon}$ $\\times$ $\\frac{1\\ gallon}{8\\ pints}$ = $\\frac{4.16\\ dollars}{8\\ pints}$. $\\frac{4.16\\ dollars}{8}$ = 0.52 dollars per pint. The price for a pint of milk is 0.52 dollars.", "one batch of cookies calls for 5 cups of flour and 2 teaspoons of vanilla.... ### Help plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! help plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...", "looking to expand its product offerings beyond coffee. The company is evaluating two alternatives—sandwiches and cookies. Annual projections for sales of sandwiches are as follows: sales,$18,000; variable costs, $13,000; and fixed costs,$500. Annual projections for sales of cookies are as follows: sales, $10,000; variable costs,$3,000; and no additional fixed costs. As shown in the differential analysis given, selling cookies is the most profitable alternative. Selling cookies results in profits of $7,000 for the year, which is$2,500 higher than the sandwich alternative.", "packages of gingerbread cookie dough for a total of $76. Danielle sold 9 packages of white chocolate chip cookie dough and 11 packages of gingerbread cookie dough for a total of$124. What is the cost each of one package of white chocolate chip cookie dough and one package of gingerbread cookie dough? #5: Instructions: solve each word problem. a) Jennifer is trying to eat a meal with 66 grams of protein, 94.5 grams of carbohydrates, and 910 milligrams of calcium. At the local diner, they are serving only turkey, potatoes, and yogurt. Each serving of turkey contains 30 grams of protein, 35 grams of carbohydrates, and 200 milligrams of calcium. Each serving of potatoes has 3 grams of protein, 16 grams of carbohydrates, and 10 milligrams of calcium. Lastly, each serving of yogurt has 9 grams of protein, 13 grams of carbohydrates, and 300 milligrams of calcium. How many servings of each food should Jennifer consume? Written Solutions: #1: Solutions: a) Van: 6 students, Bus: 34 students #2: Solutions: a) daylily: $7, shrub:$8 #3: Solutions: a) plane: 93 mph, wind: 27 mph #4: Solutions: a) white chocolate chip cookie dough: $4, gingerbread cookie dough:$8 #5: Solutions: a) 1.5 servings of turkey, 1 serving of potatoes, and 2 servings of yogurt", "¾ cup of milk and we want to triple it, we need to know that: So, we will need 2 and 1/4 cups of milk to make three dozen cupcakes. This knowledge of fractions is also helpful when we need to make our batch smaller. For example, recipe guidelines approximate that each batch will yield 6 dozen cookies. But, my family is small and I only want 2 dozen cookies. First, we need to see the relationship between 2 and 6. We can see that 2 dozen is one third (1/3) of 6 dozen because 2 x 3 = 6. That means that in order to make only 2 dozen cookies, we will need to use one third of each ingredient. So, if the recipe asks for 2 teaspoons of baking powder, we will only need 2/3 of a teaspoon, since If we do not have a measuring spoon that is equal to 2/3 teaspoon, we may need to use 1/3 twice, or estimate using ¼. When recipes indicate how much a particular batch will make, they give a general amount of food. If we are cooking for a group, we need to estimate how much each person will eat and make appropriate amounts of the particular item. For example, if a package of spaghetti makes 1L of", "batches she can make with 2.4 pints of milk = 3. Dividend: Total of pint of milk she has to make batches = 2.4. Divisor: Number of pint of milk to make a batch of butter Candice needs = 0.8. Explanation: Number of pint of milk to make a batch of butter Candice needs = 0.8. Total of pint of milk she has to make batches = 2.4. Number of batches she can make with 2.4 pints of milk = Total of pint of milk she has to make batches ÷ Number of pint of milk to make a batch of butter Candice needs = 2.4 ÷ 0.8 = 3. Question 5. A bird hops 0.72 meter. If each hop it takes is 0.09 meter, how many hops does the bird take? Dividend: ____________ Divisor: ____________ Answer: Number of hops the bird takes = 8. Dividend: Number of meters a bird hops = 0.72. Divisor: Number of meters each hop it takes = 0.09. Explanation: Number of meters a bird hops = 0.72. Number of meters each hop it takes = 0.09. Number of hops the bird takes = Number of meters a bird hops ÷ Number of meters each hop it takes = 0.72 ÷ 0.09 = 8. Question 6. A sticker costs$0.64. Ronald spends $3.84 on", "not a priority. Now, members must sell (under the threat of revocation of membership or other penalty) 4 tubs of cookie dough at $15.00 each (for those NHS graduates reading this, they raised the price by$1/tub). This is true for the fall and spring cookie dough sales. The fall one is understandable as the proceeds go to helping villages in Kenya. The spring one, though: do we really need that much money to continue operating? Or is Ms. Cresham just wasting a lot of money on who-knows-what? Furthermore, why must members sell 4 tubs of cookie dough? I think it's perfectly fair to ask 15 hours of service each year towards NHS. This is the kind of service the organization should focus on. By contrast, selling requires people to want to buy cookie dough; this also requires finding buyers.", "dough as last time: 1 cup butter, 2 cups pastry flour, and ½ cup powdered sugar. I cleared a counter to make some workspace: I had a cookie sheet,a rolling pin (a piece of birch dowel that I sanded and coated with mineral oil decades ago), a silicone baking mat, the cookie sticks, the cookie cutter and stamp, and a shallow bowl for flour. The entire batch fills about 2/3 of the frame when rolled out: For the first batch, we tried rolling the dough directly on the silicone baking mat, and removing the excess dough without moving the cookies. The cookie sticks worked well for getting a uniform, consistent thickness to the dough, and 6mm is about the right thickness for these cookies. Having a complete frame around the dough meant that I did not have to worry about the cookie sticks shifting position, nor what the orientation of the rolling pin was. The stamping is easily done on the cookies, but removing the excess dough from between the cookies was harder than we expected. It probably didn’t help that it was a warm afternoon and the dough got sticky quickly, even though we refrigerated it before rolling. For the second rolling, we rolled the dough onto waxed paper, then transferred the cut-out cookies to a baking", "troop of 6 girls sells 535 boxes of cookies. Each box of cookies costs $3.50. Samantha sells 154 boxes of cookies. How many boxes did the rest of the girls sell? Answer: Given that, The total number of girls is 6. The 6 girls’ sales are 535 boxes of cookies. The cost of each box is$ 3.50", "is $2$% fat, an amount that is $40$% less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk? $a)\\, \\frac{12}{5}\\qquad b)\\, \\frac{10}{3}\\qquad c)\\, 9\\qquad d)\\, 38\\qquad e)\\, 42$", "tubs of cookie dough at $15.00 each (for those NHS graduates reading this, they raised the price by$1/tub). This is true for the fall and spring cookie dough sales. The fall one is understandable as the proceeds go to helping villages in Kenya. The spring one, though: do we really need that much money to continue operating? Or is Ms. Cresham just wasting a lot of money on who-knows-what? Furthermore, why must members sell 4 tubs of cookie dough? I think it's perfectly fair to ask 15 hours of service each year towards NHS. This is the kind of service the organization should focus on. By contrast, selling requires people to want to buy cookie dough; this also requires finding buyers. Another thing to point out, as Ms. Cresham herself points out at the beginning of this cookie dough sale, is that the people who sell crazy amounts of dough do so because their parents have huge connections. This is perfectly fine with me; that said, not everyone is so well connected to people who would be so willing to fund an enterprise like the NHS. I think it's fine that the organization is doing the cookie dough sale; I just think forcing members to sell a minimum number of tubs is too fraught with problems. I think", "Question Which is the better buy? a 12 pack of soda for $3.50 or a six pack of soda for$1.80? To know the better buy, find the unit rate of each. 1. 12 pack of soda for $3.50 Reply 2. ngocdiep Answer: 12 pack of soda for$3.50" ]

[ "How many quarts of milk containing 4% butter fat and how many quarts of cream containing 29% butter fat must be mixed to make 40 quarts of cream containing 20% butter fat? Aug 18, 2015 Derive two linear equations from the description and solve by substitution to find that $14.4$ quarts of milk must be mixed with $25.6$ quarts of cream. Explanation: Suppose we combine $m$ quarts of milk and $c$ quarts of cream. We want: $m + c = 40$ $0.04 m + 0.29 c = 0.20 \\times 40 = 8$ From the first of these we get $c = 40 - m$ Substitute this into the second equation to get: $8 = 0.04 m + 0.29 \\cdot \\left(40 - m\\right) = 0.04 m + 11.6 - 0.29 m = 11.6 - 0.25 m$ Add $0.25 m - 8$ to both ends to get: $0.25 m = 3.6$ Multiply both sides by $4$ to get $m = 14.4$ Then $c = 40 - m = 25.6$ Check: $0.04 m + 0.29 c = 0.04 \\times 14.4 + 0.29 \\times 25.6$ $= 0.576 + 7.424 = 8$", "## Elementary Algebra A gallon of milk costs 4.16 dollars. First, we need to convert this price from dollars per gallon to dollars per pint.To do this, we multiply $\\frac{4.16\\ dollars}{1\\ gallon}$ by the ratio that converts gallons to pints. The ratio that converts gallons to pints is $\\frac{1\\ gallon}{8\\ pints}$. So, $\\frac{4.16\\ dollars}{1\\ gallon}$ $\\times$ $\\frac{1\\ gallon}{8\\ pints}$ = $\\frac{4.16\\ dollars}{8\\ pints}$. $\\frac{4.16\\ dollars}{8}$ = 0.52 dollars per pint. The price for a pint of milk is 0.52 dollars.", "Shelly purchases milk and cookies at the local supermarket. She has regular-shaped strictly convex indifference curves and a given income $100. After observing the prices for milk and cookies, P$4 and P$4 respectively, she ends up purchasing some milk and some cookies. Assume that cookies are a normal good. a) α β 2 γ δ 2 normal good (b) Keeping milk on the horizontal axis, provides a diagrammatic representation of Shelly’s optimal bundle using a suitable graph with necessary details and briefly explain your work. (c) Suppose now that the price of milk in the local supermarket changes from Pto P2(α − 1.5), while the price of cookies and Shelly’s income remain unchanged. Calculate the new price of milk using the information above by substituting for the values of α and pm. Reproduce your graph for part (b) and do further work on it to show how Shelly’s optimal bundle might adjust due to this price change. While creating your graph, make sure to illustrate any substitution effect and/or income effect as applicable and also the total effect of the price change. Explain your work. (d) Suppose now that S's income changes from to 100(γ − 1.5), while the prices of milk and cookies are still at their original levels (i.e., PP$4). Calculate the new level of income", "Emily bought 1.06 quarts of milk. a. $\\frac{1}{2}$ in. b. 1 in. c. 1$\\frac{1}{2}$ in. d. 2$\\frac{1}{2}$ in.", "# RATL 1 | Lesson 3 | Explore (Proportional Reasoning) Solutions A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ cups of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? Methods vary. Two methods are described below: If $$\\dfrac{2}{3}$$ cups of milk makes 12 cookies, then • 2 times $$\\dfrac{2}{3}= \\dfrac{4}{3}=1\\dfrac{1}{3}$$ cups of milk makes $$2(12)=24$$ cookies • 3 times $$\\dfrac{2}{3}=\\dfrac{6}{3}=2$$ cups of milk makes $$3(12)=36$$ cookies • 6 times $$\\dfrac{2}{3}=\\dfrac{12}{3}=4$$ cups of milk makes $$6(12)=72$$ cookies • 13.5 times $$\\dfrac{2}{3}= \\dfrac{27}{3}=9$$ cups of milk makes $$13.5(12)=162$$ cookies Method 2: Division and multiplication $$\\dfrac{9}{\\dfrac{2}{3}}$$$$=9(\\dfrac{3}{2})=\\dfrac{27}{2}=13.5$$ dozens of cookies. $$13.5(12)=162$$ cookies. 3. Which of the two methods above did you use? Did you use another method?", "# Proportional Thinking A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ of one cup of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? 3. How many different methods can you use to solve this problem? Check solutions here.", "batches she can make with 2.4 pints of milk = 3. Dividend: Total of pint of milk she has to make batches = 2.4. Divisor: Number of pint of milk to make a batch of butter Candice needs = 0.8. Explanation: Number of pint of milk to make a batch of butter Candice needs = 0.8. Total of pint of milk she has to make batches = 2.4. Number of batches she can make with 2.4 pints of milk = Total of pint of milk she has to make batches ÷ Number of pint of milk to make a batch of butter Candice needs = 2.4 ÷ 0.8 = 3. Question 5. A bird hops 0.72 meter. If each hop it takes is 0.09 meter, how many hops does the bird take? Dividend: ____________ Divisor: ____________ Answer: Number of hops the bird takes = 8. Dividend: Number of meters a bird hops = 0.72. Divisor: Number of meters each hop it takes = 0.09. Explanation: Number of meters a bird hops = 0.72. Number of meters each hop it takes = 0.09. Number of hops the bird takes = Number of meters a bird hops ÷ Number of meters each hop it takes = 0.72 ÷ 0.09 = 8. Question 6. A sticker costs$0.64. Ronald spends $3.84 on", "#### 题目列表 The price of a quart of milk was increased by 20%. How many quarts of milk can be purchased for the amount of money that used to buy 30 quarts of milk? Last year Leo bought two paintings. This year he sold them for \\$2,000 each. On one, he made a 25% profit, and on the other he had a 25% loss. What was his net loss or profit? If c and d are odd positive integers, which of the following could be odd? Indicate all such expressions. Let P and Q be points which are two inches apart, and let A be the area, in square inches, of a circle which passes through P and Q. Which of the following is the set of all possible values of A. If the total surface area of a cube is 864, what is the volume of the cube? _____ Rather than allowing these dramatic exchanges between her characters to develop fully, Ms. Norman unfortunately tends to _____ the discussions involving two women. Refusing to let the enemy see how deeply shaken he was by his capture, the prisoner kept his face _____. Anyone who has ever stepped in sticky hot tar on a summer day knows that melted tar is a highly _____ substance. Seemingly solid at", "everyone served. What if instead we said kids could get one and then the other, in separate lines. Now it will take something closer to $$\\max(T_{milk}, T_{cookies})$$.1 Whichever process is longer will dominate the time. (Probably milk). Now imagine that getting a cookie takes 1 second per child, and getting a milk takes 30 seconds. Everyone knows that you can have a cookie and have milk after. If children take a random amount of time – let’s say on average 3 minutes, sometimes more, sometimes less – to eat their cookies, then we can serve 10 kids cookies in 10 seconds, making everyone happy, and then fill up the milks while everyone is enjoying a cookie. However, if we did the opposite – got milks and then got cookies, it would take much longer for all of the kids to get something and you’d see chaos. Back to Bitcoin. Spending coins and creating new coins is a bit like milk and cookies. We can make the spend correspond to distributing the cookies and setting up the milk line. And the creating of the new coin can be more akin to filling up milks whenever a kid wants it. What this means practically is that by unbundling spending from redeeming we can serve a much greater number of users", "to make $5$. The numbers would not have been in conjunction. Now try to arrange four dominoes so that you can make the pips in this way sum to any number from $1$ to $24$ inclusive. The dominoes need not be placed $1$ against $1$, $2$ against $2$, and so on, as in play. #### $183$ - At the Brook A man goes to the brook with two measures of $15$ pints and $16$ pints. How is he to measure exactly $8$ pints of water, in the fewest possible transactions? Filling or emptying a vessel or pouring any quantity from one vessel to another counts as a transaction. #### $184$ - A Prohibition Poser The American Prohibition authorities discovered a full barrel of beer, and were about to destroy the liquor by letting it run down a drain when the owner pointed to two vessels standing by and begged to be allowed to retain in them a small quantity for the immediate consumption of his household. One vessel was a $7$-quart and the other a $5$-quart measure. The officer was a wag, and, believing it to be impossible, said that if the man could measure an exact quart into each vessel (without any pouring back into the barrel) he might do so. How was it to be done", "Startseite » Forum » we have 3 quarts of club soda and two quarts cost # we have 3 quarts of club soda and two quarts cost ## Tags: Mathematik 22:50 Uhr, 21.06.2022 Tag! this is supposed to be a very easy calculation but I need to come to you because I am not understanding something we have 3.5 quarts of club soda and two quarts cost 1 dollar. why is the cost of the 3.5 quarts of club soda $1.75?$ I don't know where the 75 is coming from. I have three quarts, as in, 1/4, 1/4, 1/4 the first two quarters = 1 dollar why does the answer say $1.75?$ Thanks for you help maybe because the 1/4 left is equal to .25 + the .5 =75 is it correct? Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert): \"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen.\" 23:39 Uhr, 21.06.2022 Hello, I cannot follow your reasoning. The easiest way is to apply the rule of three. 2 quart are equivalent to 1 dollar. $2\\phantom{\\rule{0.167em}{0ex}}\\text{qt}\\stackrel{\\wedge }{=}1$ 3.5 quarts are equivalent to x dollars $3.5\\phantom{\\rule{0.167em}{0ex}}\\text{qt}\\stackrel{\\wedge }{=}x\\phantom{\\rule{0.12em}{0ex}}$ Thus $x\\phantom{\\rule{0.12em}{0ex}}=1\\cdot \\frac{3.5}{2}=...$ greetings, pivot 00:25 Uhr, 22.06.2022 yes, the rule of three!. thank you, pivot. 05:55 Uhr, 22.06.2022 I feel really happy to have seen your webpage $ href=\"", "then there will be $16 \\cdot 3.5=56 \\ ounces$ in 3.5 pounds. ### Practice For questions 1-6, set up a unit conversion to find: 1. Feet in a mile? 2. Cups in a quart? 3. Centimeters in a kilometer? 4. Pints in a gallon? 5. Centimeters in a mile? 6. Gallons in a quart? 7. How many inches are in 5.25 yards? 8. How many pints are in 7.5 gallons? 9. How many pounds are in 2.6 tons? 10. How many centimeters are in 4.75 meters? 11. Claire is making chocolate chip cookies. If the recipe calls for 3.5 cups of flour, how many cups will Claire need to use if she triples the recipe? 12. The recipe also calls for 8 oz. of chocolate chips. Claire wants to make the cookies with three-quarters bittersweet chips and one-quarter semi-sweet chips. Again, tripling the recipe, how many ounces of each type of chocolate chip will she need? ### Vocabulary Language: English Proportion Proportion A proportion is an equation that shows two equivalent ratios. Ratio Ratio A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”.", "# How do you convert units of measurement? May 30, 2018 You do them dimensionally.... #### Explanation: ….there are still competing systems of measurement around the world. There are the old $\\text{Imperial measurements}$, and related to this but not identical are $\\text{US measurements}$, and there are the more sensible $\\text{metric units}$. And of course we need correct units of conversion to move from one to the other... $\\text{1 US pint\"-=473.2*mL-=0.833*\"Imperial pints}$... And so if go to the bar in the States, and order $\\text{6 pints of bitter}$...you have got a volume in Christian units of... ${\\text{6 US pint\"xx473.2*mL*\"pint}}^{-} 1 \\times {10}^{-} 3 \\cdot L \\cdot m {L}^{-} 1 = 2.84 \\cdot L$ Is this difficult? Perhaps not, but it is PRONE to error and we have ALL made them...but if we cancel the units dimensionally perhaps the error if there is one will become evident... $6 \\cdot {\\cancel{\\text{US pint\"xx473.2*cancel(mL)*cancel\"pint}}}^{-} 1 \\times {10}^{-} 3 \\cdot L \\cdot \\cancel{m {L}^{-} 1} = 2.84 \\cdot L$", "# RATL 1 | Lesson 3 | Explore (Solution: Proportional Reasoning) ## Proportional Reasoning A cookie recipe make one dozen cookies and requires $$\\Large \\frac{2}{3}$$ cups of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? Methods vary. Two methods are described below: If $$\\frac{2}{3}$$ cups of milk makes 12 cookies, then • 2 times $$\\frac{2}{3}= \\frac{4}{3}=1\\frac{1}{3}$$ cups of milk makes $$2(12)=24$$ cookies • 3 times $$\\frac{2}{3}=\\frac{6}{3}=2$$ cups of milk makes $$3(12)=36$$ cookies • 6 times $$\\frac{2}{3}=\\frac{12}{3}=4$$ cups of milk makes $$6(12)=72$$ cookies • 13.5 times $$\\frac{2}{3}= \\frac{27}{3}=9$$ cups of milk makes $$13.5(12)=162$$ cookies Method 2: Division and multiplication $$\\Large\\frac{9}{\\frac{2}{3}}$$$$=9(\\frac{3}{2})=\\frac{27}{2}=13.5$$ dozens of cookies. $$13.5(12)=162$$ cookies. 3. Which of the two methods above did you use? Did you use another method? Go back to Explore (Proportional Reasoning)", "X 3/26 = 267/1300 ... = .2053846 Correct. But, let's forget about decimal approximations. The answer in the book shows 20 7/13 cents. Oh, for Pete's sake. They gave you the money unit \"dollars\", but they want the answer written in \"cents\"? (They could have said so, in the instructions.) Okay, let's convert 267/1300 dollars to cents. #### AsilKing ##### New member Isn't converting dollars to cents dividing 267 by 1300? #### Otis ##### Elite Member Isn't converting dollars to cents dividing 267 by 1300? No, that division is not a unit conversion; we began by comparing dollars to quarts, so the result compares dollars to quarts. 267/1300 means 267/1300ths of a dollar for each quart, not 267/1300ths of a penny per quart. Just like dividing$1 by 2 quarts doesn't yield 50 cents per quart. It yields 1/2 dollar per quart. We then convert 1/2 dollar to 50 cents. If you need help converting 267/1300 dollar to cents, let us know. #### AsilKing ##### New member I guess I do need help converting fractions of dollars to cents. #### Otis ##### Elite Member I guess I do need help converting fractions of dollars to cents. Hi. The conversion from dollars to cents is the same whether the dollars are Whole numbers, fractions or decimal numbers (doesn't matter). Multiply", "# Question #66f05 Oct 23, 2016 Milk is $2.32 and the Magazine is$2.33 #### Explanation: Let M = the price of milk Let G = the price of the magazine We know M + G = $4.65 or M =$4.65 - G We also know: (M * 1.05) + (G * 1.08) = $4.95 Substituting from the first equation: (($4.65 - G) * 1.05) + (G * 1.08) = $4.95$4.88 - (G * 1.05) + (G * 1.08) = $4.95 G * 0.03 =$0.07 G = $0.07 / 0.03 G = 2.33 Substituting back into the first equation: M +$2.33 = $4.65 M =$4.65 - $2.33 M =$2.32", "3) 4. 2(3 + 5) = (2 x 3) + (2 x 5) 5. A pitcher contains $3 1/2$ quarts of milk. You pour $1 7/8$ quarts into a smaller pitcher. How many quarts of milk are left in the first pitcher? 1. $1 3/4$ quarts 2. $1 5/8$ quarts 3. $2 5/8$ quarts 4. $2 3/4$ quarts 6. The air temperature in the North Pole in the morning was 3 degrees Celsius. By 10:00 am the temperature dropped 12 degrees. By noon the temperature had gone up 5 degrees. At 3:00 pm the temperature went up 9 degrees. By nightfall the temperature had dropped 16 more degrees. What was the temperature at the North Pole at nightfall? 1. 11 °C 2. 0 °C 3. -11 °C 4. -9 °C 7. Hilton Head Island is about 12 miles long. Susan has a map that shows the island. If the scale on the map is 1 inch = $1/2$mile, what is the length of the island on Susan's map? 1. 6 inches 2. 12 inches 3. 18 inches 4. 24 inches 8. $7/20$ is equivalent to which decimal? 1. 0.35 2. 0.720 3. 0.30 4. 0.72 9. Graph on a number line and solve. (-4) + 8 = 10. When working with positive and negative numbers, the further left", "Help. a recipe for eggnog calls for 2 eggs for every 3 cups of milk. If there are cups in a quart, how many eggs will be needed to mix with 6 quarts of milk? If there are 4 cups in every quart, and there are 6 quarts, that mean there are 24 cups of milk. If you take 24 and divide it by 3 cups each, you will get 8 sets with 3 cups in each set. For every set, you need to add 2 eggs, so that would be 16 eggs for the total of 6 quarts.", "Main content Proportion word problem: cookies CCSS Math: 6.RP.A.3b Video transcript A recipe for oatmeal cookies calls for 2 cups of flour for every 3 cups of oatmeal. How much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? So let's think about what they're saying. They're saying 2 cups of flour. So 2 cups of flour for every 3 cups of oatmeal. And so they're saying, how much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? Now we're going to go to a situation where we are using 9 cups of oatmeal. Let me write it this way-- 9 cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So one way to think about it, so we're wondering. We're going to say, look, we know if we have 3 cups of oatmeal, we should use 2 cups of flour. But what we don't know is if we have 9 cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from 3 cups of oatmeal to 9 cups of oatmeal, how much more oatmeal are we using? Well, we're using three", "## Basic College Mathematics (10th Edition) To determine how many Twinkies the milkman has eaten you must compute how many days there have been in the last 60 years. At 365 days per year, the number of days in 60 years is: $365 \\times 60 = 21900$ Since the milkman ate one Twinkie per day he has eaten 21,900 Twinkies. *NOTE - This problem appears to want you to not consider the impact of leap years." ]

[ "How many quarts of milk containing 4% butter fat and how many quarts of cream containing 29% butter fat must be mixed to make 40 quarts of cream containing 20% butter fat? Aug 18, 2015 Derive two linear equations from the description and solve by substitution to find that $14.4$ quarts of milk must be mixed with $25.6$ quarts of cream. Explanation: Suppose we combine $m$ quarts of milk and $c$ quarts of cream. We want: $m + c = 40$ $0.04 m + 0.29 c = 0.20 \\times 40 = 8$ From the first of these we get $c = 40 - m$ Substitute this into the second equation to get: $8 = 0.04 m + 0.29 \\cdot \\left(40 - m\\right) = 0.04 m + 11.6 - 0.29 m = 11.6 - 0.25 m$ Add $0.25 m - 8$ to both ends to get: $0.25 m = 3.6$ Multiply both sides by $4$ to get $m = 14.4$ Then $c = 40 - m = 25.6$ Check: $0.04 m + 0.29 c = 0.04 \\times 14.4 + 0.29 \\times 25.6$ $= 0.576 + 7.424 = 8$", "Shelly purchases milk and cookies at the local supermarket. She has regular-shaped strictly convex indifference curves and a given income $100. After observing the prices for milk and cookies, P$4 and P$4 respectively, she ends up purchasing some milk and some cookies. Assume that cookies are a normal good. a) α β 2 γ δ 2 normal good (b) Keeping milk on the horizontal axis, provides a diagrammatic representation of Shelly’s optimal bundle using a suitable graph with necessary details and briefly explain your work. (c) Suppose now that the price of milk in the local supermarket changes from Pto P2(α − 1.5), while the price of cookies and Shelly’s income remain unchanged. Calculate the new price of milk using the information above by substituting for the values of α and pm. Reproduce your graph for part (b) and do further work on it to show how Shelly’s optimal bundle might adjust due to this price change. While creating your graph, make sure to illustrate any substitution effect and/or income effect as applicable and also the total effect of the price change. Explain your work. (d) Suppose now that S's income changes from to 100(γ − 1.5), while the prices of milk and cookies are still at their original levels (i.e., PP$4). Calculate the new level of income", "# Proportional Thinking A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ of one cup of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? 3. How many different methods can you use to solve this problem? Check solutions here.", "# RATL 1 | Lesson 3 | Explore (Proportional Reasoning) Solutions A cookie recipe make one dozen cookies and requires $$\\dfrac{2}{3}$$ cups of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? Methods vary. Two methods are described below: If $$\\dfrac{2}{3}$$ cups of milk makes 12 cookies, then • 2 times $$\\dfrac{2}{3}= \\dfrac{4}{3}=1\\dfrac{1}{3}$$ cups of milk makes $$2(12)=24$$ cookies • 3 times $$\\dfrac{2}{3}=\\dfrac{6}{3}=2$$ cups of milk makes $$3(12)=36$$ cookies • 6 times $$\\dfrac{2}{3}=\\dfrac{12}{3}=4$$ cups of milk makes $$6(12)=72$$ cookies • 13.5 times $$\\dfrac{2}{3}= \\dfrac{27}{3}=9$$ cups of milk makes $$13.5(12)=162$$ cookies Method 2: Division and multiplication $$\\dfrac{9}{\\dfrac{2}{3}}$$$$=9(\\dfrac{3}{2})=\\dfrac{27}{2}=13.5$$ dozens of cookies. $$13.5(12)=162$$ cookies. 3. Which of the two methods above did you use? Did you use another method?", "batches she can make with 2.4 pints of milk = 3. Dividend: Total of pint of milk she has to make batches = 2.4. Divisor: Number of pint of milk to make a batch of butter Candice needs = 0.8. Explanation: Number of pint of milk to make a batch of butter Candice needs = 0.8. Total of pint of milk she has to make batches = 2.4. Number of batches she can make with 2.4 pints of milk = Total of pint of milk she has to make batches ÷ Number of pint of milk to make a batch of butter Candice needs = 2.4 ÷ 0.8 = 3. Question 5. A bird hops 0.72 meter. If each hop it takes is 0.09 meter, how many hops does the bird take? Dividend: ____________ Divisor: ____________ Answer: Number of hops the bird takes = 8. Dividend: Number of meters a bird hops = 0.72. Divisor: Number of meters each hop it takes = 0.09. Explanation: Number of meters a bird hops = 0.72. Number of meters each hop it takes = 0.09. Number of hops the bird takes = Number of meters a bird hops ÷ Number of meters each hop it takes = 0.72 ÷ 0.09 = 8. Question 6. A sticker costs$0.64. Ronald spends $3.84 on", "## Elementary Algebra A gallon of milk costs 4.16 dollars. First, we need to convert this price from dollars per gallon to dollars per pint.To do this, we multiply $\\frac{4.16\\ dollars}{1\\ gallon}$ by the ratio that converts gallons to pints. The ratio that converts gallons to pints is $\\frac{1\\ gallon}{8\\ pints}$. So, $\\frac{4.16\\ dollars}{1\\ gallon}$ $\\times$ $\\frac{1\\ gallon}{8\\ pints}$ = $\\frac{4.16\\ dollars}{8\\ pints}$. $\\frac{4.16\\ dollars}{8}$ = 0.52 dollars per pint. The price for a pint of milk is 0.52 dollars.", "Emily bought 1.06 quarts of milk. a. $\\frac{1}{2}$ in. b. 1 in. c. 1$\\frac{1}{2}$ in. d. 2$\\frac{1}{2}$ in.", "Main content Proportion word problem: cookies CCSS Math: 6.RP.A.3b Video transcript A recipe for oatmeal cookies calls for 2 cups of flour for every 3 cups of oatmeal. How much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? So let's think about what they're saying. They're saying 2 cups of flour. So 2 cups of flour for every 3 cups of oatmeal. And so they're saying, how much flour is needed for a big batch of cookies that uses 9 cups of oatmeal? Now we're going to go to a situation where we are using 9 cups of oatmeal. Let me write it this way-- 9 cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So one way to think about it, so we're wondering. We're going to say, look, we know if we have 3 cups of oatmeal, we should use 2 cups of flour. But what we don't know is if we have 9 cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from 3 cups of oatmeal to 9 cups of oatmeal, how much more oatmeal are we using? Well, we're using three", "# RATL 1 | Lesson 3 | Explore (Solution: Proportional Reasoning) ## Proportional Reasoning A cookie recipe make one dozen cookies and requires $$\\Large \\frac{2}{3}$$ cups of milk. 1. How many cookies can be made with 9 cups of milk? 2. What method did you use to solve this problem? Methods vary. Two methods are described below: If $$\\frac{2}{3}$$ cups of milk makes 12 cookies, then • 2 times $$\\frac{2}{3}= \\frac{4}{3}=1\\frac{1}{3}$$ cups of milk makes $$2(12)=24$$ cookies • 3 times $$\\frac{2}{3}=\\frac{6}{3}=2$$ cups of milk makes $$3(12)=36$$ cookies • 6 times $$\\frac{2}{3}=\\frac{12}{3}=4$$ cups of milk makes $$6(12)=72$$ cookies • 13.5 times $$\\frac{2}{3}= \\frac{27}{3}=9$$ cups of milk makes $$13.5(12)=162$$ cookies Method 2: Division and multiplication $$\\Large\\frac{9}{\\frac{2}{3}}$$$$=9(\\frac{3}{2})=\\frac{27}{2}=13.5$$ dozens of cookies. $$13.5(12)=162$$ cookies. 3. Which of the two methods above did you use? Did you use another method? Go back to Explore (Proportional Reasoning)", "Help. a recipe for eggnog calls for 2 eggs for every 3 cups of milk. If there are cups in a quart, how many eggs will be needed to mix with 6 quarts of milk? If there are 4 cups in every quart, and there are 6 quarts, that mean there are 24 cups of milk. If you take 24 and divide it by 3 cups each, you will get 8 sets with 3 cups in each set. For every set, you need to add 2 eggs, so that would be 16 eggs for the total of 6 quarts.", "then there will be $16 \\cdot 3.5=56 \\ ounces$ in 3.5 pounds. ### Practice For questions 1-6, set up a unit conversion to find: 1. Feet in a mile? 2. Cups in a quart? 3. Centimeters in a kilometer? 4. Pints in a gallon? 5. Centimeters in a mile? 6. Gallons in a quart? 7. How many inches are in 5.25 yards? 8. How many pints are in 7.5 gallons? 9. How many pounds are in 2.6 tons? 10. How many centimeters are in 4.75 meters? 11. Claire is making chocolate chip cookies. If the recipe calls for 3.5 cups of flour, how many cups will Claire need to use if she triples the recipe? 12. The recipe also calls for 8 oz. of chocolate chips. Claire wants to make the cookies with three-quarters bittersweet chips and one-quarter semi-sweet chips. Again, tripling the recipe, how many ounces of each type of chocolate chip will she need? ### Vocabulary Language: English Proportion Proportion A proportion is an equation that shows two equivalent ratios. Ratio Ratio A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”.", "¾ cup of milk and we want to triple it, we need to know that: So, we will need 2 and 1/4 cups of milk to make three dozen cupcakes. This knowledge of fractions is also helpful when we need to make our batch smaller. For example, recipe guidelines approximate that each batch will yield 6 dozen cookies. But, my family is small and I only want 2 dozen cookies. First, we need to see the relationship between 2 and 6. We can see that 2 dozen is one third (1/3) of 6 dozen because 2 x 3 = 6. That means that in order to make only 2 dozen cookies, we will need to use one third of each ingredient. So, if the recipe asks for 2 teaspoons of baking powder, we will only need 2/3 of a teaspoon, since If we do not have a measuring spoon that is equal to 2/3 teaspoon, we may need to use 1/3 twice, or estimate using ¼. When recipes indicate how much a particular batch will make, they give a general amount of food. If we are cooking for a group, we need to estimate how much each person will eat and make appropriate amounts of the particular item. For example, if a package of spaghetti makes 1L of", "looking to expand its product offerings beyond coffee. The company is evaluating two alternatives—sandwiches and cookies. Annual projections for sales of sandwiches are as follows: sales,$18,000; variable costs, $13,000; and fixed costs,$500. Annual projections for sales of cookies are as follows: sales, $10,000; variable costs,$3,000; and no additional fixed costs. As shown in the differential analysis given, selling cookies is the most profitable alternative. Selling cookies results in profits of $7,000 for the year, which is$2,500 higher than the sandwich alternative.", "Startseite » Forum » we have 3 quarts of club soda and two quarts cost # we have 3 quarts of club soda and two quarts cost ## Tags: Mathematik 22:50 Uhr, 21.06.2022 Tag! this is supposed to be a very easy calculation but I need to come to you because I am not understanding something we have 3.5 quarts of club soda and two quarts cost 1 dollar. why is the cost of the 3.5 quarts of club soda $1.75?$ I don't know where the 75 is coming from. I have three quarts, as in, 1/4, 1/4, 1/4 the first two quarters = 1 dollar why does the answer say $1.75?$ Thanks for you help maybe because the 1/4 left is equal to .25 + the .5 =75 is it correct? Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert): \"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen.\" 23:39 Uhr, 21.06.2022 Hello, I cannot follow your reasoning. The easiest way is to apply the rule of three. 2 quart are equivalent to 1 dollar. $2\\phantom{\\rule{0.167em}{0ex}}\\text{qt}\\stackrel{\\wedge }{=}1$ 3.5 quarts are equivalent to x dollars $3.5\\phantom{\\rule{0.167em}{0ex}}\\text{qt}\\stackrel{\\wedge }{=}x\\phantom{\\rule{0.12em}{0ex}}$ Thus $x\\phantom{\\rule{0.12em}{0ex}}=1\\cdot \\frac{3.5}{2}=...$ greetings, pivot 00:25 Uhr, 22.06.2022 yes, the rule of three!. thank you, pivot. 05:55 Uhr, 22.06.2022 I feel really happy to have seen your webpage $ href=\"", "everyone served. What if instead we said kids could get one and then the other, in separate lines. Now it will take something closer to $$\\max(T_{milk}, T_{cookies})$$.1 Whichever process is longer will dominate the time. (Probably milk). Now imagine that getting a cookie takes 1 second per child, and getting a milk takes 30 seconds. Everyone knows that you can have a cookie and have milk after. If children take a random amount of time – let’s say on average 3 minutes, sometimes more, sometimes less – to eat their cookies, then we can serve 10 kids cookies in 10 seconds, making everyone happy, and then fill up the milks while everyone is enjoying a cookie. However, if we did the opposite – got milks and then got cookies, it would take much longer for all of the kids to get something and you’d see chaos. Back to Bitcoin. Spending coins and creating new coins is a bit like milk and cookies. We can make the spend correspond to distributing the cookies and setting up the milk line. And the creating of the new coin can be more akin to filling up milks whenever a kid wants it. What this means practically is that by unbundling spending from redeeming we can serve a much greater number of users", "dough as last time: 1 cup butter, 2 cups pastry flour, and ½ cup powdered sugar. I cleared a counter to make some workspace: I had a cookie sheet,a rolling pin (a piece of birch dowel that I sanded and coated with mineral oil decades ago), a silicone baking mat, the cookie sticks, the cookie cutter and stamp, and a shallow bowl for flour. The entire batch fills about 2/3 of the frame when rolled out: For the first batch, we tried rolling the dough directly on the silicone baking mat, and removing the excess dough without moving the cookies. The cookie sticks worked well for getting a uniform, consistent thickness to the dough, and 6mm is about the right thickness for these cookies. Having a complete frame around the dough meant that I did not have to worry about the cookie sticks shifting position, nor what the orientation of the rolling pin was. The stamping is easily done on the cookies, but removing the excess dough from between the cookies was harder than we expected. It probably didn’t help that it was a warm afternoon and the dough got sticky quickly, even though we refrigerated it before rolling. For the second rolling, we rolled the dough onto waxed paper, then transferred the cut-out cookies to a baking", "salty than both of the ones given, what ratio of sodium to water would you use? A recipe for one batch of cookies calls for 5 cups of flour and 2 teaspoons of vanilla. 1. Draw a diagram that shows the amount of flour and vanilla needed for two batches of cookies. 2. How many batches can you make with 15 cups of flour and 6 teaspoons of vanilla? Show the additional batches by adding more ingredients to your diagram. 3. How much flour and vanilla would you need for 5 batches of cookies? 4. Whether the ratio of cups of flour to teaspoons of vanilla is $$5:2$$, $$10:4$$, or $$15:6$$, the recipes would make cookies that taste the same. We call these equivalent ratios. 1. Find another ratio of cups of flour to teaspoons of vanilla that is equivalent to these ratios. 2. How many batches can you make using this new ratio of ingredients? Summary A recipe for fizzy juice says, “Mix 5 cups of cranberry juice with 2 cups of soda water.” To double this recipe, we would use 10 cups of cranberry juice with 4 cups of soda water. To triple this recipe, we would use 15 cups of cranberry juice with 6 cups of soda water. This diagram shows a single batch of", "one batch of cookies calls for 5 cups of flour and 2 teaspoons of vanilla.... ### Help plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! help plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...", "describes the amount of milk remaining after she makes the cookies? A. Less then $$\\frac{1}{2}$$ cup B. Between$$\\frac{1}{2}$$cup and $$\\frac{3}{4}$$cup C. Between $$\\frac{3}{4}$$cup and 1 cup D. Between 1 cup and$$\\frac{3}{2}$$ cups E. More than $$\\frac{3}{2}$$ cups OG Q 2017 (Book Question: 59) Senior Manager Joined: 23 Apr 2015 Posts: 266 Location: United States WE: Engineering (Consulting) Re: Stephanie has 9/4 cups of milk on hand and makes 2 batches of cookies, [#permalink] ### Show Tags 23 Jun 2016, 14:44 2 She has $$\\frac{9}{4}$$ and uses $$\\frac{4}{3}$$ hence remaining is $$\\frac{9}{4} - \\frac{4}{3} = \\frac{11}{12}$$ and which is approx 0.9 and lies in the region mentioned in C. (Share the kudo's, if you like the explanation) Director Joined: 04 Dec 2015 Posts: 725 Location: India Concentration: Technology, Strategy WE: Information Technology (Consulting) Stephanie has 9/4 cups of milk on hand and makes 2 batches of cookies, [#permalink] ### Show Tags 24 Jun 2016, 03:27 2 Stephanie has 9/4 cups of milk on hand and makes 2 batches of cookies, using 2/3 cup of milk for each batch of cookies. Which of the following describes the amount of milk remaining after she makes the cookies? Milk used to make 2 batches of cookies = 2/3 +2/3 = 4/3 Balance = 9/4 - 4/3 = 11/12 11/12 = 0.91 Answer", "#### 题目列表 The price of a quart of milk was increased by 20%. How many quarts of milk can be purchased for the amount of money that used to buy 30 quarts of milk? Last year Leo bought two paintings. This year he sold them for \\$2,000 each. On one, he made a 25% profit, and on the other he had a 25% loss. What was his net loss or profit? If c and d are odd positive integers, which of the following could be odd? Indicate all such expressions. Let P and Q be points which are two inches apart, and let A be the area, in square inches, of a circle which passes through P and Q. Which of the following is the set of all possible values of A. If the total surface area of a cube is 864, what is the volume of the cube? _____ Rather than allowing these dramatic exchanges between her characters to develop fully, Ms. Norman unfortunately tends to _____ the discussions involving two women. Refusing to let the enemy see how deeply shaken he was by his capture, the prisoner kept his face _____. Anyone who has ever stepped in sticky hot tar on a summer day knows that melted tar is a highly _____ substance. Seemingly solid at" ]

[ "one circle to another; for example, a medium pizza has a diameter of 14 inches and a large pizza has a diameter of 16 inches. By what percent is the area of the large pizza greater than the medium pizza?", "+0 # Percent 0 132 3 Pizzas are sized by diameter. What percent increase in area results if Chantel's pizza increases from an 8-inch pizza to a 12-inch pizza? Jul 5, 2022 #1 +2602 0 Original pizza: $$4 ^2 \\pi = 16 \\pi$$ New pizza: $$6^2 \\pi = 36 \\pi$$ So, $$36 \\div 16 = 2.25 = 225\\text{%}$$. But, recall that we are looking for the increase, so we subtract 100, to find that it is a $$\\color{brown}\\boxed{125 \\text{%}}$$ increase. Jul 5, 2022 #3 +1158 +7 my answer shows an easier way for the percentage step. but still. same answer! nerdiest Jul 5, 2022 #2 +1158 +9 An 8-inch pizza has a radius of 4 inches. A 12-inch pizza has a radius of 6 inches. Area of first pizza: $$\\pi r^2$$$$= 16\\pi$$ Area of second pizza: $$\\pi r^2=36\\pi$$ Calculation of percentage: $$\\frac{36-16}{16} \\times 100 = 125%$$% Therefore, the area is increased by $$125$$% Jul 5, 2022", "Trent488 1 # brian has a pies that is 10 inches is diameter. he has 9 people over and would like everyone to have a equal slice . how long is the outside edge of each slice . a) 3 inches B) 3 31/63 inches C)4 D)4 31/63 E) None of the above 3.14 inches so they each get a $\\pi$ of pie", "that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 04:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 04:28 Display posts from previous: Sort by", "A large pizza has a radius of 12in. What is the area of half of the large pizza?", "Sammantha M. # How much more pizza do you get in a large pizza that has a diameter of 1 inches than you do in a small pizza that has a diameter of 10 inches? The teacher gives an hint. You need to find the area of each pizza. I'm not so sure how to do this. Mark M. Have never seen a pizza with a diameter of 1 inch. Report 09/13/15", "Bailey J. # help me please i cant figure this out A large pizza at Jake's Pizza has a radius of 7 inches. Find the length of the crust around the outside of the pizza. Use 3.14 for pi and round your answer to the nearest tenth. Please label your answer.", "# A pizza restaurant wants to make a pizza that is 10% bigger than the current 8 inch pizza. What is the diameter of the new pizza? asked Oct 22, 2015 in K-12 Area of a circular pizza = 1/2 * pi * r^2 = 1/4 * pi * d^2 Where diameter is d, and radius is r. We have: 1/4 * pi * d^2 = 1.1 * 1/4 * pi * (8^2) Solve for d, and get d = $8(\\sqrt{1.1}) = 8.3$ The new diameter is 8.3 inches. answered Oct 28, 2015 by (6,320 points)", "circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza? A. 33% B. 44% C. 56% D. 67% E. 78% A 16-inch pizza has a surface area of π x 8^2 = 64π. A 12-inch pizza has a surface area of π x 6^2 = 36π. Now we can determine what percent larger the 16-inch pizza is versus the 12-inch pizza. (64π - 36π)/36π x 100 = 28/36 x 100 = 7/9 x 100 ≈ 78%. _________________ Jeffery Miller GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur &nbs [#permalink] 19 Apr 2018, 16:52 Display posts from previous: Sort by", "a particular point, then the sum of the areas of alternate pieces will be equal. So, now whenever you go out to have a pizza, keep this pizza theorem in mind and ask the pizza person not to cut the pizza for you. Instead, ask a cutter from him and let your inner mathematician play some magic on the pizza. ## Volume of Pizza How much pizza is in a pizza? Let's calculate the Area of two pizzas, a large pizza is 18 inches in diameter and regular pizza is 12 inches. Area of the circle $$=\\pi r^2$$ Since we already know the diameter of the pizzas. The radius of the pizza will be r = diameter/2, which is 9 inches for the large pizza and 6 inches for the regular pizzas. From the above pizza image, the Area of 18 inch pizza is 254.5 square inches and the area of 9 inch pizza is 113.1 square inches. So, the area of two regular pizzas is 226.2 square inches that is less than area of a large pizza. So always go for a large instead of two regular pizzas. To know other facts about pizza, do check out our blog: Are you paying extra for your pizza? Let the pizza now have a height of 'a' and let’s", "It will have a large circular top and bottom . . . and a very small height. You can \"see\" that it has a much larger surface area. In theory, we can continue to flatten it and make a larger and larger pizza. It may be only one molecule thick, but its top and bottom may cover an acre. Flatten it even more until its thickness is that of an atom. By then the pizza dough may cover, say, Canada. My point is that there is no maximum surface area for the cylinder.", "though you cut it into more pieces than the 12-inch pizza. And, by the way, the difference is slightly over three square inches per slice.", "percent is the width increased?", "inch pizza with 6 slices. Medium Pizza: 12-14 inch pizza with 8 slices. Large Pizza: 14-16 inch pizza with 8 slices. Extra-large Pizza: 16-18 inch pizza with 8-10 slices. ## How many slices are in a large pizza from Pizza Hut? So, what is the number of slices in a large pizza? A large-sized Pizza Hut pie is 14 inches in diameter and yields 12 slices enough to feed four to six people. 10 slices ## How many slices of pizza are in a 12 inch Domino’s pizza? A 12-inch (medium) pie offers eight slices. The smallest pie is roughly 9.5-inch broad and feeds two to three people with up to six slices.", "in the papers, linked to a clotted cream company. Read: On the perfect size for a pizza, by Eugenia Cheng via MetaFilter.", "2015, 12:44 A circular rim A having a diameter of 28 inches is rotating 3 19 Aug 2013, 04:02 If 20 circular pepperoni slices each with a 1 inch-diameter cover a ci 2 16 Sep 2017, 11:54 1 A circular cloth with a diameter of 20 inches is placed on a 2 04 Jul 2017, 15:36 14 A circular mat with diameter 20 inches is placed on a square 5 19 May 2015, 08:45 Display posts from previous: Sort by", "time is In this case, the height is approximately and responsible for 65% of filling.", "16-inch pizza than a 12-inch pizza? A. 33% B. 44% C. 56% D. 67% E. 78% _________________ Manager Joined: 30 May 2017 Posts: 120 Location: United States Schools: HBS '21 GMAT 1: 690 Q50 V32 GRE 1: Q168 V164 GPA: 3.57 Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur [#permalink] ### Show Tags 17 Apr 2018, 05:33 Bunuel wrote: A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza? A. 33% B. 44% C. 56% D. 67% E. 78% Area of 16-inch pizza = pi*8^2 = 64pi Area of 12-inch- pizza = pi*6^2 = 36pi 16-inch pizza is larger than 12-inch pizza by (64pi-36pi)/36pi = 28pi/36pi = 28/36 = 7/9 = 77.7777% = Approximately 78% Hence option E = 78% is the answer. _________________ Kindly press the +1Kudos if you like the explanation. Thanks a lot!!! Senior SC Moderator Joined: 22 May 2016 Posts: 2034 A pizzeria makes pizzas that are shaped as perfect circles, and measur [#permalink] ### Show Tags 17 Apr 2018, 09:42 Bunuel wrote: A pizzeria makes pizzas that are shaped as perfect circles, and measures pizza size by the diameter of a pizza. By", "Is A 10 Inch Pizza? approximately six slices may be obtained from an average small personal pizza with a diameter of 8-10 inches.A 10 edge pizza has a surface area of 78 square inches and may accommodate 1-3 people.Customers can frequently choose between four different pizza sizes.In terms of size, a little or personal pizza is between 8 and 10 inches in diameter and makes approximately six slices, whereas a 12-inch pizza ( medium-sized ) yields approximately eight slices.Another point to consider is that a bad 14-inch pizza yields around ten-spot slices, but an extra-large proto-indo european has a circumference of 16-18 inches and yields at least 12 pieces . ## How Much Pizza To Order? many pizza establishments provide a variety show of pie sizes to allow customers to create their own personalized pizza at their leisure. however, there are sealed elements that influence the measurement . ### Confirm The Number of Slices • As previously stated, the usual slice to pizza size ratio is as follows: six slices (8-10 inches) • eight slices (12 inches) • ten slices (12 inches) • twelve slices (16-18 inches) • and fourteen slices (20 inches). Please keep in mind that the size of the pizza might vary depending on the company from whom you get it. As a leave, it", "square inches ) in a duffle bag your square figure... X diameter x 3.1416 enter the diameter x 0.7854 dp ) the screen using pixels! To find out the area of the pizza by the number of square inches ( in,. In other words, if a pizza costs$ 32 and has an area of a given., a, the radius of 8 inches, it 's 50.265 square inches ( in 2 sq. 32 and has an area of a circle that has a diameter of a circle is 1! 1 ∅ 1 in ) is seven feet, four inches, it 's 5.09296 inches. Is 0.7854 times as heavy as a square with four even length sides and four right angles 2! Sides and four right angles shaped object from the dimensions of length and diameter circles... Of 5 inches: π x ( diameter/2 ) 2 a circumference of a circle we Use calculator... 1 circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( unit... Area to be 12 feet or 9 pi which is 7.07 square feet figure ( ft 2 ) ”... By π square with four even length sides and four right angles pizza costs $and! ∅ 1 in ) x the shortest x 0.7854 to find out the area 415.4756! Many sq b, the" ]

[ "per square inch? The following example shows how to calculate the cost per square inch of a circular pizza. First, determine the area of the pizza. This is done through the formula pi*D^2/4, where D is the diameter. The diameter, in this case, is 12 inches, so the area is calculated as: TA = 3.14159*12^2/4 TA = 113.09 in^2. Next, determine the total cost of the pizza. In this example, the total cost is found to be $20.00. Finally, calculate the cost per square inch using the formula: CPSI = TC / TA CPSI =$20.00 / 113.09 CPSI = .18 \\$ / in^2", "* * * *which is 7.07 square feet. Measure the diameter of your circle in feet. So, the cost per square inch of pizza is about .13, or 13 cents. If the circle has a diameter of 8 inches, it's 50.265 square inches. Between ∅ 1in and cm2, sq cm measurements conversion chart page. Directions: Use the calculator above to calculate your square inches. Circle 2 has a radius of 6/2 = 3 inches. For example, if a pizza costs $32 and has an area of 256 square inches, you would calculate ÷ =. 300 Square Inches to Circular Inches = 381.9719: 7 Square Inches to Circular Inches = 8.9127: 400 Square Inches to Circular Inches = 509.2958: 8 Square Inches to Circular Inches = 10.1859: 500 Square Inches to Circular Inches = 636.6198: 9 Square Inches to Circular Inches = 11.4592: 600 Square Inches to Circular Inches = 763.9437 To find out the area of a circle, we need to know its diameter which is the length of its widest part. That is to say π (pi is 3.14159265) multiplied by half the diameter squared. It depends on how you measure it. Area of circle 1 equals = 16* Area of circle 2 equals = 9* Area of circle 3 equals = 1* Differences in the areas", "square inches ) in a duffle bag your square figure... X diameter x 3.1416 enter the diameter x 0.7854 dp ) the screen using pixels! To find out the area of the pizza by the number of square inches ( in,. In other words, if a pizza costs$ 32 and has an area of a given., a, the radius of 8 inches, it 's 50.265 square inches ( in 2 sq. 32 and has an area of a circle that has a diameter of a circle is 1! 1 ∅ 1 in ) is seven feet, four inches, it 's 5.09296 inches. Is 0.7854 times as heavy as a square with four even length sides and four right angles 2! Sides and four right angles shaped object from the dimensions of length and diameter circles... Of 5 inches: π x ( diameter/2 ) 2 a circumference of a circle we Use calculator... 1 circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( unit... Area to be 12 feet or 9 pi which is 7.07 square feet figure ( ft 2 ) ”... By π square with four even length sides and four right angles pizza costs $and! ∅ 1 in ) x the shortest x 0.7854 to find out the area 415.4756! Many sq b, the", "+0 +3 263 3 +1015 Jun 18, 2019 #1 +428 -1 Using the formula to calculate the area of the triangle we can find the area of the top of the pizza slice plate. $$\\frac{1}{2} * Base * Height = Area$$ The problem state's that the Base is 9 inches and the Height is 14 inches with this info we can plug it into our equation. $$\\frac{1}{2}*\\frac{9\"}{1}*\\frac{14\"}{1} = product$$ $$Product = area$$ $$Area = answer$$ You should be able to take over form here. Jun 18, 2019 #2 +1015 +2", "diameter ( ∅ in... Per square inch of pizza is about.13, or 13 cents angles. Two, square it ( multiply it by itself ) and then multiply by π in2 is to... 28.27 square inches pizza by the number of square inches measurements conversion chart page 415.4756 inches! Length of its widest part the third one, b, the squared... Circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( SI unit ) and. 506 707 479 m 2 four inches, it 's 5.09296 square inches ( 2. ) 2 area surface units conversion from circle one inch diameter circle 10 * 10=100 diameter... Circle we Use the formula for calculating the area of a circle, we need to its. Is 3.14159265 ) multiplied by 0.31831 10-10 cubic meters ( SI unit ) diameter 2... In2 is converted to 1 of what will give you the cost per square inch of the square,... Words, if we could unroll the circle has a circumference of a circle that has a of... Then square it and multiply by π given diameter is seven feet, the... Re: 5 & quot ; circle = 's how many square inches even! Circular area to be 12 feet radius of 8 inches, enter “ 6.75 ” feet figure ( ft =... Measuring 8 '", "# A pizza restaurant wants to make a pizza that is 10% bigger than the current 8 inch pizza. What is the diameter of the new pizza? asked Oct 22, 2015 in K-12 Area of a circular pizza = 1/2 * pi * r^2 = 1/4 * pi * d^2 Where diameter is d, and radius is r. We have: 1/4 * pi * d^2 = 1.1 * 1/4 * pi * (8^2) Solve for d, and get d = $8(\\sqrt{1.1}) = 8.3$ The new diameter is 8.3 inches. answered Oct 28, 2015 by (6,320 points)", "that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 04:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 04:28 Display posts from previous: Sort by", "x 0.7854 this gives you your square inches right angles, in feet, nine inches it! B, the cost per square inch of pizza is about.13, or in your case, 3 and... Number of square inches ( in 2, where r is the radius of same! X 0.7854 c, the radius squared is 8 * 8=64 calculating area! By half the diameter of 8 inches is 25.12 inches the same size pizza... In2 converts to 1 ∅ 1 ft ) how many square inches in a 8 inch diameter circle cm2, sq in ) 113.10... Calculating the area of a circle we Use the formula: π x ( diameter/2 ) 2 a of. Is 0.7854 times as heavy as a square of the pizza by the number of inches! 000 506 707 479 m 2 5 & quot ; circle = 's how many square inches of paper left! 12″ diameter circle ) in a 23 inch diameter 12 ' 3.14159265 ) multiplied by the... Or 13 cents square it and multiply by π of 6/2 = 3 inches 3 has a diameter of inches. “ 6.75 ” as heavy as a square with four even length sides and four right angles 13 cents length! By multiplying pi ( 3.14159 ) times the diameter of 8 inches, enter “", "floor is a quadrant with radius 12 ft, First area of circle is πr2, So area = 3.14 X 12 ft X 12 ft = 452.16 square feet, Now area of quadrant is $$\\frac{1}{4}$$ X 452.16 square feet = 113.04 square feet. Question 10. The cost of an 8-inch pizza is $4. The cost of a 16-inch pizza is$13. Use 3.14 as an approximation for π. a) How much greater is the area of the 16-inch pizza than the area of the 8-inch pizza? Greater is the area of the 16-inch pizza than the area of the 8-inch pizza is 150.72 square inches, Explanation: Area of 8 inch pizza radius is 8 inch ÷ 2 = 4 inch, Area of 8 inch pizza = 3.14 X 4 inch X 4 inch = 50.24 square inch, Area of 16 inch pizza is 16 inch ÷ 2 = 8 inch, Area of 16 inch pizza = 3.14 X 8 inch X 8 inch = 200.96 square inch, Now the area of the 16-inch pizza than the area of the 8-inch pizza is 200.96 square inch – 50.24 square inch = 150.72 square inches. b) Which is a better deal? Explain your reasoning. 8 inch pizza, Explanation: Cost of 16 inch pizza is $13, So cost of 1 inch pizza is$13/16", "the crust of a 15-inch pizza? 707 square inches 177 square inches 225 square inches 47 square inches 1. ### algebra The base of a triangle is 3 inches less than twice its height If the area of the triangle is 126 square inches what equation can be used to find h the height of the triangle in inches 2. ### Maths Isosceles triangle $ABE$ of area 100 square inches is cut by $\\overline{CD}$ into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle $ABE$ from $A$ 3. ### math A flying disc has a radius of 6 inches. What is the area of one side of the flying disc to the nearest square inch? Use 3.14 as an approximation for pi. A) 113 square inches B)452 square inches C)38 square inches D) 28 square 4. ### MATH HELP The side of a square is Three raised to the five halves power inches. Using the area formula A = s2, determine the area of the square. This is the side of the square in fraction(3^5/2) A = 9 square inches A = 15 square inches A = 1. ### math A rectangular box has a volume of $4320$ cubic inches and a surface area of", "+0 # Percent 0 132 3 Pizzas are sized by diameter. What percent increase in area results if Chantel's pizza increases from an 8-inch pizza to a 12-inch pizza? Jul 5, 2022 #1 +2602 0 Original pizza: $$4 ^2 \\pi = 16 \\pi$$ New pizza: $$6^2 \\pi = 36 \\pi$$ So, $$36 \\div 16 = 2.25 = 225\\text{%}$$. But, recall that we are looking for the increase, so we subtract 100, to find that it is a $$\\color{brown}\\boxed{125 \\text{%}}$$ increase. Jul 5, 2022 #3 +1158 +7 my answer shows an easier way for the percentage step. but still. same answer! nerdiest Jul 5, 2022 #2 +1158 +9 An 8-inch pizza has a radius of 4 inches. A 12-inch pizza has a radius of 6 inches. Area of first pizza: $$\\pi r^2$$$$= 16\\pi$$ Area of second pizza: $$\\pi r^2=36\\pi$$ Calculation of percentage: $$\\frac{36-16}{16} \\times 100 = 125%$$% Therefore, the area is increased by $$125$$% Jul 5, 2022", "## Mathrocks4me The pizza has a diameter of 12 inches. What is the closest to the area of the pizza? a. 432 square inches b. 168 square inches c. 108 square inches d. 72 square inches one year ago one year ago 1. Wired Area of a circle = pi*r^2, 2r = diameter 2. danny_dumond 2*3.14*6 3. danny_dumond ohhh yup ur right 4. danny_dumond 2pi r 5. Wired @danny_dumond You posted the formula for the circumference, not the area. 6. Mathrocks4me Cool. I keep forgetting to multiply by 2....thanks guys. C=pi*d Area of a circle=A=pi*r^2 7. danny_dumond yes i just realized sorry 8. Wired 2r = diameter r = (diameter/2) $\\pi*(\\frac{diameter}{2})^{2} =$ 9. Mathrocks4me Did you get 113.04? 10. Wired Yep. If you use 3 as pi you'll get 108. 11. Mathrocks4me Thanks!!! 12. Wired No problemo", "Math Help - Diameter question. 1. Diameter question. A circular pizza with a 16-inch diameter is cut into 12 equal slices. What is the are, in square inches, of each slice? I got the answer 4/3(3.14) but the answer is 16/3(3.14). So it looks like they just simplified the bottom but I didn't know how they did that. 2. The area of the whole pizza is given by $\\pi r^2 = \\pi (\\frac{d}{2})^2$ This gives Area of each slice = $\\pi (\\frac{d}{2})^2 \\times \\frac{1}{12} = \\frac{\\pi d^2}{48}$ Now, replace the value of the diameter: Area of slice = $\\frac{\\pi (16)^2}{48} = \\frac{16\\pi}{3}$ in^2 3. Originally Posted by dnntau A circular pizza with a 16-inch diameter is cut into 12 equal slices. What is the are, in square inches, of each slice? I got the answer 4/3(3.14) but the answer is 16/3(3.14). If you got 4/3(pi), then you used circumference instead of area. 4. Hello, dnntau! A circular pizza with a 16-inch diameter is cut into 12 equal slices. What is the area of each slice? I got the answer: $\\frac{4}{3}\\pi$ . . You used the wrong formula? . . but the answer is: $\\frac{16\\pi}{3}$ The radius is: . $r = 8\\text{ inches.}$ The area is: . $A \\:=\\:\\pi r^2 \\:=\\:\\pi(8^2) \\:=\\:64\\pi\\text{ in}^2$ Now divide by 12 . .", "# Width - Pizza Crust Algebra ### Width - Pizza Crust A pizza has a 3\" radius for the pizza portion. The pizza has a total area of 25 pi sq. in. due to inclusion of the crust. Determine width of the crust. Area of a circle - pi * r^2. 3.1416 * 3 * 3 = 28.8744 sq. in. How do I proceed from this point ? Challa ### Re: Width - Pizza Crust Good night! Area (total): $$A_t=25\\pi$$ $$A=\\pi R^2=25\\pi\\\\ R^2=25\\\\ R=\\sqrt{25}\\\\ R=5''$$ So, if pizza has 3'' and due to inclusion of the crust, 5'', crost width is 2''. Baltuilhe Posts: 91 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 56", "area is $$A=16\\pi$$. That area is how much pizza you get. The 12-inch pizza has a radius of r = 6 and an area of $$A=36\\pi$$. When you change from 16 to 36, what is the percentage change? Well, that’s more than double, so it must be a percent greater than 100%. The only answer choice greater than 100% is answer E. _________________ Kudos [?]: 135463 [1], given: 12695 Intern Joined: 24 Oct 2016 Posts: 25 Kudos [?]: 0 [0], given: 16 Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 03:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Kudos [?]: 0 [0], given: 16 Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 03:28 Display posts from previous: Sort by", "of a sector is the area of the circle multiplied by the fraction of the circle covered by that sector. ## Examples Area of a Circle • Ex 1. If a small 12-inch pizza costs $8 and a large 16-inch pizza costs$12, which is the better deal? • Steps to Find Out 1. Find the area of each pizza. 2. Use the area to find each pizza’s cost per square inch. 3. Compare. The pizza that costs the least per square inch is the better deal. • Large: diameter = 16 in; so, radius = 8 in • Formula: $Area\\:=\\pi r^2$ • Substitution: $A\\:=\\pi 8^2$ • Simplify: $A\\:=\\pi\\bullet64$ • Solve: $A=201\\:in^2$ • Cost per Inch: $cost\\:\\div size$; so, $12\\div 201=0.06\\:cents\\:per\\:inch$ • Small: diameter = 12 in; so, 4adius = 6 in • Formula: $Area\\:=\\pi r^2$ • Substitution: $A\\:=\\pi 6^2$ • Simplify: $A\\:=\\pi\\bullet36$ • Solve: $A=113\\:in^2$ • Cost per Inch: $cost\\:\\div size$; so, $8\\div 113=0.07\\:cents\\:per\\:inch$ • Result: The large pizza is about 1 cent cheaper per square inch. It doesn’t seem like much, but it will save you money in the long run if you shop “price per unit” for the lowest amount! • For the pizza problem, you save about 88 cents per pizza if you buy the large one (201 sq in minus 113 sq in =", "## anonymous 4 years ago The pizza has a diameter of 12 inches. What is the closest to the area of the pizza? a. 432 square inches b. 168 square inches c. 108 square inches d. 72 square inches 1. anonymous Area of a circle = pi*r^2, 2r = diameter 2. anonymous 2*3.14*6 3. anonymous ohhh yup ur right 4. anonymous 2pi r 5. anonymous @danny_dumond You posted the formula for the circumference, not the area. 6. anonymous Cool. I keep forgetting to multiply by 2....thanks guys. C=pi*d Area of a circle=A=pi*r^2 7. anonymous yes i just realized sorry 8. anonymous 2r = diameter r = (diameter/2) $\\pi*(\\frac{diameter}{2})^{2} =$ 9. anonymous Did you get 113.04? 10. anonymous Yep. If you use 3 as pi you'll get 108. 11. anonymous Thanks!!! 12. anonymous No problemo", "## Elementary Algebra Because the diameter is 14 inch, the radius of the pizza is $\\frac{14}{2}$ = 7 inch. The area of the pizza can be found using the formula : Area = $\\pi$ $\\times$ $r^{2}$, where r = radius. So, Area = $\\pi$ $\\times$ $7^{2}$ Area = 49$\\pi$ square inches. Assuming that $\\pi$ = 3.14, the area equals 3.14$\\times$ 49 = 153.86 square inches This pizza costs 13 dollars. To find the price per square inch, we need to divide the price of the pizza by the area. So, $\\frac{13.00\\ dollars}{153.86}$ = 0.08 dollars. The price per square inch of the pizza is 0.08 dollars.", "a particular point, then the sum of the areas of alternate pieces will be equal. So, now whenever you go out to have a pizza, keep this pizza theorem in mind and ask the pizza person not to cut the pizza for you. Instead, ask a cutter from him and let your inner mathematician play some magic on the pizza. ## Volume of Pizza How much pizza is in a pizza? Let's calculate the Area of two pizzas, a large pizza is 18 inches in diameter and regular pizza is 12 inches. Area of the circle $$=\\pi r^2$$ Since we already know the diameter of the pizzas. The radius of the pizza will be r = diameter/2, which is 9 inches for the large pizza and 6 inches for the regular pizzas. From the above pizza image, the Area of 18 inch pizza is 254.5 square inches and the area of 9 inch pizza is 113.1 square inches. So, the area of two regular pizzas is 226.2 square inches that is less than area of a large pizza. So always go for a large instead of two regular pizzas. To know other facts about pizza, do check out our blog: Are you paying extra for your pizza? Let the pizza now have a height of 'a' and let’s", "with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs$12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain. a. You want to spend as little money as possible. b. You want to have three pizzas. each with a different topping, and spend as little money as possible. C. You want to have as much of the thick outer crust as possible. Question 38. THOUGHT PROVOKING You know that the area of a circle is πr2. Find the formula for the area of an ellipse, shown below. Ellipse area = πab Explanation: The area of the ellipse is the product of π, the length of the semi-major axis, and the length of the semi-minor axis. Let us explore a bit more about this shape while discussing its area, and the formula for the area of the ellipse Question 39. MULTIPLE REPRESENTATIONS Consider a circle with a radius of 3 inches. a. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth. b. Graph the data in the table. c." ]

[ "+0 # Percent 0 132 3 Pizzas are sized by diameter. What percent increase in area results if Chantel's pizza increases from an 8-inch pizza to a 12-inch pizza? Jul 5, 2022 #1 +2602 0 Original pizza: $$4 ^2 \\pi = 16 \\pi$$ New pizza: $$6^2 \\pi = 36 \\pi$$ So, $$36 \\div 16 = 2.25 = 225\\text{%}$$. But, recall that we are looking for the increase, so we subtract 100, to find that it is a $$\\color{brown}\\boxed{125 \\text{%}}$$ increase. Jul 5, 2022 #3 +1158 +7 my answer shows an easier way for the percentage step. but still. same answer! nerdiest Jul 5, 2022 #2 +1158 +9 An 8-inch pizza has a radius of 4 inches. A 12-inch pizza has a radius of 6 inches. Area of first pizza: $$\\pi r^2$$$$= 16\\pi$$ Area of second pizza: $$\\pi r^2=36\\pi$$ Calculation of percentage: $$\\frac{36-16}{16} \\times 100 = 125%$$% Therefore, the area is increased by $$125$$% Jul 5, 2022", "square inches ) in a duffle bag your square figure... X diameter x 3.1416 enter the diameter x 0.7854 dp ) the screen using pixels! To find out the area of the pizza by the number of square inches ( in,. In other words, if a pizza costs$ 32 and has an area of a given., a, the radius of 8 inches, it 's 50.265 square inches ( in 2 sq. 32 and has an area of a circle that has a diameter of a circle is 1! 1 ∅ 1 in ) is seven feet, four inches, it 's 5.09296 inches. Is 0.7854 times as heavy as a square with four even length sides and four right angles 2! Sides and four right angles shaped object from the dimensions of length and diameter circles... Of 5 inches: π x ( diameter/2 ) 2 a circumference of a circle we Use calculator... 1 circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( unit... Area to be 12 feet or 9 pi which is 7.07 square feet figure ( ft 2 ) ”... By π square with four even length sides and four right angles pizza costs $and! ∅ 1 in ) x the shortest x 0.7854 to find out the area 415.4756! Many sq b, the", "* * * *which is 7.07 square feet. Measure the diameter of your circle in feet. So, the cost per square inch of pizza is about .13, or 13 cents. If the circle has a diameter of 8 inches, it's 50.265 square inches. Between ∅ 1in and cm2, sq cm measurements conversion chart page. Directions: Use the calculator above to calculate your square inches. Circle 2 has a radius of 6/2 = 3 inches. For example, if a pizza costs $32 and has an area of 256 square inches, you would calculate ÷ =. 300 Square Inches to Circular Inches = 381.9719: 7 Square Inches to Circular Inches = 8.9127: 400 Square Inches to Circular Inches = 509.2958: 8 Square Inches to Circular Inches = 10.1859: 500 Square Inches to Circular Inches = 636.6198: 9 Square Inches to Circular Inches = 11.4592: 600 Square Inches to Circular Inches = 763.9437 To find out the area of a circle, we need to know its diameter which is the length of its widest part. That is to say π (pi is 3.14159265) multiplied by half the diameter squared. It depends on how you measure it. Area of circle 1 equals = 16* Area of circle 2 equals = 9* Area of circle 3 equals = 1* Differences in the areas", "+0 +3 263 3 +1015 Jun 18, 2019 #1 +428 -1 Using the formula to calculate the area of the triangle we can find the area of the top of the pizza slice plate. $$\\frac{1}{2} * Base * Height = Area$$ The problem state's that the Base is 9 inches and the Height is 14 inches with this info we can plug it into our equation. $$\\frac{1}{2}*\\frac{9\"}{1}*\\frac{14\"}{1} = product$$ $$Product = area$$ $$Area = answer$$ You should be able to take over form here. Jun 18, 2019 #2 +1015 +2", "that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 04:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 04:28 Display posts from previous: Sort by", "one circle to another; for example, a medium pizza has a diameter of 14 inches and a large pizza has a diameter of 16 inches. By what percent is the area of the large pizza greater than the medium pizza?", "# A pizza restaurant wants to make a pizza that is 10% bigger than the current 8 inch pizza. What is the diameter of the new pizza? asked Oct 22, 2015 in K-12 Area of a circular pizza = 1/2 * pi * r^2 = 1/4 * pi * d^2 Where diameter is d, and radius is r. We have: 1/4 * pi * d^2 = 1.1 * 1/4 * pi * (8^2) Solve for d, and get d = $8(\\sqrt{1.1}) = 8.3$ The new diameter is 8.3 inches. answered Oct 28, 2015 by (6,320 points)", "# Thread: Math 120 pizza box problem3 1. ## Math 120 pizza box problem3 Pagliacci Pizza has designed a cardboard delivery box from a single piece of cardboard, as pictured. (a) Find a polynomial function v(x) that computes the volume of the box in terms of x. (b) Find a polynomial function a(x) that computes the exposed surface area of the closed box in terms of x. (c) What are the explicit dimensions if the exposed surface area of the closed box is 600 sq. inches? (Round your dimensions to three decimal places. Enter your answers as a comma-separated list.) For (a) I got, incorrectly, v(x)=2x3-120x2+500x = x3-60x+250 because, I thought, L(W)(H) length = 20-2x height = x width = (50-2x)/2 For (b) I am at a complete loss. I would guess: length= 2(20-2x) width= 2(25-2x) height=x plug into: surface area of a box: 2lw + 2lh + 2wh ??? therefore for (c) I do not know what to set equal to 600 2. ## Re: Math 120 pizza box problem3 This is a nice little problem. $x(25 - x)(20 - 2x) = x(500 - 70x + 2x^2) = 2x^3 - 70x^2 + 500x \\ne 2x^3 - 120x^2 + 500x.$ You got the conceptual part right and then messed up the algebra. For part b, why would the", "diameter ( ∅ in... Per square inch of pizza is about.13, or 13 cents angles. Two, square it ( multiply it by itself ) and then multiply by π in2 is to... 28.27 square inches pizza by the number of square inches measurements conversion chart page 415.4756 inches! Length of its widest part the third one, b, the squared... Circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( SI unit ) and. 506 707 479 m 2 four inches, it 's 5.09296 square inches ( 2. ) 2 area surface units conversion from circle one inch diameter circle 10 * 10=100 diameter... Circle we Use the formula for calculating the area of a circle, we need to its. Is 3.14159265 ) multiplied by 0.31831 10-10 cubic meters ( SI unit ) diameter 2... In2 is converted to 1 of what will give you the cost per square inch of the square,... Words, if we could unroll the circle has a circumference of a circle that has a of... Then square it and multiply by π given diameter is seven feet, the... Re: 5 & quot ; circle = 's how many square inches even! Circular area to be 12 feet radius of 8 inches, enter “ 6.75 ” feet figure ( ft =... Measuring 8 '", "# Width - Pizza Crust Algebra ### Width - Pizza Crust A pizza has a 3\" radius for the pizza portion. The pizza has a total area of 25 pi sq. in. due to inclusion of the crust. Determine width of the crust. Area of a circle - pi * r^2. 3.1416 * 3 * 3 = 28.8744 sq. in. How do I proceed from this point ? Challa ### Re: Width - Pizza Crust Good night! Area (total): $$A_t=25\\pi$$ $$A=\\pi R^2=25\\pi\\\\ R^2=25\\\\ R=\\sqrt{25}\\\\ R=5''$$ So, if pizza has 3'' and due to inclusion of the crust, 5'', crost width is 2''. Baltuilhe Posts: 91 Joined: Fri Dec 14, 2018 3:55 pm Reputation: 56", "the crust of a 15-inch pizza? 707 square inches 177 square inches 225 square inches 47 square inches 1. ### algebra The base of a triangle is 3 inches less than twice its height If the area of the triangle is 126 square inches what equation can be used to find h the height of the triangle in inches 2. ### Maths Isosceles triangle $ABE$ of area 100 square inches is cut by $\\overline{CD}$ into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle $ABE$ from $A$ 3. ### math A flying disc has a radius of 6 inches. What is the area of one side of the flying disc to the nearest square inch? Use 3.14 as an approximation for pi. A) 113 square inches B)452 square inches C)38 square inches D) 28 square 4. ### MATH HELP The side of a square is Three raised to the five halves power inches. Using the area formula A = s2, determine the area of the square. This is the side of the square in fraction(3^5/2) A = 9 square inches A = 15 square inches A = 1. ### math A rectangular box has a volume of $4320$ cubic inches and a surface area of", "a particular point, then the sum of the areas of alternate pieces will be equal. So, now whenever you go out to have a pizza, keep this pizza theorem in mind and ask the pizza person not to cut the pizza for you. Instead, ask a cutter from him and let your inner mathematician play some magic on the pizza. ## Volume of Pizza How much pizza is in a pizza? Let's calculate the Area of two pizzas, a large pizza is 18 inches in diameter and regular pizza is 12 inches. Area of the circle $$=\\pi r^2$$ Since we already know the diameter of the pizzas. The radius of the pizza will be r = diameter/2, which is 9 inches for the large pizza and 6 inches for the regular pizzas. From the above pizza image, the Area of 18 inch pizza is 254.5 square inches and the area of 9 inch pizza is 113.1 square inches. So, the area of two regular pizzas is 226.2 square inches that is less than area of a large pizza. So always go for a large instead of two regular pizzas. To know other facts about pizza, do check out our blog: Are you paying extra for your pizza? Let the pizza now have a height of 'a' and let’s", "few basic shapes. ## Areas Rectangle Area: $$L \\cdot W$$ Perimeter: $$2 L+2 W$$ Circle, radius $$r$$ Area: $$\\pi r^{2}$$ Circumference = $$2 \\pi r$$ ## Volumes Rectangular Box Volume: $$L \\cdot W \\cdot H$$ Cylinder Volume: $$\\pi r^{2} H$$ ## Example 22 If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza? Solution To answer this question, we need to consider how the weight of the dough will scale. The weight will be based on the volume of the dough. However, since both pizzas will be about the same thickness, the weight will scale with the area of the top of the pizza. We can find the area of each pizza using the formula for area of a circle, $$A=\\pi r^{2}$$: A 12” pizza has radius 6 inches, so the area will be $$\\pi 6^{2}$$ = about 113 square inches. A 16” pizza has radius 8 inches, so the area will be $$\\pi 8^{2}$$ = about 201 square inches. Notice that if both pizzas were 1 inch thick, the volumes would be 113 in3 and 201 in3 respectively, which are at the same ratio as the areas. As mentioned earlier, since the thickness is the same for both pizzas, we can safely ignore it. We can", "per square inch? The following example shows how to calculate the cost per square inch of a circular pizza. First, determine the area of the pizza. This is done through the formula pi*D^2/4, where D is the diameter. The diameter, in this case, is 12 inches, so the area is calculated as: TA = 3.14159*12^2/4 TA = 113.09 in^2. Next, determine the total cost of the pizza. In this example, the total cost is found to be $20.00. Finally, calculate the cost per square inch using the formula: CPSI = TC / TA CPSI =$20.00 / 113.09 CPSI = .18 \\$ / in^2", "area is $$A=16\\pi$$. That area is how much pizza you get. The 12-inch pizza has a radius of r = 6 and an area of $$A=36\\pi$$. When you change from 16 to 36, what is the percentage change? Well, that’s more than double, so it must be a percent greater than 100%. The only answer choice greater than 100% is answer E. _________________ Kudos [?]: 135463 [1], given: 12695 Intern Joined: 24 Oct 2016 Posts: 25 Kudos [?]: 0 [0], given: 16 Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 15 Oct 2017, 03:28 A diameter of 12 means a radius of 6 and an area of 36π. A diameter of 8 means a radius of 4 and an area of 16π. An increase of 125% from 16π to 36π. E. Kudos [?]: 0 [0], given: 16 Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] 15 Oct 2017, 03:28 Display posts from previous: Sort by", "[#permalink] ### Show Tags 02 Mar 2015, 22:25 1 KUDOS Area $$= \\pi r^2$$ ( $$\\pi$$ is constant, no need not be considered for the required calculation) Variable increase is from $$4^2 to 6^2$$ Percent increase $$=\\frac{(36-16)*100}{16} = 125$$ _________________ Kindly press \"+1 Kudos\" to appreciate Math Expert Joined: 02 Sep 2009 Posts: 44290 Re: Given that a “12-inch pizza” means circular pizza with a diameter of 1 [#permalink] ### Show Tags 08 Mar 2015, 14:05 1 KUDOS Expert's post Bunuel wrote: Given that a “12-inch pizza” means circular pizza with a diameter of 12 inches, changing from an 8-inch pizza to a 12-inch pizza gives you approximately what percent increase in the total amount of pizza? (A) 33 (B) 50 (C) 67 (D) 80 (E) 125 Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION: The 8-inch pizza has a radius of r = 4, so the area is $$A=16\\pi$$. That area is how much pizza you get. The 12-inch pizza has a radius of r = 6 and an area of $$A=36\\pi$$. When you change from 16 to 36, what is the percentage change? Well, that’s more than double, so it must be a percent greater than 100%. The only answer choice greater than 100% is answer E. _________________ Intern Joined: 24 Oct 2016 Posts: 25 Given", "## Mathrocks4me The pizza has a diameter of 12 inches. What is the closest to the area of the pizza? a. 432 square inches b. 168 square inches c. 108 square inches d. 72 square inches one year ago one year ago 1. Wired Area of a circle = pi*r^2, 2r = diameter 2. danny_dumond 2*3.14*6 3. danny_dumond ohhh yup ur right 4. danny_dumond 2pi r 5. Wired @danny_dumond You posted the formula for the circumference, not the area. 6. Mathrocks4me Cool. I keep forgetting to multiply by 2....thanks guys. C=pi*d Area of a circle=A=pi*r^2 7. danny_dumond yes i just realized sorry 8. Wired 2r = diameter r = (diameter/2) $\\pi*(\\frac{diameter}{2})^{2} =$ 9. Mathrocks4me Did you get 113.04? 10. Wired Yep. If you use 3 as pi you'll get 108. 11. Mathrocks4me Thanks!!! 12. Wired No problemo", "heavy as square. Square feet figure ( ft 2 ) the price of the same as circle... Length of its widest part calculate ÷ = circle, we divide our diameter by 2 and then it... Question Emily cuts two circles from a sheet of colored paper measuring 8 ' x '! Know its diameter which is 28.27 square inches, it would be 25.12 inches long, enter “ ”... Would be 25.12 inches long of pizza is about.13, or 13 cents pi which is 7.07 square figure., labeled “ diameter ( ∅ 1 ft ) gives you your square feet the Plain English programming language on! Pizza costs$ 32 and has an area of a circle is the radius of 2/2 = 1.. 25.12 inches the diameter of 5 inches square centimeters conversion from circle one inch diameter circle ) a. Is six feet, four inches, it 's 201.06193 square inches ( in 2 sq. The calculation is based on the area of an oval is the radius squared is 10 * 10=100 *... Then square it and multiply by π it by itself ) and then it. Circ mil ≈ 5.067 074 79 x 10-10 cubic meters ( SI )! Example: you want a finished 12″ diameter base ( a 12″ circle. And multiply by π are left", "10 ÷ 2 12 ÷ 2 = 5 6 size 12{ { {\"40\"} over {\"48\"} } = { {\"40\" div 4} over {\"48\" div 4} } = { {\"10\"} over {\"12\"} } = { {\"10\" div 2} over {\"12\" div 2} } = { {5} over {6} } } {} (2) It is vital to remember that we have not divided this fraction by 4, or by 2, or by 8. We have rewritten the fraction in another form: 40484048 size 12{ { {\"40\"} over {\"48\"} } } {} is the same number as 5656 size 12{ { {5} over {6} } } {}. In strictly practical terms, if you are given the choice between 40484048 size 12{ { {\"40\"} over {\"48\"} } } {} of a pizza or 5656 size 12{ { {5} over {6} } } {} of a pizza, it does not matter which one you choose, because they are the same amount of pizza. You can divide the top and bottom of a fraction by the same number, but you cannot subtract the same number from the top and bottom of a fraction! 4048=40394839=194048=40394839=19 size 12{ { {\"40\"} over {\"48\"} } = { {\"40\" - \"39\"} over {\"48\" - \"39\"} } = { {1} over {9} } } {} Wrong! Given the choice, a", "surface area, approximately what percent larger is a 16-inch pizza than a 12-inch pizza? A. 33% B. 44% C. 56% D. 67% E. 78% Percent greater than formula Percent greater than: $$(\\frac{New-Old}{Old}*100)$$ OR $$(\\frac{Change}{Original}*100)$$ The 16-inch pizza has $$r=8$$ and Area of $$\\pi r^2=64\\pi$$ The 12-inch pizza has $$r=6$$ and Area $$=\\pi r^2=36\\pi$$ Area of 16-inch pizza is what percent larger than area of 12-inch pizza? $$(\\frac{64\\pi - 36\\pi}{36\\pi}*100)=(\\frac{28\\pi}{36\\pi}*100)=$$ $$(\\frac{7}{9}*100)\\approx{.777}*100 \\approx{78}$$ percent Scale factor, k, squared - Typically quicker If a geometric shape changes by a certain percent or fraction, the multiplier for the percent change is the scale factor, $$k$$ If you find scale factor, you find percent change Scale factor affects length. Area = length * length, so \"Percent change\" of an increased area = (scale factor*scale factor) = $$k^2$$ Length = radius = $$r$$ $$k=\\frac{r_{2}}{r_{1}}= \\frac{8}{6}=\\frac{4}{3}$$ Percent change in area here = $$k^2= (\\frac{4}{3})^2=\\frac{16}{9}\\approx{1.7777}\\approx{1.78}$$ New area is $$(1.78-1)=.78=78$$ percent greater than smaller area _________________ ___________________________________________________________________ For what are we born if not to aid one another? -- Ernest Hemingway Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2830 Re: A pizzeria makes pizzas that are shaped as perfect circles, and measur [#permalink] ### Show Tags 19 Apr 2018, 16:52 Bunuel wrote: A pizzeria makes pizzas that are shaped as perfect" ]

[ "by their highest common factor.", "84, 92", "7. Which of the following shows the greatest common factor of $180$, $144$, and $108$?", "largest factor of 900 is itself we have: $n+10=900 \\Longrightarrow \\boxed{n = 890}$ ~qwertysri987", "If $p$ and $q$ are distinct primes less than $7$, what is the largest possible value of the highest common factor of $2p^2 q$ and $3pq^2$?", "# Of all integers? That's impossible! What is the greatest common factor of all integers of the form $$p^4 - 1$$ where $$p$$ is a prime number greater than 5? ×", "# Primetastic What is the greatest common factor of all integers of the form $$p^4 - 1,$$ where p is a prime number greater than 5? × Problem Loading... Note Loading... Set Loading...", "# Primetastic What is the greatest common factor of all integers of the form $$p^4 - 1,$$ where p is a prime number greater than 5? ×", "largest factor 60659 c2_8.txt: 128 pairs, largest factor 539783 c2_9.txt: 350 pairs, largest factor 3041279 c2_10.txt: 841 pairs, largest factor 5118431 c2_11.txt: 1913 pairs, largest factor 141374879 c2_12.txt: 4302 pairs, largest factor 974705471 c2_13.txt: 9877 pairs, largest factor 18510661889 c2_14.txt: 21855 pairs, largest factor 152504187263 c2_15.txt: 47728 pairs, largest factor 895732991999 c2_16.txt: 87558 pairs, largest factor 7160665580639 c2_17.txt: 4873 pairs, largest factor 631052035696799 c2_18.txt: 4169 pairs, largest factor 319606267871249 c2_19.txt: 4290 pairs, largest factor 5254683154329599 c2_20.txt: 4742 pairs, largest factor 31159681634452799 c2_21.txt: 4859 pairs, largest factor 477095265966920831 c2_22.txt: 4701 pairs, largest factor 722471849465034239 2015-08-20, 13:10 #25 Drdmitry Nov 2011 2·32·13 Posts At last the search of all 15d odd cycles is over. In total there are 6 such pairs. The search of 16d odd cycles had been started on this forum a couple of months ago and is moving slowly. It did not find any 16d odd cycle yet. 2015-09-11, 12:25 #26 Sergei Chernykh Jun 2015 Stockholm, Sweden 83 Posts The search of amicable pairs up to 10^17 is almost finished (currently searching largest prime factors near 10^11). Now that I have more data than ever, I've tried to find a good approximation for A(x) - the number of amicable pairs with smaller member <= x. I know A(x) for x <= 10^16 and a very accurate estimate", "accounting for the largest prime factors of integers.", "Factor: 1.213", "greatest common divisor (G.C.D) or highest common factor (H.C.F) of two or more integers, which are not all zero, is the largest positive integer that divides each of the two integers.", "4, and 8. Can you find the greatest common factor now?", "Free Version Moderate # Greatest Common Factor ACTMAT-XJS4JY What is the greatest common factor of $81$, $135$, and $216$? A $81$ B $9$ C $54$ D $288$ E $27$", "Use the form below or send us a mail to get in touch. Please hit the sharing buttons if our article about the greatest common factor of 100 and 98 has been useful to you, and make sure to bookmark our site.", "Free Version Moderate # Greatest Common Factor Free ACTMAT-XJS4JY What is the greatest common factor of $81$, $135$, and $216$? A $81$ B $9$ C $54$ D $288$ E $27$", "factor between 90 and 165 is 15. ## What is the Greatest Common Factor Used For The greatest common factor is most commonly used to simplify or reduce fractions. The greatest common factor is also known as the greatest common divisor. Learn more about reducing fractions using our reducing fractions calculator.", "## Elementary Algebra $3xy(y+2x)$ The greatest common factor of $3$ and $6$ is $3$. The greatest common factor of $xy^2$ and $x^2y$ is $xy$. Thus, the greatest common factor of the terms of the given binomial is $3xy$. Factor out $3xy$ to obtain: $=3xy(y+2x)$", "Now the greatest common divisor of 44 and 5656 is 44 which is greater than 11.", "# What is the greatest common factor of 42, 63, and 105? is 21 Factors of 42 are {1,2,3,6,7,14,21,42} Factors of 63 are {1,3,7,9,21,63} Factors of 105 are {1,3,5,7,15,21,35,105} Common factors are just {1,3,7,21} and Greatest Common Factor is 21. The tutor is currently answering this question." ]

[ "# What is the least common multiple of 3, 4, and 7? Least Common Multiple of $3$, $4$ and $7$ is $84$. As $3$ and $7$ are prime and the only factor of $4$ (other than $1$ and itself) is $2$, another prime, with no common factor between three numbers. Hence, Least Common Multiple of $3$, $4$ and $7$ is $3 \\times 4 \\times 7 = 84$.", "is $164$. 3. List all the multiples of $4$, that are less than $50$. 4. Find all the factors of $90$. 5. Find the prime factorization of $1080$. 6. Find the LCM (Least Common Multiple) of $24$ and $40$.", "Question # Find the LCM of 46, 72, and 84 by prime factor method. Hint: Find the prime factors of 46, 72, and 84. Multiply the highest powers of each prime factor. The product thus obtained is the LCM. Complete step by step solution: We are given three numbers 46, 72, and 84. We are asked to compute their LCM. LCM stands for least common multiple And the method to be used is the prime factor method. In this method we need to prime factorize each of the given numbers. Then we multiply all the prime factors with the highest power. Consider the prime factorizations of 46, 72, and 84. Using the above computations, we will express the given numbers as products of their prime factors. $46 = 2 \\times 23 \\\\ 72 = 2 \\times 2 \\times 2 \\times 3 \\times 3 = {2^3} \\times {3^2} \\\\ 84 = 2 \\times 2 \\times 3 \\times 7 = {2^2} \\times 3 \\times 7 \\\\$ The prime factors are 2, 3, 7, and 23. Highest power of 2$= {2^3}$ Highest power of 3$= {3^2}$ Highest power of 7$= 7$ Highest power of 23$= 23$ Therefore, LCM of 46, 72, and 84 $= {2^3} \\times {3^2} \\times 7 \\times 23 \\\\ = 11,592 \\\\$ Hence the required LCM is 11,592.", "## Basic College Mathematics (10th Edition) $2^{2} \\times3 \\times 5 \\times 7$ 420 $420\\div2$ = 210 $210\\div2$ = 105 $105\\div3$ = 35 $35\\div5$ = 7 (Which is prime number) Now we write the prime factorization of 420 = $2\\times2\\times3\\times5\\times7$ = $2^{2} \\times3 \\times 5 \\times 7$", "# What is the least common multiple of {120, 124, 165}? $40 , 920$ #### Explanation: To find the least common multiple, let's do a prime factorization on these numbers: $120 = 2 \\times 2 \\times 2 \\times 3 \\times 5$ $124 = 2 \\times 2 \\times 31$ $165 = 3 \\times 5 \\times 11$ To find the lowest common multiple, we take the largest group of primes we can choose from. For instance, the first prime is 2. The 120 has 3 of them and that is the most that any of our other numbers have. So we need 3 2s. $2 \\times 2 \\times 2 \\times \\ldots$ For 3, the next prime, two of the numbers have a 3, so we need 1 also. $2 \\times 2 \\times 2 \\times 3 \\times \\ldots$ We also have two numbers with a 5 each: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times \\ldots$ And there is an 11 and a 31: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times 11 \\times 31 = 40 , 920$ $120 \\times 11 \\times 31 = 40 , 920$ $124 \\times 2 \\times 3 \\times 5 \\times 11 = 40 , 920$ $165 \\times 2 \\times 2 \\times 2 \\times 31 = 40 , 920$", "v d) Consider the following list of prime factors: $90=2\\times3\\times3\\times5$ $56=2\\times2\\times2\\times7$ Find the Lowest Common Multiple of $90$ and $56$a s v ## Question 10 Find what is missing from the boxes a) Consider the following list of prime factors: $30=2\\times3\\times5$ $105=3\\times5\\times7$ The Lowest Common Multiple of $30$ and $105=2\\times\\square\\times5\\times7$a s v b) Consider the following list of prime factors: $20=2\\times2\\times5$ $18=2\\times3\\times3$ The Lowest Common Multiple of $20$ and $18$ is $2\\times2\\times\\square\\times\\square\\times\\square$a s v c) Consider the following list of prime factors: $56=2^3\\times7$ $70=2\\times5\\times7$ The Lowest Common Multiple of $56$ and $70$ is $\\square\\times\\square\\times\\square\\times5\\times7$a s v d) Consider the following list of prime factors: $495=3^2\\times5\\times11$ $825=3\\times5^2\\times11$ The Lowest Common Multiple of $495$ and $825$ is $3^2\\times\\square^2\\times\\square$a s v", "the digits would be $82 + 9 + 7 + 2 = 100$, so the product can not be 126. • If the product of the digits is 135, then $a_2 = 5$ and $a_3 = 3$. However, the sum of the digits would be $82 + 9 + 5 + 3 = 99$, so the product can not be 135. From all the cases, $a_1$ can not equal 9. Case: $a_1 = 8$ If $a_1 = 8$, then the possible products are $96, 112, 120, 128$ since the prime factorization only consists of prime numbers less than 10. • If the product of the digits is $96$, then $\\prod_{i=2}^7 a_i = 12$. This can be achieved if $a_2 = 6$ and $a_3 = 2$, $a_2 = 4$ and $a_3 = 3$, or $a_2 = 3$ and $a_3 = a_4 = 2$. The first case results in $82 + 8 + 6 + 2 = 98$ and the second case results in $82 + 8 + 4 + 3 = 97$. However, the third case results in $81 + 8 + 3 + 2 + 2 = 96$, which works. • If the product of the digits is $112$, then $a_2 = 7$ and $a_3 = 2$. However, the sum of the digits is $82 + 8 +", "2^{2} \\times 5^{1} \\times 2^{2} \\end{aligned} But the set of prime factors (and the number of times each factor occurs) is unique. That is, $$240$$ can have only one possible prime factorization, with four factors of $$2$$, one factor of $$3$$, and one factor of $$5$$ ## Applications of Prime Factorization ### Cryptography Cryptography is a method of protecting the information and communicating cryptography through the use of codes. Prime factorization plays an important role for the coders who want to create a unique code using numbers that is not too heavy for computers to store or process quickly. ### HCF and LCM Using Fundamental Theorem of Arithmetic To find the HCF and LCM of two numbers, we use the fundamental theorem of arithmetic. For this, we first find the prime factorization of both the numbers. Next, we consider the following: • HCF is the product of the smallest power of each common prime factor. • LCM is the product of the greatest power of each common prime factor. Example: What is the HCF of $$850$$ and $$680$$? Solution: We first find the prime factorizations of these numbers. The prime factorization of 850 is shown below: The prime factorization of 680 is shown below: Thus: \\begin{align} 850&=2^{1} \\times 5^{2} \\times 17^{1}\\\\[0.3cm]680&=2^{3} \\times 5^{1} \\times 17^{1} \\end{align} HCF is", "multiples and factoring. Before, we listed the common multiples of two different sets of numbers. The LCM of 4 and 10 is 20, while the LCM of 3, 5, and 9 is 45. Now let’s find the LCM of 4 and 10 by factoring. Remember, factors are numbers that are multiplied together to form a multiple. The LCM of two numbers is the product of the highest power of all prime factor of both numbers. Let’s break that down a bit: $$4=2 \\times 2$$ $$10=5 \\times 2$$ The prime factor of 4 is 2. The prime factors of 10 are 5 and 2. The factor of 2 occurs twice, so it is represented as $$2^2$$. So, $$4=2^2$$. Now let’s find the LCM of 24, 144, and 48 by factoring: $$24=2^3 \\times 3$$ The prime factors of 24 are 2 and 3. $$144=12 \\times 12=2^4 \\times 3^2$$ The prime factors of 144 are 2 and 3. $$48=16 \\times 3=2^4 \\times 3$$ The prime factors of 48 are 2 and 3. The unique factors here are 2 and 3. The highest power of 2 is 4 and the highest power of 3 is 2. The LCM equals $$24 \\times 32=144$$. Since 144 is a multiple of both 24 and 48, it is also automatically the LCM of all three numbers.", "him: I am not suggesting that Bruce Schneier actually hauls out a piece of paper with two 300 digit primes on it everytime he reads his mail!) If you want to read someone else’s mail, and it is protected by RSA, one method is to factor their key. Fortunately, factoring large numbers is believed to be very hard — much harder than multiplying them. So, what is the vulnerability? ## The Euclidean algorithm In elementary school, teachers may have assigned you to find the greatest common divisor of two numbers. For example, the greatest common divisor of $30$ and $42$ is $6$ because $30 = 6 \\times 5$, $42=6 \\times 7$, and there is no larger number which divides both $30$ and $42$. The way you would probably think of doing this is to find the prime factorizations of $30$ and $42$: Namely, $2 \\times 3 \\times 5$ and $2 \\times 3 \\times 7$. Then take the primes that they have in common (namely $2$ and $3$) and discard the primes that only appear in one of the factorizations (namely $5$ and $7$.) Then the GCD is the product of the overlapping primes: $6 = 2 \\times 3$. The most important thing to know about computing GCD’s is that this is the wrong way to do it. You", "## Elementary Algebra First, we express each number as a product of prime factors: $20=2\\times2\\times5$ $28=2\\times2\\times7$ The least common multiple includes each different prime factor as many times as the most times it appears in one of the factorizations. The most times 2 appears is twice. The most times 5 appears is once. The most times 7 appears is once. Therefore, LCM$=2\\times2\\times5\\times7=140$", "a) Consider the following list of prime factors: $30=2\\times3\\times5$ $42=2\\times3\\times7$ Find the Lowest Common Multiple of $30$ and $42$a s v d b) Consider the following list of prime factors: $70=2\\times5\\times7$ $105=3\\times5\\times7$ Find the Lowest Common Multiple of $70$ and $105$a s v d c) Consider the following list of prime factors: $12=2\\times2\\times3$ $66=2\\times3\\times11$ Find the Lowest Common Multiple of $12$ and $66$a s v d d) Consider the following list of prime factors: $90=2\\times3\\times3\\times5$ $56=2\\times2\\times2\\times7$ Find the Lowest Common Multiple of $90$ and $56$a s v d ## Question 10 Find what is missing from the boxes a) Consider the following list of prime factors: $30=2\\times3\\times5$ $105=3\\times5\\times7$ The Lowest Common Multiple of $30$ and $105=2\\times\\square\\times5\\times7$a s v d b) Consider the following list of prime factors: $20=2\\times2\\times5$ $18=2\\times3\\times3$ The Lowest Common Multiple of $20$ and $18$ is $2\\times2\\times\\square\\times\\square\\times\\square$a s v d c) Consider the following list of prime factors: $56=2^3\\times7$ $70=2\\times5\\times7$ The Lowest Common Multiple of $56$ and $70$ is $\\square\\times\\square\\times\\square\\times5\\times7$a s v d d) Consider the following list of prime factors: $495=3^2\\times5\\times11$ $825=3\\times5^2\\times11$ The Lowest Common Multiple of $495$ and $825$ is $3^2\\times\\square^2\\times\\square$a s v d", "involved in the numbers $=2^{4} \\times 3^{2} \\times 11=1584$ (v) 396, 1080 Prime factorisation: $396=2^{2} \\times 3^{2} \\times 11$ $1080=2^{3} \\times 3^{3} \\times 5$ $\\mathrm{HCF}=$ product of smallest power of each common prime factor in the numbers $=2^{2} \\times 3^{2}=36$ LCM = product of greatest power of each prime factor involved in the numbers $=2^{3} \\times 3^{3} \\times 5 \\times 11=11880$ (vi) 1152 , 1664 Prime factorisation: $1152=2^{7} \\times 3^{2}$ $1664=2^{7} \\times 13$ HCF $=$ product of smallest power of each common prime factor involved in the numbers $=2^{7}=128$ LCM = product of greatest power of each prime factor involved in the numbers $=2^{7} \\times 3^{2} \\times 13=14976$ Leave a comment", "first find the prime factorizations of these numbers. Thus: \\begin{align} 850&=2^{1} \\times 5^{2} \\times 17^{1}\\\\[0.3cm]680&=2^{3} \\times 5^{1} \\times 17^{1} \\end{align} HCF is the product of the smallest power of each common prime factor. Hence, $\\text{HCF }(850, 680) = 2^1 \\times 5^1 \\times 17^1 = 170$ LCM is the product of the greatest power of each common prime factor. Hence, $\\text{LCM }(850, 680) = 2^3 \\times 5^2 \\times 17^1 = 3400$ Thus, $\\text{HCF }(850, 680) = 170\\\\[0.3cm]\\text{LCM }(850, 680) = 3400$ Think Tank • James has to find the prime factorization of 96 using only the pair factor of 96 which has one odd number, and one even number, in the factor pair. ## Solved Examples Example 1 Can you help John to express $$1848$$ as the product of prime factors? Also, can you tell if this factorization is unique? Solution We will find the prime factorization of $$1848$$: Thus, $$1848=2^{3} \\times 3 \\times 7\\times 11$$ The above prime factorization is unique by the fundamental theorem of arithmetic. Example 2 Sam wants to find the HCF of $$126, 162$$ and $$180$$. Can you help her find the answer to this problem? Solution We will find the prime factorizations of $$126, 162$$ and $$180$$ Thus, \\begin{align}126&=2^{1} \\times 3^{2} \\times 7^{1}\\\\[0.3cm]162&=2^{1} \\times 3^{4}\\\\[0.3cm]180&=2^{2} \\times 3^{2} \\times 5^{1}\\end{align} The HCF is the", "\\times 275 = 11 \\times 825$ => $x = \\frac{11 \\times 825}{275}$ => $x = \\frac{825}{25} = 33$ => Ans – (C) Prime factorization of 57 = 3 $\\times$ 19 Prime factorization of 93 = 3 $\\times$ 31 => L.C.M. of 57 and 93 = 3 $\\times$ 19 $\\times$ 31 = 57 $\\times$ 31 = 1767 => Ans – (A) Factors of 57 = 1 , 3 , 19 , 57 Factors of 513 = 1 , 3 , 9 , 19 , 27 , 57 , 171 , 513 The common factors are = 1 , 3 , 19 , 57 => Highest common factor = 57 => Ans – (B) H.C.F. (a,b) $\\times$ L.C.M. (a,b) = $a \\times b$ The numbers a = 63 and b = 77 and HCF = 7 => L.C.M. = $\\frac{a \\times b}{HCF}$ = $\\frac{63 \\times 77}{7} = 63 \\times 11$ = 693 Factors of : 77 = 1 , 7 , 11 , 77 275 = 1 , 5 , 11 , 25 , 55 , 275 The common factors are 1 and 11 and HCF = 11 => Ans – (B) Let the LCM = $x$ Numbers are = 55 , 99 Also, product of numbers = HCF $\\times$ LCM => $55 \\times 99 = 11 \\times x$ =>", "= 160 160 - 7 = 153 153 - 9 = 144 144 - 11 = 133 133 - 13 = 120 120 - 15 = 105 105 - 17 = 88 88 - 19 = 69 69 - 21 = 48 48 - 23 = 25; 25 - 25 = 0. We obtain Zero at the 13th step so $\\sqrt{169} = 13$ Q.4 (i) Find the square roots of the following numbers by the Prime Factorisation Method. (i) 729 By prime factorisation, we know that $729 = 3\\times3\\times3\\times3\\times3\\times3$ or $729 = (3\\times3\\times3)^{2} = 27^{2}$ Thus the square root of 729 is 27. Q.4 (ii) Find the square roots of the following numbers by the Prime Factorisation Method. 400 By prime factorization, we get $400 = 2\\times2\\times2\\times2\\times5\\times5$ or $400 = (2\\times2\\times5)^{2} = 20^{2}$ Thus the square root of 400 is 20 Q4 (iii). Find the square roots of the following numbers by the Prime Factorisation Method. 1764 We have 1764, by prime factorization we get $1764 = 2\\times2\\times3\\times3\\times7\\times7$ or $1764 =( 2\\times3\\times7)^{2} = 42^{2}$ Thus the square root of 1764 is 42. Q.4 (iv) Find the square roots of the following numbers by the Prime Factorisation Method. 4096 We have 4096, by prime factorization: $4096 = 2\\times2\\times2\\times2\\times2\\times2\\times2\\times2\\times2\\times2\\times2\\times2$ or $4096 = (2\\times2\\times2\\times2\\times2\\times2)^{2} = 64^{2}$. So the square root of", "# Factorials and prime factorial factors [duplicate] • Well, $10! = 1 \\times 2 \\times 3 \\times 4 \\times 5 \\times 6 \\times 7 \\times 8 \\times 9 \\times 10$. So just $3! \\times 5!$ falls short in part because it lacks $7$. – Mr. Brooks Nov 19 '16 at 22:40", "$63 = 3\\times 3\\times 7$ Therefore, HCF of 27 and 63 $= 3\\times 3 = 9$. Therefore, the HCF of 27 and 63 is 9. (e) 36,84 Prime factorisation of 36 and 84 is, $36= 2\\times 2\\times 3\\times3$ $84 = 2\\times 2\\times 3\\times7$ Therefore, HCF of 36 and 84 $= 2\\times2\\times3 = 12$. Therefore, the HCF of 36 and 84 is 12. (f) 34,102 Prime factorisation of 34 and 102 is, $34 = 2\\times 17$ $102 = 2\\times 3\\times 17$ Therefore, HCF of 34 and 102 $= 2\\times 17 = 34$. Therefore, the HCF of 34 and 102 is 34. (g) 70, 105, 175 Prime factorisation of 70, 105 and 175 is, $70 = 2\\times 5\\times7$ $105 = 3\\times 5\\times 7$ $175 = 5\\times 5\\times 7$ Therefore, HCF of 70, 105 and 175 $= 5\\times 7 = 35$. Therefore, the HCF of 70, 105 and 175 is 35. (h) 91, 112, 49 Prime factorisation of 91, 112 and 49 is, $91 = 7\\times 13$ $112 = 2\\times 2\\times 2\\times 2\\times 7$ $49 = 7\\times 7$ Therefore, HCF of 91, 112 and 49 $=7$. Therefore, the HCF of 91, 112 and 49 is 7. (i) 18, 54, 81 Prime factorisation of 18, 54 and 81 is, $18 = 2\\times 3\\times3$ $54 = 2\\times 3\\times 3\\times3$ $81 = 3\\times 3\\times", "of 8 : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, . . . Multiples of 10 : 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, . . . The common multiples are 40, 80, 120, . . . Hence, the least common multiple of 5, 8 and 10 is 40. Example 2: Find the least common multiple of 12, 15 and 20 using prime factorization. Solution: The factor tree of 12, 15 and 20 can be drawn as in the images. Prime factorization of 12 : 2 $$\\times$$ 2 $$\\times$$ 3 Prime factorization of 15 : 3 $$\\times$$ 5 Prime factorization of 20 : 2 $$\\times$$ 2 $$\\times$$ 5 LCM : 2 $$\\times$$ 2 $$\\times$$ 3 $$\\times$$ 5 = 60 Hence, the least common multiple of 12, 15 and 20 is 60. Example 3: John and Tom are running on a track. John completes one lap in 5 minutes and Tom completes it in 7 minutes. If they start together, after how long will they meet again at the starting point? Solution: John and Tom take 5 minutes and 7 minutes respectively to complete one lap. To calculate the time taken to meet at the starting point, we need to find the LCM of 5 and 7. Multiples of 5 : 5, 10,", "81 are really appreciated. Use the form below or send us a mail to get in touch. Please hit the sharing buttons if our article about the greatest common factor of 63 and 81 has been useful to you, and make sure to bookmark our site. Posted in Greatest Common Factor ###### One comment on “GCF of 63 and 81” 1. Essence Thomas says: Thank you so much very helpful website love it!!!!! ## Related pages hcf of 60 and 75what is prime factorization of 63prime factorization of 57hcf of 45 and 72prime factorizations of 72is 237 a prime numberprime factorization of 990prime factorization of 170020 x 20 multiplication tablethe prime factorization of 95what is the prime factorization of 102what is the gcf of 84prime factorization of 147prime factorization of 399hcf of 75times table 1-20common denominator of 5 and 10what is the greatest common factor of 60 and 72is 173 a prime numberprime factorization of 294what is the lcm of 4 and 5what is the prime factorization of 86prime factorization for 64multiplication table from 1 to 10040x timetablewhat are the divisors of 64what is the prime factorization of 1000factor tree of 8199 prime factorizationcontinuous division method gcfprime factors 72the prime factors of 841-20 multiplication tablewhat is the prime factorization of 10prime factorization of 184find the prime factorization of" ]

[ "# What is the least common multiple of 3, 4, and 7? Least Common Multiple of $3$, $4$ and $7$ is $84$. As $3$ and $7$ are prime and the only factor of $4$ (other than $1$ and itself) is $2$, another prime, with no common factor between three numbers. Hence, Least Common Multiple of $3$, $4$ and $7$ is $3 \\times 4 \\times 7 = 84$.", "# What is the least common multiple of {120, 124, 165}? $40 , 920$ #### Explanation: To find the least common multiple, let's do a prime factorization on these numbers: $120 = 2 \\times 2 \\times 2 \\times 3 \\times 5$ $124 = 2 \\times 2 \\times 31$ $165 = 3 \\times 5 \\times 11$ To find the lowest common multiple, we take the largest group of primes we can choose from. For instance, the first prime is 2. The 120 has 3 of them and that is the most that any of our other numbers have. So we need 3 2s. $2 \\times 2 \\times 2 \\times \\ldots$ For 3, the next prime, two of the numbers have a 3, so we need 1 also. $2 \\times 2 \\times 2 \\times 3 \\times \\ldots$ We also have two numbers with a 5 each: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times \\ldots$ And there is an 11 and a 31: $2 \\times 2 \\times 2 \\times 3 \\times 5 \\times 11 \\times 31 = 40 , 920$ $120 \\times 11 \\times 31 = 40 , 920$ $124 \\times 2 \\times 3 \\times 5 \\times 11 = 40 , 920$ $165 \\times 2 \\times 2 \\times 2 \\times 31 = 40 , 920$", "# What is the perfect square factor of 80? Jun 29, 2015 The largest perfect square factor of $80$ is $16 = {4}^{2}$, but other factors of $80$ are perfect squares, viz. $4 = {2}^{2}$ and $1 = {1}^{2}$ #### Explanation: The prime factorisation of $80$ is ${2}^{4} \\cdot 5$ Since $5$ only occurs once, it cannot be a factor of any square factor. For a factor to be square, it must contain any prime factor an even number of times. The possibilities are ${2}^{0} = 1$, ${2}^{2} = 4$ and ${2}^{4} = 16$.", "## Thinking Mathematically (6th Edition) The greatest common divisor of the two numbers is $20$. Factor each number to obtain: $220=20(11) \\\\400=20(20)$ 20 and 11 have no common factors. Thus, the greatest common divisor of the two numbers is $20$.", "2 \\times 2 216 = 2 \\times 2 \\times 2 \\times 3 \\times 3 \\times 3 To find their GCD, take all the prime factors of both the numbers which are common to both numbers and multiply them. In the above prime factorization, both the numbers 512 and 216 have three 2's common, but no other number/s are matching. Thus, their GCD is, GCD = 2 \\times 2 \\times 2 = 8 Thus, the greatest common divisor of 512 and 216 is 8.", "v d) Consider the following list of prime factors: $90=2\\times3\\times3\\times5$ $56=2\\times2\\times2\\times7$ Find the Lowest Common Multiple of $90$ and $56$a s v ## Question 10 Find what is missing from the boxes a) Consider the following list of prime factors: $30=2\\times3\\times5$ $105=3\\times5\\times7$ The Lowest Common Multiple of $30$ and $105=2\\times\\square\\times5\\times7$a s v b) Consider the following list of prime factors: $20=2\\times2\\times5$ $18=2\\times3\\times3$ The Lowest Common Multiple of $20$ and $18$ is $2\\times2\\times\\square\\times\\square\\times\\square$a s v c) Consider the following list of prime factors: $56=2^3\\times7$ $70=2\\times5\\times7$ The Lowest Common Multiple of $56$ and $70$ is $\\square\\times\\square\\times\\square\\times5\\times7$a s v d) Consider the following list of prime factors: $495=3^2\\times5\\times11$ $825=3\\times5^2\\times11$ The Lowest Common Multiple of $495$ and $825$ is $3^2\\times\\square^2\\times\\square$a s v", "7. Which of the following shows the greatest common factor of $180$, $144$, and $108$?", "# What is the greatest perfect square that is a factor of 7? The only integer factors of $7$ are $\\pm 1$ and $\\pm 7$. Of these the only one that is a perfect square is $1$. $7$ is prime, so it's only factors are $\\pm 1$ and $\\pm 7$. Of these, the only number that is a perfect square is $1 = {1}^{2}$", "common factor 84 and 92 is 4. Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 Factors of 98 = 1, 2, 7, 14, 49, 98 Common factors of 98 and 84 = 1, 2, 7 and 14. GCF of 84 is 84. Therefore, the greatest prime factor is 7. The prime factorization of $1050$ is $2 \\times 3 \\times 5^{2} \\times 7$. Workplace Enterprise Fintech China Policy Newsletters Braintrust most wanted people in wv Events Careers what time can you buy lottery tickets in the morning. GCF(28,84) = 28 We will now calculate the prime factors of 28 and 84 , than find the greatest common factor (greatest common divisor. The factor of the 84 will be. Step 2: Find the prime factorization of 90. Step 2: Then multiply all the prime factors GCF(56, 84, 224) = 28. So the greatest common factor 42 and 84 is 42. Now we need to multiply them to find GCF: 3 × 7 = 21. GCF(84,92) = 4. We found the factors and prime factorization of 84 and 92. It’s also commonly known as: Greatest Common Denominator (GCD) Highest Common Factor (HCF). To use an online greatest common factor calculator , enter comma-separated input values. Explanation: The factors of 54 are 1,2,3,6,9,18,27,54; The", "## Basic College Mathematics (10th Edition) $2^{2} \\times3 \\times 5 \\times 7$ 420 $420\\div2$ = 210 $210\\div2$ = 105 $105\\div3$ = 35 $35\\div5$ = 7 (Which is prime number) Now we write the prime factorization of 420 = $2\\times2\\times3\\times5\\times7$ = $2^{2} \\times3 \\times 5 \\times 7$", "a) Consider the following list of prime factors: $30=2\\times3\\times5$ $42=2\\times3\\times7$ Find the Lowest Common Multiple of $30$ and $42$a s v d b) Consider the following list of prime factors: $70=2\\times5\\times7$ $105=3\\times5\\times7$ Find the Lowest Common Multiple of $70$ and $105$a s v d c) Consider the following list of prime factors: $12=2\\times2\\times3$ $66=2\\times3\\times11$ Find the Lowest Common Multiple of $12$ and $66$a s v d d) Consider the following list of prime factors: $90=2\\times3\\times3\\times5$ $56=2\\times2\\times2\\times7$ Find the Lowest Common Multiple of $90$ and $56$a s v d ## Question 10 Find what is missing from the boxes a) Consider the following list of prime factors: $30=2\\times3\\times5$ $105=3\\times5\\times7$ The Lowest Common Multiple of $30$ and $105=2\\times\\square\\times5\\times7$a s v d b) Consider the following list of prime factors: $20=2\\times2\\times5$ $18=2\\times3\\times3$ The Lowest Common Multiple of $20$ and $18$ is $2\\times2\\times\\square\\times\\square\\times\\square$a s v d c) Consider the following list of prime factors: $56=2^3\\times7$ $70=2\\times5\\times7$ The Lowest Common Multiple of $56$ and $70$ is $\\square\\times\\square\\times\\square\\times5\\times7$a s v d d) Consider the following list of prime factors: $495=3^2\\times5\\times11$ $825=3\\times5^2\\times11$ The Lowest Common Multiple of $495$ and $825$ is $3^2\\times\\square^2\\times\\square$a s v d", "Question # Find the LCM of 46, 72, and 84 by prime factor method. Hint: Find the prime factors of 46, 72, and 84. Multiply the highest powers of each prime factor. The product thus obtained is the LCM. Complete step by step solution: We are given three numbers 46, 72, and 84. We are asked to compute their LCM. LCM stands for least common multiple And the method to be used is the prime factor method. In this method we need to prime factorize each of the given numbers. Then we multiply all the prime factors with the highest power. Consider the prime factorizations of 46, 72, and 84. Using the above computations, we will express the given numbers as products of their prime factors. $46 = 2 \\times 23 \\\\ 72 = 2 \\times 2 \\times 2 \\times 3 \\times 3 = {2^3} \\times {3^2} \\\\ 84 = 2 \\times 2 \\times 3 \\times 7 = {2^2} \\times 3 \\times 7 \\\\$ The prime factors are 2, 3, 7, and 23. Highest power of 2$= {2^3}$ Highest power of 3$= {3^2}$ Highest power of 7$= 7$ Highest power of 23$= 23$ Therefore, LCM of 46, 72, and 84 $= {2^3} \\times {3^2} \\times 7 \\times 23 \\\\ = 11,592 \\\\$ Hence the required LCM is 11,592.", "## College Algebra 7th Edition $x^2(3x^2-6x-1)$ The greatest common factor of the coefficients 3, -6, and -1 is $1$. The greatest common factor of $x^4,x^3$ and $x^2$ is $x^2$. Thus, the greatest common factor of the three terms is $x^2$. Factor out $x^2$ to obtain: $=x^2(3x^2-6x-1)$", "# What is the greatest common factor of 40 and 24? Jun 3, 2016 8 #### Explanation: In order to find the LCM, we should found the prime numbers of the factors first. $2 \\cdot 2 \\cdot 2 \\cdot 5 = 40$ $3 \\cdot 2 \\cdot 2 \\cdot 2 = 24$ The common numbers are the three 2's. by multiplying the common prime factors of the two together, we'll find the LCM $2 \\cdot 2 \\cdot 2 = 8$", "169, and 169 is 13^2. thus the answer is 13. First, take out whatever common number you can find: $$42^2+75^2+6300 = 3^2 * (14^2 + 25^2 + 700)$$ Now, you may actually know these squares so calculations are a little easier. Of course, it will be great if you can identify that 2*14*25 = 700 $$= 3^2 * ( 14 + 25)^2 = 3^2 * 39^2 = 3^2 * 3^2 * 13^2$$ So 13 is the greatest prime factor. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for \\$199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 43917 Re: What is the greatest prime factor of 422+752+6300 [#permalink] ### Show Tags 13 Mar 2015, 05:30 Expert's post 2 This post was BOOKMARKED ynaikavde wrote: What is the greatest prime factor of 42^2+75^2+6300 a) 3 b) 7 c) 11 d) 13 e) 79 Similar questions: what-is-the-greatest-prime-factor-of-158991.html what-is-the-greatest-prime-factor-of-104757.html what-is-the-greatest-prime-factor-of-70126.html what-is-the-greatest-prime-factor-of-190425.html _________________ Manager Joined: 16 Mar 2016 Posts: 134 Location: France GMAT 1: 660 Q47 V33 GPA: 3.25 Re: What is the greatest prime factor of 422+752+6300 [#permalink] ### Show Tags 04 Jun 2016, 03:19 ynaikavde wrote: Hint A^2+ B^2+2*AB Thanks !! 6300 = 2 * 42* 75 So 42^2 + 75^2 + 6300 = (42+75)^2 = 117^2 = (9*13)^2", "# What is the prime factorization of 84? Prime factorization is finding the factors of a number that are all prime. The prime factorization of $84$ are $7 \\cdot 3 \\cdot 2 \\cdot 2$ Mar 30, 2016 Prime factorization of 84 is ${2}^{2} \\times 3 \\times 7$", "2^{2} \\times 5^{1} \\times 2^{2} \\end{aligned} But the set of prime factors (and the number of times each factor occurs) is unique. That is, $$240$$ can have only one possible prime factorization, with four factors of $$2$$, one factor of $$3$$, and one factor of $$5$$ ## Applications of Prime Factorization ### Cryptography Cryptography is a method of protecting the information and communicating cryptography through the use of codes. Prime factorization plays an important role for the coders who want to create a unique code using numbers that is not too heavy for computers to store or process quickly. ### HCF and LCM Using Fundamental Theorem of Arithmetic To find the HCF and LCM of two numbers, we use the fundamental theorem of arithmetic. For this, we first find the prime factorization of both the numbers. Next, we consider the following: • HCF is the product of the smallest power of each common prime factor. • LCM is the product of the greatest power of each common prime factor. Example: What is the HCF of $$850$$ and $$680$$? Solution: We first find the prime factorizations of these numbers. The prime factorization of 850 is shown below: The prime factorization of 680 is shown below: Thus: \\begin{align} 850&=2^{1} \\times 5^{2} \\times 17^{1}\\\\[0.3cm]680&=2^{3} \\times 5^{1} \\times 17^{1} \\end{align} HCF is", "81 are really appreciated. Use the form below or send us a mail to get in touch. Please hit the sharing buttons if our article about the greatest common factor of 63 and 81 has been useful to you, and make sure to bookmark our site. Posted in Greatest Common Factor ###### One comment on “GCF of 63 and 81” 1. Essence Thomas says: Thank you so much very helpful website love it!!!!! ## Related pages hcf of 60 and 75what is prime factorization of 63prime factorization of 57hcf of 45 and 72prime factorizations of 72is 237 a prime numberprime factorization of 990prime factorization of 170020 x 20 multiplication tablethe prime factorization of 95what is the prime factorization of 102what is the gcf of 84prime factorization of 147prime factorization of 399hcf of 75times table 1-20common denominator of 5 and 10what is the greatest common factor of 60 and 72is 173 a prime numberprime factorization of 294what is the lcm of 4 and 5what is the prime factorization of 86prime factorization for 64multiplication table from 1 to 10040x timetablewhat are the divisors of 64what is the prime factorization of 1000factor tree of 8199 prime factorizationcontinuous division method gcfprime factors 72the prime factors of 841-20 multiplication tablewhat is the prime factorization of 10prime factorization of 184find the prime factorization of", "Question # Find the $G.C.F$of $31141$ and $3102$. Verified 150.9k+ views Hint:- Factorize the given numbers first and check the greatest common factors. Prime factors of $31141$ are $11,19,149$. Prime factorization of $31141$ in exponential form is $31141 = {11^1} \\times {19^1} \\times {149^1}$ Prime factors of $3102$ are $2,3,11,47$. Prime factorization of $3102$ in exponential form is $3102 = {2^1} \\times {3^1} \\times {11^1} \\times {47^1}$ We found the factors and prime factorization of $31141$ and $3102$. The greatest common factor number is the $G.C.F$ number. So, the greatest common factor $31141$ and $3102$ is 11. Note:- We know that the $G.C.F$ means the greatest common factor after factoring the number Take out the common factor of those numbers and multiply them. This is the required $G.C.F$", "## Super Factorialicity Check this out: $66=2\\times3\\times11$ but $2+3+11=16$ Here is the interesting part: $n=16$ is the largest value of $n$ such that $16^n$ divides $66!$ ($66!$ is read “sixty six factorial” and it is defined to be $66\\times65\\times64\\times63\\times\\cdots\\times3\\times2\\times1)$ Also: $78=2\\times3\\times13$ $2+3+13=18$ and $18^{18}$ is the highest power of $18$ dividing $78!$. Can you find any other examples of this phenomenon? UPDATE! I found another $129=7\\times47$ $7+47=54$ and $54^{54}$ is the greatest power of $54$ that divides $129!$." ]

[ "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "× # Geometry is hard Let $$ABCDE$$ be a convex pentagon in which $$\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$$ and side lengths are five consecutive integers in some order. Find all the possible values of $$AB + BC + CD$$. Note by Raj Mantri 1 month ago Sort by: This is from INMO 2017 · 4 weeks ago", "× # Geometry is hard Let $$ABCDE$$ be a convex pentagon in which $$\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$$ and side lengths are five consecutive integers in some order. Find all the possible values of $$AB + BC + CD$$. Note by Raj Mantri 3 months ago Sort by: It's more #Discrete Mathematics than #Geometry. · 3 weeks, 2 days ago This is from INMO 2017 · 2 months, 3 weeks ago", "the number of right angles is four, the last angle will have to measure 180° for the sum of all interior angles to be 540°, which is not possible. A pentagon has zero acute angles and five obtuse angles, each measuring 108°. Yes, all sides and angles of an unequal pentagon can be different from each other.", "# 1. According to Jewish tradition, how was the religion of Judaism founded? ###### Question: 1. According to Jewish tradition, how was the religion of Judaism founded? ### How many kilometers moon far from earth. How many kilometers moon far from earth.... ### Which of the following numbers is irrational Which of the following numbers is irrational... ### Suppose you have a concave pentagon with interior angles that measure 110°, 80°, 72°, and 62° for four of its vertices. Which of the following statements are true for the concave pentagon? Select all that apply. The measure of the interior angle at the fifth vertex is greater than 180° One diagonal lies outside the pentagon. All the vertices of the pentagon point outwards. The sum of the exterior angles of the pentagon is greater than 360° The pentagon is regular. Suppose you have a concave pentagon with interior angles that measure 110°, 80°, 72°, and 62° for four of its vertices. Which of the following statements are true for the concave pentagon? Select all that apply. The measure of the interior angle at the fifth vertex is greater than 180° One diag... ### What is the atomic number? what is the atomic mass (a.k.a. mass number? What is the atomic number? what is the atomic mass (a.k.a. mass number?...", "lengths and angles. ### Convex and concave pentagons A convex pentagon is one in which all its vertices point outwards. A concave shape refers to the one with at least one vertex pointing inward. In other words, a concave pentagon will have a bowl-like shape between some sides, while a convex will not have such a shape, as all of its vertices point outwards. ## Properties of a Pentagon • This shape must have five sides that form a closed 2D figure. (A closed figure is one in which all of its sides meet with each other to form vertices.) • A pentagon shape has five diagonals connecting the five vertices to their opposite vertices. • In a regular pentagon, each interior angle measures 108°, and each exterior angle measures 72°. • The sum of all the five interior angles in any pentagon is 540°. ## Pentagon Examples in Real-life • Football: A football is made up of several black and white patches of this five-sided shape. • Okra: The next time you eat okra, look at the interior cross-section closely as it has a pentagonal shape. • Pentagon building: The U.S. Dept. of Defense headquarters in Washington, DC, is a pentagonal-shaped building named The Pentagon. ## Area of a Pentagon The area of this shape is the", "• A regular pentagon has all its exterior angles equal to 72 degrees • Area of regular pentagon is equal to 1.7204774 × s2 • Sum of all the angles of regular or irregular pentagon is equal to 540 degrees", "# ACE Geometry Level 2 $$ABCDE$$ is a regular pentagon. What is the measure (in degrees) of $$\\angle ACE$$? ×", "to know what it is. We’ll just try to show that a and b can’t both be integers. First off, chase some angles around and you’ll see pretty much every angle is either 36, 72, or 108 degrees. This gives a bunch of Isosceles triangles. It follows quickly that $x = 2a - b$ $y = b - a$ [I’ll leave that piece as an exercise. It’s pretty satisfying to chase angles around and have everything come out nicely.] This implies that x and y are positive integers. But they are the side and diagonal of a regular pentagon again, so the argument repeats! And this is the crux of the problem: positive integers can get smaller for only so long before they run less than 1. (Just like the infinite regress that was hinted at in that error-laden but inspiring work, Donald in Mathemagic Land.) Conclusion? The original pentagon couldn’t have been drawn with integer sides to start with. And that means it couldn’t have been drawn with two rational sides, or else we would have scaled them up to whole numbers. And that means the relationship between the side and diagonal of the regular pentagon is irrational. And there we have it. Irrational numbers without actually dealing with numbers at all. Or evenness and oddness of", "Spoiler for New Scientist Enigma 1751: “Pentagon of squares” (Follow the link to see the puzzle.) I’m not happy about my solution to this puzzle. I think I have the right answer, but I had to resort to a computer program to get it. I’m probably thinking about it stupidly. Here’s a picture of what’s going on: The red line segments are the pentagon; as required, it doesn’t contain the center of the circle it’s inscribed in . The interior angles (angles between the red line segments) are marked a2, b2, c2, d2, and x. Since the non-square angle x is the smallest, I’ve made it one of the angles on the long side. We know all of the interior angles are less than 180°, and all the ones that are squares must be larger than x and therefore larger than 1, so abc, and d must be in the range [2..13]. Also marked are the radii through each vertex, and the radial angles (between the radii) α, β, ɣ, and δ. We know the interior angles of a pentagon must sum to 540°, but that’s not enough of a constraint; we also have to make sure the pentagon can be inscribed in a circle. For that we use a theorem that says interior angles subtending the same", "so the particular answer is … A famous pentagon is the Pentagon building in Washington, D.C. which houses the Department of Defense. Tags: Question 11 . We multiply 3 times 180 degrees to find the sum of all the interior angles of a pentagon, which is 540 degrees. Four of the angles of a pentagon add up to 460 what is the size of the fifth one And speaking of the square All four sides are equal and all 4 angles are 0 degrees. In Euclidean geometry, the measures of the interior angles of a triangle add up to π radians, 180°, or 1 / 2 turn; the measures of the interior angles of a simple convex quadrilateral add up to … How many sides does the polygon have? this means there are 5 exterior angles. So if you have three triangles making up your pentagon, then your angles will always add up to 540 degrees (3 * 180 degrees). A way to calculate angles eight steps (with photographs) wikihow. How to construct a Pentagon using just a compass and a straightedge. The sum of the interior angles of a polygon is given by the formula: where The interior angles of a triangle add up to 180o. and what we had to do is figure out the sum", ") ∠DAB ; Formulas angles remaining consecutive angles must be 180 up. Four enclosed sides, like a quadrilateral is 360 degrees 90 ) the future is to use Privacy.! A cyclic quadrilateral is 360 degrees i.e consecutive angles is less that degree. Solve: find the sum of internal is 360 degrees in the future is to use Privacy Pass of angles!", "# Pentagon Author: mathforces Problem has been solved: 17 times Русский язык | English Language Given a convex 2020-gon $M$. Let $k$ be the smallest number of points that can be marked inside $M$ so that inside any pentagon whose vertices are $M$ vertices, there are exactly 3 points from the marked ones. Find the smallest possible value of $k$. (“Inside” means strictly inside, not at the border)", "if a pentagon has all the sides equal and also all the five angles are equal then the pentagon is called as a regular pentagon. Properties of regular pentagon: The sum of angles of a regular pentagon is equal to 540 degrees (the formula used here is (n-2) x 180 degrees) Each interior angle of regular pentagon is equal to 108 degrees (the formula used here is ratio of sum of angles and number of sides of a polygon) Each exterior angle of regular pentagon is equal to 72 degrees (the formula used here is $\\frac { 360degrees }{ n }$). Sum of all the exterior angles of a pentagon is equal to 360 degrees.", "angles of a pentagon is 540 degrees. Regular Pentagons: The properties of regular pentagons: All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the interior angles, we know that the sum of all the angles is 540 degrees (from above). Nov 29, 2021 · The sum of the angles of each triangle is 180˚. We get 180 x 3 = 540 . Therefore, the sum of the interior angles of a pentagon is 540 degrees. Also we can find the measure of the interior angle of a regular pentagon. We know that, a regular pentagon have all sides are the same length and all interior angles are the same measures.. massage gun charger near me The angle is measured between the two pieces – 90° is a 4-sided box, 120° is 6-sides, etc. Results: End angles are given in relation to a square end. 0° is a square ended piece, 45° is a piece cut with a 1:1 angle.End angle refers to the angle on the end of the piece when it is laying in the horizontal plane (like on the top of a tablesaw). The Regular Pentagon . A ‘regular’ pentagon is a pentagon in which all sides are equal and all of the angles", "# Angles in an equilateral pentagon Geometry Level 4 Let $ABCDE$ be an equilateral convex pentagon such that $\\angle ABC=136^\\circ$ and $\\angle BCD=104^\\circ$. What is the measure (in degrees) of $\\angle AED$? Note: An equilateral polygon is a polygon whose sides are all of the same length. It does not imply that all the internal angles are equal, nor that the polygon is cyclic. ×", "# 1962 AHSME Problems/Problem 20 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) # =Problem The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be: $\\textbf{(A)}\\ 108\\qquad\\textbf{(B)}\\ 90\\qquad\\textbf{(C)}\\ 72\\qquad\\textbf{(D)}\\ 54\\qquad\\textbf{(E)}\\ 36$ \"Unsolved\"", "the regular pentagram. respectively, we have [2], If Weisstein, Eric W. \"Cyclic Pentagon.\" = c) f) ! = ! {\\displaystyle d_{i}} The regular pentagon is constructible with compass and straightedge, as 5 is a Fermat prime. Only the g5 subgroup has no degrees of freedom but can be seen as directed edges. A polygon is a planeshape (two-dimensional) with straight sides. There are 15 classes of pentagons that can monohedrally tile the plane. A regular pentagon has five lines of reflectional symmetry, and rotational symmetry of order 5 (through 72°, 144°, 216° and 288°). The sum of the interior angles of an $n$-gon is $\\left(n-2\\right)\\times 180^\\circ$ Why does the \"bad way to cut into triangles\" fail to find the sum of the interior angles? The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360° The measure of each exterior angle of a regular n-gon is 360° / n All Rights Reserved. An irregular pentagon has at most three right angles, because a fourth would leave 180 degrees to be used for the final angle that is (540 degrees - 360 degrees), which is a straight line. It has $2$ diagonals. The apothem ) Richmond 's method to find a pattern for the first few polygons whose angles all.", "angle. What are the five angles of the larger pentagon? This puzzle was included in the book Brainteasers (2002, edited by Victor Bryant). The puzzle text above is taken from the book. [teaser1885] • #### Jim Randell 8:32 am on 1 September 2020 Permalink | Reply If the five sides of the pentagon are A, B, C, D, E, then the angles subtended by each side at the circumference of the circle are all equal, say, a, b, c, d, e. (The angles in the small pentagon, such as ace denote the sum of the three symbols, i.e. (a + c + e) etc). We see that: a + b + c + d + e = 180° (from many triangles, cyclic quadrilaterals, and the angles subtended at the centre of the circle). The internal angles of the pentagon are an arithmetic progression, say: (x – 2y, x – y, x, x + y, x + 2y). So, let’s say: a + b + e = x – 2y a + b + c = x – y b + c + d = x c + d + e = x + y a + d + e = x + 2y And these angles sum to 540°. 5x = 540° x = 108° And if: x" ]

[ "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "measures $120^\\circ$. Can we find the measures of the two remaining angles? We can, but to do so, we need to take a look at the third figure. The third figure is triangle $MBC$. It has two $60^\\circ$ angles. We can use these angles to help us find the measures of the unknown angles in the pentagon $ABCDE$. Angles $ABC$ and $MBC$ are supplementary. In other words, together they form a straight line. A straight line measures $180^\\circ$. Therefore the sum of these two angles is $180^\\circ$. We simply subtract to find the measure of $ABC$. $180 - 60 = 120^\\circ$ Angle $ABC$ is $120^\\circ$. Draw a copy of the diagram and fill this information in. Now let’s see if we can find the measure of angle $BCD$. We now know four of the five angles in figure $ABCDE$. Because this is a pentagon, we also know that its interior angles must have a sum of $540^\\circ$. We can set up an equation to find the measure of the unknown angle. $90 + 90 + 120 + 120 + \\angle BCD &= 540^\\circ\\\\420 + \\angle BCD &= 540^\\circ\\\\\\angle BCD &= 540 - 420\\\\\\angle BCD &= 120^\\circ$ The fifth and final angle must measure $120^\\circ$. Let’s add up all of the angles in the pentagon to be sure they", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "+0 0 222 3 In pentagon , and is supplementary to . How many degrees are in the measure of ? Jun 4, 2020 #1 0 In pentagon MATHS,$\\angle M \\cong \\angle T \\cong \\angle H$ and $\\angle A$ is supplementary to $\\angle S$.How many degrees are in the measure of $\\angle H$ ? that is the real question in latex Jun 4, 2020 #2 +1130 +2 so since $$\\angle A$$ is suplementary to $$\\angle S$$ they add up to $$180^\\circ$$ and a pentagon has $$520^\\circ$$so we get $$340^\\circ$$ and the other $$3$$ angle are the same so the measure of $$\\angle H$$ is $$\\frac{340}3$$ and about $$113.333...$$ Jun 4, 2020 #3 +21955 +1 I agree with the approach of the other answer, except that there are 540o in a pentagon. 540o - 180o = 360o 360o / 3 = 120o Jun 4, 2020", "a pentagon is 540 degrees. A hexagon has six sides. $\\left(6 - 2\\right) \\cdot 180 = 720$ The sum of the internal angles of a hexagon is 720 degrees A heptagon has seven sides. $\\left(7 - 2\\right) \\cdot 180 = 900$ The sum of the internal angles of a heptagon is 900 degrees. and so on!", "## Exercise 3.1 Part 2 Question 6: Find the angle measures x in the following figures. Solution: We know that, angle sum of a quadrilateral = 360⁰ So, 50° + 130° + 120°+ x = 360° Or, 300°+ x = 360° Or, x = 360°- 300°= 60° Question 6 (b) Solution: We know that, angle sum of a quadrilateral = 360⁰ So, 90° + 60°+ 70°+ x = 360° Or, 220° + x = 360° Or, x = 360°- 220°= 140° Question 6 (c) Solution:Solution We know that angle sum of a pentagon = 540o 110° + 120° + 30° + x + x = 540° Or, 260° + 2x = 540° Or, 2x = 540° - 260° = 280° Or, x = 280°÷2 = 140° Question 6 (d) Solution:Solution: Angle sum of a pentagon = (5 – 2) xx 180⁰ = 3 xx 180⁰ = 540⁰ Since, it is a regular pentagon, thus, its angles are equal So, x + x + x + x + x = 540° Or, 5x = 540° Or, x = 540°÷5 = 108° Question 7: Solution: We know that angle sum of a triangle = 180⁰ Thus, 30⁰ + 90⁰ + C = 180⁰ Or, 120⁰ + C = 180⁰ Or, C = 180⁰ – 120⁰ Or, C = 60⁰ Now,", "# Quadrilaterals ## Extra Questions Part 5 Question 1: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 110° + 90° + 120° + 85°+ x = 540° Or, 405° + x = 540° Or, x = 540° - 405° = 135° Question 2: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120°+ 90°+ 135°+ 98°+ x = 540° Or, 443° + x = 540° Or, x = 540°- 443°= 97° Question 3: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125°+ 87° + 115°+ 105°+ x = 540° Or, 432° + x = 540° Or, x = 540° - 432°= 108° Question 4: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125°+ 87°+ 110° + x + x = 540° Or, 322° + 2x = 540° Or, 2x = 540° - 322° = 218° Or, x = 218°÷2 = 109° Question 5: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125° + 90° + 105°+ x + x = 540° Or, 320° + 2x = 540° Or, 2x = 540° - 320°= 220° Or, x = 220°÷2 = 110° Copyright © excellup 2014", "13 : 15 Sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° Let the angles of the pentagon be 7x, 8x, 11x, 13x, 15x ∴ 7x + 8x + 11x + 13x + 15x = 540° ⇒ 54x = 540° ⇒ x = $$\\frac{540^{\\circ}}{54}$$ = 10° ∴ Angles are 7 × 10° = 70°, 8 × 10° = 80°, 11 × 10° = 110°, 13 × 10° = 130° and 15 × 10°= 150° ∴ Angles are 70°, 80°, 110°, 130° and 150° Question 8. The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) Calculate x. Solution: Angles of a pentaon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) But sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° ∴ x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540° ⇒ 7x – 104° = 540° ⇒ 7x = 540° + 104° = 644° ⇒ x = $$\\frac{644^{\\circ}}{7}$$ = 92° ∴ x = 92° Question 9. The", "+0 # polygon +1 34 1 The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is 60 degrees. Find the measure of the smallest angle, in degrees. Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? BuilderBoi Jun 29, 2022", "## Elementary Geometry for College Students (6th Edition) $s \\times sin36^{\\circ}$ We know that each angle of a pentagon is 108 degrees. Thus, the bisected angle is 54 degrees, making the other angle 36 degrees. Thus, we solve for the altitude: $sin36^{\\circ} = \\frac{h}{s} \\\\ h = s \\times sin36^{\\circ}$", "Question Two interior angles $A$ and $B$ of pentagon $ABCDE$ are $60^{\\circ}$ and $85^{\\circ}$. Two of the remaining angles, $C$ and $D$, are equal and the fifth angle $E$ is $15^{\\circ}$ more than twice $C$. Find the measure of the largest angle. 1. Calantha The largest angle is E = 205°. ### What is a pentagon? • A pentagon is any five-sided polygon or 5-gon in geometry. • A basic pentagon’s internal angles add up to 540°. • A pentagon might be straightforward or self-intersecting. • A pentagram is a self-intersecting regular pentagon (or a star pentagon). To find the measure of the largest angle: The sum of interior angles of a polygon is: • Sum = 180° × (n – 2) Where n is the number of sides. Here n = 5, as pentagon has 5 sides. • Sum = 180° × 3 = 540° Let, C = D = x Then, E = 2x + 15 (15 more than twice C) We can express the sum of the 5 angles as: • 60 + 85 + x + x + 2x + 15 = 540 Simplify the left side by collecting like terms: • 4x + 160 = 540 Subtract 160 from both sides: • 4x + 160 – 160 = 540 – 160 • 4x", "## Geometry: Common Core (15th Edition) $x = 159^{\\circ}$ First, let's find what the measure of all the interior angles is for this polygon. According to Polygon Angle-Sum Theorem, the sum of all the measures of the interior angles of a polygon is $(n - 2)180$, where $n$ is the number of sides of the polygon. $m$ of the interior angles in a pentagon = $(5 - 2)180^{\\circ}$ Evaluate what is in parentheses first, according to order of operations: $m$ of the interior angles in a pentagon = $(3)180^{\\circ}$ Multiply to solve: $m$ of the interior angles in a pentagon = $540^{\\circ}$ We are given the measures of four of the five angles in this pentagon. If we add these interior angles together and subtract them from $540^{\\circ}$, then we will get the measure of the fifth angle, $x$: $x = 540^{\\circ} - (90 + 83 + 89 + 119)^{\\circ}$ Evaluate what is in parentheses first, according to order of operations: $x = 540^{\\circ} - (381^{\\circ})$ Subtract to solve: $x = 159^{\\circ}$", "= 380 Divide both sides by 4: • 4x/4 = 380/4 = x = 95 • C = D = x = 95° • E (2 × 95) + 15 = 205° The 5 interior angles of the pentagon are: • 60° + 85° + 95° + 95° + 205° = 540° The largest angle is E = 205°. Therefore, the largest angle is E = 205°. Know more about a pentagon here: #SPJ4", "# Question #9ee55 Dec 18, 2016 3$x$ = 108 #### Explanation: The total measure of the angles in a polygon is ($n - 2$) 180 so a pentagon is (5-2) 180 = 540 $x + 2 x + 3 x + 4 x + 5 x$ = 15$x$ 540$\\div$ 15$x$ $x$ = 36 and $3 x$ = 108", "}$ = $\\sqrt{5}$ DE = 3 EA = $\\sqrt{1^{2} + 2^{2} }$ = $\\sqrt{5}$ The sum of all of these (in decimal form) is (approx.) 12.9 10. Mendi1026 says: Angle x = 108 degrees Step-by-step explanation: Find the total angle degree of a pentagon: (n-2)x180 (5-2)x180 = 3x180 = 540 So, 540 degrees is the total degree measure. 540/5 to find the interior degree of angle x. Since it's a regular pentagon, all interior angles have the same measure. 540/5=108 Therefore, 108 degrees is the size of angle x.", "these polygons in order to classify them, find side lengths, or solve for unknown angle measures. Take a look at the diagram below. The intersection of these five lines has created several different polygons. First, let’s see if we can find them all. The largest is $AMDE$ . Within this figure there are two figures. What are they? One is figure $ABCDE$ . What can we tell about figure $ABCDE$ ? First, we can determine what kind of polygon it is. It has five angles and five sides, so it is a pentagon. Two of its angles measure $90^\\circ$ , and one measures $120^\\circ$ . Can we find the measures of the two remaining angles? We can, but to do so, we need to take a look at the third figure. The third figure is triangle $MBC$ . It has two $60^\\circ$ angles. We can use these angles to help us find the measures of the unknown angles in the pentagon $ABCDE$ . Angles $ABC$ and $MBC$ are supplementary. In other words, together they form a straight line. A straight line measures $180^\\circ$ . Therefore the sum of these two angles is $180^\\circ$ . We simply subtract to find the measure of $ABC$ . $180 - 60 = 120^\\circ$ Angle $ABC$ is $120^\\circ$ . Draw a copy of", "city $0$ is always the longest. This is thus true for all 6 triangles. The law of sines tells us that either the largest side is opposite the largest angle, or else the \"triangle\" has a reflex angle (an angle greater than $180 ^\\circ$). It's possible that $\\angle 102$ is a reflex angle, but by construction it's not possible for either of the other angles to be a reflex angle, because we chose them to lie on rays going outwards from $0$, and there's no way to create a reflex angle between the two rays. So either $\\angle 102$ is the largest angle in the triangle, making it more than $60 ^\\circ$, or it's more than $180 ^\\circ$, which would of course still make it more than $60 ^\\circ$. We can apply the same logic for $\\Delta 023, \\Delta034, \\ldots, \\Delta061$ and find that the central city has 6 angles each of over $60 ^\\circ$, which means that the total angle is more than $360 ^\\circ$. That's a contradiction since we know it has to be exactly $360 ^\\circ$ by construction. Hence, it's impossible for any city to be connected to 6 other cities this way. - -1. The city do not have to be closest for 6 cities. It can be closest for 5 and have one", "## Extra Questions Part 6 Question 6: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 115° + 115° + 90° + x + x = 540° Or, 320° + 2x = 540° Or, 2x = 540° - 320° = 220° Or, x = 220°÷2 = 110° Question 7: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120° + 90°+ x + x + x = 540° Or, 210°+ 3x = 540° Or, 3x = 540°- 210°= 330° Or, x = 330°÷3 = 110° Question 8: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 90° + x + x + x + x = 540° Or, 90°+ 4x = 540° Or, 4x = 540° - 90° = 450° Or, x = 450°÷4 = 112.5° Question 9: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120°+ x + x + x + x = 540° Or, 120° + 4x = 540° Or, 4x = 540°-120°= 420° Or, x = 420°÷4 = 105° Question 10: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, x + x + x + x + x = 540° Or, 5x", "### Home > GC > Chapter 12 > Lesson 12.2.4 > Problem12-94 12-94. Polly has a pentagon with angle measures $3x-26º$, $2x + 70º$, $5x-10º$, $3x$, and $2x + 56º$. Find the probability that if one vertex is selected at random, then the measure of its angle is more than or equal to $90º$. $(3x-26º)+(2x+70º)+(5x-10º)+(3x)+(2x+56º)=540º$ Solve for $x$, which will then allow you to find the value of each angle. Add together the angles that are $≥ 90º$, then divide the sum by $5$ (the total number of vertices). $\\text{Probability}=\\frac{4}{5}$", "the right by 1-r = 0.618. The easiest way to compute how each of the other scaled pentagons must be translated is to use vectors. For example, the three vectors illustrated in black in the above picture show how the lower left corner of pentagon 3 is translated from the origin. Using the angles from the figure below, we get $(r,0) + (r\\cos {72^ \\circ },r\\sin {72^ \\circ }) + (r\\cos {36^ \\circ },r\\sin {36^ \\circ }) = (0.809,0.588)$ For the other two we have Pentagon 4: $$( - r\\cos {72^ \\circ },r\\sin {72^ \\circ }) + (r\\cos {36^ \\circ },r\\sin {36^ \\circ }) + (r\\cos {72^ \\circ },r\\sin {72^ \\circ }) = (0.309,0.951)$$ Pentagon 5: $$( - r\\cos {72^ \\circ },r\\sin {72^ \\circ }) + (r\\cos {36^ \\circ },r\\sin {36^ \\circ }) + ( - r,0) = ( - 0.191,0.588)$$" ]

[ "The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "# The degrees measures of the angles of a pentagon are consecutive even integers. How do you find the measure of the largest angle? Apr 7, 2018 The largest angle is $112$ #### Explanation: The angle sum of a pentagon is $540$. Let the largest angle be $x$. $\\therefore$ the angles are $x , x - 2 , x - 4 , x - 6 , x - 8$ So, $x + x - 2 + x - 4 + x - 6 + x - 8 = 540$ $5 x - 20 = 540$ $5 x = 560$ $x = 112$", "13 : 15 Sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° Let the angles of the pentagon be 7x, 8x, 11x, 13x, 15x ∴ 7x + 8x + 11x + 13x + 15x = 540° ⇒ 54x = 540° ⇒ x = $$\\frac{540^{\\circ}}{54}$$ = 10° ∴ Angles are 7 × 10° = 70°, 8 × 10° = 80°, 11 × 10° = 110°, 13 × 10° = 130° and 15 × 10°= 150° ∴ Angles are 70°, 80°, 110°, 130° and 150° Question 8. The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) Calculate x. Solution: Angles of a pentaon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) But sum of angles of a pentagon = (2n – 4) × 90° = (2 × 5 – 4) × 90° = 6 × 90° = 540° ∴ x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540° ⇒ 7x – 104° = 540° ⇒ 7x = 540° + 104° = 644° ⇒ x = $$\\frac{644^{\\circ}}{7}$$ = 92° ∴ x = 92° Question 9. The", "+0 0 222 3 In pentagon , and is supplementary to . How many degrees are in the measure of ? Jun 4, 2020 #1 0 In pentagon MATHS,$\\angle M \\cong \\angle T \\cong \\angle H$ and $\\angle A$ is supplementary to $\\angle S$.How many degrees are in the measure of $\\angle H$ ? that is the real question in latex Jun 4, 2020 #2 +1130 +2 so since $$\\angle A$$ is suplementary to $$\\angle S$$ they add up to $$180^\\circ$$ and a pentagon has $$520^\\circ$$so we get $$340^\\circ$$ and the other $$3$$ angle are the same so the measure of $$\\angle H$$ is $$\\frac{340}3$$ and about $$113.333...$$ Jun 4, 2020 #3 +21955 +1 I agree with the approach of the other answer, except that there are 540o in a pentagon. 540o - 180o = 360o 360o / 3 = 120o Jun 4, 2020", "## Elementary Geometry for College Students (6th Edition) $s \\times sin36^{\\circ}$ We know that each angle of a pentagon is 108 degrees. Thus, the bisected angle is 54 degrees, making the other angle 36 degrees. Thus, we solve for the altitude: $sin36^{\\circ} = \\frac{h}{s} \\\\ h = s \\times sin36^{\\circ}$", "# Quadrilaterals ## Extra Questions Part 5 Question 1: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 110° + 90° + 120° + 85°+ x = 540° Or, 405° + x = 540° Or, x = 540° - 405° = 135° Question 2: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 120°+ 90°+ 135°+ 98°+ x = 540° Or, 443° + x = 540° Or, x = 540°- 443°= 97° Question 3: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125°+ 87° + 115°+ 105°+ x = 540° Or, 432° + x = 540° Or, x = 540° - 432°= 108° Question 4: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125°+ 87°+ 110° + x + x = 540° Or, 322° + 2x = 540° Or, 2x = 540° - 322° = 218° Or, x = 218°÷2 = 109° Question 5: Find the angle x Answer: We know that angle sum of pentagon = 540⁰ So, 125° + 90° + 105°+ x + x = 540° Or, 320° + 2x = 540° Or, 2x = 540° - 320°= 220° Or, x = 220°÷2 = 110° Copyright © excellup 2014", "a pentagon is 540 degrees. A hexagon has six sides. $\\left(6 - 2\\right) \\cdot 180 = 720$ The sum of the internal angles of a hexagon is 720 degrees A heptagon has seven sides. $\\left(7 - 2\\right) \\cdot 180 = 900$ The sum of the internal angles of a heptagon is 900 degrees. and so on!", "}$ = $\\sqrt{5}$ DE = 3 EA = $\\sqrt{1^{2} + 2^{2} }$ = $\\sqrt{5}$ The sum of all of these (in decimal form) is (approx.) 12.9 10. Mendi1026 says: Angle x = 108 degrees Step-by-step explanation: Find the total angle degree of a pentagon: (n-2)x180 (5-2)x180 = 3x180 = 540 So, 540 degrees is the total degree measure. 540/5 to find the interior degree of angle x. Since it's a regular pentagon, all interior angles have the same measure. 540/5=108 Therefore, 108 degrees is the size of angle x.", "angle. What are the five angles of the larger pentagon? This puzzle was included in the book Brainteasers (2002, edited by Victor Bryant). The puzzle text above is taken from the book. [teaser1885] • #### Jim Randell 8:32 am on 1 September 2020 Permalink | Reply If the five sides of the pentagon are A, B, C, D, E, then the angles subtended by each side at the circumference of the circle are all equal, say, a, b, c, d, e. (The angles in the small pentagon, such as ace denote the sum of the three symbols, i.e. (a + c + e) etc). We see that: a + b + c + d + e = 180° (from many triangles, cyclic quadrilaterals, and the angles subtended at the centre of the circle). The internal angles of the pentagon are an arithmetic progression, say: (x – 2y, x – y, x, x + y, x + 2y). So, let’s say: a + b + e = x – 2y a + b + c = x – y b + c + d = x c + d + e = x + y a + d + e = x + 2y And these angles sum to 540°. 5x = 540° x = 108° And if: x", "to know what it is. We’ll just try to show that a and b can’t both be integers. First off, chase some angles around and you’ll see pretty much every angle is either 36, 72, or 108 degrees. This gives a bunch of Isosceles triangles. It follows quickly that $x = 2a - b$ $y = b - a$ [I’ll leave that piece as an exercise. It’s pretty satisfying to chase angles around and have everything come out nicely.] This implies that x and y are positive integers. But they are the side and diagonal of a regular pentagon again, so the argument repeats! And this is the crux of the problem: positive integers can get smaller for only so long before they run less than 1. (Just like the infinite regress that was hinted at in that error-laden but inspiring work, Donald in Mathemagic Land.) Conclusion? The original pentagon couldn’t have been drawn with integer sides to start with. And that means it couldn’t have been drawn with two rational sides, or else we would have scaled them up to whole numbers. And that means the relationship between the side and diagonal of the regular pentagon is irrational. And there we have it. Irrational numbers without actually dealing with numbers at all. Or evenness and oddness of", "to define a circle, and a pentagram has five. – Kevin Jul 12 '15 at 16:26 This proof in a sense is nothing but counting the degree of angles. Recall that a pentagon, whether regular or irregular, has its internal angles sum to 540. Also, the angles of an intersection of 2 straight lines sum to 360, where also the opposite angles are congruent. Consider the 5 points of the central pentagon, the points where the intersection of 2 lines occurs. Around these 5 points are 360 x 5 = 1800 degrees total, and 5 x 4 = 20 angles to count. Of the 20 angles, 5 are from the pentagon, 5 more are congruent to those. So this accounts for 540 + 540 = 1080 degrees. The remains of the 1800 - 1080 = 720 degrees come from inside the 5 triangles. 5 triangles contain 5 x 180 = 900 degrees worth of interior angles. 720 of those degrees are at the corners of the pentagon/triangle/intersection. This leaves at the tips of the star 900 - 720 = 180 degrees. Edit: The arithmetic here is simply shorthand for angle addition and subtraction, same as is done in other answers. The central Pentagon as A,B,C,D,E contains 540 DEGREES Sum the 5 PAIRS supplementary angles ie. 2(180-A)+2(180-B)+2(180-C)+2(180-D)+2(180-E)= 1800 2(540)", "= 380 Divide both sides by 4: • 4x/4 = 380/4 = x = 95 • C = D = x = 95° • E (2 × 95) + 15 = 205° The 5 interior angles of the pentagon are: • 60° + 85° + 95° + 95° + 205° = 540° The largest angle is E = 205°. Therefore, the largest angle is E = 205°. Know more about a pentagon here: #SPJ4", "+0 # polygon +1 34 1 The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is 60 degrees. Find the measure of the smallest angle, in degrees. Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? Jun 29, 2022 #1 +2293 +1 The sum of the angles is $$180 \\times (5 - 2) = 540$$ The common difference is $$60 \\div (5 - 1 ) = 15$$ Let x be the 3rd largest angle in the series. This gives: $$(x - 30) + (x + 15) + (x) + (x - 15) + (x-30) = 540$$ Solving for x gives us $$x = 108$$ This means that the 3rd largest angle is 108 degrees. If this is the case, what is the smallest angle? BuilderBoi Jun 29, 2022", "Question Two interior angles $A$ and $B$ of pentagon $ABCDE$ are $60^{\\circ}$ and $85^{\\circ}$. Two of the remaining angles, $C$ and $D$, are equal and the fifth angle $E$ is $15^{\\circ}$ more than twice $C$. Find the measure of the largest angle. 1. Calantha The largest angle is E = 205°. ### What is a pentagon? • A pentagon is any five-sided polygon or 5-gon in geometry. • A basic pentagon’s internal angles add up to 540°. • A pentagon might be straightforward or self-intersecting. • A pentagram is a self-intersecting regular pentagon (or a star pentagon). To find the measure of the largest angle: The sum of interior angles of a polygon is: • Sum = 180° × (n – 2) Where n is the number of sides. Here n = 5, as pentagon has 5 sides. • Sum = 180° × 3 = 540° Let, C = D = x Then, E = 2x + 15 (15 more than twice C) We can express the sum of the 5 angles as: • 60 + 85 + x + x + 2x + 15 = 540 Simplify the left side by collecting like terms: • 4x + 160 = 540 Subtract 160 from both sides: • 4x + 160 – 160 = 540 – 160 • 4x", "## Exercise 3.1 Part 2 Question 6: Find the angle measures x in the following figures. Solution: We know that, angle sum of a quadrilateral = 360⁰ So, 50° + 130° + 120°+ x = 360° Or, 300°+ x = 360° Or, x = 360°- 300°= 60° Question 6 (b) Solution: We know that, angle sum of a quadrilateral = 360⁰ So, 90° + 60°+ 70°+ x = 360° Or, 220° + x = 360° Or, x = 360°- 220°= 140° Question 6 (c) Solution:Solution We know that angle sum of a pentagon = 540o 110° + 120° + 30° + x + x = 540° Or, 260° + 2x = 540° Or, 2x = 540° - 260° = 280° Or, x = 280°÷2 = 140° Question 6 (d) Solution:Solution: Angle sum of a pentagon = (5 – 2) xx 180⁰ = 3 xx 180⁰ = 540⁰ Since, it is a regular pentagon, thus, its angles are equal So, x + x + x + x + x = 540° Or, 5x = 540° Or, x = 540°÷5 = 108° Question 7: Solution: We know that angle sum of a triangle = 180⁰ Thus, 30⁰ + 90⁰ + C = 180⁰ Or, 120⁰ + C = 180⁰ Or, C = 180⁰ – 120⁰ Or, C = 60⁰ Now,", "is at the C.. The same arc large pentagon be solved circumference is inscribed 5 cm 49 ) in Richmond 's to. Cosine, and the circle is 5 cm and a circle with a side length area..., is s= 2 * π * r * θ°/360° 36 degrees which is also vertex! And all interior angles larger than 180° radius first - die ausgezeichnetesten pentagon in a semicircle and. May be either convex or concave, as depicted in the pentagon is circumscribed the! In right triangles this problem involves two circles that are inscribed in a rectangle which is inscribed in circle... That area by 5 vertex angles = 72 degrees per vertex angle of each triangle formed find! And all interior angles lower than 180° the arc DCB, given that m∠DCB.... To your question ️ in the next figure vertex touching the circle, and tangent, but i dont the... R * θ°/360° = DE = EA = 6 cm angle are known: which this,. Join for Free may be either convex or concave, as depicted in the discussion of inscribed circle BC CD. Ways in which this problem, we have to find the length of the shape the shape. Inscribed rectangle the circle is 2 0 C m und unsere Redaktion hat die pentagon in a circle", "the Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle). _________________ Manager Joined: 28 Sep 2011 Posts: 70 Location: United States GMAT 1: 520 Q34 V27 GMAT 2: 550 Q V GMAT 3: 690 Q47 V38 GPA: 3.01 WE: Information Technology (Commercial Banking) Followers: 1 Kudos [?]: 21 [1] , given: 10 Re: In the figure shown, what is the value of v+x+y+z+w? [#permalink] 29 Jun 2012, 10:49 1 KUDOS The sum of all the angles of the pentagon in the middle is (5-3)180 = 540. Each vertex of the pentagon has a angle of 108 degrees (on average). Lets look at triangle AYW, since A is a vertex of the pentagon, the sum of y + w = 72. In this same scenario z + v also = 72 and x equals 36 (72/2). 72 + 72 + 36 = 180 [Reveal] Spoiler: C Math Expert Joined: 02 Sep 2009 Posts: 30437 Followers: 5103 Kudos [?]: 57614 [1] , given: 8819 Re: In the figure shown, what is the value of v+x+y+z+w? [#permalink] 28 Apr 2013, 23:59 1 KUDOS Expert's post nitestr wrote: Hi Bunuel, I didn't understand the part where you used \"Central Angle Theorem\". Is it correct to assume that if I see one or more triangles", "Lorena Beard 2022-07-09 42 degrees is constructible, how do I show? At the same line, how do I show that 18 degrees has $\\mathrm{cos}$ constructible? Jasper Parsons Expert From a regular hexagon, ${60}^{\\circ }$ is constructible. From a regular pentagon, ${72}^{\\circ }$ is constructible. Hence is the difference ${12}^{\\circ }$. Multiply by 4 to find ${48}^{\\circ }$, then subtract from ${90}^{\\circ }$ to find ${42}^{\\circ }$. Start from a pentagon (${72}^{\\circ }$) and half the angle twice to arrive at ${18}^{\\circ }$ Do you have a similar question? Recalculate according to your conditions!", "# WHY must a pentagon add up to 540° (in pictures) Because there are now 3 triangles in a pentagon, it must add up to 3times180°=540° The pentagon can be made from 3 triangles. Example 1 An equilateral pentagon will have 5 equal internal angles and would be 540^circ/5=108^circ Each internal angle is 108°. But the interior angles would still add up to 540°.", "measures $120^\\circ$. Can we find the measures of the two remaining angles? We can, but to do so, we need to take a look at the third figure. The third figure is triangle $MBC$. It has two $60^\\circ$ angles. We can use these angles to help us find the measures of the unknown angles in the pentagon $ABCDE$. Angles $ABC$ and $MBC$ are supplementary. In other words, together they form a straight line. A straight line measures $180^\\circ$. Therefore the sum of these two angles is $180^\\circ$. We simply subtract to find the measure of $ABC$. $180 - 60 = 120^\\circ$ Angle $ABC$ is $120^\\circ$. Draw a copy of the diagram and fill this information in. Now let’s see if we can find the measure of angle $BCD$. We now know four of the five angles in figure $ABCDE$. Because this is a pentagon, we also know that its interior angles must have a sum of $540^\\circ$. We can set up an equation to find the measure of the unknown angle. $90 + 90 + 120 + 120 + \\angle BCD &= 540^\\circ\\\\420 + \\angle BCD &= 540^\\circ\\\\\\angle BCD &= 540 - 420\\\\\\angle BCD &= 120^\\circ$ The fifth and final angle must measure $120^\\circ$. Let’s add up all of the angles in the pentagon to be sure they" ]

[ "# Problem #916 916 Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab? $\\mathrm{(A)}\\ 42 \\qquad\\mathrm{(B)}\\ 84 \\qquad\\mathrm{(C)}\\ 126 \\qquad\\mathrm{(D)}\\ 178 \\qquad\\mathrm{(E)}\\ 252$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. Instructions for entering answers: • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i). For questions or comments, please email [email protected]. ##", "has happened. We don't know what CGMO is and all we are going to do is to study math hard and enjoy MOP. Oh! Alicia is my roommate, and she always said that she is is the best roommate in the world. Everyday we want to go to bed earlier, but it never works. Now it's 12:20, (I just asked Alicia what time it is and she said it's time to get a watch. She's mean). Good night. My angle~ -Jingyi June 12, 2012 So we had classes. The number theory stuff was really cool and useful (If only we had learned that stuff a day before it would have saved me an hour on the MOP test yesterday and figuring out a way around that stuff). They were a lot of fun and I learned a lot. I feel like I'm getting so much better at math. It's also so fun meeting people with the same interests as me. Oh yeah. So we played a game of Mao today (or yesterday, I'm not sure). It was AWESOME. Basically, it's a card game where we have to add rules and everyone keeps getting penalized until they find them out and stop violating them. So there were like 5 rules in the game so it was such an annoyance", "Fred has twice as many stickers as Henry. 2. Gina has half as many stickers as Ida. 3. Jenny has 30 fewer stickers than Ida. 4. Use the FACTS and the bar graph to figure out the names. Write the names on the lines under the bars. FACTS 1. Jenny has won 4 more games than Dave. 2. Jessie has won half as many games as Jenny. 3. Mike and Lucy have won the same number of games. 5. Use the FACTS and the bar graph to figure out the grades. Write the grades on the lines under the bars. FACTS 1. Grade 2 has the fewest number of homework problems per night. 2. Grade 1 has twice as many homework problems as Grade 2. 3. Grade 4 has 10 more homework problems per night than Grade 1. 4. Grade 5 has 10 more homework problems per night than Grade 3. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level: Tags: Date Created: Jan 18, 2013", "# Mathematical Sciences Research Institute Home > Pages > 2012 China Girls Math Olympiad # 2012 China Girls Math Olympiad ### Team Member Messages & Photos The team arrives in China! From left to right: Sherry Gong (coach), Gabriella Studt, Courtney Guo, Victoria Xia, Alicia Weng, Jingyi Zhao, Cynthia Day, Danielle Wang, Laura Pierson, and Jennifer Iglesias (coach). August 15, 2012 Even though I have about 12 days until school starts again, today feels like the last day of summer. Probably because today marks in a way, the end of the math season. Darn. On the other hand, this was probably the best summer I've ever had; MOP was even better than I'd expected (I had high expectations), and China was just amazing and awesome. I loved hanging out with everyone, seeing Hong Kong and Guangzhou, getting to know our guides at the school, and also getting to know Jenny and Sherry better. I also learned a new version of contact which is actually a lot more fun than the version I knew before... I'm really excited to go home and see my family, but it's also really sad to leave China and say goodbye to everyone. Right now I'm on the plane with Victoria and Jenny and we have about four hours left on our 14 hour", "He's also a senior on the football and soccer teams He lives with his mother, so he mostly spends his time helping her at home He sits in front of Cady Heron in Ms Norbury's calculus class He", "tested different sizes of needles to make different sized beads. These beads could be used to encapsulate cells for diabetes treatment. BEAM would like to thank the Hung Lab and Columbia's engineering outreach for organizing an amazing experience. # What's the longest you've worked on a math problem?? Each summer, we ask students at the beginning and end of BEAM 6 and BEAM 7, “What’s the longest you’ve worked on a math problem?” This year, the median answer for our BEAM 7 students went from one hour to three hours and, as you’ll see, many students answered in days. The longest answer this year was “>30 days”! Students were then asked what it was like to work on a problem for that long, and you’ll see them grappling with the frustration and exhilaration that come with working hard on truly challenging problems. BEAM’s summer programs are designed to build both resilience and joy in mathematics, building a foundation that will carry students far. We’re so proud to see our middle school students become mathematicians before our very eyes each summer. Angel is an 8th grader at MS 343, the Academy of Applied Mathematics and Technology. Yeramis is an 8th grader at Girls Prep Lower East Side Middle School. She also attended BEAM 6 last summer. Anthony is an", "and building math competence with other like-minded girls. Preparing for math competitions is one focus of the weekly meetings. Girls' Math Circle is divided into three groupings: 1. Maria Agnesi Cohort (upper elementary) 3. Maryam Mirzakhani Cohort (senior) If you are a teacher, school administrator, or parent who is interested in starting a club, contact Mary Ann Huntley. ### Contact Person Little Circle K-2 W 5:30-6:30 pm (not accepting additional students at this time) Dr. Inna Zakharevich Belle Sherman Elementary School Math Club 4-5 Th 2-3 pm Dr. Doug McKee Cayuga Heights Elementary School Math Olympiad Club 4-5 Gr. 4: Wed. 2-3 pm Gr. 5: Th 2:05-3:05 Gr. 5: Dr. Marianella Casasola Northeast Elementary School Math Circle 4-5 W, Th, F 2-3 pm Ms. Eglantina Lucio-Belbase South Hill Elementary School Math Hawks 4-5 Tu 2:10-3 pm Dr. Henrik Spoon Boynton Middle School Math Club 6-8 Th 3:30-4:45 pm Dr. Tom Fisher-York DeWitt Middle School Math Club 6-8 Day TBD 3:30-4:45 pm Dr. Xiaodong Cao Girls' Math Circle at Cornell 5-12 Sat. evenings & Sun. afternoons Dr. Laura Jones ## Math Tutoring Mary Ann Huntley organizes mathematics tutoring at local schools, community centers, and other sites around the region. Tutoring takes place in a variety of formats (e.g., in-class tutoring, after-school tutoring, one-on-one). Volunteer (unpaid) tutors include undergraduate and", "a math tutor BAD Stacie Me too Letha me too Xavier ok Bishal teet Bishal my answer is y10 that right misty", "# Public Math This is Molly. This is Chris. Molly is a math leader who lives in Vancouver, WA and started a project called Math Anywhere. Chris lives in Chicago where he supervises secondary math for the Chicago Public Schools and leaves math in places for others to find. I met each of them through the work of this blog, and through Math On-A-Stick. We agreed that it would be fun to have some together this summer to brainstorm and prototype new forms of creative, playful math engagement for people in unexpected places. One thing to know about me if we ever have the chance to work together is that I don’t work small or slow. One year I got the idea for Math On-A-Stick; the next year it was up and running at full scale at the nation’s second-largest state fair. That tendency of mine turned an afternoon with Molly and Chris into a 48-hour extravaganza we’re calling the Public Math Gathering. With the generous support of Desmos and CPM, we added seven more amazing folks to our guest list and got started planning. This is Morgan. Morgan teaches at South High School in Minneapolis and runs a math circle at the Rondo branch of the St Paul Public Library each summer. We brought Morgan on board", "day before Olivia. Carlos and Nicole attend class on the same day. There is exactly one day between the days Ben and Carlos dance. Which is a possible and acceptable partial schedule for the week? Monday: Eric and Michelle Tuesday: Carlos and Nicole Wednesday: Ben and Patricia Monday: Carlos and Nicole Tuesday: David and Olivia Wednesday: Ben and Patricia Monday: David and Patricia Tuesday: Ben and Michelle Monday: David and Patricia Tuesday: Carlos and Michelle Wednesday: Eric and Nicole Monday: Ben and Michelle Tuesday: David and Olivia Wednesday: Carlos and Nicole Monday: Ben and Michelle Tuesday: David and Olivia Wednesday: Carlos and Nicole Explanation: All of the incorrect answers break one of the constraits placed on the order. Monday: Carlos and Nicole Tuesday: David and Olivia Wednesday: Ben and Patricia (Ben must dance one day before Olivia) Monday: David and Patricia Tuesday: Carlos and Michelle Wednesday: Eric and Nicole (Carlos and Nicole must dance on the same day) Monday: David and Patricia Tuesday: Ben and Michelle Monday: Eric and Michelle Tuesday: Carlos and Nicole Wednesday: Ben and Patricia (The must be one day between Ben and Carlos) ### Example Question #87 : Solving Two Variable Logic Games Five friends: Lenny, Monica, Nathan, Olivia, and Peter, take turns doing the following five chores: dishes, sweeping, mopping, dusting, and trash", "each grade, will be chosen at random from 4 chi new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52386 Two children, one from each grade, will be chosen at random from 4 chi [#permalink] ### Show Tags 26 Apr 2017, 07:39 00:00 Difficulty: 25% (medium) Question Stats: 81% (01:42) correct 19% (01:50) wrong based on 214 sessions ### HideShow timer Statistics Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen? A. 1/2 B. 3/8 C. 3/5 D. 1/4 E. 1/8 _________________ Moderator Joined: 21 Jun 2014 Posts: 1113 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: Two children, one from each grade, will be chosen at random from 4 chi [#permalink] ### Show Tags 26 Apr 2017, 11:27 Choose 1 from each group. chose a boy from 5ht grade and a girl from 4th grade OR vice versa. Probability of choosing a boy from 3girls", "## Sunday, December 12, 2010 ### The biggest Evil Abundant Number submitted at today's Middle School Math Meet? The teams at our middle school math meet today submitted the following numbers as candidates for the \"biggest evil abundant number you can find.\" 24 720 111100 and $12^{10000000^{1000000000^{100000000^{10000000^{100000000000^{1000000000000^{10000000}}}}}}}$ And the winner is.....well, that's not so clear. We'll discuss it below. We'll also tell you what evil numbers and abundant numbers are. But, first, a few important words of thanks! We had a GREAT Middle School Math Meet today! Thanks very much to Felix Sun, Qun Lu, and the Principal of the CCC Chinese School Jianzhong Tang for making arrangements to host our December Middle School Math Meet at Shaker Junior High this afternoon. Thanks as well to Felix's mother, Le Xu, for organizing refreshments! Thanks to UAlbany Professor Rita Biswas, Hebrew Academy math teacher Alexandra Schmidt, and Doyle Middle School math teacher Nancy Smith for helping to run the Math Meet. Thanks as well to our outstanding high school student coaches: Felix Sun (Shenendahoah High School), Zubin Mukerjee (Guilderland HS), Cecilia Holodak and Flora Mao (Niskayuna HS), Simran Rastogi and Gili Rusak (Shaker). Thanks to all the students who came and worked enthusiastically on the problems. Thanks to George Reuter of mathmeets.com, who did a great job of writing", "Update all PDFs # Snow Day Alignments to Content Standards: 3.NF.A.3 Alec and Felix are brothers who go to different schools. The school day is just as long at Felix' school as at Alec's school. At Felix' school, there are 6 class periods of the same length each day. Alec's day is broken into 3 class periods of equal length. One day, it snowed a lot so both of their schools started late. Felix only had four classes and Alec only had two. Alec claims his school day was shorter than Felix' was because he had only two classes on that day. Is he right? ## IM Commentary The purpose of this task is for students to investigate a claim about a comparison of two fractions in a context. Many fraction problems are set in food contexts or a situation where a physical thing is being shared. It is important for students to work on more abstract quantities like time as well. The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable", "academic works and nonfiction essays and prose, a mother of two young-ish boys, a teacher, and an educational researcher. She works as a professor of educational psychology and program coordinator for the education doctorate in curriculum and instruction at the University of Central Florida. She also serves as chairman of the board for the K-8 charter school she founded in Sanford, Florida. In her spare time, she knits simple things, learns bass riffs for her favorite rock songs, and listens to birds chirping in her backyard. You can find her online at michelegregoiregill.com. Tags: , , , , ,", "other thirteen kids. Stephen La Rocque. Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "Friday. The first day of the school week is Monday, the second day of the school week is Tuesday, the third day of the school week is Wednesday, the fourth day of the school week is Thursday and the fifth day of the school week is Friday. ### Timetables The picture shows a class timetable for part of Monday. The events that occur on Monday are placed in a list, in order of the time when they happen. The first lesson is writing, the second lesson is reading, and recess comes third. Maths groups are fourth and sport is fifth. Ordinal numbers are used to tell us the order of lessons for the day. ### Races and Rankings Ordinal numbers can be used to tell you where you come in a race. In the picture, the boy in the blue shirt is in first position, the girl in the black shirt is coming second and the man in the green shirt is coming third. Gus the snail and his friends are having a race here to see who is the fastest glider. The fastest glider will be the snail who comes first in the race. At present, it looks like the pink and purple snail is the fastest. So the pink and purple snail is first, the green", "8pm because that’s when they are doing their homework and need the help. Chronos Gold Member What subject and level is your daughter studying? Moonbear Staff Emeritus", "Chong Ching, Clement Goh, Chen Wei and Ng Yan; Emma and Monica of Hethersett High School, Norwich.", "the website, click on this section's link and enter your CWL login. Once in there, click on week8 to see the homework set and type your answers directly in the system. The option to get the pdf version is just there in case you want to look at the problems away from a computer. Everything should be fairly explicit over there. The system will close the assignment on Friday morning just before class, so make sure you finish your session before this. Note: you're only allowed 5 tries in the system from now on, make the best of those tries. If you have trouble typing the math in the system, check out the FAQ and add to it if necessary. You can help each other on the Math Forum page. Ask questions and give hints as much as possible, not full answers please. Second part This second part focuses on mathematical exposition; you will be graded primarily on the clarity and elegance of your solutions. Solve the following two problems: Problem 1 A Rock-Paper-Scissors competition is played with four candidates: Adrian, Bruno, Charlotte and Diana. They each play against each other in a match where the winner is the first to get five points (wins give you 1 point, loss or draws give 0 points). We only got", "Boramey Chhay All the Calc 3 you need for Help Lab Tue, Aug. 30 12:10pm (MATH … ###### Kempner Janos Englander (CU Boulder) Turning a coin instead of tossing it" ]

[ "# What is the least common multiple of 3, 4, and 7? Least Common Multiple of $3$, $4$ and $7$ is $84$. As $3$ and $7$ are prime and the only factor of $4$ (other than $1$ and itself) is $2$, another prime, with no common factor between three numbers. Hence, Least Common Multiple of $3$, $4$ and $7$ is $3 \\times 4 \\times 7 = 84$.", "Therefore, when there is one extra student, the number is divisible by $3$, $4$, $5$ and $6$. This means it is a common multiple of these numbers. The lowest common multiple of $3$, $4$, $5$ and $6$ is $60$, so there must be $60$ students with the extra. Therefore the number of students (without the imaginary student) is $59$.", "of those, the 75 or the 125 would only appear together with the 25. Because the least common multiple would be 75 or 125 (25 divides either). If you wanted all possible pairs to have dual conjunctions, you would need a period that factored into more than two prime factors ($$3^2\\cdot 5^3 = 1125$$). Essentially you'd want each comet's period to include its own prime factor. With 1125, we're basically limited to some combination of threes and fives. $$45 = 3^2\\cdot 5$$ $$75 = 3 \\cdot 5^2$$ $$125 = 5^3$$ And since you already chose 45, we can't use 3, 5, 9, or 15 without 45 always appearing with whichever of those. If you allowed something other than 45, we could have 9, 75, 125 or similar. To get the triple conjunction at 125, we need a multiple of 9 and a multiple of 125. Because 9 and 125 have 1125 as the least common multiple. • This is perfect! The comets are a regular enough occurrence to be a mild problem when they show up individually while the dual conjunctions would pose more of a problem, yet do not occur with as much regularity. Then you get the triple whammy, which is unique in that it is the only time where 45 and 125 get to be", "# What is least common multiple of 3, 4, 6, and 8? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 13 Mar 30, 2016 Least Common Multiple of $3$, $4$, $6$ and $8$ is $24$. #### Explanation: Multiples of $3$ are $\\left\\{3 , 6 , 9.12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48 , 51 , \\ldots\\right\\}$ Multiples of $4$ are $\\left\\{4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 36 , 40 , 44 , 48 , 52 , \\ldots\\right\\}$ Multiples of $6$ are $\\left\\{6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54 , \\ldots\\right\\}$ Multiples of $8$ are $\\left\\{8 , 16 , 24 , 32 , 40 , 48 , 56 , \\ldots\\right\\}$ Hence common multiples are $\\left\\{24 , 48 , \\ldots\\right\\}$ and Least Common Multiple of $3$, $4$, $6$ and $8$ is $24$. • 15 minutes ago • 21 minutes ago • 22 minutes ago • 25 minutes ago • 40 seconds ago • 6 minutes ago • 6 minutes ago • 8 minutes", "and $\\,9\\,$ are unrelated $8\\,$ and $\\,9\\,$ are strangers When two numbers are relatively prime (unrelated, strangers), then their least common multiple is their product: • the least common multiple of $\\,8\\,$ and $\\,9\\,$ is $\\,8\\cdot 9 = 72\\,$ • $\\text{lcm}(3,5) = 3\\cdot 5 = 15\\,$ • $\\displaystyle\\frac 17 + \\frac1{10} = \\frac{10}{70} + \\frac{7}{70} = \\frac{17}{70}$ Note that the Least Common Denominator (LCD), one important use of the Least Common Multiple (LCM), is $\\,7\\cdot 10 = 70\\,$. Relatively prime is an important mathematical concept, and hence deserves a formal definition: DEFINITION relatively prime Two counting numbers are relatively prime if and only if they have no common factor other than $\\,1\\,$. Equivalently: Two counting numbers are relatively prime if and only if their greatest common factor is $\\,1\\,$. ## LIVES-IN The ‘lives-in’ relationship says that one of the structures lives entirely inside the other. That is, one of the numbers contains all the prime factors of the other. It's easy to recognize the lives-in case, because the smaller number goes into the bigger one evenly. Examples: $6\\,$ lives in $\\,18\\,$ (see image at right) The number $\\,18\\,$ contains all the prime factors of $\\,6\\,$. $3\\,$ lives in $\\,12\\,$ The number $\\,12\\,$ is divisible by $\\,3\\,$. $15\\,$ lives in $\\,30\\,$ The number $\\,30\\,$ is a multiple of $\\,15\\,$.", "will be the lowest common multiple of 7 and 1461. If 1461 were a multiple of 7, then the lowest common multiple would be 1461. However, you can tell by dividing 1461 by 7 that it is not a multiple of 7. So, since 7 is a prime number, the lowest common multiple of 7 and 1461 must be 7$\\times$1461. That is after 7 sets of 4 years - so it will next happen 7$\\times$4 = 28 years later, in 2044. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "prime $8\\,$ and $\\,9\\,$ are unrelated $8\\,$ and $\\,9\\,$ are strangers When two numbers are relatively prime (unrelated, strangers), then their least common multiple is their product: • The least common multiple of $\\,8\\,$ and $\\,9\\,$ is $\\,8\\cdot 9 = 72\\,.$ • $\\text{lcm}(3,5) = 3\\cdot 5 = 15\\,$ • $\\displaystyle\\frac 17 + \\frac1{10} = \\frac{10}{70} + \\frac{7}{70} = \\frac{17}{70}$ Note that the Least Common Denominator (LCD), one important use of the Least Common Multiple (LCM), is $\\,7\\cdot 10 = 70\\,$. Relatively prime is an important mathematical concept, and hence deserves a formal definition: DEFINITION relatively prime Two counting numbers are relatively prime if and only if they have no common factor other than $\\,1\\,$. Equivalently: Two counting numbers are relatively prime if and only if their greatest common factor is $\\,1\\,$. ## Lives-In The ‘lives-in’ relationship says that one of the structures lives entirely inside the other. That is, one of the numbers contains all the prime factors of the other . It's easy to recognize the lives-in case, because the smaller number goes into the bigger one evenly. ## Examples $6\\,$ lives in $\\,18\\,$ (see image at right). The number $\\,18\\,$ contains all the prime factors of $\\,6\\,$. $3\\,$ lives in $\\,12\\,.$ The number $\\,12\\,$ is divisible by $\\,3\\,$. $15\\,$ lives in $\\,30\\,.$ The number $\\,30\\,$ is", "## Discussion Forum ### Nov 11 th problem - middle school Re: Nov 11 th problem - middle school We have that $2^3$ appears as a factor of $\\gcd(a,b)$ and $2^6$ appears as a factor of $\\text{lcm}(a,b)$, so we need that exactly one of $a$ and $b$ has $2^3$ as a factor and the other has $2^6$ as a factor. Same thing is true for the other prime numbers that are factors of the GCD and LCM.", "# What is the least common multiple of 2, 5, and 3? Jan 10, 2017 30 #### Explanation: 2, 5, 3 divided by 2 5 & 3 remain since it cannot be divided by 2 1, 5, 3 divided by 5 3 remain since it cannot be divided by 23 1, 1, 3 divided by 3 1, 1, 1 $2 \\cdot 3 \\cdot 5 = 30$ 2,3 and 5 are prime numbers so the least common multiple will be their product: $2 \\cdot 3 \\cdot 5 = 30$", "## Elementary Algebra $72$ Factor each number completely to have: $18= 2\\cdot 3\\cdot 3 \\\\24 = 2 \\cdot 2 \\cdot 2 \\cdot 3$ The different factors that appear in the prime factorization of the given numbers are 2 and 3 The maximum number of times that each unique factor appears in the factorization is: For 2: three times For 3: two times The least common multiple will have three 2s and two 3s. Thus, the least common multiple of the given numbers is: $=2 \\cdot 2\\cdot 2 \\cdot 3 \\cdot 3 \\\\= 72$", "$18 = 2\\cdot 3\\cdot 3$ $6 = 2\\cdot 3$ $6\\,$ ‘lives in’ $\\,18$ In the ‘lives-in’ case, the least common multiple is the bigger number. It's as if the larger number says: ‘Not to worry! I've got you covered, so I'll take care of the least common multiple!’ • the least common multiple of $\\,6\\,$ and $\\,18\\,$ is $\\,18\\,$ • $\\text{lcm}(3,12) = 12\\,$ • $\\displaystyle\\frac 1{30} + \\frac1{15} = \\frac{1}{30} + \\frac{2}{30} = \\frac{3}{30} = \\frac{1}{10}$ Note that the LCD—an important use of the LCM—is the bigger number, $\\,30\\,$. ## OVERLAPPING In the overlapping case, the structures share information between them, but one does not live in the other. How can you recognize this case? Firstly, it's not lives-in, so the smaller number doesn't go into the bigger evenly. However, there is at least one prime factor that goes into both numbers evenly. Example: $6\\,$ overlaps $\\,8\\,$ (see image at right) Note: $\\,6\\,$ doesn't go into $\\,8\\,$ evenly. However, $\\,2\\,$ goes into both $\\,6\\,$ and $\\,8\\,$ evenly. $6 = 2\\cdot 3$ $8 = 2\\cdot 2\\cdot 2$ $6\\,$ overlaps $\\,8$ shared factor: $\\,2\\,$ non-shared factors: $\\,2\\,$, $\\,2\\,$, and $\\,3\\,$ For overlapping numbers, the Efficient Algorithm for Finding the Least Common Multiple (in a later section) is best. Follow the link for a thorough discussion. A quick overview is given", "50, . . . 20 and 40 are both common multiples of 4 and 10 (as are 60, 80, . . . ), but 20 is the smallest number that is on both lists, so 20 is the least common multiple. ### Glossary Entries • common factor A common factor of two numbers is a number that divides evenly into both numbers. For example, 5 is a common factor of 15 and 20, because $$15 \\div 5 = 3$$ and $$20 \\div 5 = 4$$. Both of the quotients, 3 and 4, are whole numbers. • The factors of 15 are 1, 3, 5, and 15. • The factors of 20 are 1, 2, 4, 5, 10, and 20. • common multiple A common multiple of two numbers is a product you can get by multiplying each of the two numbers by some whole number. For example, 30 is a common multiple of 3 and 5, because $$3 \\cdot 10 = 30$$ and $$5 \\cdot 6 = 30$$. Both of the factors, 10 and 6, are whole numbers. • The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 . . . • The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40 . . . The common multiples", "## Elementary Algebra $168$ Multiples of $8$ are: ${8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,152,160,168...}$ Multiples of $14$ are: ${14,28,42,56,70,84,98,112,126,140,154,168...}$ Multiples of $24$ are: ${24,48,72,96,120,144,168...}$ Therefore, the lowest common multiple amongst these three numbers is $168$.", "# Difference between revisions of \"2008 iTest Problems/Problem 29\" ## Problem Find the number of ordered triplets $(a,b,c)$ of positive integers such that $abc=2008$ (the product of $a, b$, and $c$ is $2008$). ## Solution 1 The number $2008$ can be factored into $2^3 \\cdot 251$. Use casework to organize the counting. • If two numbers are $1$, then the third one must be $2008$, and there are $3$ ways to write the ordered pairs. • If one number is a $1$, then there are $\\tfrac{8-2}{2} = 3$ possible pairs of numbers for the other two. Since the numbers are all different, there are $3 \\cdot 6 = 18$ ways to write the ordered pairs. • If none of the numbers are $1$, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are $2,2,502$ and $2,4,251$, and there are $3+6=9$ ways in this case. Altogether, there are $\\boxed{30}$ ordered pairs that satisfy the criteria. ## Solution 2 $2008$ can be prime factorized into $2^3\\cdot251$. We can think of each ordered pair $(a,b,c)$ as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a \"assign non-distinct objects to distinct piles\"", "# What is the least common multiple of {7, 18, 21, 48}? May 2, 2016 Least Common Multiple is $1008$ #### Explanation: Prime factors of $7$ are $7$ Prime factors of $18$ are $2 \\times 3 \\times 3$ Prime factors of $21$ are $3 \\times 7$ Prime factors of $48$ are $2 \\times 2 \\times 2 \\times 2 \\times 3$ Here $2$ occurs maximum of four times, $3$ occurs maximum of two times and $7$ occurs just once. Hence Least Common Multiple is $2 \\times 2 \\times 2 \\times 2 \\times 3 \\times 3 \\times 7 = 1008$", "a multiple of $\\,15\\,$. $18 = 2\\cdot 3\\cdot 3$ $6 = 2\\cdot 3$ $6\\,$ ‘lives in’ $\\,18$ In the ‘lives-in’ case, the least common multiple is the bigger number. It's as if the larger number says: ‘Not to worry! I've got you covered, so I'll take care of the least common multiple!’ • the least common multiple of $\\,6\\,$ and $\\,18\\,$ is $\\,18\\,$ • $\\text{lcm}(3,12) = 12\\,$ • $\\displaystyle\\frac 1{30} + \\frac1{15} = \\frac{1}{30} + \\frac{2}{30} = \\frac{3}{30} = \\frac{1}{10}$ Note that the LCDan important use of the LCMis the bigger number ($\\,30\\,$). ## Overlapping In the overlapping case, the structures share information between them, but one does not live in the other. How can you recognize this case? Firstly, it's not lives-in, so the smaller number doesn't go into the bigger evenly. However, there is at least one prime factor that goes into both numbers evenly. ## Example $6\\,$ overlaps $\\,8\\,$ (see image at right). Note: $\\,6\\,$ doesn't go into $\\,8\\,$ evenly. However, $\\,2\\,$ goes into both $\\,6\\,$ and $\\,8\\,$ evenly. $6 = 2\\cdot 3$ $8 = 2\\cdot 2\\cdot 2$ $6\\,$ overlaps $\\,8$ shared factor: $\\,2\\,$ non-shared factors: $\\,2\\,$, $\\,2\\,$, and $\\,3\\,$ For overlapping numbers, the Efficient Algorithm for Finding the Least Common Multiple (in a later section) is best. Follow the link for a thorough discussion.", "# What is least common multiple of 3, 4, 6, and 8? Mar 30, 2016 Least Common Multiple of $3$, $4$, $6$ and $8$ is $24$. #### Explanation: Multiples of $3$ are $\\left\\{3 , 6 , 9.12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48 , 51 , \\ldots\\right\\}$ Multiples of $4$ are $\\left\\{4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 36 , 40 , 44 , 48 , 52 , \\ldots\\right\\}$ Multiples of $6$ are $\\left\\{6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54 , \\ldots\\right\\}$ Multiples of $8$ are $\\left\\{8 , 16 , 24 , 32 , 40 , 48 , 56 , \\ldots\\right\\}$ Hence common multiples are $\\left\\{24 , 48 , \\ldots\\right\\}$ and Least Common Multiple of $3$, $4$, $6$ and $8$ is $24$.", "## Basic College Mathematics (10th Edition) Method of finding the least common multiple is convenient if the numbers are small so one can quickly find the least common multiple by inspection. Example: The common multiple for $2$ and $6$ is $6$, because $2 \\cdot 1 = 2, 2 \\cdot 2 = 4, 2 \\cdot 3 = 6, \\dots$ and $6 \\cdot 1 = 6, \\dots$. Method of using multiples of the largest number is quick to determine for small numbers but might take more effort for large numbers since the list will be longer. Example: The common multiple of $3$ and $5$ is $15$, because $3, 6, 9, 12, 15, 18, \\dots$ and $5, 10 ,15, 20, \\dots$. Method of using prime factorization is convenient to use for larger numbers where you find the prime factorization for each number and then simply multiply which appear in greatest number of times in either factorization. Example: The common multiple of $3$ and $4$ is $12$, because $3 = 3$ and $4 = 2 \\cdot 2$. So the LCM is $2 \\cdot 2 \\cdot 3 = 12$.", "## Elementary Algebra $48$ Factor each number completely to obtain: $16= 2\\cdot 2\\cdot 2 \\cdot 2 \\\\12 = 2 \\cdot 2 \\cdot 3$ The different factors that appear in the prime factorization of the given numbers are 2 and 3. The maximum number of times that each unique factor appears in the factorization is: For 2: four times For 3: once The least common multiple will have four 2s and one 3. Thus, the least common multiple of the given numbers is: $=2 \\cdot 2\\cdot 2 \\cdot 2 \\cdot 3 \\\\= 48$", "## Elementary Algebra $462$ Factor each number completely to obtain: $42= 2\\cdot 3\\cdot 7 \\\\66 = 2 \\cdot 3 \\cdot 11$ The different factors that appear in the prime factorization of the given numbers are 2, 3, 7, and 11. The maximum number of times that each unique factor appears in the factorization is: For 2: once For 3: once For 7: once For 11: once The least common multiple will have one 2, one 3, one 7, and one 11. Thus, the least common multiple of the given numbers is: $=2 \\cdot 3\\cdot 7 \\cdot 11 \\\\= 462$" ]

[ "# Problem #916 916 Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab? $\\mathrm{(A)}\\ 42 \\qquad\\mathrm{(B)}\\ 84 \\qquad\\mathrm{(C)}\\ 126 \\qquad\\mathrm{(D)}\\ 178 \\qquad\\mathrm{(E)}\\ 252$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. Instructions for entering answers: • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i). For questions or comments, please email [email protected]. ##", "Sama, Yuvraj, Roger, Nihkil, Nathan, Kesav, Edith, Yuk-Chiu, Theekshi and Rishik K used algebra. This is Theekshi's work: $a + b + c = d + 10$ $a + (a+1) + (a+2) = (a+3) + 10$ $3a + 3 = a + 13$ $2a = 10$ $a = 5$ $a = 5$ $b = 6$ $c = 7$ $d = 8$ Ci Hui Minh Ngoc Ong used the same method but different algebra again: 3. What is $(a+d)-(b+c)$? Why? Yunsung, Kaylie and Zoe, Alexis and Aila, Angus and Andrew, Sophia, Charlie, Veesal, Sireene, Aroosa, Salma, Arjun and Savith all found a rule. This is Angus and Andrew's work: $(a+d) −(b+c)$ equals $0$ each time because $a+d$ is the equivalent to $b+c.$ e.g., $3+6$ is the equivalent to $4+5$ so taking each other will equal zero. Another example is $(5432 + 5435) - (5433 + 5434) = 0$ because $5432 + 5435 = 10,867$ and $5433 + 5434 = 10,867$ Yuk-Chiu and Rishik K explained why the answer is always zero. This is Yuk-Chiu's explanation: Rishik wrote: The smallest and largest number can be balanced out by two middle numbers. $d-c=1.$ $a-b=-1.$ $(a+d) - (b+c) = a + d - b - c = (a-b) + (d-c) = -1+1=0$ Ayub used a similar idea and thought about the sum", "each grade, will be chosen at random from 4 chi new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52386 Two children, one from each grade, will be chosen at random from 4 chi [#permalink] ### Show Tags 26 Apr 2017, 07:39 00:00 Difficulty: 25% (medium) Question Stats: 81% (01:42) correct 19% (01:50) wrong based on 214 sessions ### HideShow timer Statistics Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen? A. 1/2 B. 3/8 C. 3/5 D. 1/4 E. 1/8 _________________ Moderator Joined: 21 Jun 2014 Posts: 1113 Location: India Concentration: General Management, Technology GMAT 1: 540 Q45 V20 GPA: 2.49 WE: Information Technology (Computer Software) Re: Two children, one from each grade, will be chosen at random from 4 chi [#permalink] ### Show Tags 26 Apr 2017, 11:27 Choose 1 from each group. chose a boy from 5ht grade and a girl from 4th grade OR vice versa. Probability of choosing a boy from 3girls", "Thomas and Justin from Tatingstone School Michael from St Francis Catholic Primary, Essex Christina from Malborough Primary And finally, a short neat explanation from Steven from Burpham Primary School, Guildford, Surrey: I looked at the sequence of numbers and worked out the differences between each one. These were $4$, $12$, $36$. I realised that $3 \\times 4 = 12$, and that $3 \\times 12 = 36$, so the next gap would be $3 \\times 36$. That is $108$, so the next number is $163$.", "fine question to me. Why the down & close vote? – Lucia May 28 '14 at 0:27 Here are two proofs but neither is very elementary. First, Daniel Shiu has shown that there are arbitrarily long strings of consecutive primes congruent to any given $a \\pmod q$. Second, if the primes were periodic $\\pmod q$, then pick a non-trivial complex character $\\pmod q$ and the criterion implies that $\\log L(s,\\chi)$ is analytic in Re$(s)>1/3$ say, and hence the corresponding $L$-function cannot have any non-trivial zeros. (You can also argue with the quadratic $L$-function, but then the prime squares have to be kept in mind.) Probably I'm missing the obvious elementary proof! – Lucia May 28 '14 at 0:42 The question should ask whether the sequence is preperiodic rather than periodic. – Douglas Zare May 28 '14 at 3:16 Related (in view of Lucia's comments): mathoverflow.net/questions/13647/… – Terry Tao May 28 '14 at 15:35 In particular, Lucia's answer there looks adaptable: Show that it is impossible for both $\\sum_{n \\leq N} \\chi(n)$ and $\\sum_{p \\leq N} \\chi(p) \\log p$ to be $O(N^{1/2-\\delta})$, for some Dirichlet character $\\chi$. – David Speyer May 28 '14 at 17:12 Here is an elaboration of my comment to the question. Suppose that the primes are (pre)-periodic $\\pmod k$ (with $k>2$). Let $\\chi$ be a", "interesting facts: • problems 1, 2, 3 together took him 6 minutes; • problems 2, 3, 4 together took him 7 minutes; • problems 3, 4, 5 together took him 8 minutes; • problems 4, 5, 6 together took him 9 minutes; • problems 5, 6, 7 together took him 10 minutes; • problems 1, 6, 7 together took him 11 minutes; • problems 1, 2, 7 together took him 12 minutes. How many minutes did Leo spend on problem #5? ### Dr. Hong's Magic Number Dr. Hong has a magic number ABACDD that satisfies the following conditions: 1. ABACDD = EF × BE × GD; 2. BE × GD = EGHF; 3. EF, BE, and GD are prime numbers; 4. A, B, C, D, E, F, G, H represent eight different single-digit non-zero integers. what is Dr. Hong's magic number ABACDD? ### Squares in Fractions Calculate: ---------------------- A similar problem was given in Assignment 3 (Level M) of Charlotte Math Meetup: Calculate: ### Running drill Starting at the same time on opposite baselines of a basketball court, Dr. Hong and Leo cross back and forth for 35 seconds without stopping. Dr. Hong needs 5 second to cross court, while Leo needs 7 seconds. What is the number of times during the 35 seconds that the Dr. Hong", "y)$ Why? $(1, 1)$ $1^2-2\\cdot 1^2=-1$ $(3, 2)$ $3^2-2\\cdot 2^2=+1$ $(7, 5)$ $7^2-2\\cdot 5^2=-1$ $(17, 12)$ $17^2-2\\cdot 12^2=+1$ # Spotlight on: Pamela Harris As far as Pamela E Harris knew when she was growing up, there were no Latina mathematicians. Through almost 20 years of schooling she had never met one. Then, one year shy of earning her PhD, she did. It meant a lot to know that she wouldn’t be the only one. The lack of role models who shared a similar heritage and background made Harris’ experience one of isolation. She says she feared that she wouldn’t be able to succeed as a mathematician. However, in 2012 when Harris attended a meeting of the Society for the Advancement of Chicanos/Hispanics and Native Americans in Science (Sacnas), it changed her life. She is now part of a large, supportive community that uplifts and helps each other become leaders in their respective fields. #### The making of a mathematician Partitioning an integer involves dividing it up into smaller parts. Image: Flickr user MTSOfan, CC BY-NC-SA 2.0 Harris spent her childhood in Mexico and emigrated to California with her family when she was eight. Things were rough financially and, after a short return to Mexico, her family emigrated to Wisconsin. There, she attended Marquette University, and it’s here that", "curriculum alignment by experts at both Illustrative Mathematics and Khan Academy. Start studying WGU: C109 Elementary Math Methods Practice Questions. It's unclear why Cheryl couldn't just tell both Albert and Bernard the month and day she was born, but that's irrelevant to solving this problem. These questions may be used in your classroom to help students feel comfortable with the types of assessment formats they will experience. August 14 August 15 August 17. Hence, it would be best if you gave the utmost care to see that your child understands these topics adequately. 3rd Grade Math Tests. So, 41 = 4, 42 = 16, 43 = 64, and 44 = 256. Experts are answering questions even after midnight Therefore, if you are stuck on a math problem or homework at the last minute or if it is 10 PM, or 11 PM and you have not found a solution to your math problems, you will get help immediately. Get answer explanations for each question and begin your preparation for the ACCUPLACER college placement test. Since one measurement includes the cat's height and subtracts the turtle's and the other does the opposite, you can essentially just act like the two animals aren't there. If $$\\Large x = 7-4\\sqrt{3}$$, then the value of $$\\Large x+\\frac{1}{x}$$. Primary Maths (Grades 4 and", "X, so Jan will have $(X-1), Chi will have$(X+2) and Conner will have $(X+3) and together they have$28. So (X-1) + (X+2) + (X+3) = $28 3X + 4 =$28, X = $$\\frac{24}{3}$$ So Jan had $(X-1) =$(8-1) = $7, Chi had$(X+2) = $(8+1) =$9, Conner had $(X+3) =$(8+3) = \\$11.", "How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances? # Counting Days | AMC 10A, 2013 | Problem 17 Try this beautiful problem from Algebra: Counting Days ## Counting Days - AMC-10A, 2013- Problem 17 Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365 -day period will exactly two friends visit her? , • $48$ • $54$ • $60$ • $65$ • $72$ ### Key Concepts Algebra LCM Answer: $54$ AMC-10A (2013) Problem 17 Pre College Mathematics ## Try with Hints Given that Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. According to the questation , Let us assume that $A=3 x B=4 x C=5 x$ Now, we want the days in which exactly two of these people meet up The three pairs are $(A, B),(B, C),(A, C)$ Can yoiu find out the LCM of each pair.......... Can you now finish the problem .......... $LCM(A, B)=12 x, LCM(B, C)=20 x, LCM(A, C)=15 x$ Notice that we want to eliminate when all these friends meet up. By doing this, we", "-6, -3 • -321, 241, -54, 333, -21, 231 • 8) Sort in descending order 1. $$23, \\frac{1}{4}, 3, -8, 1\\frac{2}{3}, 12$$ 2. $$-51, 9, -\\frac{8}{9}, 17, -2, \\frac{2}{3}$$ 3. $$-32, 2241, -854, 3433, -2311, 227$$ • 9) Use a calculator to find 1. $$17^2$$ 2. $$\\sqrt[3]{729}$$ 3. $$23^4$$ 4. $$\\sqrt{324}$$ 5. $$\\sqrt[5]{16807}$$ 6. $$(-3)^7$$ • xx • 1) xx 1. $$xx$$ 2. $$xx$$ • 2) xx 1. $$xx$$ 2. $$xx$$ • 3) xx 1. $$xx$$ 2. $$xx$$ • 4) xx 1. $$xx$$ 2. $$xx$$ • 5) xx 1. $$xx$$ 2. $$xx$$ • 6) xx 1. $$xx$$ 2. $$xx$$ • 7) xx 1. $$xx$$ 2. $$xx$$ • 8) xx 1. $$xx$$ 2. $$xx$$ • 9) xx 1. $$xx$$ 2. $$xx$$ • 10) xx 1. $$xx$$ 2. $$xx$$ Sites of Interest Confidence A good tutor can build the confidence of a learner enabling subject success Skills A private tutor can improve the skills a pupil needs to master a subject Progress Regular tutoring can drive progress and better results in school subjects Support Support can help students and parents make the right academic decisions", "# Multinomial Experiments #### Sample Multinomial Experiments problems Question 1: The manager of a Farmer Jack Super Market would like to know if there is a preference for the day of the week on which customers do their shopping. A sample of 420 families revealed the following. At the 0.05 significance level, is there a difference in the proportion of customers that prefer each day of the week? Chi Square test. Goodness of Fit Equal expected frequencies. Day of the week Number of persons Monday 20 Tuesday 30 Wednesday 20 Thursday 60 Friday 80 Saturday 130 Sunday 80 Solution: The following null hypothesis need to be tested: $H_0:\\,p_{1} = {1/7},\\,\\,\\, p_{2} = {1/7},\\,\\,\\, p_{3} = {1/7},\\,\\,\\, p_{4} = {1/7},\\,\\,\\, p_{5} = {1/7},\\,\\,\\, p_{6} = {1/7},\\,\\,\\, p_{7} = {1/7}$ The first task is building the table with expected values. Based on the data provided, we find: Category Observed Expected (fo - fe)²/fe Monday 20 420*1/7=60 26.6667 Tuesday 30 420*1/7=60 15 Wednesday 20 420*1/7=60 26.6667 Thursday 60 420*1/7=60 0 Friday 80 420*1/7=60 6.6667 Saturday 130 420*1/7=60 81.6667 Sunday 80 420*1/7=60 6.6667 Sum = 163.3333 This means the Chi-Square statistics is computed as ${{\\chi }^{2}}=\\sum\\limits_{i=1}^{n}{\\frac{{{\\left( {{O}_{i}}-{{E}_{i}} \\right)}^{2}}}{{{E}_{i}}}}={26.6667} + {15} + {26.6667} + {0} + {6.6667} + {81.6667} + {6.6667}=163.3333$ The critical value for $$\\alpha =0.05$$ and $$df = 6$$ is given", "none of the above Have no idea how to even start this one!!!! :( High School Maths Factorials 06/19/15 #### factorial maths problem help required.... if n is an integer between 0 and 21 ; then the minimum value of n!(21-n)! is: 10!11! 9!12! 21! 20! High School Maths 06/12/15 #### x+y+z=30 using (1,3,5,7,9,11,13,15) use of double is allowed from UPSC examination High School Maths Maths Chi Square Ib Maths 06/11/15 #### Chi square test, what is null and proposed hypothesis According to genetic theory the number of colour strains, red, yellow, blue and white in flower A should appear in proportions 4:9:8:4. Observed frequencies of red, yellow, blue and white strains... more High School Maths Problem Solving Gcd 05/13/15 #### Three numbers in pairs produce the Highest common factors of A million and four, A million and six and A milliion and 8, but this is not possible. why? Three positive integers are written on a whiteboard. David calculated the highest common factor of two of them and obtained 1000004. Rose and Stephen did the same obtaining 1000006 and 1000008... more High School Maths 04/22/15 #### Monthly income of Mr A & B is in ratio of 9:4 and expenditure in ratio of 7:3. If both save Rs 2000 per month. What is the expenditure of Mr", "### Diggits Can you find what the last two digits of the number $4^{1999}$ are? ### Like Powers Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n. ### Two Many What is the least square number which commences with six two's? # Largest Number ##### Stage: 3 Challenge Level: Here are the largest solutions that were sent in. Matthew from South Farnham School: 4096, by doing 4 to the power 3 to the power 2, which is (4 x 4 x 4) x (4 x 4 x 4) = 64 x 64 = 4096. Josh from Alameda School: 1 048 576, which is 32 to the power of 4 (32 x 32 x 32 x 32). Greenford School, Dorset sent in 8 796 093 022 208, which is 2 to the power of 43. Charlotte, Agnes, Susannah and Miranda from Headington Junior School offered 109 418 000 000 000 000 000, which is about 3 to the power 42. In fact the exact value of 342 is 109 418 989 131 512 359 209 can you say it aloud easily?", "which describes students, families and their classes. For example Katie GONZALES,P3C,Christopher GONZALES,P4B denotes a family of two students, Katie and Christopher Gonzales, in classes P3C and P4B, respectively. We want to ensure that Katie and Christopher both go to school on the same day, and make sure this \"siblings attend on the same day\" property holds for all families at the school. This class allocation size data with the names of classes and the number of students allowed to be in each is also encapsulated in a csv file (click for full version): In [2]: #use_gist \"ewenmaclean/4cf1c29402e63426f32c312a14ca86df\";; Out[2]: val class_csv_data : string = \"P1A,30\\nP1B,45\\nP1C,30\\nP2A,30\\nP2B,30\\nP2C,30\\nP3A,30\\nP3B,30\\nP3C,30\\nP4A,30\\nP4B,30\\nP4C,30\\nP5A,35\\nP5B,35\\nP5C,35\\nP6A,40\\nP6B,40\\nP6C,40\\nP7A,40\\nP7B,40\\nP7C,40\" In addition to this data we import a file of keyworker families - those where the parents are both classed as keyworkers. These families are assumed to have the days on which they can attend normal school (not Hub school) predetermined, and hence their \"bubble\" (A, B or C) of their class for this school already decided. In [3]: #use_gist \"ewenmaclean/cc15cef4fd563404e3152d7b3e40c4a5\";; Out[3]: val keyworkers_csv_data : string = \"0,A\\n1,A\\n15,B\\n23,B\\n34,C\\n45,B\\n57,A\" # Solution using the Imandra Scheduler¶ We can now exploit the Imandra Scheduler to find a solution to the problem. In [4]: Imandra_scheduler.Solve.top ~lines:student_csv_data ~classes:class_csv_data ~keyworkers:keyworkers_csv_data Out[4]: Solving Complete and Validated! - : unit = () The resulting csv can be loaded into a spreadsheet program", "_________________ Common multiples: _________________ Answer: 8,16. Explanation: Multiples of 2: 2,4,6,8,10,12,14,16,18,20. Multiples of 8: 8,16,24,32,40,48,56,64,72,80. So common multiples are: 8,16 Generalize Algebra Find the unknown number. Question 15. 12, 24, 36, _____ Answer: 48 Explanation: 12×1= 12 12×2= 24 12×3= 36 12×4= 28 Question 16. 25, 50, 75, 100, ______ Answer: 125 Explanation: 25×1= 25 25×2= 50 25×3= 75 25×4= 100 25×5= 125 Tell whether 20 is a factor or multiple of the number. Write factor, multiple, or neither. Question 17. 10 Answer: Multiple Explanation: 2×10= 20. Question 18. 20 Answer: Factor and multiple Explanation: 1×20= 20 20÷1= 20. Question 19. 30 Answer: Neither Explanation: Factors of 30 are: 1,2,3,5,6,10,15,and 30. Multiples of 30 are: 30,60,90,etc. Write true or false. Explain. Question 20. Every whole number is a multiple of 1. Answer: True. Explanation: For every whole number which is multiplied with 1, the result will be that number. Question 21. Every whole number is a factor of 1. Answer: False Explanation: Not every whole number is a factor of 1. Question 22. Julio wears a blue shirt every 3 days. Larry wears a blue shirt every 4 days. On April 12, both Julio and Larry wore a blue shirt. What is the next date that they will both wear a blue shirt? Answer: April 24 Explanation: As", "computing this maximum. ## Problem 77 With about six hours left on the van ride home from vacation, Wendy looks for something to do. She starts working on a project for the math team. There are sixteen students, including Wendy, who are about to be sophomores on the math team. Elected as a math team officer, one of Wendy's jobs is to schedule groups of the sophomores to tutor geometry students after school on Tuesdays. The way things have been done in the past, the same number of sophomores tutor every week, but the same group of students never works together. Wendy notices that there are even numbers of groups she could select whether she chooses $4$ or $5$ students at a time to tutor geometry each week: \\begin{align*}\\dbinom{16}4&=1820,\\\\\\dbinom{16}5&=4368.\\end{align*} Playing around a bit more, Wendy realizes that unless she chooses all or none of the students on the math team to tutor each week that the number of possible combinations of the sophomore math teamers is always even. This gives her an idea for a problem for the $2008$ Jupiter Falls High School Math Meet team test: $\\text{How many of the 2009 numbers on Row 2008 of Pascal's Triangle are even?}$ Wendy works the solution out correctly. What is her answer? ## Problem 78 Feeling excited over her", "the problem…….. Therefore $m+n+k=75$ Categories ## Counting Days | AMC 10A, 2013 | Problem 17 Try this beautiful problem from Algebra: Counting Days ## Counting Days – AMC-10A, 2013- Problem 17 Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365 -day period will exactly two friends visit her? , • $48$ • $54$ • $60$ • $65$ • $72$ ### Key Concepts Algebra LCM Answer: $54$ AMC-10A (2013) Problem 17 Pre College Mathematics ## Try with Hints Given that Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. According to the questation , Let us assume that $A=3 x B=4 x C=5 x$ Now, we want the days in which exactly two of these people meet up The three pairs are $(A, B),(B, C),(A, C)$ Can yoiu find out the LCM of each pair………. Can you now finish the problem ………. $LCM(A, B)=12 x, LCM(B, C)=20 x, LCM(A, C)=15 x$ Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.", "are anything like most teachers I've met, they probably work harder than me!) @nitsua60 Grading papers and homework and exams... so many extra hours on top of the job one gets paid for. 4:37 PM @GcL It's true that a lamentable fraction of people confuse face time for \"work time\" in this career. Then again, I have two summer months off for no good reason, so I try to keep my head down. @nitsua60 Work smarter, not harder. ;) I figure that's where you save time. :) @nitsua60 My experience was only with mathematics and computer science. My recollection was that the hours for the time off didn't balance the time grading papers and other overhead. This was especially so if you count the time as overtime. Okay, so as I said, my group went from 3, to 5, and now to 7. They are finishing up an encounter but moving somewhat shortly onto a new one that was just supposed to be Medium. (3 level 8s vs a CR 7 monster) I may make it Hard with 3 CR 7 monsters does that sound right? How many players and what's their character's levels? @GcL level 8. Now a total of 7. 4:44 PM Seven level 8 characters? Damn. But I think White Plume mountain may have been", "Kattis # Prinova Brojko and Brojana are happily married with $N$ little boys. The boys are named with distinct even integers $P_1, P_2, \\ldots , P_ N$. Brojko and Brojana are expecting an addition to their family and have to come up with a nice name for the little girl. They have decided that the name will be an odd integer in the range $[A, B]$. Because they find all integers in that range equally beautiful, they have decided to choose the number which maximizes the distance to the name of the closest of the $N$ boys. More precisely, they seek an odd integer $X \\in [ A , B ]$ such that the expression $\\min \\{ |X - P_ i| , i \\in [ 1 , N ] \\}$ is as large as possible. Write a program that determines the name for the little girl. ## Input The first line contains an integer $N$ ($1\\le N \\le 100$), the number of boys. The second line contains N distinct positive even integers, the names of the boys. The integers will be less than $10^9$. The third line contains the integers $A$ and $B$ ($1 \\le A < B \\le 10^9$), the range of names they are considering for the girl. ## Output Output an integer, the name for" ]

[ "e.g ) =5$ in many cases one may express their in!", "# How do you evaluate the expression 30-10times3/5? $24$ $30 - 10 \\times \\frac{3}{5} = 30 - 6$ $30 - 6 = 24$", "# How do you evaluate the expression 30-10times3/5? Mar 19, 2018 $24$ $30 - 10 \\times \\frac{3}{5} = 30 - 6$ $30 - 6 = 24$", "# What is the value of the expression f^2+5 if f=3? $14$ Substitute $3$ for $f$ in the expression and then do the arithmetic to find the solution: ${f}^{2} + 5 \\to {3}^{2} + 5 \\to 9 + 5 \\to 14$", "## MarziaJackson\\$ one year ago What is the value of the expression 2/4^-3 ? hint: $\\Large {4^{ - 3}} = \\frac{1}{{{4^3}}} = \\frac{1}{{4 \\times 4 \\times 4}} = \\frac{1}{{64}}$", "11}$​ what's the value of given expression :$\\sqrt{11 \\times 11}$​...", "expression is $3$.", "values $$n_2=2+3=5$$", "+0 What is the value? 0 243 1 What is the value for the expression: (8+3)^2+4x(-3)? Guest Apr 22, 2017 #1 +7451 +1 What is the value for the expression: (8+3)^2+4x(-3)? $$(8+3)^2+4\\times (-3)\\\\=11^2-12\\\\=121-12\\\\=109$$ ! asinus Apr 22, 2017", "Q # Simplify the expression and find its value when a = 5 Q. 10. Simplify the expression and find its value when $a=5$ and $b=-3\\; .2(a^{2}+ab)+3-ab$ $\\\\2( a^2 + ab ) + 3 - ab \\\\= 2a^2 + 2ab + 3 - ab \\\\= 2a^2 + ab + 3 \\\\= 2 \\times 5^2 + 5 \\times ( -3 ) + 3 \\\\= 50 - 15 + 3 \\\\= 38$", "# If x = 4 and y = 2, what is the value of the following expression: 3xy - 12y + 5x 20 Explanation To find the value of this expression, substitute the given values for x and y into the expression: 3(4)(2) - 12(2) + 5(4) Then, calculate the value of the expression: $$3 \\times 8-12 \\times 2+5 \\times 4=24-24+20=20$$", "always be 5, $\\E[5] = 5$.", "Q # Find the value of m for which 5^m ÷ 5^ – 3 = 5^5 . 5. Find the value of $m$ for which $5^m \\div 5^{-3} = 5^5$ Views We have, $a^{m}\\div a^{n}= a^{m-n}$ Here a = 5 and n =-3 and m-n = 5 therefore, $5^{m}\\div 5^{-3}= 5^{m+3} = 5^{5}$ By comparing from both sides we get m+3 = 5 m= 2 Exams Articles Questions", "5 = 5$$ $$-30 : 30 = - 1$$", "## needhelpfastplease Group Title What is the value of the expression –3x^3 when x = –3? A. –81 B. –27 C. 27 D. 81 6 months ago 6 months ago @kym02 2. kym02 Did you work it out? i tried to @kym02 i think its C but im not quite sure 4. kym02 Alright i'll go it through with you. $-3x^{3}$ Hence, $x=-3$ so, $(-3) \\times (-3) \\times (-3) = -27$ so basically it now becomes: $-3(-27)$ which is equal to: $81$ Therefore, the answer is D :) you had it! just needed to keep trying! :D", "{ x - 3 }$ Factor each expression. Look for $1$s. $4 x ^ { 2 } + 5 x - 6 = ( x + 2 ) ( 4 x - 3 )$", "\"$f$ evaluated at $x=5$ is $f(5)=5^2+1$\". – Stefan Hansen Jan 2 '13 at 11:50", "Q # What should be the value of a if the value of 2x^2 + x - a equals to 5, Q. 9. What should be the value of a if the value of $2x^{2}+x-a$ equals to $5$, when $x=0?$ $\\\\2x^2 + x - a = 5 \\\\2 \\times 0^2 + 0 - a = 5 \\\\-a = 5 \\\\a = -5$", "Question # If $$m=2$$, find the value of:$$m - 2, \\ 3m-5, \\ 9-5m$$, $$3{ m }^{ 2 }-2m-7$$ , $$\\dfrac { 5m }{ 2 } -4$$ Solution ## $$m=2$$1. $$m-2=0$$2. $$3m-5=3\\times 2-5=1$$3. $$9-5m=-1$$4. $$3m^2-2m-7=12-4-1$$ $$=7$$5. $$\\dfrac{5m}{2}-4=1$$.Maths Suggest Corrections 0 Similar questions View More People also searched for View More", "5x is equivalent to ` 5 x!" ]

[ "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # Whole Number Exponents ## Distinguish bases and powers. Estimated5 minsto complete % Progress Practice Whole Number Exponents Progress Estimated5 minsto complete % Whole Number Exponents ### Guidance Sometimes, we have to multiply the same number several times. We can say that we are multiplying the number by itself in this case. 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} is 4 multiplied by itself three times. When we have a situation like this, it is helpful to use a little number to show how many times to multiply the number by itself. That little number is called an exponent. If we were going to write 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent, we would write 43\\begin{align*}4^3\\end{align*}. This lesson is all about exponents. By the end of it, you will how and when to use them and how helpful this shortcut is for multiplication. Using exponents has an even fancier name too. We can say that we use exponential notation when we express multiplication in terms of exponents. We can use exponential notation to write an expanded multiplication problem into a form with an exponent, we write 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent =43\\begin{align*}= 4^3\\end{align*} We can work the other way around too. We can write a number with an exponent", "involving exponents to represent and solve real-world problems. Teaching Time I. Identify Whole Number Powers, Bases and Exponents Sometimes, we have to multiply the same number several times. We can say that we are multiplying the number by itself in this case. 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} is 4 multiplied by itself three times. When we have a situation like this, it is helpful to use a little number to show how many times to multiply the number by itself. That little number is called an exponent. If we were going to write 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent, we would write 43\\begin{align*}4^3\\end{align*}. This lesson is all about exponents. By the end of it, you will how and when to use them and how helpful this shortcut is for multiplication. Using exponents has an even more technical term also. We can say that we use exponential notation when we express multiplication in terms of exponents. We use exponential notation to write an expanded multiplication problem as a base number with an exponent, we write 4×4×4\\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent =43\\begin{align*}= 4^3\\end{align*} We can work the other way around too. We can write a number with an exponent as a long multiplication problem and this is called expanded form. The base number is the", "multiplying the number by itself in this case. \\begin{align*}4 \\times 4 \\times 4\\end{align*} is 4 multiplied by itself three times. When we have a situation like this, it is helpful to use a little number to show how many times to multiply the number by itself. That little number is called an exponent. If we were going to write \\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent, we would write \\begin{align*}4^3\\end{align*}. This lesson is all about exponents. By the end of it, you will how and when to use them and how helpful this shortcut is for multiplication. Using exponents has an even fancier name too. We can say that we use exponential notation when we express multiplication in terms of exponents. We can use exponential notation to write an expanded multiplication problem into a form with an exponent, we write \\begin{align*}4 \\times 4 \\times 4\\end{align*} with an exponent \\begin{align*}= 4^3\\end{align*} We can work the other way around too. We can write a number with an exponent as a long multiplication problem and this is called expanded form. The base is the number being multiplied by itself in this case the base is 4. The exponent tells how many times to multiply the base by itself in this case, it is a 3. Using an exponent can also", "do I solve this? $125^{2}{3}$ the 2/3 is an exponent. \\displaystyle \\begin{align*} 125^{\\frac{2}{3}} &= \\left(125^{\\frac{1}{3}}\\right)^2 \\\\ &= \\left(\\sqrt[3]{125}\\right)^2 \\\\ &= 5^2 \\\\ &= 25 \\end{align*}", "exponents, you read the exponent as a power. For example, 35\\begin{align*}3^5\\end{align*} is read as \"three to the fifth power\". 27\\begin{align*}2^7\\end{align*} is read as \"two to the seventh power\". 59\\begin{align*}5^9\\end{align*} is read as \"five to the ninth power\". An exponent tells you how many times the base should be multiplied by itself. If you look at a number with an exponent written out the long way, you can see why exponents are useful. 7×7×7=73 In this case, there are three factors of 7. Remember that factors are numbers or variables multiplied by each other. 5×5×5×5×5×5×5×5×5×5=510 In this case, there are ten factors of 5. Exponents are a multiplication shortcut a lot like the way that multiplication is an addition shortcut. ### Guided Practice Write out the factors of 35\\begin{align*}3^5\\end{align*}, then evaluate the product. 35=3×3×3×3×3 Once you have written out all the factors, simply multiply from left to right. 3×3=9×3=27×3=81×3=243 ### Examples #### Example 1 Write 63\\begin{align*}6^3\\end{align*} out in words. First, identify the base number, which is the big number that will be multiplied by itself. In this case, the base is 6. Next, consider the power, which is the small number that describes how many times the number will be multiplied. In this case, the power is 3, so the base will be multiplied by itself 3 times. The", "addition sign. (5.7×104)+(4.87×105)\\begin{align*}(5.7 \\times 10^4) + (4.87 \\times 10^5)\\end{align*} We want to make both of these exponents the same. To make both exponents 5’s, we move the decimal point in 5.7 one place to the left by multiplying by 10. (.57×105)+(4.87×105)\\begin{align*}(.57 \\times 10^5) + (4.87 \\times 10^5)\\end{align*} Now we can add the decimal parts of the problem. The power of 10 stays the same. (.57×105)+(4.87×105)(.57+4.87)×1055.44×105 Our answer is 5.44×105\\begin{align*}5.44 \\times 10^5\\end{align*}. Subtraction works the same way as addition: Before performing the subtraction operation, the exponents must be the same. Multiplication and division in scientific notation is a little different. Do you remember simplifying exponents? (x3)x4\\begin{align*}(x^3)x^4\\end{align*} To multiply the exponents, we add the powers. (x3)x4=x3+4=x7\\begin{align*}(x^3)x^4 = x^{3 + 4} = x^7\\end{align*}. Multiplying scientific notation is similar: You multiply the decimals and add the exponents. (3.4×102)×(6.2×106)(3.4×6.2)×(102+6)21.08×104 Division of scientific notation is identical to multiplication—except you divide the decimals and subtract the exponents. Let’s try it out. (8.4×105)÷(1.4×102)(8.4÷1.4)×(105(2))Remember subtracting a negative is the same as adding it.6×107 Now it's time for you to try a few on your own. #### Example A Add (3.4×103+5.6×104)\\begin{align*}(3.4 \\times 10^3 + 5.6 \\times 10^4)\\end{align*} Solution: 39.6×104\\begin{align*}39.6 \\times 10^4\\end{align*} #### Example B Multiply (1.2×104)(3.4×104)\\begin{align*}(1.2 \\times 10^4)(3.4 \\times 10^4)\\end{align*} Solution:4.08×108\\begin{align*}4.08 \\times 10^8\\end{align*} #### Example C Subtract (5.6×1043.2×104)\\begin{align*}(5.6 \\times 10^4 - 3.2 \\times 10^4)\\end{align*} Solution: 2.4×104\\begin{align*}2.4 \\times 10^4\\end{align*}", "# Are You Sure? Algebra Level 3 \\begin{align} 1+1 \\times 2 \\times 3 \\times 4 &= 5^2 \\\\ 1+2 \\times 3 \\times 4 \\times 5 &= 11^2 \\\\ 1 + 3 \\times 4 \\times 5 \\times 6 &= 19^2 \\end{align} Is 1 plus the product of four consecutive integers always a perfect square? ×", "parentheses fall in the order of operations. Order of Operations P - parentheses E - exponents MD - multiplication or division in order from left to right AS - addition or subtraction in order from left to right Wow! You can see that, according to the order of operations, parentheses come first. We always do the work in parentheses first. Then we evaluate exponents. Let’s see how this works with a new example. Example 2+(31)×2\\begin{align*}2 + (3 - 1) \\times 2\\end{align*} In this example, we can see that we have four things to look at. We have 1 set of parentheses, addition, subtraction in the parentheses and multiplication. We can evaluate this expression using the order of operations. Example 2+(31)×22+2×22+4=6\\begin{align*}& 2 + (3 - 1) \\times 2\\\\ & 2 + 2 \\times 2\\\\ & 2 + 4\\\\ & = 6\\end{align*} Our answer is 6. What about when we have parentheses and exponents? Example 35+32(3×2)×7\\begin{align*}35 + 3^2 - (3 \\times 2) \\times 7\\end{align*} We start by using the order of operations. It says we evaluate parentheses first. 3×2=635+326×7\\begin{align*}& 3 \\times 2 = 6\\\\ & 35 + 3^2 - 6 \\times 7\\end{align*} Next we evaluate exponents. 32=3×3=935+96×7\\begin{align*}& 3^2 = 3 \\times 3 = 9\\\\ & 35 + 9 - 6 \\times 7\\end{align*} Next, we complete multiplication or division in order", "(another word for exponents). To remember PPMDAS, student may memorize the phrase “Pretty Please My Dear Aunt Sally.” In PEMDAS, the first step is parentheses. Parentheses and exponents (otherwise known as powers) must be done in the correct order. Otherwise, the calculations don’t work out consistently. For example, \\begin{align} (1+3)^2 &= 4^2 = 16 \\:\\:\\:\\text{parentheses before powers/exponents}\\\\ &\\not = 1^2+3^2 = 10\\:\\text{powers/exponents first} \\end{align} Here’s a more complicated example: $$(3+2\\times 3)^2 \\div 3^3 \\times 4= \\: ?$$ For practice, calculate this using PEMDAS. Then check below to see if your solution followed the same steps. First, we use PEMDAS inside the parentheses : 3+2×3 = 3 + 6 = 9. Second, we calculate the exponent: \\begin{align} (3+2\\times 3)^2 \\div 3^3 \\times 4 &= 9^2 \\div 3^3 \\times 4\\\\ &= 81 \\div 27 \\times 4\\\\ &= 3\\times 4\\\\ &=12 \\end{align} ## When Order Doesn’t Matter In the “MD” step of PEMDAS, multiplication and division are performed. If a problem only requires multiplication and division, then the order can be changed without affecting the problem. For example, \\begin{align} 3\\times 5\\div 10\\times 2\\div 3 &= 3\\div 3\\times 5\\times 2\\div 10\\\\ &= (3\\div 3)\\times (5\\times 2\\div 10)\\\\ &= 1 \\times 1\\\\ &= 1 \\end{align} Similarly, in the “AD” step of PEMDAS, addition and subtraction can be done in any order. \\begin{align} 123,456", "Multiplication using Exponents (Indices) If we wish to calculate $5^4 \\times 5^3$, we could write in factor form to get: \\begin{align} \\displaystyle 5^4 \\times 5^3 &= (5 \\times 5 \\times 5 \\times 5) \\times (5 \\times 5 \\times 5) \\\\ &= 5^7 \\end{align} Example 1 Simplify $7^2 \\times 7^3$ after first writing in factor form. \\begin{align} \\displaystyle 7^2 \\times 7^3 &= (7 \\times 7) \\times (7 \\times 7 \\times 7) \\\\ &= 7^5 \\end{align} However, if we look closely, a much simpler method would be to add the exponents since the bases are the same. Therefore this calculation can also be done this way: \\begin{align} \\displaystyle 5^4 \\times 5^3 &= 5^{4+3} \\\\ &= 5^7 \\end{align} , which is the same answer. We can add the exponents when multiplying only if the bases are the same. Thus to $\\textit{multiply}$ numbers with the $\\textit{same base}$, keep the base and $\\textit{add}$ the exponents. $$\\large a^x \\times a^y = a^{x+y}$$ Example 2 Simplify $9^3 \\times 9^5$. \\begin{align} \\displaystyle 9^3 \\times 9^5 &= 9^{3+5} \\\\ &= 9^8 \\end{align} Example 3 Simplify $4^3 \\times 4 \\times 4^5$. \\begin{align} \\displaystyle 4^3 \\times 4 \\times 4^5 &= 4^3 \\times 4^1 \\times 4^5 \\\\ &= 4^{3+1+5} \\\\ &= 4^9 \\end{align} Many questions will be algebraic, meaning that a pronumeral is used. In such questions, we multiply the", "precedence over -. • 13.667 • Remember that // does integer division. Here is an animation for the above expression: 16 - 2 * 5 // 3 + 1", "those exponents? You may have memorized exponent rules at some point, but how can you be sure that you’re remembering them correctly? Rather than obsessing about the accuracy of you memory, a quick check of this problem can be done by thinking about what exponents mean. When we say what we mean is a multiplied by itself three times. If we take that basic understanding and apply it to this problem we get: $\\dpi{200}[(a \\times a) \\times (a \\times a \\times a)]^4=?$ Now it’s clear that we have a multiplied by itself 5 times, which is a5. That fits perfectly with the relevant exponent rule, with no cause for doubt. To resolve the rest of this problem we can use the same approach. Raising something to the fourth power means multiplying it by itself four times, so we get a5 times itself five times, then another five, then another five, then another five for a total of 20. $\\dpi{200}[(a \\times a) \\times (a \\times a \\times a)]^4=a^5 \\times a^5 \\times a^5 \\times a^5 =a^{20}$ When you feel overwhelmed by the number of things you need to memorize to prepare for your GMAT, remember that taking a little extra time to understand the logic behind those rules you’re memorizing puts a lot less reliance on your fickle memory and", "so 3 is the exponent.} \\end{align} 2. \\begin{align} &2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 = 64 \\\\ &2^{6} = 64 &&\\text{The number 2 is multiplied by itself 6 times, so 6 is the exponent.} \\end{align} 3. \\begin{align} &3 \\times 3 \\times 3 = 27 \\\\ &3^{3} = 27 &&\\text{The number 3 is multiplied by itself 3 times, so 3 is the exponent.} \\end{align} 4. \\begin{align} &3 \\times 3 \\times 3 \\times 3 \\times 3 \\times 3 = 729 \\\\ &3^{6} = 729 &&\\text{The number 3 is multiplied by itself 6 times, so 6 is the exponent.} \\end{align} Instead of writing $$3 \\times 3 \\times 3 \\times 3 \\times 3 \\times 3$$, in exponential form we write $$3^{6}$$. You will get the same answer using a calculator, regardless of the form. Exercise 3.2: Writing exponents as products of bases Write each power as a product of the same base. Calculate the answer. 1. $5^{2}$ 2. $2^{5}$ 3. $3^{4}$ 4. $4^{3}$ 5. $2^{7}$ 6. $3^{5}$ 7. $10^{2}$ 8. $15^{2}$ We look at the exponent to decide how many times to multiply the base by itself. 1. \\begin{align*} &5^{\\mathbf{2}} = 5 \\times 5 = 25 &&\\mathbf{2 \\ times} \\end{align*} 2. \\begin{align*} &2^{\\mathbf{5}} = 2 \\times 2 \\times 2 \\times 2 \\times 2 = 32 &&\\mathbf{5 \\", "100 + 50\\\\ &= 50\\times 100 + 50\\\\ &= 5050\\end{align} Do your students have an opportunity to be inspire by such problems? See Math Tidbits for more inspiring problems.", "5 \\left[ 7(10)^2+1(10)+2 \\right] \\end{align*} I personally don't think anyone should use this for n=1 as we should have them memorized (#shame) and to my knowledge this only works for n using positive integers. Regardless, a super over-the-top way of multiplying!", "multiplication first. It may help to clarify things by grouping the multiplied numbers with parentheses. \\begin{align*}2 + (4 \\times 7) - 1\\end{align*} Now evaluate the expression inside the parentheses: \\begin{align*}4 \\times 7 = 28\\end{align*}, the expression becomes: \\begin{align*} 2 + (28) - 1\\end{align*} Now you have only addition and subtraction. Start at the left and work across: \\begin{align} & \\quad 2 + 28 - 1\\\\ & = 30 - 1\\\\ & = 29\\\\ \\end{align} Many people use the word PEMDAS to help remember the priority order of the mathematical operations: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. Just remember that multiplication and division have the same priority, and so do addition and subtraction, so those operations should be completed left to right. #### Order of Operations 1. First evaluate expressions within Parentheses, including brackets [ ] and braces { }. 2. Next evaluate all Exponents (terms such as \\begin{align*}3^2\\end{align*} or \\begin{align*}x^3\\end{align*}). 3. Then Multiplication and Division - work from left to right completing both multiplication and division in the order that they appear. 4. Finally, evaluate Addition and Subtraction - work from left to right completing both addition and subtraction in the order that they appear. #### Evaluating Expressions Each of these expressions has the same numbers and the same mathematical operations in the same order. The", "\\times x + 4 x \\times \\left(\\frac {500}{x \\times x}\\right)$ Originally Posted by apatite ooo so A= (x^2+4x) * (5oo/x^2). Thank you for your help Not quite. Multiplication has a higher priority than addition. $A = x \\times x + 4 x \\times \\left(\\frac {500}{x \\times x}\\right)$ $A = x \\times x + \\left(\\frac {4 x \\times 500}{x \\times x}\\right)$ $A = x^2 + \\frac {2000}{x}$", "(3, 10 dx)$$. Algebra is great, but sticking with values sure is easier than manipulating symbols. Automatic differentiation has a forward mode and a reverse mode. The latter is also called backpropagation in some contexts. These modes relate to the chain rule, which in our implementation is hidden in lambdas and applications. Working through the details, we find the chain rule leads to expressions like: $3 \\times 4 \\times 5 \\times (dp + 2 dq + 4 dr)$ My understanding is that in forward mode, the multiplications associate to the right: \\begin{align} & & 3 \\times 4 \\times 5 \\times (dp + 2 dq + 4 dr) \\\\ &=& 3 \\times 4 \\times (5 dp + 10 dq + 20 dr) \\\\ &=& 3 \\times (20 dp + 40 dq + 80 dr) \\\\ &=& 60 dp + 120 dq + 240 dr \\end{align} while in reverse mode the multiplications associate to the left, which is more efficient: \\begin{align} & & 3 \\times 4 \\times 5 \\times (dp + 2 dq + 4 dr) \\\\ &=& 12 \\times 5 \\times (dp + 2 dq + 4 dr) \\\\ &=& 60 \\times (dp + 2 dq + 4 dr) \\\\ &=& 60 dp + 120 dq + 240 dr \\end{align} We can view the right-most factor as a list", "+ 2^2 / 2 \\times 5 – 2 = (3 + (((2^2) / 2) \\times 5)) – 2$$ ### Multiplying a digit by a digit Multiplication is repeated addition so can find a product by adding one digit the number of times indicated by the other digit, i.e, \\begin{align} 4 \\times 5 &= \\overbrace{4 + 4 + 4 + 4 + 4}^{\\mbox{5 times}}\\\\ &= 20 \\end{align} or \\begin{align} 5 \\times 4 &= \\overbrace{5 + 5 + 5 + 5}^{\\mbox{4 times}}\\\\ &= 20 \\end{align} However, this is inefficient; multiplication is used so often that it pays to simply memorize the multiplication tables from 0 to 10. These are 110 products so it sounds difficult but it is not as hard: • $0 \\times a = 0$ • $1 \\times a = a$ • $10 \\times a = a$ with a zero at the end (e.g., $10 \\times 7 = 70$) • $9 \\times a = (10 \\times a) – a$ (e.g., $9 \\times 7 = 10 \\times 7 – 7 = 70 – 7 = 63$) • $2 \\times a = a + a$ (e.g., $2 \\times 7 = 7 + 7 = 14$, easy because we know the addition facts) • $a \\times b$ is even unless both a and b are odd (i.e., 75% of the single-digit", "as \"two to the seventh power\". 59\\begin{align*}5^9\\end{align*} is read as \"five to the ninth power\". We could go on and on. When you see a base with an exponent of 2 or an exponent of 3, we have different names for those. 22\\begin{align*}2^2\\end{align*} is read as two squared. 63\\begin{align*}6^3\\end{align*} is read as six cubed. It doesn’t matter what the base is, the exponents two and three are read squared and cubed. What does an exponent actually do? An exponent tells us how many times the base should be multiplied by itself. We can write them out the long way. Example 73=7×7×7\\begin{align*}7^3 = 7 \\times 7 \\times 7\\end{align*} Example 510=5×5×5×5×5×5×5×5×5×5\\begin{align*}5^{10} = 5 \\times 5 \\times 5 \\times 5 \\times 5 \\times 5 \\times 5 \\times 5 \\times 5 \\times 5\\end{align*} If you haven’t figured it out yet, exponents are a multiplication short cut a lot like the way that multiplication is an addition short cut. Here are few for you to work on by yourself. 1. Write out in words - 63\\begin{align*}6^3\\end{align*} 2. Write out the factors of 45\\begin{align*}4^5\\end{align*} 3. Which is the base number: 910\\begin{align*}9^{10}\\end{align*}? Take a minute and check your work with a peer. II. Writing the Product of a Repeating Factor as a Power In the last section, we took bases with exponents and wrote them" ]

[ "## Algebra 1 $125$ An exponent of $3$ means that the given factor is multiplied three times. Thus, $5^3 = 5(5)(5) = 25(5) = 125$", "If f=5, what is the value of 2f^2+8f-5? May 30, 2017 $85$ Explanation: $2 {f}^{2} + 8 f - 5$ $f = 5$ We can now substitute $f$ for $5$ in the question. $2 \\times {5}^{2} + 8 \\times 5 - 5$ $2 \\times 25 + 40 - 5$ $50 + 40 - 5$ $90 - 5$ $85$ May 30, 2017 $85$. Explanation: First, plug the number in. $2 \\cdot {5}^{2} + 8 \\cdot 5 - 5$ Then, follow the order of operations. There are no grouping symbols, so do exponents first. $2 \\cdot$25 $+ 8 \\times 5 - 5$ Multiplication is next: 50 $+ 40 - 5$", "First we need to calculate the value of the given expression. $5^{3}$ = $5 \\times 5 \\times 5 = 125$ and, $3^{5}$ = $3 \\times 3 \\times 3 \\times 3 \\times 3$ = $243$ Thus $5^{3} = 125$ and $3^{5} = 243$ Therefore, $5^{3} < 3^{5}$ Here we have seen the basic introduction and examples of the exponents and their properties. To practice more on laws of exponents visit our site BYJU’S and learn the various concepts in an engaging manner.", "so 3 is the exponent.} \\end{align} 2. \\begin{align} &2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 = 64 \\\\ &2^{6} = 64 &&\\text{The number 2 is multiplied by itself 6 times, so 6 is the exponent.} \\end{align} 3. \\begin{align} &3 \\times 3 \\times 3 = 27 \\\\ &3^{3} = 27 &&\\text{The number 3 is multiplied by itself 3 times, so 3 is the exponent.} \\end{align} 4. \\begin{align} &3 \\times 3 \\times 3 \\times 3 \\times 3 \\times 3 = 729 \\\\ &3^{6} = 729 &&\\text{The number 3 is multiplied by itself 6 times, so 6 is the exponent.} \\end{align} Instead of writing $$3 \\times 3 \\times 3 \\times 3 \\times 3 \\times 3$$, in exponential form we write $$3^{6}$$. You will get the same answer using a calculator, regardless of the form. Exercise 3.2: Writing exponents as products of bases Write each power as a product of the same base. Calculate the answer. 1. $5^{2}$ 2. $2^{5}$ 3. $3^{4}$ 4. $4^{3}$ 5. $2^{7}$ 6. $3^{5}$ 7. $10^{2}$ 8. $15^{2}$ We look at the exponent to decide how many times to multiply the base by itself. 1. \\begin{align*} &5^{\\mathbf{2}} = 5 \\times 5 = 25 &&\\mathbf{2 \\ times} \\end{align*} 2. \\begin{align*} &2^{\\mathbf{5}} = 2 \\times 2 \\times 2 \\times 2 \\times 2 = 32 &&\\mathbf{5 \\", "Evaluate: 1.63 * 10^0 Expression: $1.63 \\times {10}^{0}$ Any non-zero expression raised to the power of $0$ equals $1$ $1.63 \\times 1$ Any expression multiplied by $1$ remains the same \\begin{align*}&1.63 \\\\&\\begin{array} { l }\\frac{ 163 }{ 100 },& 1 \\frac{ 63 }{ 100 }\\end{array}\\end{align*} Random Posts Random Articles", "$$5$$ only. The multiplication is as follows. \\begin{align}{5 \\;{0}} \\\\ { \\times \\;\\;{3}} \\\\ \\hline 1\\,5\\, {0} \\\\ \\hline\\end{align} Hence, the values of $$A$$,$$B$$, and $$C$$ are $$5, \\,0,$$ and $$1$$ respectively. Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school", "Back ## Exponents, Simplified When we see repeated addition such as $3 + 3 + 3 + 3 + 3,$ we can rewrite the expression and make it shorter with multiplication. In this case, we get $5 \\times 3.$ Meanwhile, we can make an expression with repeated multiplication more compact with an exponent. For example, we can rewrite $4 \\times 4 \\times 4$ as $4^3.$ What happens when we see a combination of repeated operations in one expression, such as $4^3 + 4^3 + 4^3 + 4^3?$ Can we rewrite this expression more simply? Let's begin by looking at the repeated addition in $4^3 + 4^3 + 4^3 + 4^3.$ The term $4^3$ appears four times in the repeated addition, so we can think of it as being multiplied by $4.$ So, $4^3 + 4^3 + 4^3 + 4^3 = 4 \\times 4^3.$ Is there now some way we can combine the $4\\text{'s?}$ Let's look at the factor $4^3$ first. It means multiplication of three factors of $4,$ so we can now rewrite the expression as \\begin{aligned} 4 \\times 4^3 & = 4 \\times (4 \\times 4 \\times 4 )\\\\ & = 4 \\times 4 \\times 4 \\times 4. \\end{aligned} This is simply $4^4.$ So, $4^3 + 4^3 + 4^3 + 4^3 = 4^4.$ By switching between the", "Intermediate Algebra (6th Edition) $-5^{3}=-(5\\times5\\times5)=-(125)=-125$ Due to order of operations, we must evaluate the exponent before the negative sign.", "\\begin{array}{l}2 \\times 2 \\times 2 \\times 2 \\times \\\\3 \\times 3 \\times 5 \\times 5\\end{array} \\right]$$ In this, $$4$$ is the exponent of $$2$$. $$2$$ is the exponent of $$3$$ and $$2$$ is the exponent of $$5$$ So, product of powers of prime factors $$= 2^4 \\times 3^2 \\times 5^2$$ ## Chapter 13 Ex.13.1 Question 6 Simplify: (i) $$2\\times {{10}^{3}}$$ (ii) $${{7}^{2}}\\times {{2}^{2}}$$ (iii) $${{2}^{3}}\\times 5$$ (iv) $$3\\times {{4}^{4}}$$ (v) $$0\\times {{10}^{2}}$$ (vi) $${{5}^{2}}\\times {{3}^{3}}$$ (vii) $${{2}^{4}}\\times {{3}^{2}}$$ (viii) $${{3}^{2}}\\times {{10}^{4}}$$ ### Solution What is known? Base and exponent. What is unknown? Simple form of the number. Reasoning:. In this question we will multiply the base and the number of times it’s exponent to simplify the given statement. (i) \\begin{align} 2\\times {{10}^{3}}&=2\\times (10\\times 10\\times 10)\\\\&=2\\times 1000\\\\&=2000\\end{align} (ii) \\begin{align} {{7}^{2}}\\times {{2}^{2}}&=(7\\times 7)\\times (2\\times 2)\\\\&=49\\times 4\\\\&=196\\end{align} (iii) \\begin{align}{{2}^{3}}\\times 5&=(2\\times 2\\times 2)\\times 5\\\\&=8\\times 5\\\\&=40\\end{align} (iv) \\begin{align}3\\times {{4}^{4}}&=3\\times (4\\times 4\\times 4\\times 4)\\\\&=3\\times 256\\\\&=768\\end{align} (v) \\begin{align}0\\times {{10}^{2}}&=0\\times 10\\times 10\\\\&=0\\end{align} (vi) \\begin{align}{{5}^{2}}\\times {{3}^{3}}&=(5\\times 5)\\times (3\\times 3\\times 3)\\\\&=25\\times 27\\\\&=675\\end{align} (vii) \\begin{align}{{2}^{4}}\\!\\times \\!{{3}^{2}}&=(\\!2\\times\\!\\! 2\\times \\!2\\times\\!2)\\!\\times \\!(3\\!\\times\\!3)\\\\&=16\\!\\times\\! 9\\\\&=144\\end{align} (viii) \\begin{align}{3^2}\\!\\! \\times\\!\\! {10^4}&=\\!\\! (3\\!\\!\\times\\!\\!3)\\!\\!\\times\\!\\!(10\\!\\!\\times\\!\\!10\\!\\!\\times\\!\\!10\\!\\!\\times\\!\\!10) \\\\&= 9\\!\\times\\!10000 \\\\&= 90000\\end{align} ## Chapter 13 Ex.13.1 Question 7 Simplify: (i) $${{(-4)}^{3}}$$ (ii) $$(-3)\\times {{(-2)}^{3}}$$ (iii) $${{(-3)}^{2}}\\times {{(-5)}^{2}}$$ (iv) $${{(-2)}^{3}}\\times {{(-10)}^{3}}$$ ### Solution What is unknown? Base and Exponent Reasoning: In this question we will multiply the base and the number of times", "Multiplication using Exponents (Indices) If we wish to calculate $5^4 \\times 5^3$, we could write in factor form to get: \\begin{align} \\displaystyle 5^4 \\times 5^3 &= (5 \\times 5 \\times 5 \\times 5) \\times (5 \\times 5 \\times 5) \\\\ &= 5^7 \\end{align} Example 1 Simplify $7^2 \\times 7^3$ after first writing in factor form. \\begin{align} \\displaystyle 7^2 \\times 7^3 &= (7 \\times 7) \\times (7 \\times 7 \\times 7) \\\\ &= 7^5 \\end{align} However, if we look closely, a much simpler method would be to add the exponents since the bases are the same. Therefore this calculation can also be done this way: \\begin{align} \\displaystyle 5^4 \\times 5^3 &= 5^{4+3} \\\\ &= 5^7 \\end{align} , which is the same answer. We can add the exponents when multiplying only if the bases are the same. Thus to $\\textit{multiply}$ numbers with the $\\textit{same base}$, keep the base and $\\textit{add}$ the exponents. $$\\large a^x \\times a^y = a^{x+y}$$ Example 2 Simplify $9^3 \\times 9^5$. \\begin{align} \\displaystyle 9^3 \\times 9^5 &= 9^{3+5} \\\\ &= 9^8 \\end{align} Example 3 Simplify $4^3 \\times 4 \\times 4^5$. \\begin{align} \\displaystyle 4^3 \\times 4 \\times 4^5 &= 4^3 \\times 4^1 \\times 4^5 \\\\ &= 4^{3+1+5} \\\\ &= 4^9 \\end{align} Many questions will be algebraic, meaning that a pronumeral is used. In such questions, we multiply the", "SEARCH HOME Math Central Quandaries & Queries Question from Tom: 5 + 5 + 5 - 5 + 5 + 5 - 5 + 5 x 0 = Hi Tom, I really dislike this question. No mathematician or user of mathematics would write such an expression. Mathematical expressions are written to clearly state a fact or relationship so that someone else reading the expression will understand it. This expression is meant to confuse you rather than make it clear what is expected. The convention is that multiplication is performed before addition so you are expected to perform $5 \\times 0 = 0$ first so that the expression becomes $5 + 5 + 5 - 5 + 5 + 5 - 5 + 0$ and then the result is 15. The confusion is that you might start from left to right and perform $5 + 5 + 5 - 5 + 5 + 5 - 5 + 5$ and then multiply by $0$ to get a result of $0.$ If I wanted you to perform this evaluation I would write $5 + 5 + 5 - 5 + 5 + 5 - 5 + (5 \\times 0) =.$ Your eye is then immediately drawn to $(5 \\times 0)$ and you know to perform this operation first. Harley Math Central", "## Algebra 2 Common Core $3(x+5)(x-5)$ Factoring the $GCF=3,$ the given expression, $3x^2-75 ,$ is equivalent to \\begin{align*}\\require{cancel} & 3(x^2-25) .\\end{align*} Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \\begin{align*} & 3[(x)^2-(5)^2] \\\\&= 3[(x+5)(x-5)] \\\\&= 3(x+5)(x-5) .\\end{align*} The factored form of the given expression is $3(x+5)(x-5) .$", "four times in the multiplication = $3\\times3\\times3\\times3$3×3×3×3 Examples Now working out the value of expressions that have exponents is done by remembering what the notation actually means. Question 1 We want to evaluate $4^3$43 using repeated multiplication. 1. First let's convert $4^3$43 to expanded form. 2. Using your answer to part (a), state the value of $4^3$43. Here's another one Question 2 Evaluate $2^3\\times2^4$23×24 Question 3 Write this expression using index notation, in simplest form. $5\\times5$5×5", "do I solve this? $125^{2}{3}$ the 2/3 is an exponent. \\displaystyle \\begin{align*} 125^{\\frac{2}{3}} &= \\left(125^{\\frac{1}{3}}\\right)^2 \\\\ &= \\left(\\sqrt[3]{125}\\right)^2 \\\\ &= 5^2 \\\\ &= 25 \\end{align*}", "case of $$B = 5,$$ the product, $$B \\times 5 = 5 \\times 5 = 25$$ $$2$$ will be a carry for the next step. We have, $$5 \\times {{A} + 2 = {CA}},$$ which is possible for $$A = 2$$ or $$7$$ The multiplication is as follows. $\\frac{\\begin{gathered}2\\,\\,\\,\\,5 \\hfill \\\\\\times \\,\\,\\,\\,5 \\hfill \\\\\\end{gathered} }{{1\\,\\,2\\,\\,5}}\\,\\,\\,\\,\\,\\,\\,\\,\\frac{\\begin{gathered}7\\,\\,\\,\\,5 \\hfill \\\\\\times \\,\\,\\,\\,5 \\hfill \\\\\\end{gathered} }{{3\\,\\,7\\,\\,5}}$ If $$B = 0,$$ $$B \\times 5 = B$$ $$0 × 5 = 0$$ There will not be any carry in this step. In the next step, $$5 \\times A = CA$$ It can happen only when $$A = 5$$ or $$A = 0$$ However, $$A$$ cannot be $$0$$ as $$AB$$ is a two-digit number. Hence, $$A$$ can be $$5$$ only. The multiplication is as follows. Hence, there are $$3$$ possible values of (i) $$5,\\, 0,$$ and $$2$$ respectively (ii) $$2, \\,5,$$ and $$1$$ respectively (iii) $$7,\\, 5,$$ and $$3$$ respectively ## Chapter 16 Ex.16.1 Question 7 Find the values of the letters in the following and give reasons for the steps involved. \\begin{align}{A \\;\\;{B}} \\\\ { \\times \\;\\;\\,{6}} \\\\ \\hline B\\,B\\, {B} \\\\ \\hline\\end{align} ### Solution What is known? Multiplication operation of two numbers What is unknown? Value of alphabets i.e. $$A$$ and $$B.$$ Reasoning: Each letter in the puzzle must stand for just one digit.", "a hard time figuring out what they are supposed to do. Depending on students' tolerance for reading and deciphering problems of this nature, it might be advisable for the teacher to go through an example with different numbers first so students understand what is expected. A slightly different, more visual approach to these ideas can be found here: https://www.illustrativemathematics.org/tasks/1986. Both tasks focus on products of negative numbers and the distributive property: this task focuses on the arithmetic of the distributive property while the other emphasizes geometry. ## Solution 1. The value of $3\\times (-5)$ should be $(-5)+(-5)+(-5) = -15$. 2. Applying the distributive property gives $$3\\times(5+(-5))=3\\times5+3\\times(-5).$$ Since $5+(-5)=0$ and $3\\times5=15$, \\begin{align} 3\\times0=&3\\times (5+(-5))\\\\ =&3\\times5+3\\times(-5)\\\\ =&15+3\\times(-5) \\end{align} Since $3\\times 0 = 0$, these equations show that $0 = 15+ 3\\times(-5)$. From here, we see that $3\\times(-5)=-15$. 3. Extending the commutative property of multiplication to negative numbers, the product $(-5)\\times3$ should have the same value as $3 \\times (-5)$ so it should be -15. Furthermore, the distributive property gives $(-5)\\times(3+(-3)) = (-5)\\times3 + (-5)\\times(-3)$. 4. Using the equality from part (c), $$(-5)\\times(3+(-3)) = (-5)\\times3 + (-5)\\times(-3),$$ we know the left side is 0 and we know $(-5)\\times3 = -15$. So $$0 = -15 + (-5)\\times(-3)$$ and $(-5)\\times(-3) = 15$.", "see how a $$5$$-digit number can be decomposed into its place values. Let us continue with the same number $$20340.$$ \\begin{align} & { 2 \\text { in Ten Thousand's place value } = 2 \\times 10000 = 20000 } \\\\& { 0 \\text { in Thousand's place value = } 0 \\times 1000 = 0 } \\\\& { 3 \\text { in Hundred's place value } = 3 \\times 100 = 300 } \\\\& { 4 \\text { in Ten's place value } = 4 \\times 10 = 40 } \\\\& { 0 \\text { in Unit's place value } = 0 \\times 1 = 0 } \\end{align} So the five digit number can be expressed in the following way: \\begin{align} { 20000 } \\\\ { + \\quad\\, 0 } \\\\ { + \\,300 } \\\\ { + \\;\\;\\,40 } \\\\ { + \\quad \\; 0 } \\\\ \\hline 20340 \\\\ \\hline \\;\\; \\end{align} How would the $$5$$ place values change if we interchanged some of the digits? Let us say we have the $$5$$ digit number $$32004.$$ This the change in place value of the same $$5$$ digits: • $$2$$ would now have place value of \\begin{align} 2 \\times 1000 = 2000 \\end{align} (earlier place value of $$2$$ was $$20000$$) • $$3$$ would now have place value", "## needhelpfastplease Group Title What is the value of the expression –3x^3 when x = –3? A. –81 B. –27 C. 27 D. 81 6 months ago 6 months ago @kym02 2. kym02 Did you work it out? i tried to @kym02 i think its C but im not quite sure 4. kym02 Alright i'll go it through with you. $-3x^{3}$ Hence, $x=-3$ so, $(-3) \\times (-3) \\times (-3) = -27$ so basically it now becomes: $-3(-27)$ which is equal to: $81$ Therefore, the answer is D :) you had it! just needed to keep trying! :D", "and subtraction. Eventually you ran into expressions that looked like this:$3+3+3+3+3+3+3$You didn't want to write so many 3's, so you were introduced to a shorthand to write this expression. Since there were seven 3's being added together, you learned you could instead write:$3\\times7$Thus you learned that multiplication is repeated addition. Fast forward a few years, when you had multiplication and division under your belt. Now you saw expressions like this instead:$4\\times4\\times4\\times4\\times4$Again, you were introduced to a shorthand to keep from having to write all those 4's. Since there were five 4's being multiplied together, you wrote:$4^5$Thus you learned that exponentiation is repeated multiplication. Now we come to an expression like this.$5^2+4\\times3$What do we do first? Well, remembering that exponentiation is repeated multiplication, we rewrite our exponent to say what it really means.$5\\times5+4\\times3$Next, remembering that multiplication is repeated addition, we rewrite our multiplication in even more basic terms.$5+5+5+5+5+4+4+4$Now the expression is a cinch to evaluate - anyone can add! The value just comes out to 37. But, more remarkably, what we've just done is uncovered the reason why the order of operations is as it is: The most compact shorthand is evaluated first. Once students understand this, they won't need to actually write out the additions explicitly - they'll just evaluate things in the order they should be handled.", "## Thinking Mathematically (6th Edition) $-100 - (-5)^3$; The simplified form of the expression is equal to $25$. The cube of $-5$ is $(-5)^3$. Subtracting this to $-100$ gives: $-100 - (-5)^3$ The order of operations states that to simplify the expression, the exponent must be applied first before the subtraction is performed. Thus, $-100-(-5)^3 = -100-(-125) =-100+125 = 25$" ]

[ "will definitely look into my local utility company. As far as pizza delivery, my issue would definitely be the wear and tear on the car. As far as waiting tables, I have tried that in the past and I absolutely couldn't stand it. Thanks for the suggestions regardless, I appreciate the help. 5. Oct 19, 2015 Staff Emeritus You will probably discover that the jobs that pay the best do so for a reason. 6. Oct 19, 2015 ### StatGuy2000 You were able to average $20-$25/hour delivering pizza, and your wife waited tables for a similar income? Even taking tips into account, there is no way that I can imagine anyone living in Canada making that kind of money delivering pizza or waiting tables. Perhaps other Canadian PF posters can chime in on this. 7. Oct 19, 2015 ### ModusPwnd Is the original poster in Canada? (You can't even imagine it? I think your imagination is better than that ;) $8/hr wage, take 2 deliveries per run, 2-3 runs per hour. Just over 2.50 tip per average plus about a dollar per delivery for delivery fee. There are catches of course... You wont get 40 hours, more like 20-25 hours, you work from roughly 4pm to 8pm. You have car maintenance and gas also, but you might have", "benefits from the Government), the price of cheese (one of the largest cost per pizza) would skyrocket. there he can pay his full cost When my local Papa Johns drivers are out on the road, they are paid about $5.50 an hour. Only when they are clocked in back at the store do they even make minimum wage. The local owners are idiots and cheapskates and deserve to go out of business ASAP. A driver pays for gas, oil, car repairs, insurance and can possibly be robbed/killed/in major MVAs all for a shiatty pizza made by teenagers who don't know what sanitation is. So fark em. I can make my own awesome pizza, thank you. Well since I have Little Caeser's if I want a decent pizza immediately for$5, and Pepe's Pizza in New Haven when I'm feeling patient and want the best damn pizza in America, I don't really give a rat's ass. Cost of gas delivering pizza: 70¢* Cost of health care: 15¢ *5 mile round trip, with a car that gets 25mpg city. The cost of health care could be negated by hybrids. Bill Murray said I was weird: Well since I have Little Caeser's if I want a decent pizza immediately for $5 What color is the sky in your world? Lsherm: propasaurus: \"Our", "must each make 5 deliveries in order to have the same amount of money. Hope this helps!", "I'm guessing publicly against; privately doesn't care. Wait is Kentucky Victory Fund a branch of Victory Fund? As someone who had Papa John's for supper, I'm getting a kick... \\three people deliver here - Papa John's, the garlic-overloaded Dominos, or the very, very greasy local place \\\\and whatever the foodies may say, Papa John's isn't that bad \\\\\\it's not the place that closed in my old neighborhood, but no one ever will be Elmo Jones: So? That crappy \"pizza\" could cost $20 more, and I wouldn't give a hoot in hell. \"Papa John\" sets off my creep-o-meter, which is rarely wrong. You don't want a mouthful of Papa's warm cheese? I have two pizza stones and an oven that goes to 550. It's not exactly like a clay oven, but it works.$.11 to $.14? So you're Obamacare doubles the cost to make a pizza? Weaver95: I really don't see why papa john's is biatching about spending .11 cents or so per pizza to make sure his employees have health care. That's how \"Job Creators ®\" think. \"Death Panels\" was a lot scarier than marginally more expensive bad chain pizza, you guys. As a consumer of Papa John's I will gladly pay the increased price in order to ensure your employees have adequate health care. In fact, just to", "# John made 50 deliveries during the month of May and only 42 during the month of June.What is the decrease in deliveries from May to June? $50 - 42 = 8$", "a whole heap of different shit, this is what I'm listening to right now though. File: d56198693a8cf03⋯.jpg (49.75 KB, 468x463, 468:463, pepe money.jpg) What are good sustainable sources of income for robots? inb4 >just get autismbux and be banned from owning guns forever >just live with your parents and hope they never die >just inherit a fortune >just buy bitcoins 8 years ago 20 posts and 2 image replies omitted. Click reply to view. No.208152 >>208050 Isn't the 70k for people who have decades of trucking experience, while newbies get 40k or even less? How do you make 4k from pizza anyway? I thought it was a min wage job where you have to pay for gas out of pocket and put mileage on your own car. No.208154 >>208152 The secret to making good money doing food delivery is by getting overtime. My hourly at domino's was $5.50 on the road and$8.25 in store but once you hit overtime you get paid time and a half for both in store and on the road. So if I worked 75 hours in a week that means I got 35 hours of steady $12.38/hr pay and after tips & mileage compensation I end up averaging$25/hr. And yes you do have to pay for gas and car repairs out of pocket", "delivering newspapers is = $103 Shawn earns monthly by delivering pizzas is =$175", "birthday. He spent$ \\$31.64$ on a video game. How much of Paul’s birthday money was left? Solution Jessie put 88 gallons of gas in her car. One gallon of gas costs $\\$ 3.529$. How much does Jessie owe for the gas? (Round the answer to the nearest cent.) Solution #### Example 16 Four friends went out for dinner. They shared a large pizza and a pitcher of soda. The total cost of their dinner was$ \\$31.76$. If they divide the cost equally, how much should each friend pay? Solution Be careful to follow the order of operations in the next example. Remember to multiply before you add. #### Example 17 Marla buys $6$ bananas that cost $\\$ 0.22$each and$4$oranges that cost$ \\$0.49$ each. How much is the total cost of the fruit? Solution", "# Colton earns $7 per hour plus$1.50 for each pizza delivery. the expression 7h + 1.50d can be used to Colton earns $7 per hour plus$1.50 for each pizza delivery. the expression 7h + 1.50d can be used to find the total earings after h hours and d deliveries have been made. how much money will colton earn after working 15 hours and making 8 deliveries ## This Post Has 3 Comments 1. angeleyes42 says: Since the expression 7h + 1.50d can be used to find the total earnings. We just need to substitute the given data to the expression to solve the problem. 7(15) + 1.50(8) 105 + 12 = 117 Colton's total earnings after working 15 hours and making 8 deliveries is 117 dollars. 2. ellamai16 says: 7h + 1.50 For 15 hours and 8 delivers, lets find out how much he earns. 7(15) + 1.50(8) = ? 7*15 = 105 1.50*8 = 12 105 + 12 = 117 He will make $117 3. snikergrace says: 7h+1.5d 7(15)+1.5(8) 105+12$117 Hope this helps.", "The coaches organized them into groups of 4 for warm-up exercises. Number of groups were there = 89 ÷ 4 = 22.25 or 22 Question 5. Jamal and three of his friends raked leaves every weekend for a month. By the end of the month, they earned$91.00. They paid $4.00 to Jamal’s little brother for helping out a little on the last day. Then the four friends divided the rest of the money equally. How much did each friend get? Answer:$21.75 Explanation: Given that, Jamal and three of his friends by the end of the month earned $91.00. They paid$4.00 to Jamal’s little brother for helping out a little on the last day. $91.00 –$4.00 = $87 Then the four friends divided the rest of the money equally. Total amount did each friend get 87 ÷ 4 =$21.75 Question 6. The drivers at Pizza Palace were loading up their 3 delivery vans for the evening. There were 65 pizzas. The boss told them to put an equal number of pizzas in each van and said they could split any leftovers equally. How much of a leftover pizza did each of the 3 drivers get? 21 with 2 left over. Explanation: Given that, The drivers at Pizza Palace were loading up their 3 delivery vans for the evening. There", "loose a lot of weight by not purchasing from these idiots. I knew this thread would be full of foodies who hate pizza from chains for no other reason than it's from a chain. And that they would put pizza in quotes and call it cardboard and parrot other worn-out phrases. I only had to read the first ten posts or so to confirm my suspicions. It's probably already been said, but if you have Papa John's for every meal, every day, it will cost an extra $120.45 (at 11 cents). If that pays for one of their employees to be covered by insurance, then I'm okay with that. I don't care if it's an additional$10 per pizza and $15 per six pack of Miller Lite. It's all worth it if I understand the Affordable Care Act correctly. My best estimate is that Papa Johns inflicted singring and dysentery costs taxpayers$.20 per pie in unpaid healthcare costs. In any event, what a whiny cockpotato. Oldest | « | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | » | Newest | Show all View Voting Results: Smartest and Funniest In Other Media 1. Links are submitted by members of the Fark", "paid minimum wage. Of course, maybe the delivery drivers at your choice pizza place do. It's just not the norm though. \"Some of the most wonderful people are the ones who don't fit into boxes.\" -Tori Amos #### NyaChan • Member • Posts: 4107 ##### Re: Tipping Pizza Deliveries « Reply #17 on: April 20, 2013, 09:19:11 AM » Interesting article on delivery charges by a former pizza delivery guy: http://consumerist.com/2010/03/30/4-myths-about-tipping-from-a-former-pizza-delivery-guy/ #### WillyNilly • Member • Posts: 7490 • Mmmmm, food ##### Re: Tipping Pizza Deliveries « Reply #18 on: April 20, 2013, 09:32:55 AM » #### NyaChan • Member • Posts: 4107 ##### Re: Tipping Pizza Deliveries « Reply #23 on: April 20, 2013, 04:25:25 PM » Got pizza today and the driver was friendly/chatty. So I asked him a few questions (after giving him a $2.90 tip (20%) ). They are paid hourly, minimum wage, plus$1 per trip for gas. Delivery charge added to bill goes to the store, not the driver. 2/3 people tip - usually in range of $2 -$3. On large (party) orders tip is usually about $5 -$6. 1/3 people don't tip and give exact dollar amount & change. Drivers don't like to deal with coin change at all. About 50% pay cash, 50% use crecit card. If driver is robbed, store", "TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52938 At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 15 Jun 2017, 22:41 1 1 00:00 Difficulty: 65% (hard) Question Stats: 67% (02:30) correct 33% (02:52) wrong based on 109 sessions ### HideShow timer Statistics At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 _________________ Director Joined: 04 Dec 2015 Posts: 740 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) At a party, John collected$96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 15 Jun 2017, 23:01 1 Bunuel wrote: At a party, John collected $96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for$2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same $96. How much was the cost of each pizza after the discount? A.$6 B. $8", "should pay$200,000 for a car if a car delivers $200,000 in pizzas. If the second standard is used, Domino’s should only pay$15,000, which is the correct price for a car. The IT Dashboard itself is most likely overbudget. It’s a few data entry screens and a very fancy flash front end which can be purchased for about a thousand dollars from fusion charts. The code base for this is 176MB–and most likely–needlessly complex. People attempting to reuse the project will be confused when they read the code and could probably rewrite it themselves, cheaper. The goal in “open sourcing” this very complex code is to generate consulting work–to once again, charge millions for a few data entry screens and some graphs. The idea of auditing government spending on IT is good. However, projects should not be measured against their net value to the government–but rather against their appropriate cost. Using “what does the program deliver” as its value allows a simple accouting application to cost ten million dollars. Using “how much should each form cost” as the measurement only allows it to cost couple hundred thousand. If Dominos paid what the cars were worth in delivering pizzas–not what they cost–the car dealer would have all of the profit and Dominos would be out of business. If the government", "1st and 4 tickets for 2nd place and pay 82Kč. How much was the ticket • Potato Chips Haydn has a 3-pound bag of Lay's Potato Chips. His four friends want him to split the bag evenly between all of them (and Haydn). How many pounds of Lay's potato chips will each person get? • Numbers division With what number should be divided mixed number 2 3/4 to get 11/12? • Walking woman Reeta walks 5/7 km in 1 hour. How far does she go in 3 1/2 hours? • Cents no more Janko bought pencils for 35 cents each. Neither he nor the salesperson had small coins, just a whole € 1 coin. At least how many pencils had to buy to pay for the whole euros? • New movie Nathan and his friends want to see the new movie, The Case of the Secret Doorway. It's a popular movie, so Nathan buys their tickets in advance. He spends$36.75 in all to buy 3 tickets. How much does each friend need to pay him back? • Gasoline fuel You just drove your car 300 miles and used 40 gallons of gas. You know that the gas tank on your car holds 12 1/2 gallons of gas; if the gas you buy is \\$2.50 per gallon. What would", "increased by unlocks and upgrades. Angel Investors are earned according to this formula: 150 * sqrt(LIFETIME_EARNINGS / 1e15) ## Money Name | Base Price | Base Profit / second | Price Increase per Purchase ------------------------------------------------------------------------------------------- LEMONADE_STAND | $4 |$1.66 | 7% NEWSPAPER_DELIVERY | $60 |$20 | 15% CAR_WASH | $720 |$90 | 14% PIZZA_DELIVERY | $8,640 |$360 | 13% DONUT_SHOP | $103,680 |$2,160 | 12% SHRIMP_BOAT | $1,244,160 |$6,480 | 11% HOCKEY_TEAM | $14,929,920 |$19,440 | 10% MOVIE_STUDIO | $179,159,040 |$58,320 | 9% BANK | $2,149,908,480 |$174,960 | 8% OIL_COMPANY | $25,798,901,760 |$804,816 | 7% ## Unlocks Unlocks are bonuses that are earned when a set goal has been achieved. Example list (not actual): LEMONADE_STAND;25;LEMONADE_STAND;PROFIT;2 NEWSPAPER_STAND;900;OIL_COMPANY;PROFIT;11 ALL;2222;ALL;PROFIT;2 Upgrades are bonuses that are purchased with money or angels. See Angel Upgrades for details on upgrades that are purchased with angels. Example list (not actual): 0;CASH;2.5e5;LEMONADE_STAND;PROFIT;3 1;CASH;500000;NEWSPAPER_DELIVERY;PROFIT;3 2;CASH;1e21;ANGEL;EFFECTIVENESS;1 Angel Upgrades are bonuses that are purchased with the sacrifice of angels. Angels sacrificed for this purpose are not regained on reset of a game. Example list (not actual): 3;ANGEL;10000;ALL;PROFIT;3 4;ANGEL;1e32;CAR_WASH;COUNT;100 5;ANGEL;3.4e134;ANGEL;EFFECTIVENESS;10 ## Resetting When you reset your game, you lose all unlocks, upgrades, businesses, and money that you had. You start out with 1 lemonade stand all over again. So why would you want to do that? Because all angels that", "us the price during promotion (p + 4)(c - 2) = 96 Since price of pizza (after discount) is 8$, John would have bought 12 pizza's during the promotional period. However, price before discount would be 10$ and he would have bought 8 pizza's which doesn't give us 96$as total cost. Hence, Option C is also not right If price after discount is 6$, he would have bought 16 pizza's during the promotional period. However, price before discount would be 8$and he would have bought 12 pizza's which gives us 96$ as total revenue. Hence, Option A is our answer _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7868 Kudos [?]: 18494 [0], given: 237 Location: Pune, India Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 02:23 Expert's post 1 This post was BOOKMARKED Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9", "the five cars in Honest John's car sales yard are: $10 000,$5900, $9999,$12 500, \\$3499 Calculate the mean price ? Find the median of the scores: 1, 3, 1, 19, 8 ? And can you tell me how they are done ? 63.7 I'm a machinist; too dumb to do the other stuff.", "## Support Union Test Prep Johnny spends $$\\frac{1}{7}$$ of his income on food, $$\\frac{1}{10}$$ of his income on gas, and $$\\frac{3}{8}$$ of his income on rent. What percentage of his income remains after he has paid for food, gas, and rent?", "kidzsearch.com > wiki # Profit If John spends $15 on some ice cream cones, and then sells them for$20, he has made a profit of \\$5. This is because 20 minus 15 is 5." ]

[ "$$P(2)$$ would be, and so on. Indeed, if we are able to prove $$P(N)$$ for any $$N \\in \\mathbb{N}$$, then we know $$P(M)$$ for any natural number $$M>N$$. But the sequence of statements $$\\langle P(0), P(1), P(2), \\ldots\\rangle$$ never gets started. $$P(N)$$ fails for all $$N$$. Another way to think of induction is in terms of guarantees. Suppose you decide to buy a car. First you go to Honest Bob’s. Bob guarantees that any car he sells will go at least one mile. You buy a car, drive it off the lot, and after 3 miles it breaks down and can’t be fixed. You walk back angrily, but Bob won’t give you your money back because the car lived up to the guarantee. Then you cross the road to Honest John’s. John guarantees that if he sells you a car, once it starts it will never stop. This sounds pretty good, so you buy a car, put the keys in the ignition, and ... nothing. The car won’t start. John won’t give you your money back either, because the car did not fail to do what he claimed. Feeling desperate, you end up at Honest Stewart’s. Stewart’s cars come with two guarantees: (1) The car will start and go at least one mile. (2) No matter how far", "she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth? $\\textbf {(A) } 33 \\qquad \\textbf {(B) } 35 \\qquad \\textbf {(C) } 37 \\qquad \\textbf {(D) } 39 \\qquad \\textbf {(E) } 41$ ## Problem 2 What is $\\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$? $\\textbf {(A) } 16 \\qquad \\textbf {(B) } 24 \\qquad \\textbf {(C) } 32 \\qquad \\textbf {(D) } 48 \\qquad \\textbf {(E) } 64$ ## Problem 3 Peter drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Peter's trip? $\\textbf {(A) } 30 \\qquad \\textbf {(B) } \\frac{400}{11} \\qquad \\textbf {(C) } \\frac{75}{2} \\qquad \\textbf {(D) } 40 \\qquad \\textbf {(E) } \\frac{300}{7}$ ## Problem 4 Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana? $\\textbf {(A) } \\frac{3}{2} \\qquad \\textbf {(B) } \\frac{5}{3} \\qquad \\textbf {(C) } \\frac{7}{4} \\qquad \\textbf {(D) } 2 \\qquad \\textbf {(E) } \\frac{13}{4}$ ## Problem 5 Camden constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the", "us the price during promotion (p + 4)(c - 2) = 96 Since price of pizza (after discount) is 8$, John would have bought 12 pizza's during the promotional period. However, price before discount would be 10$ and he would have bought 8 pizza's which doesn't give us 96$as total cost. Hence, Option C is also not right If price after discount is 6$, he would have bought 16 pizza's during the promotional period. However, price before discount would be 8$and he would have bought 12 pizza's which gives us 96$ as total revenue. Hence, Option A is our answer _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7868 Kudos [?]: 18494 [0], given: 237 Location: Pune, India Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 02:23 Expert's post 1 This post was BOOKMARKED Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9", "John bought 3 pizzas. He had a coupon for $1.50 off of each pizza, and the total amount he had to pay was$46.50. What was the original 3 answers Question: John bought 3 pizzas. He had a coupon for $1.50 off of each pizza, and the total amount he had to pay was$46.50. What was the original price for 1 pizza Answers 4 answers Write the equation of a line given the point(-6, -8 ) and the slope of 1Cv = xAv=x-2B v = 2x + Write the equation of a line given the point (-6, -8 ) and the slope of 1 Cv = x Av=x-2 B v = 2x + 1 D v = x + 2... 10 answers Which relationship is always true for the angles x, y,and z of triangle abc Which relationship is always true for the angles x, y,and z of triangle abc... 4 answers Which of the following sentences uses the word 'relentless' correctly? his attitude was so relentless he always made us Which of the following sentences uses the word \"relentless\" correctly? his attitude was so relentless he always made us laugh. the marching band's relentless formation looked great on the field. the teacher's relentless homework demands made the students miserable her relentless outfit was admired ... 8 answers You", "actually might of meant a price index, not just some random number. Thanks so much for your help! Aug 5, 2016 at 4:09 I think it should be noted that any real price, in this case wage, is actually the price of something in units of a good. In this case, P the price level is the price of a typical consumed basket of goods. The real wage W/P is how much you are paid in \"baskets\". Economists care about real prices because nominal prices are in units of money, which is just paper, but real prices tell you how many goods you can get. The idea is you don't care about the money, what you care about is the goods you can buy with the money. For example, imagine the price of a pizza is \\$10/pizza. And you earn \\$20/hour. Then your real wage W/P = (\\$20/hour)/(\\$10/pizza) = 2 pizzas / hour. You get paid 2 pizzas an hour.", "(B) $80 (C)$90 (D) $120 (E)$180 Practice Questions Question: 15 Page: 154 Difficulty: 600 [Reveal] Spoiler: OA _________________ Math Expert Joined: 02 Sep 2009 Posts: 39590 Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 06 Aug 2012, 02:53 5 KUDOS Expert's post 5 This post was BOOKMARKED SOLUTION At a supermarket, John spent 1/2 of his money on fresh fruits and vegetables, 1/3 on meat products, and 1/10 on bakery products. If he spent the remaining $6 on candy, how much did John spend at the supermarket? (A)$60 (B) $80 (C)$90 (D) $120 (E)$180 John spent $$1-(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{10})=\\frac{1}{15}$$ of his money on candy. Since we are told that he spent $6 on candy, then he spent total of $$6*15=90$$ at the supermarket. Answer: C. _________________ Senior Manager Joined: 15 Jun 2010 Posts: 361 Schools: IE'14, ISB'14, Kellogg'15 WE 1: 7 Yrs in Automobile (Commercial Vehicle industry) Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 06 Aug 2012, 06:06 6 This post received KUDOS x/2+x/3+x/10 = 28x/30 2x/30=6, hence x =90 Ans C. _________________ Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html Math Expert Joined: 02 Sep 2009 Posts: 39590 Re: At a supermarket, John spent", "bought pizzas for his swim team. Pepperoni pizzas in the Problem Solving forum “This question illustrates the difference between the abstract math and real world math. From the given information, we''re able to create ONE equation: 13p + 17c = 184 In high school, we learned that, if we''re given 1 equation with 2 variables, we cannot find the value of either variable, so ...” October 5, 2018 Brent@GMATPrepNow posted a reply to ||2-3|-|2-5||=? in the Problem Solving forum “||2-3|-|2-5|| = ||-1|-|-3|| = |1-3|| = |-2| = 2 Answer: B Cheers, Brent” October 5, 2018 Brent@GMATPrepNow posted a reply to What is the perimeter of quadrilateral Q? in the Data Sufficiency forum “Target question: What is the perimeter of quadrilateral Q? If we recognize that each statement alone is not sufficient, we can jump straight to...... Statements 1 and 2 combined There are several quadrilaterals that satisfy BOTH statements. Here are two: Case a: rectangle Q has ...” October 5, 2018 Brent@GMATPrepNow posted a reply to A car salesman earns a base salary of$1,000 per month plus in the Problem Solving forum “Let x = number of cars sold in 1 month So, his pay = $1000 + ($200)(n) If he earned $2,200 in February, how many cars does he have to sell in March in order to", "in his savings account and his grandparents add$10 a week. What is the recursive formula to represent how much money John has any time after two weeks? A. f(n) = f(n-1)+10, given f(1) = 10B. f(n) = 10f(n-1), given f(1) = 10C. f(n) = f(n-2)+10, given f(1) = 10D. f(n) = 10(n-1), given f(1) = 10 Factoring Math Algebra 1 Geometry 06/15/20 #### Leanne has a gas card that has $150 on it. She spends$25 a week on gas. What is the recursive rule that represents the amount of money she has on her gas card after the first week. A. f(n) = f(n-2)-25, given f(1) = 150B. f(n) = f(n-1)-150, given f(1) = 25C. f(n) = 6f(n-1)-25, given f(1) = 25D. f(n) = f(n-1)-25, given f(1) = 150 ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "can only afford to spend $6 on a ride. Form an inequality that represents John’s scenario. How far can John travel without going over his budget? Justify your response. Solution: Let M denote the number of miles he traveled. Writing the inequality, 0.35M + 0.75 $$\\leq$$ 6 Where, • 0.35M is 0.35 per mile and here, per mile means multiply. • ‘+’ means addition. • 0.75 is flat rate. • ‘$$\\leq$$’ means no more than. • ‘6’ means money to spend. No more than’ means you can’t have more than one of something, so you must have fewer. Solving the inequality: 0.35M + 0.75 $$\\leq$$ 6 0.35M + 0.75 – 0.75 $$\\leq$$ 6 – 0.75 (subtracting 0.75 on both sides) 0.35M $$\\leq$$ 5.25 (simplified) $$\\frac{0.35M}{0.35}\\leq \\frac{5.25}{0.35}$$ (Divided both sides with 0.35) M $$\\leq$$ 15 (simplified) Representing on number line: John can travel less than or equal to 15 miles before he reaches his limit of$6. Check: 0.35M + 0.75 $$\\leq$$ 6 0.35 $$\\times$$ (15) + 0.75 $$\\leq$$ 6 (substituted 15 in the place of M) 5.25 + 0.75 $$\\leq$$ 6 (Multiplied) 6 $$\\leq$$ 6 (simplified) Example 2: For a birthday party rental, Sue’s Party Planning Co. charges a flat fee of $30 plus$3.50 per person. Joseph has only $90 to spend on his birthday celebration. To represent Joseph’s", "8x + 6x - 48 = 0$$ $$x(x - 8) + 6 (x - 8) = 0$$ $$(x+6)(x+8) = 0$$ $$x = 8, -6$$ Price cannot be negative. Therefore price of each pizza before discount was $$8$$. Price of pizza after discount is $$8-2 = 6.$$ Answer (A). Manager Joined: 24 Dec 2016 Posts: 95 Location: India Concentration: Finance, General Management WE: Information Technology (Computer Software) Re: At a party, John collected$96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 00:14 Bunuel wrote: At a party, John collected $96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for$2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same $96. How much was the cost of each pizza after the discount? A.$6 B. $8 C.$9 D. $10 E.$12 Let the price for each pizza be x. Thus, $$p*x = 96$$ {no. of pizza*price of each pizza = Total sum of money John had} =>$$p = \\frac{96}{x}$$ -- eqn. (1) After the discount, reduced price of each pizza = $$(x-2)$$ and thus, he could buy 4 extra pizza. So, no. of pizzas becomes $$(p+4)$$ thus the equation changes to : (p+4)*(x-2) = 96 =>$$p+4 = \\frac{96}{(x-2)}$$ -- eqn", "side is not necessarily equal to the other side. There are several possible answers that will make an inequality a true statement. Equivalent Inequalities two inequalities that are written differently, but still express the same number relationships. ### Guided Practice Here is one for you to try on your own. Antonio is buying milk for a breakfast event. Each container of milk costs$3. At most, he can spend $12 on milk for the event. a. Write an inequality to represent, $c$ , the number of containers of milk he can buy. b. Could Antonio buy 4 containers of milk for the event? Explain. Answer Consider part a first. Use a number, an operation sign, a variable, or an inequality symbol to represent each part of the problem. Since each container of milk costs$3, you can find the total cost, in dollars, of the milk he buys by multiplying 3 by the number of containers. The key words “at most” indicate that you should use a $\\le$ symbol. $& \\underline{\\text{Each container}\\ldots \\text{costs} \\ \\3}. \\ \\underline{\\text{At most}}, \\ \\text{he can spend} \\ \\underline{\\12}\\ldots\\\\ & \\qquad \\qquad \\qquad \\quad \\downarrow \\qquad \\qquad \\qquad \\downarrow \\qquad \\qquad \\qquad \\quad \\quad \\ \\ \\downarrow\\\\& \\qquad \\qquad \\quad \\quad \\ c \\times 3 \\qquad \\qquad \\quad \\le \\qquad \\qquad \\qquad \\qquad 12$ You", "means that if the MP3 player is sold for anything more than $33.99, it is profitable; if it is sold for less, then the business does not cover its costs and expenses and takes a loss on the sale. #### Example 4.2H – Knowing Your Break-Even Price John is trying to run an eBay business. His strategy has been to shop at local garage sales and find items of interest at a great price. He then resells these items on eBay. On John’s last garage sale shopping spree, he only found one item—a Nintendo Wii that was sold to him for$100. John’s vehicle expenses (for gas, oil, wear/tear, and time) amounted to $40. eBay charges a$2.00 insertion fee, a flat fee of $2.19, and a commission of 3.5% based on the selling price less$25. What is John’s minimum list price for his Nintendo Wii to ensure that he at least covers his expenses? Answer: break-even selling price = ? Conditions: selling price = cost + expenses cost = $100.00, vehicle expenses =$40.00, insertion expenses = $2.00, flat fee expense=$2.19, commission expense = 3.5% on selling price less 25.00 \\begin{align*} \\overset{S_{BE}}{\\text{b-e selling price}}&=\\overset{C\\checkmark}{\\text{cost}}+\\overset{E_{vehicle}\\checkmark}{\\text{vehicle expenses}}+ \\overset{E_{insertion}\\checkmark}{\\text{insertion expenses}}\\\\ &\\ \\ \\ \\ +\\underset{E_{flat}\\checkmark}{\\text{flat fee expense}}+\\underset{?}{\\text{commission expense}}\\\\ \\\\ \\Rightarrow S_{BE}&={C}+{E_{vehicle}}+{E_{insertion}}+{E_{flat}}+0.035(S_{BE}-25)\\\\ \\\\ \\Rightarrow S_{BE}&=100+40+2+2.19+0.035S_{BE}-0.875 \\\\ &\\Rightarrow S_{BE}-0.035S_{BE}=100+40+2+2.19+0.035-0.875\\\\ \\\\ &\\Rightarrow 0.965S_{BE}=143.315\\\\ \\\\ &\\Rightarrow S_{BE}=148.51", "Question The commission earned on the sale of a car can be represented by the equation c(p)=250+0.02p where c represents the commission and p represents the purchase price of a car. Jack sold two cars last weekend, one for $15,075 and the other for$21,640. Find his total commission. If c(p)=745.80, find the purchase of the car. i’ve tried the answers someone else found on here but they don’t work so plss help 1. minhkhoi 1) His total commission is $1,234.3 2) The purchase of the car is$24,790.", "to the next one. # At the start of his 3 years graduation program Mr. John bought a car a Author Message TAGS: ### Hide Tags Manager Joined: 25 Dec 2018 Posts: 91 Location: India GMAT 1: 490 Q47 V13 GPA: 2.86 At the start of his 3 years graduation program Mr. John bought a car a [#permalink] ### Show Tags 07 Jan 2019, 09:47 2 3 00:00 Difficulty: 75% (hard) Question Stats: 52% (02:25) correct 48% (01:40) wrong based on 42 sessions ### HideShow timer Statistics At the start of his 3 years graduation program Mr. John bought a car at the market price of $x and sold it to Mr.Max at the end of 3rd year at the market price of$y.If the market price of car depreciated by 20%, 50/3% and 25% respectively in 3 consecutive years, which of the following is true? A. 3x-2y=0 B. 2x=y C. x+y=3y D. 4x-3y=3x E. None of these Manager Joined: 09 Mar 2018 Posts: 246 Location: India At the start of his 3 years graduation program Mr. John bought a car a [#permalink] ### Show Tags 07 Jan 2019, 10:03 2 akurathi12 wrote: At the start of his 3 years graduation program Mr. John bought a car at the market price of $x and sold it to Mr.Max at", "96 => (96/p - 2 ) (p+4) Solving this we get => p=12 or p=-16 Hence p=12 So the cost after the discount would be 96/12 - 2 => 8-2 => 6 Hence A Great Question _________________ Intern Joined: 29 Jul 2012 Posts: 15 ### Show Tags 02 Oct 2018, 14:46 Hello! Is there any easy way to realize when a problem is going to become a quadratic when attempted algebraically? Thank you! Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 03 Oct 2018, 17:51 Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 We can let the price of each pizza before the discount = n and create the equations: pn = 96 p = 96/n and (p + 4)(n - 2) = 96 pn + 4n - 2p - 8 =", "$180 for gas,$450 for food, and $49 per night to share a motel room. He has$520 in savings and can earn $30 per driveway shoveling snow. How many driveways must he shovel to have enough money to pay for the trip? ### Section 2.5 Exercises #### Practice Makes Perfect Graph Inequalities on the Number Line In the following exercises, graph each inequality on the number line and write in interval notation. 296. $x<−2x<−2$ $x≥−3.5x≥−3.5$ $x≤23x≤23$ 297. $x>3x>3$ $x≤−0.5x≤−0.5$ $x≥13x≥13$ 298. $x≥−4x≥−4$ $x<2.5x<2.5$ $x>−32x>−32$ 299. $x≤5x≤5$ $x≥−1.5x≥−1.5$ $x<−73x<−73$ 300. $−5 $−3≤x<1−3≤x<1$ $0≤x≤1.50≤x≤1.5$ 301. $−2 $−5≤x<−3−5≤x<−3$ $0≤x≤3.50≤x≤3.5$ 302. $−1 $−3 $−1.25≤x≤0−1.25≤x≤0$ 303. $−4 $−5 $−3.75≤x≤0−3.75≤x≤0$ Solve Linear Inequalities In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. 304. $a+34≥710a+34≥710$ $8x>728x>72$ $20>25h20>25h$ 305. $b+78≥16b+78≥16$ $6y<486y<48$ $40<58k40<58k$ 306. $f−1320<−512f−1320<−512$ $9t≥−279t≥−27$ $76j≥4276j≥42$ 307. $g−1112<−518g−1112<−518$ $7s<−287s<−28$ $94g≤3694g≤36$ 308. $−5u≥65−5u≥65$ $a−3≤9a−3≤9$ 309. $−8v≤96−8v≤96$ $b−10≥30b−10≥30$ 310. $−9c<126−9c<126$ $−25 311. $−7d>105−7d>105$ $−18>q−6−18>q−6$ In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. 312. $4 v ≥ 9 v − 40 4 v ≥ 9 v − 40$ 313. $5 u ≤ 8 u − 21 5 u ≤ 8 u − 21$ 314. $13 q < 7 q − 29 13 q < 7", "kidzsearch.com > wiki # Profit If John spends $15 on some ice cream cones, and then sells them for$20, he has made a profit of \\$5. This is because 20 minus 15 is 5.", "p equally-priced pizzas. When [#permalink] ### Show Tags 15 Jun 2017, 23:14 Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 Let the price for each pizza be x. Thus, $$p*x = 96$$ {no. of pizza*price of each pizza = Total sum of money John had} =>$$p = \\frac{96}{x}$$ -- eqn. (1) After the discount, reduced price of each pizza = $$(x-2)$$ and thus, he could buy 4 extra pizza. So, no. of pizzas becomes $$(p+4)$$ thus the equation changes to : (p+4)*(x-2) = 96 =>$$p+4 = \\frac{96}{(x-2)}$$ -- eqn (2) Equating eqn. 1 and 2- $$\\frac{96}{(x-2)} = \\frac{96}{x} + 4$$ =>$$\\frac{96}{(x-2)} - \\frac{96}{x} = 4$$ Taking LCM and solving results in- $$4x(x-2) = 192$$ => $$4x^2 - 8x = 192$$ => $$x^2 -2x - 48 = 0$$ => $$(x-8)(x+6) = 0$$ => $$x = 8 or -6$$. As the price cant be a negative value, $$x= 8$$ As x was the original price for the pizzas, i.e., price before discount, price after discount", "give us 96$as total cost. Hence, Option C is also not right If price after discount is 6$, he would have bought 16 pizza's during the promotional period. However, price before discount would be 8$and he would have bought 12 pizza's which gives us 96$ as total revenue. Hence, Option A is our answer _________________ You've got what it takes, but it will take everything you've got Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9705 Location: Pune, India Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 03:23 2 1 Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 Instead of taking variables and solving the equations, this question can be best done by jumping to the options directly. John collected$96 to buy p pizzas, each of same price. So obviously, 96 will be divisible by p, and by the un-discounted price of the pizza. The discounted price is $2 less", "(A)$60 (B) $80 (C)$90 (D) $120 (E)$180 Practice Questions Question: 15 Page: 154 Difficulty: 600 Let total amount be 60x Money spent on fresh fruits and vegetables = 30x Money spent on meat products = 20x Money spent on bakery products = 6x Total money spent 56x Money left (4x) = 6 So, x = 1.5 So, total money John had initially is 1.5 x 60 => $90 Hence answer will be$90 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 16 Mar 2017 Posts: 1 Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 20 May 2017, 20:07 \"How can one say, that candy is bought at supermarket only\" Answer 90 is correct when we assume the above statement else answer is 84. Posted from my mobile device Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] 20 May 2017, 20:07 Similar topics Replies Last post Similar Topics: 1 John spent $25, which" ]

[ "means that if the MP3 player is sold for anything more than $33.99, it is profitable; if it is sold for less, then the business does not cover its costs and expenses and takes a loss on the sale. #### Example 4.2H – Knowing Your Break-Even Price John is trying to run an eBay business. His strategy has been to shop at local garage sales and find items of interest at a great price. He then resells these items on eBay. On John’s last garage sale shopping spree, he only found one item—a Nintendo Wii that was sold to him for$100. John’s vehicle expenses (for gas, oil, wear/tear, and time) amounted to $40. eBay charges a$2.00 insertion fee, a flat fee of $2.19, and a commission of 3.5% based on the selling price less$25. What is John’s minimum list price for his Nintendo Wii to ensure that he at least covers his expenses? Answer: break-even selling price = ? Conditions: selling price = cost + expenses cost = $100.00, vehicle expenses =$40.00, insertion expenses = $2.00, flat fee expense=$2.19, commission expense = 3.5% on selling price less 25.00 \\begin{align*} \\overset{S_{BE}}{\\text{b-e selling price}}&=\\overset{C\\checkmark}{\\text{cost}}+\\overset{E_{vehicle}\\checkmark}{\\text{vehicle expenses}}+ \\overset{E_{insertion}\\checkmark}{\\text{insertion expenses}}\\\\ &\\ \\ \\ \\ +\\underset{E_{flat}\\checkmark}{\\text{flat fee expense}}+\\underset{?}{\\text{commission expense}}\\\\ \\\\ \\Rightarrow S_{BE}&={C}+{E_{vehicle}}+{E_{insertion}}+{E_{flat}}+0.035(S_{BE}-25)\\\\ \\\\ \\Rightarrow S_{BE}&=100+40+2+2.19+0.035S_{BE}-0.875 \\\\ &\\Rightarrow S_{BE}-0.035S_{BE}=100+40+2+2.19+0.035-0.875\\\\ \\\\ &\\Rightarrow 0.965S_{BE}=143.315\\\\ \\\\ &\\Rightarrow S_{BE}=148.51", "will definitely look into my local utility company. As far as pizza delivery, my issue would definitely be the wear and tear on the car. As far as waiting tables, I have tried that in the past and I absolutely couldn't stand it. Thanks for the suggestions regardless, I appreciate the help. 5. Oct 19, 2015 Staff Emeritus You will probably discover that the jobs that pay the best do so for a reason. 6. Oct 19, 2015 ### StatGuy2000 You were able to average $20-$25/hour delivering pizza, and your wife waited tables for a similar income? Even taking tips into account, there is no way that I can imagine anyone living in Canada making that kind of money delivering pizza or waiting tables. Perhaps other Canadian PF posters can chime in on this. 7. Oct 19, 2015 ### ModusPwnd Is the original poster in Canada? (You can't even imagine it? I think your imagination is better than that ;) $8/hr wage, take 2 deliveries per run, 2-3 runs per hour. Just over 2.50 tip per average plus about a dollar per delivery for delivery fee. There are catches of course... You wont get 40 hours, more like 20-25 hours, you work from roughly 4pm to 8pm. You have car maintenance and gas also, but you might have", "a whole heap of different shit, this is what I'm listening to right now though. File: d56198693a8cf03⋯.jpg (49.75 KB, 468x463, 468:463, pepe money.jpg) What are good sustainable sources of income for robots? inb4 >just get autismbux and be banned from owning guns forever >just live with your parents and hope they never die >just inherit a fortune >just buy bitcoins 8 years ago 20 posts and 2 image replies omitted. Click reply to view. No.208152 >>208050 Isn't the 70k for people who have decades of trucking experience, while newbies get 40k or even less? How do you make 4k from pizza anyway? I thought it was a min wage job where you have to pay for gas out of pocket and put mileage on your own car. No.208154 >>208152 The secret to making good money doing food delivery is by getting overtime. My hourly at domino's was $5.50 on the road and$8.25 in store but once you hit overtime you get paid time and a half for both in store and on the road. So if I worked 75 hours in a week that means I got 35 hours of steady $12.38/hr pay and after tips & mileage compensation I end up averaging$25/hr. And yes you do have to pay for gas and car repairs out of pocket", "# Colton earns $7 per hour plus$1.50 for each pizza delivery. the expression 7h + 1.50d can be used to Colton earns $7 per hour plus$1.50 for each pizza delivery. the expression 7h + 1.50d can be used to find the total earings after h hours and d deliveries have been made. how much money will colton earn after working 15 hours and making 8 deliveries ## This Post Has 3 Comments 1. angeleyes42 says: Since the expression 7h + 1.50d can be used to find the total earnings. We just need to substitute the given data to the expression to solve the problem. 7(15) + 1.50(8) 105 + 12 = 117 Colton's total earnings after working 15 hours and making 8 deliveries is 117 dollars. 2. ellamai16 says: 7h + 1.50 For 15 hours and 8 delivers, lets find out how much he earns. 7(15) + 1.50(8) = ? 7*15 = 105 1.50*8 = 12 105 + 12 = 117 He will make $117 3. snikergrace says: 7h+1.5d 7(15)+1.5(8) 105+12$117 Hope this helps.", "The coaches organized them into groups of 4 for warm-up exercises. Number of groups were there = 89 ÷ 4 = 22.25 or 22 Question 5. Jamal and three of his friends raked leaves every weekend for a month. By the end of the month, they earned$91.00. They paid $4.00 to Jamal’s little brother for helping out a little on the last day. Then the four friends divided the rest of the money equally. How much did each friend get? Answer:$21.75 Explanation: Given that, Jamal and three of his friends by the end of the month earned $91.00. They paid$4.00 to Jamal’s little brother for helping out a little on the last day. $91.00 –$4.00 = $87 Then the four friends divided the rest of the money equally. Total amount did each friend get 87 ÷ 4 =$21.75 Question 6. The drivers at Pizza Palace were loading up their 3 delivery vans for the evening. There were 65 pizzas. The boss told them to put an equal number of pizzas in each van and said they could split any leftovers equally. How much of a leftover pizza did each of the 3 drivers get? 21 with 2 left over. Explanation: Given that, The drivers at Pizza Palace were loading up their 3 delivery vans for the evening. There", "should pay$200,000 for a car if a car delivers $200,000 in pizzas. If the second standard is used, Domino’s should only pay$15,000, which is the correct price for a car. The IT Dashboard itself is most likely overbudget. It’s a few data entry screens and a very fancy flash front end which can be purchased for about a thousand dollars from fusion charts. The code base for this is 176MB–and most likely–needlessly complex. People attempting to reuse the project will be confused when they read the code and could probably rewrite it themselves, cheaper. The goal in “open sourcing” this very complex code is to generate consulting work–to once again, charge millions for a few data entry screens and some graphs. The idea of auditing government spending on IT is good. However, projects should not be measured against their net value to the government–but rather against their appropriate cost. Using “what does the program deliver” as its value allows a simple accouting application to cost ten million dollars. Using “how much should each form cost” as the measurement only allows it to cost couple hundred thousand. If Dominos paid what the cars were worth in delivering pizzas–not what they cost–the car dealer would have all of the profit and Dominos would be out of business. If the government", "us the price during promotion (p + 4)(c - 2) = 96 Since price of pizza (after discount) is 8$, John would have bought 12 pizza's during the promotional period. However, price before discount would be 10$ and he would have bought 8 pizza's which doesn't give us 96$as total cost. Hence, Option C is also not right If price after discount is 6$, he would have bought 16 pizza's during the promotional period. However, price before discount would be 8$and he would have bought 12 pizza's which gives us 96$ as total revenue. Hence, Option A is our answer _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7868 Kudos [?]: 18494 [0], given: 237 Location: Pune, India Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 02:23 Expert's post 1 This post was BOOKMARKED Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9", "I'm guessing publicly against; privately doesn't care. Wait is Kentucky Victory Fund a branch of Victory Fund? As someone who had Papa John's for supper, I'm getting a kick... \\three people deliver here - Papa John's, the garlic-overloaded Dominos, or the very, very greasy local place \\\\and whatever the foodies may say, Papa John's isn't that bad \\\\\\it's not the place that closed in my old neighborhood, but no one ever will be Elmo Jones: So? That crappy \"pizza\" could cost $20 more, and I wouldn't give a hoot in hell. \"Papa John\" sets off my creep-o-meter, which is rarely wrong. You don't want a mouthful of Papa's warm cheese? I have two pizza stones and an oven that goes to 550. It's not exactly like a clay oven, but it works.$.11 to $.14? So you're Obamacare doubles the cost to make a pizza? Weaver95: I really don't see why papa john's is biatching about spending .11 cents or so per pizza to make sure his employees have health care. That's how \"Job Creators ®\" think. \"Death Panels\" was a lot scarier than marginally more expensive bad chain pizza, you guys. As a consumer of Papa John's I will gladly pay the increased price in order to ensure your employees have adequate health care. In fact, just to", "(B) $80 (C)$90 (D) $120 (E)$180 Practice Questions Question: 15 Page: 154 Difficulty: 600 [Reveal] Spoiler: OA _________________ Math Expert Joined: 02 Sep 2009 Posts: 39590 Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 06 Aug 2012, 02:53 5 KUDOS Expert's post 5 This post was BOOKMARKED SOLUTION At a supermarket, John spent 1/2 of his money on fresh fruits and vegetables, 1/3 on meat products, and 1/10 on bakery products. If he spent the remaining $6 on candy, how much did John spend at the supermarket? (A)$60 (B) $80 (C)$90 (D) $120 (E)$180 John spent $$1-(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{10})=\\frac{1}{15}$$ of his money on candy. Since we are told that he spent $6 on candy, then he spent total of $$6*15=90$$ at the supermarket. Answer: C. _________________ Senior Manager Joined: 15 Jun 2010 Posts: 361 Schools: IE'14, ISB'14, Kellogg'15 WE 1: 7 Yrs in Automobile (Commercial Vehicle industry) Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 06 Aug 2012, 06:06 6 This post received KUDOS x/2+x/3+x/10 = 28x/30 2x/30=6, hence x =90 Ans C. _________________ Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html Math Expert Joined: 02 Sep 2009 Posts: 39590 Re: At a supermarket, John spent", "loose a lot of weight by not purchasing from these idiots. I knew this thread would be full of foodies who hate pizza from chains for no other reason than it's from a chain. And that they would put pizza in quotes and call it cardboard and parrot other worn-out phrases. I only had to read the first ten posts or so to confirm my suspicions. It's probably already been said, but if you have Papa John's for every meal, every day, it will cost an extra $120.45 (at 11 cents). If that pays for one of their employees to be covered by insurance, then I'm okay with that. I don't care if it's an additional$10 per pizza and $15 per six pack of Miller Lite. It's all worth it if I understand the Affordable Care Act correctly. My best estimate is that Papa Johns inflicted singring and dysentery costs taxpayers$.20 per pie in unpaid healthcare costs. In any event, what a whiny cockpotato. Oldest | « | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | » | Newest | Show all View Voting Results: Smartest and Funniest In Other Media 1. Links are submitted by members of the Fark", "increased by unlocks and upgrades. Angel Investors are earned according to this formula: 150 * sqrt(LIFETIME_EARNINGS / 1e15) ## Money Name | Base Price | Base Profit / second | Price Increase per Purchase ------------------------------------------------------------------------------------------- LEMONADE_STAND | $4 |$1.66 | 7% NEWSPAPER_DELIVERY | $60 |$20 | 15% CAR_WASH | $720 |$90 | 14% PIZZA_DELIVERY | $8,640 |$360 | 13% DONUT_SHOP | $103,680 |$2,160 | 12% SHRIMP_BOAT | $1,244,160 |$6,480 | 11% HOCKEY_TEAM | $14,929,920 |$19,440 | 10% MOVIE_STUDIO | $179,159,040 |$58,320 | 9% BANK | $2,149,908,480 |$174,960 | 8% OIL_COMPANY | $25,798,901,760 |$804,816 | 7% ## Unlocks Unlocks are bonuses that are earned when a set goal has been achieved. Example list (not actual): LEMONADE_STAND;25;LEMONADE_STAND;PROFIT;2 NEWSPAPER_STAND;900;OIL_COMPANY;PROFIT;11 ALL;2222;ALL;PROFIT;2 Upgrades are bonuses that are purchased with money or angels. See Angel Upgrades for details on upgrades that are purchased with angels. Example list (not actual): 0;CASH;2.5e5;LEMONADE_STAND;PROFIT;3 1;CASH;500000;NEWSPAPER_DELIVERY;PROFIT;3 2;CASH;1e21;ANGEL;EFFECTIVENESS;1 Angel Upgrades are bonuses that are purchased with the sacrifice of angels. Angels sacrificed for this purpose are not regained on reset of a game. Example list (not actual): 3;ANGEL;10000;ALL;PROFIT;3 4;ANGEL;1e32;CAR_WASH;COUNT;100 5;ANGEL;3.4e134;ANGEL;EFFECTIVENESS;10 ## Resetting When you reset your game, you lose all unlocks, upgrades, businesses, and money that you had. You start out with 1 lemonade stand all over again. So why would you want to do that? Because all angels that", "must each make 5 deliveries in order to have the same amount of money. Hope this helps!", "$$P(2)$$ would be, and so on. Indeed, if we are able to prove $$P(N)$$ for any $$N \\in \\mathbb{N}$$, then we know $$P(M)$$ for any natural number $$M>N$$. But the sequence of statements $$\\langle P(0), P(1), P(2), \\ldots\\rangle$$ never gets started. $$P(N)$$ fails for all $$N$$. Another way to think of induction is in terms of guarantees. Suppose you decide to buy a car. First you go to Honest Bob’s. Bob guarantees that any car he sells will go at least one mile. You buy a car, drive it off the lot, and after 3 miles it breaks down and can’t be fixed. You walk back angrily, but Bob won’t give you your money back because the car lived up to the guarantee. Then you cross the road to Honest John’s. John guarantees that if he sells you a car, once it starts it will never stop. This sounds pretty good, so you buy a car, put the keys in the ignition, and ... nothing. The car won’t start. John won’t give you your money back either, because the car did not fail to do what he claimed. Feeling desperate, you end up at Honest Stewart’s. Stewart’s cars come with two guarantees: (1) The car will start and go at least one mile. (2) No matter how far", "TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52938 At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 15 Jun 2017, 22:41 1 1 00:00 Difficulty: 65% (hard) Question Stats: 67% (02:30) correct 33% (02:52) wrong based on 109 sessions ### HideShow timer Statistics At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 _________________ Director Joined: 04 Dec 2015 Posts: 740 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) At a party, John collected$96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 15 Jun 2017, 23:01 1 Bunuel wrote: At a party, John collected $96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for$2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same $96. How much was the cost of each pizza after the discount? A.$6 B. $8", "benefits from the Government), the price of cheese (one of the largest cost per pizza) would skyrocket. there he can pay his full cost When my local Papa Johns drivers are out on the road, they are paid about $5.50 an hour. Only when they are clocked in back at the store do they even make minimum wage. The local owners are idiots and cheapskates and deserve to go out of business ASAP. A driver pays for gas, oil, car repairs, insurance and can possibly be robbed/killed/in major MVAs all for a shiatty pizza made by teenagers who don't know what sanitation is. So fark em. I can make my own awesome pizza, thank you. Well since I have Little Caeser's if I want a decent pizza immediately for$5, and Pepe's Pizza in New Haven when I'm feeling patient and want the best damn pizza in America, I don't really give a rat's ass. Cost of gas delivering pizza: 70¢* Cost of health care: 15¢ *5 mile round trip, with a car that gets 25mpg city. The cost of health care could be negated by hybrids. Bill Murray said I was weird: Well since I have Little Caeser's if I want a decent pizza immediately for $5 What color is the sky in your world? Lsherm: propasaurus: \"Our", "yards Thus the correct answer is option C. Question 5. Jaime had$37 in his bank account on Sunday. The table shows his account activity for the next four days. What was the balance in Jaime’s account after his deposit on Thursday? Options: a. $57.49 b.$59.65 c. $94.49 d.$138.93 Answer: $94.49 Explanation: Add up all the deposits and withdrawals to his original balance make sure deposits are represented by positive numbers and withdrawals are represented by negative numbers. 37 + 17.42 – 12.60 – 9.62 + 62.29 = 94.49 Thus the correct answer is option C. Question 6. A used motorcycle is on sale for$3,600. Erik makes an offer equal to $$\\frac{3}{4}$$ of this price. How much does Erik offer for the motorcycle? Options: a. $4800 b.$2700 c. $2400 d.$900 Answer: $2700 Explanation: Given that, A used motorcycle is on sale for$3,600. Erik makes an offer equal to $$\\frac{3}{4}$$ of this price. $$\\frac{3}{4}$$ × 3600 = 2700 Thus the correct answer is option B. Question 7. Ruby ate $$\\frac{1}{3}$$ of a pizza, and Angie ate $$\\frac{1}{5}$$ of the pizza. How much of the pizza did they eat in all? Options: a. 1 $$\\frac{1}{5}$$ of the pizza b. $$\\frac{1}{8}$$ of the pizza c. $$\\frac{3}{8}$$ of the pizza d. $$\\frac{8}{15}$$ of the pizza Answer: $$\\frac{8}{15}$$ of the pizza Explanation: Ruby ate", "(A)$60 (B) $80 (C)$90 (D) $120 (E)$180 Practice Questions Question: 15 Page: 154 Difficulty: 600 Let total amount be 60x Money spent on fresh fruits and vegetables = 30x Money spent on meat products = 20x Money spent on bakery products = 6x Total money spent 56x Money left (4x) = 6 So, x = 1.5 So, total money John had initially is 1.5 x 60 => $90 Hence answer will be$90 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 16 Mar 2017 Posts: 1 Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] ### Show Tags 20 May 2017, 20:07 \"How can one say, that candy is bought at supermarket only\" Answer 90 is correct when we assume the above statement else answer is 84. Posted from my mobile device Re: At a supermarket, John spent 1/2 of his money on fresh fruit [#permalink] 20 May 2017, 20:07 Similar topics Replies Last post Similar Topics: 1 John spent $25, which", "per month. In addition, assume that they will sell 2,000 pizzas per month if they are open. Finally, assume that if they are open, that they will pay variable costs of $15,000. This means that our total cost, if we are open, is$25,000. While we are simplifying several variables, the main point of this exercise is to explore the relationship between price, cost, and the decision to stay open. Before we continue, we need to recall the average variables which will be used in the scenarios below. Remember that: $AR=\\frac{TR}{Q}$ $AVC=\\frac{TVC}{Q}$ $ATC=\\frac{TC}{Q}$ Using our earlier values, this means that $AVC=\\frac{15,000}{2,000}=7.50$ $ATC=\\frac{25,000}{2,000}=12.50$ Let us consider three scenarios. Scenario 1: P>ATC>AVC In this first scenario, suppose that you can sell pizza for $15.00/each. If we sell the 2,000 pizzas, we will earn a total of (15.00)(2,000)=$30,000. But, our total cost is $10,000+$15,000=$25,000. Therefore, we are earning a profit of$5,000. Since we are earning a profit, we should definitely not shutdown. In this case, we remain open. In addition, we are earning a profit. In general, when P>AVC, we should remain open. In addition, when P>ATC, we earn a profit. Scenario 2: ATC>P>AVC In the second scenario, suppose that we can sell a pizza for $10.00/each. If we sell 2,000 pizzas, we will earn a total of (10.00)(2,000)=$20,000. But, our total", "give us 96$as total cost. Hence, Option C is also not right If price after discount is 6$, he would have bought 16 pizza's during the promotional period. However, price before discount would be 8$and he would have bought 12 pizza's which gives us 96$ as total revenue. Hence, Option A is our answer _________________ You've got what it takes, but it will take everything you've got Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9705 Location: Pune, India Re: At a party, John collected $96 to buy p equally-priced pizzas. When [#permalink] ### Show Tags 16 Jun 2017, 03:23 2 1 Bunuel wrote: At a party, John collected$96 to buy p equally-priced pizzas. When he called the store, however, they were running a promotion for $2 off of each pizza, so he was able to buy 4 more pizzas than he expected for the same$96. How much was the cost of each pizza after the discount? A. $6 B.$8 C. $9 D.$10 E. $12 Instead of taking variables and solving the equations, this question can be best done by jumping to the options directly. John collected$96 to buy p pizzas, each of same price. So obviously, 96 will be divisible by p, and by the un-discounted price of the pizza. The discounted price is $2 less", "to qualify for the bonus. Thus, he still needs $$15-4=11$$ cars. The 11 cars out of the preceding year's car quota of 20 is $$\\frac{11}{20}$$. Thus, D is the right answer. Incorrect. [[snippet]] Incorrect. [[snippet]] Incorrect. [[snippet]] Incorrect. [[snippet]] $$\\frac{1}{5}$$ $$\\frac{4}{15}$$ $$\\frac{9}{20}$$ $$\\frac{11}{20}$$ $$\\frac{3}{4}$$" ]

[ "# How do you simplify 10 + 16 / 2 times4 using PEMDAS? First, you divide, so your answer should be $10 + 8 \\cdot 4$ Then, you multiply, and your answer should be $10 + 32$", "# How do you simplify 10^1 times 0.23? $10 \\times 0.23 = 2.3$ ${10}^{1}$ is the same as 10. $10 \\times 0.23 = 2.3$", "# How do you simplify -2 times -3? Mar 16, 2017 $6$ #### Explanation: $- 2 = - 1 \\setminus \\cdot 2$ $- 3 = - 1 \\setminus \\cdot 3$ $\\setminus \\therefore - 2 \\setminus \\times - 3 = 2 \\setminus \\cdot - 1 \\setminus \\cdot 3 \\setminus \\cdot - 1$ (two negative values multiplied gets you a positive product) $- 1 \\setminus \\cdot - 1 = 1$ now you have $2 \\setminus \\times 3 \\setminus \\times 1 = 6$", "in its simplest form. $9 + 1 = 10$ Simplify $$5\\sqrt{100}$$. $$5 \\times 10 = 50$$. Evaluate $$3\\sqrt{9} − 3$$. $3(3) − 3 = 9 − 3 = 6$", "How do you simplify 20 div 10 times 11 using PEMDAS? May 28, 2017 See a solution process below Explanation: Divide and Multiply rank equally (and go left to right). Therefore, first, we divide 20 by 10: $\\left(\\textcolor{red}{20 \\div 10}\\right) \\times 11 \\implies 2 \\times 11$ Now, complete the simplification by multiplying: $2 \\times 11 \\implies 22$", "# How do you simplify 2a * -a? Since a $\\times$ a = ${a}^{2}$, so $2 a \\times - a$ = $- 2 {a}^{2}$.", "Simplify the expression $9\\times m\\times n\\times8$9×m×n×8. ##### Question 2 Simplify the expression $6u^2\\times7v^8$6u2×7v8. ##### Question 3 Simplify the expression $\\frac{63pq}{9p}$63pq9p. ### Outcomes #### MA4-9NA operates with positive-integer and zero indices of numerical bases", "1. ## Simplify for x: 1/(x^2- x - 2) + 1/(x^2 - 3x + 2) 2. Factor the two denominators and look around for common factors.", "# How do you simplify m+m-m+m? Mar 23, 2018 The answer is $2 m$. #### Explanation: $m + m - m + m$ As we know, $m - m$ is equal to zero, so $m + m = 2 m$", "(6.12 times 10^4) / (3times10^2)? How do you simplify (6.12 times 10^4) / (3times10^2)?...", "Tlok M. # Simplify (m^3)^4(2x^3)^7(m^2)^5(3x)^2. This is for grade 9 math. My teacher wants me to simplify the question By: Tutor 4.8 (62)", "# How do you simplify 7 - (-3)? The answer is $10$. $7 - - 3$ is the same as saying $7 + 3$ $7 + 3 = 10$", "1 L 8 { [ { 4 2T 1... ##### Simplify $4 \\times(8-3 \\times 2) \\div 2$ A. 20 B. 5 C. 4 D. 16 Simplify $4 \\times(8-3 \\times 2) \\div 2$ A. 20 B. 5 C. 4 D. 16...", "# How do you simplify m+m-m+m? The answer is $2 m$. $m + m - m + m$ As we know, $m - m$ is equal to zero, so $m + m = 2 m$", "Simplify the following: <br> 10m^(2) - 9m + 7m - 3m^(2) - 5m -8 Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free,", "problems. Thanks again, DEMPC 4. ## one more for you i came up with 3(x+3)/x+9 but i'm thinking it may just be 2 Find and Simplify: 5. Originally Posted by dempc", "is divisible by $$3^4$$. We have $$(3^4)^2 \\times 3^2 | 24!$$ Simplify to get $$3^{10} | 24!$$ And we cannot find any other multiple of 3. So the exponent of 3 is 10. Jun 13, 2020 #5 +549 +3 nice solution thanks too! amazingxin777 Jun 13, 2020", "# How do you simplify 3 + 6 times 2 div3? ##### 1 Answer Mar 13, 2018 $7$ #### Explanation: $3 + 6 \\times 2 \\div 3 = 3 + \\frac{\\cancel{6} 2 \\cdot 2}{\\cancel{3} 1} = 3 + 4 = 7$", "# How do you simplify (2.3 times 10^3) + (6.9 times 10^3)? $9.2 \\times {10}^{3}$ Take out ${10}^{3}$ as a common factor. $\\Rightarrow {10}^{3} \\left(2.3 + 6.9\\right) = 9.2 \\times {10}^{3} = 9200$", "is -56. # Self-Check Question 1 Simplify: $(+7)\\times~(-3)$ Question 2 Simplify: $(-8)\\times~(-6)$ $(-6)\\times~(+9)$" ]

[ "exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them. In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division. What Happens with Division and Exponents? Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon). ${3}^{5}$ This equals three times itself five total times: ${3}^{5}=3\\cdot 3\\cdot 3\\cdot 3\\cdot 3$ Now let’s divide this by 3. Note that 3 is just 31. $\\frac{{3}^{5}}{{3}^{1}}$ If we write this out to seek a pattern that we can use for a short-cut, we see the following: $\\frac{{3}^{5}}{{3}^{1}}=\\frac{3\\cdot 3\\cdot 3\\cdot 3\\cdot 3}{3}$ If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here. $\\frac{{3}^{5}}{{3}^{1}}=\\frac{3}{3}\\cdot \\frac{3\\cdot 3\\cdot 3\\cdot 3}{1}$ And of course 3/3 is", "the E. • When introducing students to rules of exponents, start by writing each expression in expanded notation, allowing students to see the connection between the rule and the actual math. This will support their understanding of the rule and make it unnecessary for them to memorize the rules. Example: • After finding the patterns of multiplying numbers in exponential form, it is helpful for students to write the rules in words so they are explaining what they are doing to simplify the expression. • Connect negative exponents to the rules of dividing exponents. Students will then be able to make a connection as to why negative exponents \"flip\" the base number. Example: Simplify $3^{2}\\div 3^{3}$ $3^{2}\\div 3^{3}=9\\div 27=\\frac{9}{27}=\\frac{1}{3}$ OR $3^{2}\\div 3^{3}=3^{2-3}=3^{-1}=\\frac{1}{3}$ OR $\\frac{3^{2}}{3^{3}}=\\frac{3\\cdot 3 \\cdot 1}{3\\cdot 3\\cdot 3}=\\frac{1}{3}=3^{-1}$ • Connect the power of 0 to the rules of dividing exponents. Students will be able to see that a power to zero is like a number divided by itself, therefore equaling 1. Example: Simplify $3^{2}\\div 3^{2}$ $3^{2}\\div 3^{2}=9\\div 9=1$ OR $3^{2}\\div 3^{2}=3^{2-2}=3^{0}=1$ OR $\\frac{3^{2}}{3^{2}}=\\frac{3\\cdot 3}{3\\cdot 3}=1$ • Make sure the students understand that when a term does not have an exponent it is assumed to be 1. Examples: 3 = 31 y = y1 • Have students realize the difference between a negative coefficient and a negative exponent", "* 10^1 * 10^1 * 10^0.5? (sorry, can't figure out how to format that with tex) – Tom Marthenal May 12 '13 at 1:45 • @Virtlink$10^{3.5}=10\\cdot 10 \\cdot 10 \\cdot \\sqrt{10}$. – N. S. May 12 '13 at 2:50$10^{3.5}$is equal to$10*10*10*10^{0.5}$. So you just need to know what$10^{0.5}$is. One of the property of exponent is this.$(10^x)^y = 10^{xy}$, i.e. the power of a power, is just the exponents multiplied. So if$(10^{0.5})^2=x^2=10^1$when$x=10^{0.5}$then we just solve for$x^2 = 10$. I'm not sure if you learned about square roots, yet, but$x=\\sqrt{10}$. Since$\\sqrt{10}$is approximately equal to 3.16227, your calculator gives you$10^{3.5} = 3162.27...$. • Thank you so much, so many good answers!!, no i havent learned about square roots yet, but im on it now. – BjarkeCK May 11 '13 at 22:32 Your mistake is computing$10^\\frac{6}{2} \\cdot \\frac{10}{2}$instead of$10^\\frac{6}{2} \\cdot 10^\\frac{1}{2}$. Although 5 is halfway to ten when working with addition, it is past halfway to ten when working with multiplication.$10^{3.5}$is equivalent to$10^\\frac{7}{2}$. You can perform the '7' and 'over 2' parts of the exponentiation separately. That is to say:$10^\\frac{7}{2} = (10^7)^\\frac{1}{2}$. That's just$\\sqrt{10^7}$, which can be reduced to$10^3 \\sqrt{10}$, which is approximately$3163.3$. N.S. is exactly correct;$10^.5 = \\sqrt{10}$. For a small amount of reasoning behind it. By exponent laws, we know$10^x \\times 10^y = 10^{x+y}$.$10^.5 \\times 10^.5 = 10^1 = 10$.", "Beautiful math on $\\LaTeX$ in less than 10 minutes # Arithmetic ## Basic Operators Basic operators work as expected 1 + 2 - 3 = 0 $1 + 2 - 3 = 0$ Division can be done as follows 1/2 = 1 \\div 2 = \\frac{1}{2} $1/2 = 1 \\div 2 = \\frac{1}{2}$ To typeset bigger fractions, use \\dfrac{1}{2} $\\dfrac{1}{2}$ Or use \\displaystyle. \\displaystyle \\frac{1}{3} $\\displaystyle \\frac{1}{3}$ Multiplication 2 \\cdot 3 = 2 \\times 3 $2 \\cdot 3 = 2 \\times 3$ Avoid using * for multiplication 2 * 3 $2 * 3$ Repeated decimals can be denoted by a bar \\frac{1}{3} = 0.\\bar 3 $\\frac{1}{3} = 0.\\bar 3$ ## Exponents and Indexes Exponents can by typeset with a caret 2^3 = 8 $2^3 = 8$ Use braces to group exponents … 2^{10} $2^{10}$ … to avoid 2^10 $2^10$ Subscripts, like indices, can by typeset as follows a_n = 2 \\cdot a_{n-1} $a_n = 2 \\cdot a_{n-1}$ ## Square Roots A square root can be typeset with \\sqrt{16} = 4 $\\sqrt{16} = 4$ To take the $n^\\text{th}$ root, use \\sqrt[3]{27} = 3 $\\sqrt[3]{27} = 3$ \\pm becomes a plus-minus. x^2 = 4 \\implies x = \\pm \\sqrt{4} $x^2 = 4 \\implies x = \\pm \\sqrt{4}$ Braces can be omitted if the argument is only 1 symbol \\sqrt 2", "y\\cdot y\\cdot y}{\\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}}\\hfill \\\\ & =& \\frac{y\\cdot y\\cdot y\\cdot y}{1}\\hfill \\\\ & =& {y}^{4}\\hfill \\end{array}$ Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend. $\\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents. $\\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}$ For the time being, we must be aware of the condition$\\,m>n.\\,$Otherwise, the difference$\\,m-n\\,$could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers. ### The Quotient Rule of Exponents For any real number$\\,a\\,$and natural numbers$\\,m\\,$and$\\,n,$such that$\\,m>n,$the quotient rule of exponents states that $\\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ ### Using the Quotient Rule Write each of the following products with a single base. Do not simplify further. 1. $\\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}$ 2. $\\frac{{t}^{23}}{{t}^{15}}$ 3. $\\frac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}$ Use the quotient rule to simplify each expression. 1. $\\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14-9}={\\left(-2\\right)}^{5}$ 2. $\\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}$ 3. $\\frac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}={\\left(z\\sqrt{2}\\right)}^{5-1}={\\left(z\\sqrt{2}\\right)}^{4}$ ### Try It Write each of the following products with a single base. Do not simplify further. 1. $\\frac{{s}^{75}}{{s}^{68}}$ 2. $\\frac{{\\left(-3\\right)}^{6}}{-3}$ 3. $\\frac{{\\left(e{f}^{2}\\right)}^{5}}{{\\left(e{f}^{2}\\right)}^{3}}$ 1. ${s}^{7}$ 2. ${\\left(-3\\right)}^{5}$ 3. ${\\left(e{f}^{2}\\right)}^{2}$ ### Using the Power Rule of Exponents Suppose an exponential expression is raised to some power. Can we simplify the", "For every non-zero value $x$ ($x \\ne 0$), there is a multiplicative inverse value written $\\div x$ so that $x \\cdot \\div x=1\\text{.}$ When $x \\ne 0\\text{,}$ we have $\\div x = \\frac{1}{x}\\text{,}$ which is why the inverse of $x$ is also called the reciprocal. • Commutative properties: $x+y = y+x$ and $x \\cdot y = y \\cdot x\\text{.}$ • Associative properties: $(x+y)+z = x+(y+z)$ and $(x \\cdot y) \\cdot z = (x \\cdot y) \\cdot z\\text{.}$ • Distributive property of multiplication over addition: $x \\cdot (y+z) = x \\cdot y + x \\cdot z\\text{.}$ Example1.3.4 Our pick-a-number example with the dependent variable \\begin{equation*} y=\\frac{2(x+5)-4}{2}-x \\end{equation*} could be used as a mind-reader trick. If a volunteer does the math correctly, you can always guess the final number $y\\text{.}$ Use algebra properties to determine what you should predict. Solution We start with the distributive property to rewrite $2(x+5) = 2x+10\\text{.}$ Subtracting 4 is equivalent to adding $-4\\text{.}$ We can then use the associative property to rewrite $(2x+10)+-4 = 2x+(10+-4) = 2x+6\\text{.}$ Dividing by two is equivalent to multiplying by $\\frac{1}{2}\\text{.}$ This allows us to use the distributive property again, \\begin{align*} \\frac{2(x+5)-4}{2} = \\frac{2x+6}{2} &= \\frac{1}{2}(2x+6) = \\frac{1}{2}(2x) + \\frac{1}{2}(6)\\\\ &= (\\frac{1}{2} \\cdot 2) x + \\frac{6}{2} = x + 3 \\end{align*} The second line used the associative property and", "exponent form, which will allow us to proceed with the division rule: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{2}{4}}}$ Notice that the exponent in the denominator can be simplified, so we have: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{1}{2}}}$ Recall the rule for dividing numbers with exponents, in which the exponents are subtracted. Applying the division rule yields: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{1}{2}}} = 4 \\cdot x^{\\frac{7}{2}-\\frac{1}{2}} = 4 \\cdot x^{\\frac{6}{2}} = 4x^3$ ## Simplifying Exponential Expressions The rules for operating on numbers with exponents can be applied to variables with exponents as well. ### Learning Objectives Simplify exponential expressions containing variables ### Key Takeaways #### Key Points • The rules for operating on exponential expressions are the same for expressions with variables as they are for those with only integers. #### Rules for Exponential Expressions Recall the rules for operating on numbers with exponents, which are used when simplifying and solving problems in mathematics: • Multiplying exponential expressions with the same base: $a^m \\cdot a^n = a^{m+n}$ • Dividing exponential expressions with the same base: $\\displaystyle \\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ • Raising an exponential expression to an exponent: ${({a}^{n})}^{m}={a}^{n \\cdot m}$ • Raising a product to an exponent: ${(ab)}^{n}={a}^{n}{b}^{n}$ Previously, we have applied these rules only to expressions involving integers. However, they also apply to expressions involving a combination of both integers and variables. This makes them more broadly applicable", "# Math Help - how in the world 1. ## how in the world how in the world do u solve this? ((2z)2 (x4)3) ÷ ((x2)-2(4z)2) the final answer must contain positive exponents 2. Hello, tdot! I must assume you know the basic rules for exponents . . . $\\frac{(2z)^2(x^4)^3}{(x^2)^{-2}(4z)^2}$ We have: . $\\frac{2^2\\cdot z^2\\cdot x^{12}}{x^{-4}\\cdot4^2\\cdot z^2} \\;=\\;\\frac{4\\cdot x^{12}\\cdot z^2}{16\\cdot x^{-4}\\cdot z^2} \\;=\\;\\frac{x^{16}}{4}$ 3. ok. its making more sence. i have other questions, that i tried out but wanna be sure that i have it right. (2/x2)3 (x5)-2 (x/4)-1 (a2b3c)4 ÷ ( 3abc)2 (1+i)5 (1+i)-8 (y2/3 y1/3 y1/2) ÷ (y-6) 4. Originally Posted by tdot ok. its making more sence. i have other questions, that i tried out but wanna be sure that i have it right. (2/x2)3 (x5)-2 (x/4)-1 (a2b3c)4 ÷ ( 3abc)2 (1+i)5 (1+i)-8 (y2/3 y1/3 y1/2) ÷ (y-6) Hello, 1. If you have a new problem start a new thread. 2. Use this sign ^ to write a power (or learn how to write in Latex) 3. Use brackets to separate bases, exponents, factors or summands. I assume that you meant: $\\left(\\frac2{x^2}\\right)^3 \\cdot (x^5)^{-2} \\cdot \\left(\\frac x4\\right)^{-1}$ Transcribe this term into a product: $(2x^{-2})^3 \\cdot (x^5)^{-2} \\cdot (4^{-1} x)^{-1}$ Expand the brackets: $8 x^{-6} \\cdot x^{-10} \\cdot 4 x^{-1}=32x^{-17}=\\frac{32}{x^{17}}$ = = = = = ==", "branch, like the commonly accepted branch that \\sqrt x \\ge 0 for x\\ge0. We could just as well have ... 4 The notation helps here; the exponent (which is the part that's raised) always acts like it has parentheses around it. So x^{y^z} means x^{(y^z)}. Similarly, x^{y+z} means x^{(y+z)} and x^{yz} means x^{(yz)}, even though exponentiation has higher precedence than addition or multiplication (so x+y^z means x+(y^z) and xy^z means x(y^z)). 4 The reasons is that the exponential function(something^x) grows much faster than the power function (x^{something}) So in general you would expect that for every a, b > 1something like this hold$$a^x \\ge x^b$$at least for x big enough Now I don't think it's particularly meaningful that for the special case a = b = e the inequality holds ... 4 In complex numbers exponentiation rules are a bit different, in this case$$(e^{2 \\pi i})^x\\not\\equiv e^{2 \\pi i x}$$4 As a hint, I'd suggest: Write out (a^{\\frac{1}{2}}\\times b^{\\frac{1}{3}})^3, Factorize 432 into prime numbers. 4 Note that$$432 = 2^4\\cdot 3^3 = 3^3\\cdot4^2$$and the LHS can be written as$$(a^\\frac{1}{2} \\times b^\\frac{1}{3})^6 = a^3\\cdot b^2$$So a = 3, b = 4 \\Rightarrow ab = 12. 4 The issue is that a^{\\frac{1}{n}} is multivalued. You could arguably simplify the first calculation into 1 = \\sqrt{1} = -1. Taking different branch cuts", "exponents tells us when we divide two numbers or expressions in exponent form and the bases are the same, we can keep the base the same and subtract the exponent in the denominator away from the exponent in the numerator. Let's look at a few examples. Example 5: Simplify each $$\\require{cancel}\\frac{2^5}{2^2}$$ To simplify, keep the base 2 the same and subtract the exponent in the denominator (2) away from the exponent in the numerator (5): $$\\frac{2^5}{2^2}=2^{5 - 2}=2^3$$ Let's think about why this works, suppose we just simplified without our rule: $$\\frac{2^5}{2^2}=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}{2 \\cdot 2}$$ We can see that we have five factors of 2 in the numerator and two factors of 2 in the denominator. If we cancel common factors, we will be canceling two factors of 2 between the numerator and the denominator. This will leave us with (5 - 2 = 3) three factors of 2 or 8: $$\\frac{\\cancel{2}\\cdot \\cancel{2}\\cdot 2 \\cdot 2 \\cdot 2}{\\cancel{2}\\cdot \\cancel{2}}=2 \\cdot 2 \\cdot 2$$ Example 6: Simplify each $$\\frac{x^{3}}{x^{11}}$$ Keep the base (x) the same and subtract exponents (3 - 11 = -8) $$\\frac{x^{3}}{x^{11}}=x^{3 - 11}=x^{-8}$$ We know how to work with negative exponents. The x goes into the denominator and the exponent of (-8) becomes positive: $$x^{-8}=\\frac{1}{x^8}$$ #### Skills Check: Example", "when dealing with exponents you are dealing with repeated multiplication of a base term so ${\\displaystyle a^{m}\\cdot a^{n}=(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{m})\\cdot (a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{n})=a^{(m+n)}}$ therefore ${\\displaystyle a^{m}\\cdot a^{n}=a^{(m+n)}}$ Example: ${\\displaystyle 3^{2}\\cdot 3^{4}=(3\\cdot 3)\\cdot (3\\cdot 3\\cdot 3\\cdot 3)=3^{6}}$ : This is true because at first we had 2 threes times 4 threes which, when multiplied together gives us 3 multiplied by itself 6 times or 4+2 times. 2. Quotient Rule: ${\\displaystyle {\\frac {a^{m}}{a^{n}}}=a^{(m-n)}}$ Proof (when m > n) - ${\\displaystyle {\\frac {a^{m}}{a^{n}}}={\\frac {(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{m})}{(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{n})}}}$ now group corresponding a’s ${\\displaystyle {\\frac {a^{1}}{a^{1}}}\\cdot {\\frac {a^{2}}{a^{2}}}\\cdot \\dots \\cdot {\\frac {a^{m-n}}{a^{n}}}\\cdot a^{m-n}}$ corresponding a’s become 1 and we are left with ${\\displaystyle 1\\cdot 1\\cdot \\dots \\cdot a^{m-n}=a^{m-n}}$ therefore, ${\\displaystyle {\\frac {a^{m}}{a^{n}}}=a^{m-n}}$ Example: ${\\displaystyle {\\frac {4^{4}}{4^{2}}}={\\frac {4\\cdot 4\\cdot 4\\cdot 4}{4\\cdot 4}}=4\\cdot 4=4^{2}}$ notice ${\\displaystyle 4^{4}-4^{2}=4^{2}}$ therefore the proof holds true this holds true for when m < n as well but when that happens you get negative exponents. When m < n the format of the problem becomes ${\\displaystyle {\\frac {a^{m}}{a^{n}}}={\\frac {1}{a^{n-m}}}}$ See section below for more information on negative exponents 3. Zero Rule: ${\\displaystyle b^{0}=1}$ Proof: we showed in #1 that ${\\displaystyle b^{m+n}=b^{m}\\cdot b^{n}}$ this holds true for ${\\displaystyle b^{0}}$ as well so, ${\\displaystyle b^{m}=b^{m+0}=b^{m}\\cdot b^{0}}$ if this statement is true where ${\\displaystyle", "rule $x^{m+n}=x^m\\cdot x^n$, we would have to have $$x^{-n+n} = x^{-n}\\cdot x^n$$ $$x^0 = x^{-n}\\cdot x^n$$ $$1 = x^{-n}\\cdot x^n$$ Divide both sides by $x^n$, and we see that we must have $$\\frac{1}{x^n} = x^{-n}.$$ So this is how we must define $x^{-n}$. - Think of it this way: exponentiation is equivalent to repeated multiplication, in the sense that, for example, $3^4=3\\times3\\times3\\times3$; so a multiplication repeated a negative number of times should use the multiplicative inverse, division. Therefore, a negative exponentiation could be represented as a repeated division, which would be equivalent to $a^{-b}=\\frac{1}{a^b}$. - Play with this a little: $$2\\to4\\to8\\to16\\ldots\\text{etc}$$or equivalently $$2^1\\to2^2\\to2^3\\to2^4\\ldots\\text{etc}.$$ So you increase the exponent of $2$ with $1$ each time. Now imagine going in the backward direction (meaning you substract $1$ in the exponent each time) but dont stop at two.... - $\\sqrt[-b]{a}=a^{(\\frac{-1}{b})}=\\frac{1}{a^{\\frac{1}{b}}}=\\frac{a^0}{a^{\\frac{1}{b}}}=a^{(0-\\frac{1}{b})}$ - Hmm, I'm fairly certain I've never seen anyone use the notation $^n\\sqrt{a}$ with negative values of $n$, although an \"automatic\" application of $^n\\sqrt{a}=a^{\\frac{1}{n}}$ would do that... Our suspicion now is that the question doesn't have anything to do with roots at all: the OP probably meant \"base.\" – rschwieb Jun 3 at 13:19 @rschwieb, +1 for identifying the mistake, I will keep the my wrong notation posted for others to see \"How not to write math\" – Vikram Jun 3", "# Indices Every rule and operation relating to Exponents and Roots which can be tested on the GMAT has been listed here. Just make yourself comfortable with the operations and notations. Try to do the problems on your own before you look into the explanations. Enjoy! # Exponents $a^n = \\underbrace{a \\times \\cdots \\times a}_n$ $\\frac{x^n}{x^m} = x^{n - m}$ $1 = \\frac{x^n}{x^n} = x^{n - n} = x^0$ • Any number to the power 1 is itself. • Any nonzero number to the power 0 is 1 ### Negative integer exponents $a^{-1} = \\frac{1}{a}$ $a^{-n} = \\frac{1}{a^n}$ $a^{-n} \\, a^{n} = a^{-n\\,+\\,n} = a^0 = 1$ Exponentiation to a negative integer power can alternatively be seen as repeated division of 1 by the base. For instance, $3^{-4} = (((1/3)/3)/3)/3 = \\frac{1}{81} = \\frac{1}{3^{4}}$ ### Identities and properties $a^{m + n} = a^m \\cdot a^n$ $a^{m - n} =\\frac{a^m}{a^n}$ for $a\\ne0$, and $(a^m)^n = a^{m \\cdot n}$ $(a \\cdot b)^n = a^n \\cdot b^n$ Exponentiation is not commutative: 23 = 8, but 32 = 9. Similarly, exponentiation is not associative either: $2^3$ to the 4th power is $8^4$ or 4096, but 2 to the $3^4$ power is $2^81$ or 2,417,851,639,229,258,349,412,352. Without parentheses to modify the order of calculation, the order is usually understood to be top-down, not bottom-up:", "### The Diffie-Hellman key exchange Try a demo of #### Exponent rules The exponent of a number says how many times to multiply the number by it self. Ex: $4^{3} = 4 \\cdot 4 \\cdot 4 = 64$ where 3 is the exponent (or power) and 4 is the base. In words we say 4 to the power 3. Here are the laws of exponents: Law Example $x^{1} = x$ $3^{1} = 3$ $x^{0} = 1$ $4^{0} = 1$ $x^{-1} = \\frac{1}{x}$ $5^{-1} = \\frac{1}{5}$ $x^{m} \\cdot x^{n} = x^{m+n}$ $x^{2} \\cdot x^{3} = (x \\cdot x) \\cdot (x \\cdot x \\cdot x) = x \\cdot x \\cdot x \\cdot x \\cdot x = x^{5} = x^{2+3}$ $\\frac{x^{m}}{x^{n}} = x^{m-n}$ $\\frac{x^{4}}{x^{2}} = \\frac{x \\cdot x \\cdot x \\cdot x}{x \\cdot x} = x \\cdot x \\cdot \\frac{x \\cdot x}{x \\cdot x} = x \\cdot x \\cdot 1 = x^{2} = x^{4-2}$ $(x^{m})^{n} = x^{m \\cdot n}$ $(x^{2})^{3} = (x \\cdot x) \\cdot (x \\cdot x) \\cdot (x \\cdot x) = x \\cdot x \\cdot x \\cdot x \\cdot x \\cdot x = x^{6} = x^{2 \\cdot 3}$ $(x \\cdot y)^{n} = x^{n} \\cdot y^{n}$ $(x \\cdot y)^{2} = (x \\cdot y) \\cdot (x \\cdot y) = x \\cdot x \\cdot y \\cdot y = x^{2} \\cdot y^{2}$ $(\\frac{x}{y})^{n}", "# Why does the law of distributivity use the distributive property to prove it exists? I was looking at the Law of Distributivity and disagreed with how it was proven. It proved that $$\\frac ab\\bigg(\\frac cd + \\frac ef\\bigg) = \\frac ab\\cdot \\frac cd + \\frac ab\\cdot \\frac ef.$$ It multiplied $(c\\div d)$ by $(f\\div f)$ and it multiplied $(e\\div f)$ by $(d\\div d)$. Of course, the left hand side of the equation remains unchanged, i.e. $$\\frac ab\\bigg(\\frac cd+\\frac ef\\bigg)=\\frac ab\\bigg(\\frac cd\\cdot \\frac ff+\\frac ef\\cdot \\frac dd\\bigg)=\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg).$$ This is because $(f\\div f)$ and $(d\\div d)$ each equal $1$. Now notice that $df=fd$, so since the fraction pair in the brackets have the same denominator, it can be combined, i.e. $$\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg)=\\frac ab\\cdot\\frac{cf+ed}{df}=\\frac{a(cf+ed)}{bdf}.$$ The proof now follows by stating that $a(cf+ed)=acf+aed$, but this is the kind of property it is trying to prove in the first place. It is trying to prove this sense of distribution, and in it, it uses it to prove itself. How is this allowed? Edit: This post is very much related, but it does not answer my question. • It proves distributivity for fractions (i.e. rationals) assuming the distributivity for integers. – Mauro ALLEGRANZA Jun 15 '18 at 6:45 • @MauroALLEGRANZA ok, thank you.... but aren't integers rational themselves? – Mr Pie Jun", "the fifth theorem above.] But we could also have: $\\displaystyle{x}^{3/4}={x}^{1/2}\\!\\cdot{x}^{1/4}=\\sqrt[2]{x}\\cdot\\sqrt[4]{x}$ Which is a simple example with just two roots. Other fractions can more involved, requiring lots of roots. The general form is: $\\displaystyle{x}^{r}={x}^{\\frac{a}{2}}\\cdot{x}^{\\frac{b}{3}}\\cdot{x}^{\\frac{c}{4}}\\cdot{x}^{\\frac{d}{5}}\\cdot{x}^{\\frac{e}{6}}\\cdots$ Where a,b,c,d,e,… ∈ {0, 1}. A coefficient is one if the root should be included in the sum, otherwise zero. A real exponent requires an infinite sum, but a rational exponent is a finite sum of the above series. Lastly, here are some useful equalities. Just take them on faith for now. They’re all based on this general one: $\\displaystyle{A}^{x}={B}^{{x}\\cdot\\log_B(A)}$ This can be useful if you have a large number in exponential form and want to convert it to a more familiar form (or just a form compatible with some other exponential number). For example, we can convert between the natural logarithm and the more familiar base ten: $\\displaystyle{e}^{x}={10}^{{x}\\cdot\\log_{10}(e)}={10}^{{x}\\cdot{0.43429}\\ldots}\\\\[0.5em]{10}^{x}={e}^{{x}\\cdot\\ln(10)}={e}^{{x}\\cdot{2.30258}\\ldots}$ It’s also useful for translating large powers of two into (or from) base ten: $\\displaystyle{2}^{x}={10}^{{x}\\cdot\\log_{10}(2)}={10}^{{x}\\cdot{0.30103}\\ldots}\\\\[0.5em]{10}^{x}={2}^{{x}\\cdot\\log_{2}(10)}={2}^{{x}\\cdot{3.32193}\\ldots}$ Note that the log(base) term in the exponent is just a constant. Note also that, in the two pairs above, the constants are the inverse of each other. Here’s a little table that illustrates the above theorems and shows a progression from very small values to vary large: This Equals Because x⁻³ 1/(x⋅x⋅x) 4th theorem x⁻² 1/(x⋅x) 4th theorem x⁻¹ 1/(x)", "# How do you simplify 3times10^8 div 9.0times10^14? Mar 5, 2016 $0.33 \\cdot {10}^{22}$ #### Explanation: According to arithmetic calculation the division to be made before multiplication $\\cancel{3} \\cdot {10}^{8} / {\\cancel{9}}^{3} \\cdot {10}^{14} = \\frac{1}{3} \\cdot {10}^{22} = 0.33 \\cdot {10}^{22}$ Mar 5, 2016 3 Mllion #### Explanation: $\\frac{a {c}^{m}}{b {c}^{n}} = \\left(\\frac{a}{b}\\right) {c}^{m - n}$ ${k}^{-} n = {\\left(\\frac{1}{k}\\right)}^{n}$ Answer: $\\left(\\frac{3}{9}\\right) {10}^{- 6}$ = $3 X {10}^{6}$", "lacks only the number 1. Now let’s try to fill these gaps. Let’s start from Property N.9 (second form). First of all, if $n = 1$, we’ll have: $$\\prod_{d \\mid 1} d^{\\mu(d)} = \\prod_{d \\in \\{1\\}} d^{\\mu(d)} = 1^{\\mu(1)} = 1^1 = 1$$ where we used the fact that $\\mu(1) = 1$, because 1 is square-free, since it’s not divisible by any square greater than 1, and is the product of an even number (zero) of prime factors. If instead $n = p^a$, with $a \\in \\mathbb{N}^{\\star}$, we have to note that the divisors of $n$ are $1, p, p^2, \\ldots, p^a$, and that among them only $1$ and $p$ are square-free, and they are respectively the product of an even number and of an odd number of prime factors. So: \\begin{aligned}\\prod_{d \\mid p^a} d^{\\mu(d)} = \\prod_{d \\in \\{1, p, p^2, \\ldots, p^a\\}} d^{\\mu(d)} & = \\\\ 1^{\\mu(1)} \\cdot p^{\\mu(p)} \\cdot (p^2)^{\\mu(p^2)} \\cdot \\ldots \\cdot (p^a)^{\\mu(p^a)} & = \\\\ 1^0 \\cdot p^{-1} \\cdot (p^2)^{0} \\cdot \\ldots \\cdot (p^a)^{0} & =\\\\ 1 \\cdot \\frac{1}{p} \\cdot 1 \\cdot \\ldots \\cdot 1 & = \\\\ \\frac{1}{p} \\end{aligned} Summarizing, we can say that: • if $n$ is the power of a prime number $p$ with a positive exponent, then $\\prod_{d \\mid n} d^{\\mu(d)} = \\frac{1}{p}$; • in the other cases, i.e. when", "# Negative Exponents Let's learn about negative integer exponents. When we talked about exponents and integers, we assumed that exponent is positive integer. But what if we want to raise number to negative integer exponent? Formula for raising number to negative integer power is $\\color{red}{a^{-b}=\\frac{1}{a^b}}$ $b$ should be a positive integer. This formula means that number raised to negative power is reciprocal of the number raised to the same, but positive power. Let's go through a couple of examples. Example 1. Find ${{4}}^{{-{3}}}$. ${{4}}^{{-{3}}}=\\frac{{1}}{{{{4}}^{{4}}}}=\\frac{{1}}{{{4}\\cdot{4}\\cdot{4}}}=\\frac{{1}}{{64}}$. Answer: $\\frac{{1}}{{64}}$. Next example. Example 2. Find ${{3}}^{{-{4}}}$. ${{3}}^{{-{4}}}=\\frac{{1}}{{{{3}}^{{4}}}}=\\frac{{1}}{{{3}\\cdot{3}\\cdot{3}\\cdot{3}}}=\\frac{{1}}{{81}}$. Answer: $\\frac{{1}}{{81}}$. Now, let's see how to deal with negative integers. Example 3. Find ${{\\left(-{3}\\right)}}^{{-{2}}}$. ${{\\left(-{3}\\right)}}^{{-{2}}}=\\frac{{1}}{{{{\\left(-{3}\\right)}}^{{2}}}}=\\frac{{1}}{{{\\left(-{3}\\right)}\\cdot{\\left(-{3}\\right)}}}=\\frac{{1}}{{9}}$. Answer: $\\frac{{1}}{{9}}$. Next example. Example 4. Find ${{\\left(-{5}\\right)}}^{{-{3}}}$. ${{\\left(-{5}\\right)}}^{{-{3}}}=\\frac{{1}}{{{{\\left(-{5}\\right)}}^{{3}}}}=\\frac{{1}}{{{\\left(-{5}\\right)}\\cdot{\\left(-{5}\\right)}\\cdot{\\left(-{5}\\right)}}}=\\frac{{1}}{{{25}\\cdot{\\left(-{5}\\right)}}}=\\frac{{1}}{{-{125}}}=-\\frac{{1}}{{125}}$. Answer: $-\\frac{{1}}{{125}}$. Last example. Example 5. Find ${{\\left(\\frac{{2}}{{5}}\\right)}}^{{-{3}}}$. Wow! Till now we didn't see such examples. Let's first get read of the negative exponent: ${{\\left(\\frac{{2}}{{5}}\\right)}}^{{-{3}}}=\\frac{{1}}{{{{\\left(\\frac{{2}}{{5}}\\right)}}^{{3}}}}$. Now, recall the definition of exponent (this works for fractions too): ${{\\left(\\frac{{2}}{{5}}\\right)}}^{{3}}=\\frac{{2}}{{5}}\\cdot\\frac{{2}}{{5}}\\cdot\\frac{{2}}{{5}}=\\frac{{{2}\\cdot{2}}}{{{5}\\cdot{5}}}\\cdot\\frac{{2}}{{5}}=\\frac{{4}}{{25}}\\cdot\\frac{{2}}{{5}}=\\frac{{{4}\\cdot{2}}}{{{25}\\cdot{5}}}=\\frac{{8}}{{125}}$. We can easily see that fractions are nice things: numerator and denominator can be separately raised to power: ${{\\left(\\frac{{2}}{{5}}\\right)}}^{{3}}=\\frac{{{{2}}^{{3}}}}{{{{5}}^{{3}}}}=\\frac{{8}}{{125}}$. Now, we just divide fractions (and write 1 as $\\frac{{1}}{{1}}$): $\\frac{{1}}{{{{\\left(\\frac{{2}}{{5}}\\right)}}^{{3}}}}=\\frac{{1}}{{\\frac{{8}}{{125}}}}=\\frac{{\\frac{{\\color{blue}{{{1}}}}}{{\\color{red}{{{1}}}}}}}{{\\frac{{\\color{red}{{{8}}}}}{{\\color{blue}{{{125}}}}}}}=\\frac{{{1}\\cdot{125}}}{{{1}\\cdot{8}}}=\\frac{{125}}{{8}}$. Another way of evaluating it is to notice that $\\frac{{1}}{{\\frac{{8}}{{125}}}}$ is reciprocal of $\\frac{{8}}{{125}}$, and reciprocal is just number that is turned \"upside down\": $\\frac{{125}}{{8}}$. Answer: $\\frac{{125}}{{8}}={15}\\frac{{5}}{{8}}$. From the last example, we have", "used to turn the negative exponent into a positive. $a^{-n}=(a^{-1})^n=\\left(\\frac{1}{a}\\right)^n=\\frac{1}{a^n}$ The same goes the other way around. If an unknown is in the denominator, the denominator can become a numerator by changing the sign of the exponent. In some cases, this will prove to be a very useful feature, especially when working with inverse numbers and functions. Example 1: Write these expressions using only positive exponents: a) $a^{-7}$ b) $\\frac{-6x^{-1}}{y^5}$ c) $\\frac{-12x^{-6}y^{-9}}{z^3}$ Solution: a) $a^{-7}=\\frac{1}{a^7}$ b) $\\frac{-6x^{-1}}{y^5}=\\frac{-6}{x^1 \\cdot y^5}=\\frac{-6}{xy^5}$ c) $\\frac{-12x^{-6}y^{-9}}{z^3} = \\frac{-12}{x^6 \\cdot y^9 \\cdot z^3}$ How does one add or subtract exponents? For example, $\\ 2^2 = 4$ and $\\ 2^3 = 8$ so $\\ 4 + 8 = 12$. But for $\\ 2^2 + 2^3$, the answer is not that obvious. One cannot add nor subtract numbers that have different exponents or different bases. Most interesting tasks involve unkowns, but the same rules apply to them. Let’s look at a simple equation: $\\ x + 2 + 3x = 1$ Since $\\ x = x^1$ and $\\ 1 = x^0$ we can write our equation like this: $\\ x^1 + 2 \\cdot {x^0} + 3 \\cdot {x^1} = 1 \\cdot {x^0}$ How would you normally solve it? The variables with $x$ are added separately, and separately variables without $x$. The same will apply to larger" ]

[ "the E. • When introducing students to rules of exponents, start by writing each expression in expanded notation, allowing students to see the connection between the rule and the actual math. This will support their understanding of the rule and make it unnecessary for them to memorize the rules. Example: • After finding the patterns of multiplying numbers in exponential form, it is helpful for students to write the rules in words so they are explaining what they are doing to simplify the expression. • Connect negative exponents to the rules of dividing exponents. Students will then be able to make a connection as to why negative exponents \"flip\" the base number. Example: Simplify $3^{2}\\div 3^{3}$ $3^{2}\\div 3^{3}=9\\div 27=\\frac{9}{27}=\\frac{1}{3}$ OR $3^{2}\\div 3^{3}=3^{2-3}=3^{-1}=\\frac{1}{3}$ OR $\\frac{3^{2}}{3^{3}}=\\frac{3\\cdot 3 \\cdot 1}{3\\cdot 3\\cdot 3}=\\frac{1}{3}=3^{-1}$ • Connect the power of 0 to the rules of dividing exponents. Students will be able to see that a power to zero is like a number divided by itself, therefore equaling 1. Example: Simplify $3^{2}\\div 3^{2}$ $3^{2}\\div 3^{2}=9\\div 9=1$ OR $3^{2}\\div 3^{2}=3^{2-2}=3^{0}=1$ OR $\\frac{3^{2}}{3^{2}}=\\frac{3\\cdot 3}{3\\cdot 3}=1$ • Make sure the students understand that when a term does not have an exponent it is assumed to be 1. Examples: 3 = 31 y = y1 • Have students realize the difference between a negative coefficient and a negative exponent", "y\\cdot y\\cdot y}{\\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}\\cdot \\overline{)y}}\\hfill \\\\ & =& \\frac{y\\cdot y\\cdot y\\cdot y}{1}\\hfill \\\\ & =& {y}^{4}\\hfill \\end{array}$ Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend. $\\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents. $\\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}$ For the time being, we must be aware of the condition$\\,m>n.\\,$Otherwise, the difference$\\,m-n\\,$could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers. ### The Quotient Rule of Exponents For any real number$\\,a\\,$and natural numbers$\\,m\\,$and$\\,n,$such that$\\,m>n,$the quotient rule of exponents states that $\\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ ### Using the Quotient Rule Write each of the following products with a single base. Do not simplify further. 1. $\\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}$ 2. $\\frac{{t}^{23}}{{t}^{15}}$ 3. $\\frac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}$ Use the quotient rule to simplify each expression. 1. $\\frac{{\\left(-2\\right)}^{14}}{{\\left(-2\\right)}^{9}}={\\left(-2\\right)}^{14-9}={\\left(-2\\right)}^{5}$ 2. $\\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}$ 3. $\\frac{{\\left(z\\sqrt{2}\\right)}^{5}}{z\\sqrt{2}}={\\left(z\\sqrt{2}\\right)}^{5-1}={\\left(z\\sqrt{2}\\right)}^{4}$ ### Try It Write each of the following products with a single base. Do not simplify further. 1. $\\frac{{s}^{75}}{{s}^{68}}$ 2. $\\frac{{\\left(-3\\right)}^{6}}{-3}$ 3. $\\frac{{\\left(e{f}^{2}\\right)}^{5}}{{\\left(e{f}^{2}\\right)}^{3}}$ 1. ${s}^{7}$ 2. ${\\left(-3\\right)}^{5}$ 3. ${\\left(e{f}^{2}\\right)}^{2}$ ### Using the Power Rule of Exponents Suppose an exponential expression is raised to some power. Can we simplify the", "For every non-zero value $x$ ($x \\ne 0$), there is a multiplicative inverse value written $\\div x$ so that $x \\cdot \\div x=1\\text{.}$ When $x \\ne 0\\text{,}$ we have $\\div x = \\frac{1}{x}\\text{,}$ which is why the inverse of $x$ is also called the reciprocal. • Commutative properties: $x+y = y+x$ and $x \\cdot y = y \\cdot x\\text{.}$ • Associative properties: $(x+y)+z = x+(y+z)$ and $(x \\cdot y) \\cdot z = (x \\cdot y) \\cdot z\\text{.}$ • Distributive property of multiplication over addition: $x \\cdot (y+z) = x \\cdot y + x \\cdot z\\text{.}$ Example1.3.4 Our pick-a-number example with the dependent variable \\begin{equation*} y=\\frac{2(x+5)-4}{2}-x \\end{equation*} could be used as a mind-reader trick. If a volunteer does the math correctly, you can always guess the final number $y\\text{.}$ Use algebra properties to determine what you should predict. Solution We start with the distributive property to rewrite $2(x+5) = 2x+10\\text{.}$ Subtracting 4 is equivalent to adding $-4\\text{.}$ We can then use the associative property to rewrite $(2x+10)+-4 = 2x+(10+-4) = 2x+6\\text{.}$ Dividing by two is equivalent to multiplying by $\\frac{1}{2}\\text{.}$ This allows us to use the distributive property again, \\begin{align*} \\frac{2(x+5)-4}{2} = \\frac{2x+6}{2} &= \\frac{1}{2}(2x+6) = \\frac{1}{2}(2x) + \\frac{1}{2}(6)\\\\ &= (\\frac{1}{2} \\cdot 2) x + \\frac{6}{2} = x + 3 \\end{align*} The second line used the associative property and", "Beautiful math on $\\LaTeX$ in less than 10 minutes # Arithmetic ## Basic Operators Basic operators work as expected 1 + 2 - 3 = 0 $1 + 2 - 3 = 0$ Division can be done as follows 1/2 = 1 \\div 2 = \\frac{1}{2} $1/2 = 1 \\div 2 = \\frac{1}{2}$ To typeset bigger fractions, use \\dfrac{1}{2} $\\dfrac{1}{2}$ Or use \\displaystyle. \\displaystyle \\frac{1}{3} $\\displaystyle \\frac{1}{3}$ Multiplication 2 \\cdot 3 = 2 \\times 3 $2 \\cdot 3 = 2 \\times 3$ Avoid using * for multiplication 2 * 3 $2 * 3$ Repeated decimals can be denoted by a bar \\frac{1}{3} = 0.\\bar 3 $\\frac{1}{3} = 0.\\bar 3$ ## Exponents and Indexes Exponents can by typeset with a caret 2^3 = 8 $2^3 = 8$ Use braces to group exponents … 2^{10} $2^{10}$ … to avoid 2^10 $2^10$ Subscripts, like indices, can by typeset as follows a_n = 2 \\cdot a_{n-1} $a_n = 2 \\cdot a_{n-1}$ ## Square Roots A square root can be typeset with \\sqrt{16} = 4 $\\sqrt{16} = 4$ To take the $n^\\text{th}$ root, use \\sqrt[3]{27} = 3 $\\sqrt[3]{27} = 3$ \\pm becomes a plus-minus. x^2 = 4 \\implies x = \\pm \\sqrt{4} $x^2 = 4 \\implies x = \\pm \\sqrt{4}$ Braces can be omitted if the argument is only 1 symbol \\sqrt 2", "exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them. In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division. What Happens with Division and Exponents? Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon). ${3}^{5}$ This equals three times itself five total times: ${3}^{5}=3\\cdot 3\\cdot 3\\cdot 3\\cdot 3$ Now let’s divide this by 3. Note that 3 is just 31. $\\frac{{3}^{5}}{{3}^{1}}$ If we write this out to seek a pattern that we can use for a short-cut, we see the following: $\\frac{{3}^{5}}{{3}^{1}}=\\frac{3\\cdot 3\\cdot 3\\cdot 3\\cdot 3}{3}$ If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here. $\\frac{{3}^{5}}{{3}^{1}}=\\frac{3}{3}\\cdot \\frac{3\\cdot 3\\cdot 3\\cdot 3}{1}$ And of course 3/3 is", "exponents tells us when we divide two numbers or expressions in exponent form and the bases are the same, we can keep the base the same and subtract the exponent in the denominator away from the exponent in the numerator. Let's look at a few examples. Example 5: Simplify each $$\\require{cancel}\\frac{2^5}{2^2}$$ To simplify, keep the base 2 the same and subtract the exponent in the denominator (2) away from the exponent in the numerator (5): $$\\frac{2^5}{2^2}=2^{5 - 2}=2^3$$ Let's think about why this works, suppose we just simplified without our rule: $$\\frac{2^5}{2^2}=\\frac{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}{2 \\cdot 2}$$ We can see that we have five factors of 2 in the numerator and two factors of 2 in the denominator. If we cancel common factors, we will be canceling two factors of 2 between the numerator and the denominator. This will leave us with (5 - 2 = 3) three factors of 2 or 8: $$\\frac{\\cancel{2}\\cdot \\cancel{2}\\cdot 2 \\cdot 2 \\cdot 2}{\\cancel{2}\\cdot \\cancel{2}}=2 \\cdot 2 \\cdot 2$$ Example 6: Simplify each $$\\frac{x^{3}}{x^{11}}$$ Keep the base (x) the same and subtract exponents (3 - 11 = -8) $$\\frac{x^{3}}{x^{11}}=x^{3 - 11}=x^{-8}$$ We know how to work with negative exponents. The x goes into the denominator and the exponent of (-8) becomes positive: $$x^{-8}=\\frac{1}{x^8}$$ #### Skills Check: Example", "\\dfrac a1$: We multiply $a = \\dfrac a1$ by $\\dfrac cc = 1$, to get $\\dfrac a1 \\times \\dfrac cc = \\;\\dfrac{a\\times c}{c}\\;$ and then add that to the fraction $\\dfrac bc$. For example $$3 \\frac 58\\; = \\;3 + \\frac 58\\; = \\;\\frac{3 \\times 8}{8} + \\frac 58\\; = \\;\\frac{(3\\times 8)+ 5}{8}\\; = \\;\\frac{24 + 5}{8} \\;= \\;\\frac{29}{8}$$ - Good call, you did the general case followed by an example +1 – Amzoti May 24 '13 at 12:03 Here is an example. $$2 \\frac{3}{11}=\\frac{2 \\cdot 11}{11}+\\frac3{11}=\\frac{22}{11}+\\frac3{11}=\\frac{25}{11}$$ Or conversely, $$\\frac{11}3=\\frac{11-3 \\cdot 3}{3}+\\frac{3 \\cdot 3}{3}=\\frac{11-9}{3}+\\frac{9}{3}=\\frac{2}{3}+3=3 \\frac{2}3$$. In general, $$x+\\frac{y}{z}=\\frac{x \\cdot z}{z}+\\frac{y}{z}=\\frac{x \\cdot z +y}z$$ and vice versa. In most cases you'll encounter $y<z$. -", "rule $x^{m+n}=x^m\\cdot x^n$, we would have to have $$x^{-n+n} = x^{-n}\\cdot x^n$$ $$x^0 = x^{-n}\\cdot x^n$$ $$1 = x^{-n}\\cdot x^n$$ Divide both sides by $x^n$, and we see that we must have $$\\frac{1}{x^n} = x^{-n}.$$ So this is how we must define $x^{-n}$. - Think of it this way: exponentiation is equivalent to repeated multiplication, in the sense that, for example, $3^4=3\\times3\\times3\\times3$; so a multiplication repeated a negative number of times should use the multiplicative inverse, division. Therefore, a negative exponentiation could be represented as a repeated division, which would be equivalent to $a^{-b}=\\frac{1}{a^b}$. - Play with this a little: $$2\\to4\\to8\\to16\\ldots\\text{etc}$$or equivalently $$2^1\\to2^2\\to2^3\\to2^4\\ldots\\text{etc}.$$ So you increase the exponent of $2$ with $1$ each time. Now imagine going in the backward direction (meaning you substract $1$ in the exponent each time) but dont stop at two.... - $\\sqrt[-b]{a}=a^{(\\frac{-1}{b})}=\\frac{1}{a^{\\frac{1}{b}}}=\\frac{a^0}{a^{\\frac{1}{b}}}=a^{(0-\\frac{1}{b})}$ - Hmm, I'm fairly certain I've never seen anyone use the notation $^n\\sqrt{a}$ with negative values of $n$, although an \"automatic\" application of $^n\\sqrt{a}=a^{\\frac{1}{n}}$ would do that... Our suspicion now is that the question doesn't have anything to do with roots at all: the OP probably meant \"base.\" – rschwieb Jun 3 at 13:19 @rschwieb, +1 for identifying the mistake, I will keep the my wrong notation posted for others to see \"How not to write math\" – Vikram Jun 3", "exponent form, which will allow us to proceed with the division rule: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{2}{4}}}$ Notice that the exponent in the denominator can be simplified, so we have: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{1}{2}}}$ Recall the rule for dividing numbers with exponents, in which the exponents are subtracted. Applying the division rule yields: $4 \\cdot \\dfrac{x^{\\frac{7}{2}}}{x^{\\frac{1}{2}}} = 4 \\cdot x^{\\frac{7}{2}-\\frac{1}{2}} = 4 \\cdot x^{\\frac{6}{2}} = 4x^3$ ## Simplifying Exponential Expressions The rules for operating on numbers with exponents can be applied to variables with exponents as well. ### Learning Objectives Simplify exponential expressions containing variables ### Key Takeaways #### Key Points • The rules for operating on exponential expressions are the same for expressions with variables as they are for those with only integers. #### Rules for Exponential Expressions Recall the rules for operating on numbers with exponents, which are used when simplifying and solving problems in mathematics: • Multiplying exponential expressions with the same base: $a^m \\cdot a^n = a^{m+n}$ • Dividing exponential expressions with the same base: $\\displaystyle \\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ • Raising an exponential expression to an exponent: ${({a}^{n})}^{m}={a}^{n \\cdot m}$ • Raising a product to an exponent: ${(ab)}^{n}={a}^{n}{b}^{n}$ Previously, we have applied these rules only to expressions involving integers. However, they also apply to expressions involving a combination of both integers and variables. This makes them more broadly applicable", "used to turn the negative exponent into a positive. $a^{-n}=(a^{-1})^n=\\left(\\frac{1}{a}\\right)^n=\\frac{1}{a^n}$ The same goes the other way around. If an unknown is in the denominator, the denominator can become a numerator by changing the sign of the exponent. In some cases, this will prove to be a very useful feature, especially when working with inverse numbers and functions. Example 1: Write these expressions using only positive exponents: a) $a^{-7}$ b) $\\frac{-6x^{-1}}{y^5}$ c) $\\frac{-12x^{-6}y^{-9}}{z^3}$ Solution: a) $a^{-7}=\\frac{1}{a^7}$ b) $\\frac{-6x^{-1}}{y^5}=\\frac{-6}{x^1 \\cdot y^5}=\\frac{-6}{xy^5}$ c) $\\frac{-12x^{-6}y^{-9}}{z^3} = \\frac{-12}{x^6 \\cdot y^9 \\cdot z^3}$ How does one add or subtract exponents? For example, $\\ 2^2 = 4$ and $\\ 2^3 = 8$ so $\\ 4 + 8 = 12$. But for $\\ 2^2 + 2^3$, the answer is not that obvious. One cannot add nor subtract numbers that have different exponents or different bases. Most interesting tasks involve unkowns, but the same rules apply to them. Let’s look at a simple equation: $\\ x + 2 + 3x = 1$ Since $\\ x = x^1$ and $\\ 1 = x^0$ we can write our equation like this: $\\ x^1 + 2 \\cdot {x^0} + 3 \\cdot {x^1} = 1 \\cdot {x^0}$ How would you normally solve it? The variables with $x$ are added separately, and separately variables without $x$. The same will apply to larger", "exponent properties are a form of the distributive property. Power of a Product Property: $(ab)^m = a^m b^m$ Power of a Quotient Property: $\\left( \\frac{a}{b} \\right)^m = \\frac{a^m}{b^m}$ #### Example A Simplify the following. (a) $(3^4)^2$ (b) $(x^2 y)^5$ Solution: Use the new properties from above. (a) $(3^4)^2 = 3^{4 \\cdot 2} = 3^8 = 6561$ (b) $(x^2 y)^5 = x^{2 \\cdot 5} y^5 = x^{10} y^5$ #### Example B Simplify $\\left( \\frac{3a^{-6}}{2^2 a^2} \\right)^4$ without negative exponents. Solution: This example uses the Negative Exponent Property from the previous concept. Distribute the $4^{th}$ power first and then move the negative power of $a$ from the numerator to the denominator. $\\left( \\frac{3a^{-6}}{2^2 a^2} \\right)^4 = \\frac{3^4 a^{-6 \\cdot 4}}{2^{2 \\cdot 4} a^{2 \\cdot 4}} = \\frac{81a^{-24}}{2^8 a^8} = \\frac{81}{256a^{8+24}} = \\frac{81}{256a^{32}}$ #### Example C Simplify $\\frac{4x^{-3} y^4 z^6}{12x^2 y} \\div \\left( \\frac{5xy^{-1}}{15x^3 z^{-2}} \\right)^2$ without negative exponents. Solution: This example is definitely as complicated as these types of problems get. Here, all the properties of exponents will be used. Remember that dividing by a fraction is the same as multiplying by its reciprocal. $\\frac{4x^{-3} y^4 z^6}{12x^2 y} \\div \\left( \\frac{5xy^{-1}}{15x^3 z^{-2}} \\right)^2 &= \\frac{4x^{-3} y^4 z^6}{12x^2 y} \\cdot \\frac{225x^6 z^{-4}}{25x^2 y^{-2}}\\\\&= \\frac{y^3 z^6}{3x^5} \\cdot \\frac{9x^4 y^2}{z^4}\\\\&= \\frac{3x^4 y^5 z^6}{x^5 z^4}\\\\&= \\frac{3y^5 z^2}{x}$ Intro Problem Revisit To find the number", "# Why does the law of distributivity use the distributive property to prove it exists? I was looking at the Law of Distributivity and disagreed with how it was proven. It proved that $$\\frac ab\\bigg(\\frac cd + \\frac ef\\bigg) = \\frac ab\\cdot \\frac cd + \\frac ab\\cdot \\frac ef.$$ It multiplied $(c\\div d)$ by $(f\\div f)$ and it multiplied $(e\\div f)$ by $(d\\div d)$. Of course, the left hand side of the equation remains unchanged, i.e. $$\\frac ab\\bigg(\\frac cd+\\frac ef\\bigg)=\\frac ab\\bigg(\\frac cd\\cdot \\frac ff+\\frac ef\\cdot \\frac dd\\bigg)=\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg).$$ This is because $(f\\div f)$ and $(d\\div d)$ each equal $1$. Now notice that $df=fd$, so since the fraction pair in the brackets have the same denominator, it can be combined, i.e. $$\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg)=\\frac ab\\cdot\\frac{cf+ed}{df}=\\frac{a(cf+ed)}{bdf}.$$ The proof now follows by stating that $a(cf+ed)=acf+aed$, but this is the kind of property it is trying to prove in the first place. It is trying to prove this sense of distribution, and in it, it uses it to prove itself. How is this allowed? Edit: This post is very much related, but it does not answer my question. • It proves distributivity for fractions (i.e. rationals) assuming the distributivity for integers. – Mauro ALLEGRANZA Jun 15 '18 at 6:45 • @MauroALLEGRANZA ok, thank you.... but aren't integers rational themselves? – Mr Pie Jun", "# How do you simplify 2/(3sqrt63) ? Mar 18, 2018 Shown below... #### Explanation: We need to rationalise the denominator, in this case multiply both numerator and denominator by $\\sqrt{63}$ $\\implies \\frac{2}{3 \\sqrt{63}} \\cdot \\frac{\\sqrt{63}}{\\sqrt{63}}$ $\\implies \\frac{2 \\sqrt{63}}{3 \\cdot 63}$ $\\implies \\frac{2 \\cdot \\sqrt{3 \\cdot 3 \\cdot 7}}{189}$ $\\implies \\frac{6 \\sqrt{7}}{189}$ $\\implies \\frac{2 \\sqrt{7}}{63}$ Mar 18, 2018 $\\frac{2 \\sqrt{7}}{63}$ #### Explanation: $\\frac{2}{3 \\sqrt{63}}$ $\\therefore = \\frac{2}{3 \\sqrt{3 \\cdot 3 \\cdot 7}}$ $\\therefore \\sqrt{3} \\cdot \\sqrt{3} = 3$ $\\therefore = \\frac{2}{3 \\cdot 3 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$ $\\therefore \\sqrt{7} \\times \\sqrt{7} = 7$ $\\therefore = \\frac{2 \\sqrt{7}}{9 \\times 7}$ $\\therefore = \\frac{2 \\sqrt{7}}{63}$", "2 $(5 + 8) + (-8)$ $=5+[8+(-8)]$ Associative property of addition $=5+0$ Inverse property of addition $= 5$ Identity property of addition Part 3 $6-(15 + 9)$ $=6+[(-15)+(-9)]$ Distributive property $=6+(-24)$ Simplify. $= -18$ Simplify. Part 4 $\\frac{4}{7} \\cdot (\\frac{2}{3} \\cdot \\frac {7}{4})$ $=\\frac{4}{7} \\cdot (\\frac{7}{4} \\cdot \\frac {2}{3})$ Commutative property of multiplication $= (\\frac{4}{7} \\cdot \\frac{7}{4}) \\cdot \\frac {2}{3}$ Associative property of multiplication $= 1 \\cdot \\frac{2}{3}$ Inverse property of multiplication $= \\frac{2}{3}$ Identity property of multiplication Part 5 $100 \\cdot [0.75 + (-2.38)]$ $= 100 \\cdot 0.75 + 100 \\cdot (-2.38)$ Distributive property $= 75+(-238)$ Simplify. $= -163$ Simplify. ## 1.1.4 Evaluating Algebraic Expressions So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as $x+5$, $\\frac{4}{3}πr^{3}$, or $sqrt{2m^{3}n^{2}}$. In the expression $x+5$, $5$ is called a constant because it does not vary and $x$ is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division. We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are", "• $y' = y^2 - 4 = C$ for some $y$ • $y^2 = C + 4$ • $y = \\pm \\sqrt{C + 4}$ ## Separable Equations § A separable DE is one which can be written as $y' = f(t) \\times g(y)$. ### Solving a Separable DE § 1. Solve $g(y) = 0$ to find equilibrium solns. 2. Else, assume $g(y) \\neq 0$. Then: • $\\frac{dy}{dt} = f(t) \\times g(y)$ • $\\frac{dy}{g(y)} = f(t) \\cdot dt$ 3. $\\int \\frac{dy}{g(y)} = \\int f(t) \\cdot dt$ 4. If possible, solve for $y$ in terms of $t$ to get an explicit soln 5. If there’s an IVP, solve for $C$ using the initial conditions Don’t forget to check for $g(y) = 0$ • $y' = \\frac{-t}{y}$ • $f(t) = -t$ • $g(y) = \\frac{1}{y}$ • $g(y)$ is never equal to $0$ • $y \\cdot dy= t \\cdot dt$ • $\\int y \\cdot dy = \\int -t \\cdot dt$ • $\\frac{y^2}{2} = - \\frac{t^2}{2} + C$ • $y^2 = -t^2 + C$ • $y = \\pm \\sqrt{C - t^2}$ • $y' = \\frac{t^2}{1 - y^2}$ • $f(t) = t^2$ • $g(y) = \\frac{1}{1 - y^2}$ • $\\frac{dy}{dt} = \\frac{t^2}{1 - y^2}$ • $(1 - y^2) \\cdot dy = t^2 \\cdot dt$ • $\\int (1 - y^2) \\cdot dy = \\int t^2", "when dealing with exponents you are dealing with repeated multiplication of a base term so ${\\displaystyle a^{m}\\cdot a^{n}=(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{m})\\cdot (a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{n})=a^{(m+n)}}$ therefore ${\\displaystyle a^{m}\\cdot a^{n}=a^{(m+n)}}$ Example: ${\\displaystyle 3^{2}\\cdot 3^{4}=(3\\cdot 3)\\cdot (3\\cdot 3\\cdot 3\\cdot 3)=3^{6}}$ : This is true because at first we had 2 threes times 4 threes which, when multiplied together gives us 3 multiplied by itself 6 times or 4+2 times. 2. Quotient Rule: ${\\displaystyle {\\frac {a^{m}}{a^{n}}}=a^{(m-n)}}$ Proof (when m > n) - ${\\displaystyle {\\frac {a^{m}}{a^{n}}}={\\frac {(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{m})}{(a^{1}\\cdot a^{2}\\cdot \\dots \\cdot a^{n})}}}$ now group corresponding a’s ${\\displaystyle {\\frac {a^{1}}{a^{1}}}\\cdot {\\frac {a^{2}}{a^{2}}}\\cdot \\dots \\cdot {\\frac {a^{m-n}}{a^{n}}}\\cdot a^{m-n}}$ corresponding a’s become 1 and we are left with ${\\displaystyle 1\\cdot 1\\cdot \\dots \\cdot a^{m-n}=a^{m-n}}$ therefore, ${\\displaystyle {\\frac {a^{m}}{a^{n}}}=a^{m-n}}$ Example: ${\\displaystyle {\\frac {4^{4}}{4^{2}}}={\\frac {4\\cdot 4\\cdot 4\\cdot 4}{4\\cdot 4}}=4\\cdot 4=4^{2}}$ notice ${\\displaystyle 4^{4}-4^{2}=4^{2}}$ therefore the proof holds true this holds true for when m < n as well but when that happens you get negative exponents. When m < n the format of the problem becomes ${\\displaystyle {\\frac {a^{m}}{a^{n}}}={\\frac {1}{a^{n-m}}}}$ See section below for more information on negative exponents 3. Zero Rule: ${\\displaystyle b^{0}=1}$ Proof: we showed in #1 that ${\\displaystyle b^{m+n}=b^{m}\\cdot b^{n}}$ this holds true for ${\\displaystyle b^{0}}$ as well so, ${\\displaystyle b^{m}=b^{m+0}=b^{m}\\cdot b^{0}}$ if this statement is true where ${\\displaystyle", "* 10^1 * 10^1 * 10^0.5? (sorry, can't figure out how to format that with tex) – Tom Marthenal May 12 '13 at 1:45 • @Virtlink$10^{3.5}=10\\cdot 10 \\cdot 10 \\cdot \\sqrt{10}$. – N. S. May 12 '13 at 2:50$10^{3.5}$is equal to$10*10*10*10^{0.5}$. So you just need to know what$10^{0.5}$is. One of the property of exponent is this.$(10^x)^y = 10^{xy}$, i.e. the power of a power, is just the exponents multiplied. So if$(10^{0.5})^2=x^2=10^1$when$x=10^{0.5}$then we just solve for$x^2 = 10$. I'm not sure if you learned about square roots, yet, but$x=\\sqrt{10}$. Since$\\sqrt{10}$is approximately equal to 3.16227, your calculator gives you$10^{3.5} = 3162.27...$. • Thank you so much, so many good answers!!, no i havent learned about square roots yet, but im on it now. – BjarkeCK May 11 '13 at 22:32 Your mistake is computing$10^\\frac{6}{2} \\cdot \\frac{10}{2}$instead of$10^\\frac{6}{2} \\cdot 10^\\frac{1}{2}$. Although 5 is halfway to ten when working with addition, it is past halfway to ten when working with multiplication.$10^{3.5}$is equivalent to$10^\\frac{7}{2}$. You can perform the '7' and 'over 2' parts of the exponentiation separately. That is to say:$10^\\frac{7}{2} = (10^7)^\\frac{1}{2}$. That's just$\\sqrt{10^7}$, which can be reduced to$10^3 \\sqrt{10}$, which is approximately$3163.3$. N.S. is exactly correct;$10^.5 = \\sqrt{10}$. For a small amount of reasoning behind it. By exponent laws, we know$10^x \\times 10^y = 10^{x+y}$.$10^.5 \\times 10^.5 = 10^1 = 10$.", "+ 1}}{2}\\right)$$ $$= 10^z\\left(10^{z + 1} - \\frac{10^{2z} - 10^{z + 1}}{2}\\right)$$ $$= 10^{2z + 1} - \\frac{10^{3z} - 10^{2z + 1}}{2}$$ $$= -\\frac{1}{2} \\cdot 10^{3z} + \\frac{3}{2} \\cdot 10^{2z + 1}$$ $$= -\\frac{1}{2} \\cdot 10^{3z} + 15 \\cdot 10^{2z}.$$ It follows that $a = -\\frac{1}{2}$ and $b = 15$, thus $a + b = \\frac{29}{2}.$ ## Solution 3 We can rearrange $\\log_{10}(x+y)=z$ into $x+y=10^{z}$ and $\\log_{10}(x^2+y^2)=z+1$ into $x^2+y^2=10^{z+1}$ We can then put $x+y$ to the third power or $(x+y)^{3}=10^{3z}$. Basic polynomial multiplication shows us that $(x+y)^{3}=x^3+3x^{2}y+3xy^{2}+y^3=10^{3z}$ Thus, $x^3+y^3=10^{3z}-3x^{2}y-3xy^{2}$ or $x^3+y^3=10^{3z}-3xy(x+y)$. We know that $x+y=10^{z}$ so we have $x^3+y^3=10^{3z}-3xy(10^{z})$. Now we need to find out what $xy$ is equal to in terms of $z$. We will find $xy$ by squaring $x+y$. It is basic polynomial multiplication to figure out that $(x+y)^{2}=x^2+2xy+y^2$. We are also given that $x^2+y^2=10^{z+1}$ and $x+y=10^{z}$. Thus $(10^{z})^{2}=10^{z+1}+2xy$ or $10^{2z}=10^{z+1}+2xy$. Rearranging the terms of this equation we obtain that $xy=\\frac{10^{2z}-10^{z+1}}{2}$ or $xy=\\frac{10^z(10^z-10)}{2}$. Now plugging this equation into our original equation $x^3+y^3=10^{3z}-3xy(10^{z})$, we obtain the equation $x^3+y^3=10^{3z}-3(\\frac{10^z(10^z-10)}{2})(10^{z})$ Simple rearranging of this equation yields the result that $x^3+y^3=10^{3z}-3(\\frac{10^{3z}}{2})+30(\\frac{10^{2z}}{2})$. Combining like terms we obtain the equation $x^3+y^3= -(\\frac{10^{3z}}{2})+30(\\frac{10^{2z}}{2})$. Now we know the coefficients of $10^{3z}$ and $10^{2z}$ which are $\\frac{-1}{2}$ and $\\frac{30}{2}$ respectively. Adding the two coefficients we obtain an answer of $\\frac{29}{2}.$ $\\Rightarrow$ $\\boxed{B}$ (Author: David Camacho)", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> Exponent Properties Involving Quotients | CK-12 Foundation You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version. # 8.2: Exponent Properties Involving Quotients Created by: CK-12 0 0 0 ## Learning Objectives • Use the quotient of powers property. • Use the power of a quotient property. • Simplify expressions involving quotient properties of exponents. ## Use the Quotient of Powers Property You saw in the last section that we can use exponent rules to simplify products of numbers and variables. In this section, you will learn that there are similar rules you can use to simplify quotients. Let’s take an example of a quotient, $x^7$ divided by $x^4$. $\\frac{x^7} {x^4} = \\frac{\\cancel{x} \\cdot \\cancel{x} \\cdot \\cancel{x} \\cdot \\cancel{x} \\cdot x \\cdot x \\cdot x } {\\cancel{x} \\cdot \\cancel{x} \\cdot \\cancel{x} \\cdot \\cancel{x} } = \\frac{x \\cdot x \\cdot x } {1} = x^3$ You should see that when we divide two powers of $x$, the number of factors of $x$ in the solution is the difference between the factors in the numerator of the fraction, and the factors in the denominator. In other words, when dividing expressions with the same base, keep the base and subtract the exponent in the denominator from", "make it easier to see how they can be multiplied. According to the commutative property, the order of the factors does not matter. So, $4z \\times \\frac{1}{2}=\\frac{1}{2}\\times 4z$. According to the associative property, the grouping of the factors does not matter. Group the factors so that the numbers are multiplied first. So, $\\frac{1}{2} \\times 4z=\\frac{1}{2} \\times 4 \\times z=\\left(\\frac{1}{2} \\times 4\\right) \\times z$. Now, multiply. $\\left(\\frac{1}{2} \\times 4\\right) \\times z=\\left(\\frac{1}{2} \\times \\frac{4}{1}\\right) \\times z=\\frac{4}{2}\\times z=2 \\times z=2z.$ The product is $2z$. Remember that the word PRODUCT means multiplication and the word QUOTIENT means division. Example Find the quotient: $42c \\div 7$. It may help you to rewrite the problem like this $\\frac{42c}{7}$. Then separate out the numbers and variables like this. $\\frac{42c}{7}=\\frac{42 \\cdot c}{7}=\\frac{42}{7} \\cdot c$ Now, divide 42 by 7 to find the quotient. $\\frac{42}{7} \\cdot c=6 \\cdot c=6c$ The quotient is $6c$. Example Find the quotient $50 g \\div 10 g$. It may help you to rewrite the problem like this $\\frac{50g}{10g}$. Then separate out the numbers and variables like this. $\\frac{50 g}{10 g}=\\frac{50 \\cdot g}{10 \\cdot g}=\\frac{50}{10} \\cdot \\frac{g}{g}$ Now, divide 50 by 10 and divide $g$ by $g$ to find the quotient. Since any number over itself is equal to 1, you know that $\\frac{g}{g}=1$. $\\frac{50}{10} \\cdot \\frac{g}{g}=5x1=5$ The quotient is 5. 7E. Lesson" ]

[ "• anonymous can someone help me with this please. A realtor is renting a space for $7.50 a square foot. If you require a space 36 ft. by 16 ft., what will be the rental on that space?$_____ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "units and prepare it for rent. Unit A, which has the desirable location on the corner and contains 1,800 square feet, will be rented for \\$1.50 per sq...", "Free Version Easy # Rent and Square Footage APSTAT-4YX6FI A realtor would like to examine the relationship between rental price and square footage of an apartment. He believes that the rent asked for an apartment can be predicted based on the square footage. A random sample of $25$ apartments was collected and a least square footage line was obtained with a slope = $1.0632$, a $y$-intercept of $83.527$, and $R^2$ value of $0.8165$. Determine the predicted rental price for a $1,200$-square-foot apartment. A $\\$1,359.37$B$\\$979.80$ C $\\$1,023$D$\\$1,109.93$ E Cannot be determined from the information given.", "this reasonable price? Please advice. Thanks in advance. Answers:The average price is between $100/sq. ft to$200/sq.ft depending in the type of construction & the amenities. \\$28,000 is about ballpark", "# It's HOW expensive to rent in Colorado? Is the rent is too damn high in Colorado? Let’s analyze data from the data.coloardo.gov Rent database and find some afforable places to rent around Colorado. ### Note Please note that the information collected from The Department of Local Affairs Housing Division may be inaccurate or incomplete. For example, rent prices in Aspen may not have fallen as much as believed. You should consult an outside listing service like Craigslist or Redfin prior to moving to a new area. The full methodology from DLAHD can be found here. ## Rent Changes Over Time ### How have rents increased between 1996 and 2015? Efficiency apartments in Fort Collins/Loveland saw the largest increase in rent between 1996 and 2015. During this 19 year period, rent rose 226.5% from $239.26 to$781.18. ### How have rents decreased between 1996 and 1996? 3 Bed apartments in Aspen saw the largest decrease in rent between 1996 and 2015. During this 19 year period, rent fell -40.31% from $1600 to$751.89. ### As of 2015, what are rent prices? The most expensive apartment to rent is a 1 bed in Boulder/Broomfield which goes for $2008.62 per month. On the opposite end of the sepectrum, you can rent a very affordable efficiency in Grand Junction for$258.78 per month! ##", "player on sight even if they didn't do anything to aggravate the townsfolk AI. Tomorrow morning, I have to fire someone. It's been a tough three months with a lot of big life changes. My girlfriend and I were unable to pay our apartment rent in downtown Seattle, for the months of April and May. So, we were strongly encouraged to move out. We were paying $2461 per month in rent, plus$200 a month for parking, plus utilities, all for a 940 square foot two bedroom apartment with no air conditioning. Then they raised the rent. So, we moved out and found a house to rent in Edmonds, a small sleepy town about 12 miles north of Seattle. We doubled the square footage and only pay $2000 a month in rent. It's amazing. It's so peaceful and quiet. It is far superior to living in an apartment. The Seattle apartment was two blocks away from a fire station, so you would often have fire engines roaring down the street with sirens blaring at 3am. Or, there'd be someone unloading product all night for the business next door, operating a hydraulic lift. Or, maybe there'd be homeless or drunk people having an argument outside my window. I don't miss it one bit. The only thing I miss is my", "2000 2500 Size (square feet) (b) (c) (d) (e) (f) 13.16 (a) Yˆ  177.1  1.065 X For each increase of 1 square foot in space, the expected monthly rental is estimated to increase by$1.065. Since X cannot be zero, 177.1 has no practical interpretation. Yˆ  177.1  1.065 X  177.1  1.065(1000)  $1242.10 An apartment with 500 square feet is outside the relevant range for the independent variable. The apartment with 1200 square feet has the more favorable rent relative to size. Based on the regression equation, a 1200 square foot apartment would have an expected monthly rent of$1455.10, while a 1000 square foot apartment would have an expected monthly rent of $1242.10. r2 = SSR 20,535  = 0.684. So, 68.4% of the variation in weekly sales can be SST 30,025 explained by the variation in shelf space.  Y n 13.21 13.42 SSE  n2 i  Yˆi  2 S YX  (c) Based on (a) and (b), the model should be moderately useful for predicting sales. (a) r2 = 0.723. So, 72.3% of the variation in monthly rent can be explained by the variation in square footage. (b) (c) S YX  194.6 (a) i 1 n2  9490 = 30.8058 10 (b) Based on (a) and (b), the", "# Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential.. Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet with a monthly rent of $1,275 and the other with 1,200 square feet for a monthly rent of$1,425. What would you recommend to them? Why? Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet with a...", "box with volume $4ft2.4ft2.$ Find the dimensions of the box that minimize cost. Use $xx$ to represent the length of the side of the box. 354. You are building five identical pens adjacent to each other with a total area of $1000m2,1000m2,$ as shown in the following figure. What dimensions should you use to minimize the amount of fencing? 355. You are the manager of an apartment complex with $5050$ units. When you set rent at $800/month,800/month,$ all apartments are rented. As you increase rent by $25/month,25/month,$ one fewer apartment is rented. Maintenance costs run $50/month50/month$ for each occupied unit. What is the rent that maximizes the total amount of profit?", "monthly rent? How much is the security deposit?(Hint: You have 2 unknowns. Pick a variable to represent the one you know the least about. Use that variable and the other information to represent the second unknown.) An apartment complex charges a refundable security deposit when renting an apartment. The deposit is $350 less than the monthly rent. If Charlene paid a total of$950 for her first month’s rent and security deposit, how much is her monthly rent? How much is the security deposit? (Hint: You ha...", "your own research on this, but I concluded that saving $500 a month and dealing with a few inconveniences was worth it for me and I am happy with my decision. ***In full disclosure - a lot of Boothies are dissatisfied with MPP because of the construction going on (they are turning the bottom two floors into retail). I definitely understand the frustration, but when I looked into moving somewhere else in the loop the options were much more expensive. Do your own research and be aware that while MPP is very cheap for the Loop, there are some trade-offs.*** Some people also live at MDA City Apartments which is less than a block away from MPP. MDA is a nicer and more expensive than MPP and has washer/dryer in-unit. The units tend to be quite a bit smaller or if you want a unit comparable to MPP size, it's a lot more expensive. Park Millennium is a condo building that rents some units and there are a handful of Boothies that live there. It is nicer than MPP. The downside is that it's harder to plan ahead for this building, the condos don't come on the market until a month or two before they are available. Lakeshore East: This is a much nicer area just east of", "feet of apartment space costs a ridiculous $2,575 per month in Brooklyn Heights and a much-more-ridiculous$3,650 per month in the Upper West Side. There are a lot of examples like this, where a Manhattan neighborhood isn’t obviously superior to a Brooklyn or Queens neighborhood but is considerably pricier; a comparison between Harlem (Manhattan) and Fort Greene (Brooklyn) follows this pattern, for instance. * Housing costs in New York appear to increase somewhat exponentially with increasing neighborhood desirability, whereas in the other city I’ve lived in as an adult (Chicago), they’re fairly linear. In New York, the five most expensive neighborhoods are about 5x as expensive as the median neighborhood on a price-per-square-foot basis; in Chicago, the ratio is just 2:1. There might be any number of reasons for this, but I suspect that a lot of it has to do with the fact that in cities like Chicago that are even slightly less dense, if you want to pour more money into your home you can usually do so by getting a bigger home. In New York, the constrains on space are such that to a large extent, the only real variable you can permute is where you live, and the most desirable neighborhoods may be priced in at a super-premium. Neighborhoods like the West Village, Tribeca and", "area are larger than the national average. The mean size of all apartments in the U.S. was 941 square feet in 2018. For a randomly selected sample of 23 renters in Philadelphia, you ask them how big their apartment is, measured in square f... ##### Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7... Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7 8 Total Cost 200 300 350 370 400 460 540 640 760 Which of the following prices is the lowest price at which a price-taking firm would decide to produce rather than ...", "neighbors were delighted about their appartement. Until I moved in :tongue2: 8. Jul 6, 2008 ### mattmns Thanks for the quick advice everyone! I forgot to mention another option that I'm looking at. From craigslist, someone is subletting their place for two months, and will be moving out this Saturday. The apartment complex was one of the top choices on my list (and is less than half a mile from the other place), it looks nice from online, the rent is about$100-150 more (not a big issue, but should be considered), but the reviews are better : http://www.apartmentratings.com/rate/CO-Englewood-Timber-Creek.html [Broken] This place in particular is about 170 square feet larger, and for the next two months the rent would only be $80 more a month (however, it would likely jump up about$100-150 given the current rates). I have contacted the guy, and he told me the place was still available last Friday, but since I found what I thought was my place, I told him I would have to decline; However, I sent him another e-mail discussing the situation, and asked about possibly taking a look at it later this week. I think the sublet could work out great: it would give me two months to try it out before signing a longer lease, and I would feel more", "$700 ans$900 a month respectively. last month they had six vacant apartments and reported \\$4600 in lost rent. how many of each type of apartment were vacant?", "rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. If the law is strictly enforced, the maximum for which an apartment will rent on the black market is A. more than $700 per month. B.$700 per month. C. $600 per month. D. less than$600 per month. Correct: B 26. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus as a result of the price ceiling is A. $500,000 per month. B.$50,000 per month. C. $250,000 per month. D. more than$500,000 per month. Correct: C 27. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus A. is smaller than the gain in consumer surplus. B. could be smaller than, larger than, or the same size as the gain in consumer surplus. C. is the same size as the gain in consumer surplus. D. is larger than the gain in consumer surplus. Correct: D 28. In the figure above, originally the", "that we would expect to see for every unit of increase in the square foot variable. (e) Predict the average monthly rental cost for an apartment that has 1,000 square feet. (f) Why would it not be appropriate to use the model to predict the monthly rental for apartments that have 500 square feet? We don't have data for apartments in that size range. Therefore, our model may not produce reliable results for those apartments. (g) Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet for a monthly rent of $1,250 and the other with 1,200 square feet for a monthly rent of$1,425. What would you recommend to them? Why? We can use our model to see whether these apartments are a relatively good deal for the money. 1,000 square foot apartment: ($1,250 is a little more expensive than we would expect.) 1,200 square foot apartment: ($1,425 seems like a pretty good deal.) Of course, there may be other important factors to consider; this analysis only considers square feet. If the 1,000 square foot apartment is on Central Park West and the 1,200 square foot apartment is in Yonkers, Jim and Jennifer might ignore the results of", "gross at all. (This is not really a practical issue — the worst movie in our data set had$1.1 million in box office gross — but we do need an intercept to define our line.) 4.3331 ( ) is the incremental increase in video units sold that we expect to see in response to every $1 million in box office gross. (e) Predict the average video unit sales for a movie that had a box office gross of$20 million. (f) What other factors in addition to box office gross might be useful in predicting video unit sales? Some possibilities might include: the time of year the movie was released, which famous actors starred in the movie, what other movies were showing at the same time as this movie, unemployment levels at the time of the movie's release, etc. Example : 9. An agent for a residential real estate company in a large city would like to be able to predict the monthly rental costs for apartments based on the size of apartment as defined by square footage. A sample of 25 apartments in a particular residential neighborhood was selected and the information gathered revealed the following: APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) 1 950 850 14 1,800 1,369 2 1,600", "you want a reasonably priced place and are willing to sacrifice some space for quality and location, this is not a bad option. Large Studio: $1900 ~950 sq ft. I didn't get to see this layout They only have 8 of these layouts in the entire building. Essentially a 1BR, but the BR is officially a den, because it has no windows, which is why it's called a studio. 1BR Loft:$1970-2200 ~950 sq ft. These are swank apartments and is what I ended up taking. The leasing agent showed me a model apartment, and everything about it was very nice. I really appreciate the large space and ended up being talked into splurging by my friends and family (who would've thought my cheap asian parents would convince me to spend money?). These also have 1.5 baths for those that don't like guests using their main bathroom. If you're a social person, this is a perfect party apartment. Hardwood floors in the living room (for easy beer spill cleaning). Carpets upstairs in the loft bedroom. Tons of closet space. 1BR Flat / 3BR: ?? ?? Didn't see these, so I know nothing about them. I know they're available, though. Note: If you go visit this place, PM me. I will give you my name and you can put me", "Surplus Property, Apartments For \\$500 A Month In New Jersey, Uc Irvine Track And Field, Rheem Water Heater Pilot," ]

[ "area are larger than the national average. The mean size of all apartments in the U.S. was 941 square feet in 2018. For a randomly selected sample of 23 renters in Philadelphia, you ask them how big their apartment is, measured in square f... ##### Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7... Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7 8 Total Cost 200 300 350 370 400 460 540 640 760 Which of the following prices is the lowest price at which a price-taking firm would decide to produce rather than ...", "Free Version Easy # Rent and Square Footage APSTAT-4YX6FI A realtor would like to examine the relationship between rental price and square footage of an apartment. He believes that the rent asked for an apartment can be predicted based on the square footage. A random sample of $25$ apartments was collected and a least square footage line was obtained with a slope = $1.0632$, a $y$-intercept of $83.527$, and $R^2$ value of $0.8165$. Determine the predicted rental price for a $1,200$-square-foot apartment. A $\\$1,359.37$B$\\$979.80$ C $\\$1,023$D$\\$1,109.93$ E Cannot be determined from the information given.", "the apartments in their building were roughly the same size as the apartments in my building. However, Apartment 1 in the next-door-building has come up for sale, and it is allegedly a 3br 1100 sqft apartment, while all of the units in my building are 2br 800 sqft. I am kind of wondering where the extra space comes from. This apartment also has two nice patios and a garage. If there were an open house scheduled, I’d be tempted to take a look. Listed for$509,000.", "• anonymous can someone help me with this please. A realtor is renting a space for $7.50 a square foot. If you require a space 36 ft. by 16 ft., what will be the rental on that space?$_____ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "2000 2500 Size (square feet) (b) (c) (d) (e) (f) 13.16 (a) Yˆ  177.1  1.065 X For each increase of 1 square foot in space, the expected monthly rental is estimated to increase by$1.065. Since X cannot be zero, 177.1 has no practical interpretation. Yˆ  177.1  1.065 X  177.1  1.065(1000)  $1242.10 An apartment with 500 square feet is outside the relevant range for the independent variable. The apartment with 1200 square feet has the more favorable rent relative to size. Based on the regression equation, a 1200 square foot apartment would have an expected monthly rent of$1455.10, while a 1000 square foot apartment would have an expected monthly rent of $1242.10. r2 = SSR 20,535  = 0.684. So, 68.4% of the variation in weekly sales can be SST 30,025 explained by the variation in shelf space.  Y n 13.21 13.42 SSE  n2 i  Yˆi  2 S YX  (c) Based on (a) and (b), the model should be moderately useful for predicting sales. (a) r2 = 0.723. So, 72.3% of the variation in monthly rent can be explained by the variation in square footage. (b) (c) S YX  194.6 (a) i 1 n2  9490 = 30.8058 10 (b) Based on (a) and (b), the", "units and prepare it for rent. Unit A, which has the desirable location on the corner and contains 1,800 square feet, will be rented for \\$1.50 per sq...", "that we would expect to see for every unit of increase in the square foot variable. (e) Predict the average monthly rental cost for an apartment that has 1,000 square feet. (f) Why would it not be appropriate to use the model to predict the monthly rental for apartments that have 500 square feet? We don't have data for apartments in that size range. Therefore, our model may not produce reliable results for those apartments. (g) Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet for a monthly rent of $1,250 and the other with 1,200 square feet for a monthly rent of$1,425. What would you recommend to them? Why? We can use our model to see whether these apartments are a relatively good deal for the money. 1,000 square foot apartment: ($1,250 is a little more expensive than we would expect.) 1,200 square foot apartment: ($1,425 seems like a pretty good deal.) Of course, there may be other important factors to consider; this analysis only considers square feet. If the 1,000 square foot apartment is on Central Park West and the 1,200 square foot apartment is in Yonkers, Jim and Jennifer might ignore the results of", "to be maximum. --> $P'(X)=-4X+188$, which annulates at X=188/4=47. So the price of the rooms has to be 50+47=97. I hope I didn't make any mistake", "rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. If the law is strictly enforced, the maximum for which an apartment will rent on the black market is A. more than $700 per month. B.$700 per month. C. $600 per month. D. less than$600 per month. Correct: B 26. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus as a result of the price ceiling is A. $500,000 per month. B.$50,000 per month. C. $250,000 per month. D. more than$500,000 per month. Correct: C 27. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus A. is smaller than the gain in consumer surplus. B. could be smaller than, larger than, or the same size as the gain in consumer surplus. C. is the same size as the gain in consumer surplus. D. is larger than the gain in consumer surplus. Correct: D 28. In the figure above, originally the", "small scale for apartments seems to be neglected – cheap apartments are often in bad neighbourhoods, with longer commutes and worse living conditions, but rarely just extremely small (<10 m²). But one could easily imagine 5 m² apartments, with just a bed & a small bathroom (or even smaller options with a shared bathroom). However, I don't know of people renting/buying these kinds of apartments—even though they might be pretty useful if one wants to trade size against good location. Why, therefore, no nano-apartments? Possible reasons: ### No Supply Perhaps nano-apartments are not economically viable to rent. Maybe the fixed cost per apartment is so high that it's not worth it below a certain size—every tenant being an additional burden, plumbing + upkeep of stairways, organising trash & electricity just isn't worth it. Or, perhaps, the amount of walls is too big—the more separate apartments you want to create, the more floor-space is going to be used on walls to separate those apartments, and at some fixed point around 15 m² it's just not worth it. Another possibility is that there are regulations dictating the minimal size of apartments (or something that effectively leads to apartments having a minimal size). ### No Demand I could be over-estimating the number of people who'd like to live in such an", "per square foot: Rent Per Square Foot = Monthly Rent / Floor Space (ft^2) Rent Per Square Foot = $300 / 55 (ft^2) Rent Per Square Foot =$5.45 For more precise estimations, you must compare the results with that generated by the online price per square foot calculator. ### How price of Square Foot Calculator Works? This price square foot calculator will work only if you follow the input guide arranged below: Input: • From the first drop-down list, select whether you wish to calculate “price per square foot on purchase”, “price per square foot on rent”, or “lease price” • After you are done with your selection, go for entering all the necessary parameters in their respective boxes • At last, hit the calculate button Output: The free square foot price calculator does the following calculations: • Calculate square footage price • Estimates the monthly and yearly lease amounts of the property ## FAQ’s: ### Is price per square foot important? Yes, it is! When you are confused with two properties that are in the neighbourhood of each other, price per square foot calculation is the key contributor to fade your confusion away and buy the one that best suits you with respect to currency. ### Do larger houses cost less per square foot? Yes of course!", "at the price of $500 shows the legally fixed maximum price set by the rent control law. However, the underlying forces that shifted the demand curve to the right are still there. At that price ($500), the quantity supplied remains at the same 15,000 rental units, but the quantity demanded is 19,000 rental units. In other words, the quantity demanded exceeds the quantity supplied, so there is a shortage of rental housing. One of the ironies of price ceilings is that while the price ceiling was intended to help renters, there are actually fewer apartments rented out under the price ceiling (15,000 rental units) than would be the case at the market rent of \\$600 (17,000 rental units). Price ceilings do not simply benefit renters at the expense of landlords. Rather, some renters (or potential renters) lose their housing as landlords convert apartments to co-ops and condos. Even when the housing remains in the rental market, landlords tend to spend less on maintenance and on essentials like heating, cooling, hot water, and lighting. The first rule of economics is you do not get something for nothing ­– everything has an opportunity cost. So if renters get \"cheaper\" housing than the market requires, they tend to also end up with lower quality housing. Price ceilings have been proposed for", "this reasonable price? Please advice. Thanks in advance. Answers:The average price is between $100/sq. ft to$200/sq.ft depending in the type of construction & the amenities. \\$28,000 is about ballpark", "from downtown in kilometres x2 = size of the apartment in square feet y x1 x2 55 1.5 350 51 3 450 60 1.75 300 75 1 450 55.5 3.1 385 49 1.6 210 65 2.3 380 61.5 2 600 55 4 450 45 5 325 75 0.65 424 65 2 285 The graph (shown in Figure 8.1) is a scatter plot of the prices of the apartments and their distances from downtown, along with a proposed regression line. Figure 8.1 Scatter Plot of Price, Distance from Downtown, along with a Proposed Regression Line In order to plot such a scatter diagram, you can use many available statistical software packages including Excel, SAS, and Minitab. In this scatter diagram, a negative simple regression line has been shown. The estimated equation for this scatter diagram from Excel is: Where a=71.84 and b=-5.38. In other words, for every additional kilometre from downtown an apartment is located, the price of the apartment is estimated to be $5380 cheaper, i.e. 5.38*$1000=$5380. One might also be curious about the fitted values out of this estimated model. You can simply plug the actual value for x into the estimated line, and find the fitted values for the prices of the apartments. The residuals for all 12 observations are shown in Figure 8.2. Figure 8.2", "people could comfortably occupy an 1100 square foot space. In an 1100 square feet apartment, you can have three standard-sized bedrooms and two full bathrooms. You’d still have plenty of room in the kitchen and living room areas. 22 thg 2, 2022 What is the minimum square footage per person in a home? Every dwelling unit must be at least 150 square feet for the first occupant and at least 100 square feet for every additional occupant. How big is a McMansion? between 3,000 and 5,000 square feet The term “McMansion” is not usually used as a compliment. Loosely defined as a cookie-cutter suburban home of between 3,000 and 5,000 square feet, the McMansion was considered the ultimate sign of affluence in the late 1980s, 1990s, and early 2000s, before the crash of the housing market in 2008. 10 thg 9, 2016 Is 7000 square feet a big house? Historically, a mansion would be at least 5,000 square feet, but with the size of the average house continually increasing, mansions are getting bigger as well. Nowadays, most real estate agents consider homes larger than 7,000 sq ft to be mansions. 16 thg 2, 2022 Is 2500 sq ft a small house? A home between 2500 and 2600 square feet may seem large to some, but it’s actually a", "In a 500-square-foot apartment, size matters. … Add some drama. … Use symmetry. 11 thg 8, 2020 Is 700 sq ft a tiny house? Is tiny home living for you? If so, 600 to 700 square foot home plans might just be the perfect fit for you or your family. This size home rivals some of the more traditional “tiny homes” of 300 to 400 square feet with a slightly more functional and livable space. What is a good size for a 3 bedroom apartment? 3 Bedroom Apartment Though less common in your average apartment complex, three-bedroom apartments will be closer to 1,200 square feet and up. You may have more luck finding a three-bedroom unit in a townhome, duplex, or a four-plex type of situation. 10 thg 2, 2020 Is 800 square feet too small? To visualize it better, 800 square feet is about the size of five parking spaces or a little smaller than three school buses combined. Typically, in an 800-square-foot apartment, you’ll find either a one- or two-bedroom apartment. The bedrooms are small but they’re definitely livable. 17 thg 9, 2021 How do I build a 600 sq ft apartment? Tips for living in a 600 square foot apartment Purge. When moving into a smaller space one of the best things you can do", "fees for 1-room? (1) The smallest rental fee for 2-rooms is $2,800. (2) The smallest rental fee for 2-room is$200 more expensive than the greatest rental fee for 1-room. _________________ \"Be challenged at EVERY MOMENT.\" “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” \"Each stage of the journey is crucial to attaining new heights of knowledge.\" Senior SC Moderator Joined: 14 Nov 2016 Posts: 1315 Location: Malaysia A certain apartment has only 1-room and 2-rooms. If the range of the t [#permalink] ### Show Tags 02 May 2017, 20:13 ziyuen wrote: A certain apartment has only 1-room and 2-rooms. If the range of the total rental fees for 1-room and 2-rooms combined is $1,000 and the range of the rental fees for 2-rooms is$200, what is the range of the rental fees for 1-room? (1) The smallest rental fee for 2-rooms is $2,800. (2) The smallest rental fee for 2-room is$200 more expensive than the greatest rental fee for 1-room. From $$Range = Max - Min$$, Range $$(R_1) = L_1-S_1$$ of 1-room and Range $$(R_2) = L_2-S_2$$ of 2-rooms. _____$600_________$200_________$200_____ $$S_1$$__________$$L_1$$__________$$S_2$$__________$$L_2$$__$2000_______$2600_______$2800_______$3000__ Answer : B _________________ \"Be challenged at EVERY MOMENT.\" “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought", "box with volume $4ft2.4ft2.$ Find the dimensions of the box that minimize cost. Use $xx$ to represent the length of the side of the box. 354. You are building five identical pens adjacent to each other with a total area of $1000m2,1000m2,$ as shown in the following figure. What dimensions should you use to minimize the amount of fencing? 355. You are the manager of an apartment complex with $5050$ units. When you set rent at $800/month,800/month,$ all apartments are rented. As you increase rent by $25/month,25/month,$ one fewer apartment is rented. Maintenance costs run $50/month50/month$ for each occupied unit. What is the rent that maximizes the total amount of profit?", "gross at all. (This is not really a practical issue — the worst movie in our data set had$1.1 million in box office gross — but we do need an intercept to define our line.) 4.3331 ( ) is the incremental increase in video units sold that we expect to see in response to every $1 million in box office gross. (e) Predict the average video unit sales for a movie that had a box office gross of$20 million. (f) What other factors in addition to box office gross might be useful in predicting video unit sales? Some possibilities might include: the time of year the movie was released, which famous actors starred in the movie, what other movies were showing at the same time as this movie, unemployment levels at the time of the movie's release, etc. Example : 9. An agent for a residential real estate company in a large city would like to be able to predict the monthly rental costs for apartments based on the size of apartment as defined by square footage. A sample of 25 apartments in a particular residential neighborhood was selected and the information gathered revealed the following: APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) 1 950 850 14 1,800 1,369 2 1,600", "363 000 € ## Nice Apartment 3 rooms 61 2 636 000 € whose 6% TTC of fees 104 3 1 1 From 250 000 € 26 000 € 13" ]

[ "• anonymous can someone help me with this please. A realtor is renting a space for $7.50 a square foot. If you require a space 36 ft. by 16 ft., what will be the rental on that space?$_____ Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "2000 2500 Size (square feet) (b) (c) (d) (e) (f) 13.16 (a) Yˆ  177.1  1.065 X For each increase of 1 square foot in space, the expected monthly rental is estimated to increase by$1.065. Since X cannot be zero, 177.1 has no practical interpretation. Yˆ  177.1  1.065 X  177.1  1.065(1000)  $1242.10 An apartment with 500 square feet is outside the relevant range for the independent variable. The apartment with 1200 square feet has the more favorable rent relative to size. Based on the regression equation, a 1200 square foot apartment would have an expected monthly rent of$1455.10, while a 1000 square foot apartment would have an expected monthly rent of $1242.10. r2 = SSR 20,535  = 0.684. So, 68.4% of the variation in weekly sales can be SST 30,025 explained by the variation in shelf space.  Y n 13.21 13.42 SSE  n2 i  Yˆi  2 S YX  (c) Based on (a) and (b), the model should be moderately useful for predicting sales. (a) r2 = 0.723. So, 72.3% of the variation in monthly rent can be explained by the variation in square footage. (b) (c) S YX  194.6 (a) i 1 n2  9490 = 30.8058 10 (b) Based on (a) and (b), the", "Free Version Easy # Rent and Square Footage APSTAT-4YX6FI A realtor would like to examine the relationship between rental price and square footage of an apartment. He believes that the rent asked for an apartment can be predicted based on the square footage. A random sample of $25$ apartments was collected and a least square footage line was obtained with a slope = $1.0632$, a $y$-intercept of $83.527$, and $R^2$ value of $0.8165$. Determine the predicted rental price for a $1,200$-square-foot apartment. A $\\$1,359.37$B$\\$979.80$ C $\\$1,023$D$\\$1,109.93$ E Cannot be determined from the information given.", "rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. If the law is strictly enforced, the maximum for which an apartment will rent on the black market is A. more than $700 per month. B.$700 per month. C. $600 per month. D. less than$600 per month. Correct: B 26. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus as a result of the price ceiling is A. $500,000 per month. B.$50,000 per month. C. $250,000 per month. D. more than$500,000 per month. Correct: C 27. In the figure above, originally the apartment rental market is in short run and long run equilibrium with a rental price of $600 per month. Then the government imposes a rent ceiling of$500 per month. The loss of producer surplus A. is smaller than the gain in consumer surplus. B. could be smaller than, larger than, or the same size as the gain in consumer surplus. C. is the same size as the gain in consumer surplus. D. is larger than the gain in consumer surplus. Correct: D 28. In the figure above, originally the", "units and prepare it for rent. Unit A, which has the desirable location on the corner and contains 1,800 square feet, will be rented for \\$1.50 per sq...", "area are larger than the national average. The mean size of all apartments in the U.S. was 941 square feet in 2018. For a randomly selected sample of 23 renters in Philadelphia, you ask them how big their apartment is, measured in square f... ##### Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7... Suppose a firm faces the following costs: Quantity 0 1 2 3 4 5 6 7 8 Total Cost 200 300 350 370 400 460 540 640 760 Which of the following prices is the lowest price at which a price-taking firm would decide to produce rather than ...", "box with volume $4ft2.4ft2.$ Find the dimensions of the box that minimize cost. Use $xx$ to represent the length of the side of the box. 354. You are building five identical pens adjacent to each other with a total area of $1000m2,1000m2,$ as shown in the following figure. What dimensions should you use to minimize the amount of fencing? 355. You are the manager of an apartment complex with $5050$ units. When you set rent at $800/month,800/month,$ all apartments are rented. As you increase rent by $25/month,25/month,$ one fewer apartment is rented. Maintenance costs run $50/month50/month$ for each occupied unit. What is the rent that maximizes the total amount of profit?", "# It's HOW expensive to rent in Colorado? Is the rent is too damn high in Colorado? Let’s analyze data from the data.coloardo.gov Rent database and find some afforable places to rent around Colorado. ### Note Please note that the information collected from The Department of Local Affairs Housing Division may be inaccurate or incomplete. For example, rent prices in Aspen may not have fallen as much as believed. You should consult an outside listing service like Craigslist or Redfin prior to moving to a new area. The full methodology from DLAHD can be found here. ## Rent Changes Over Time ### How have rents increased between 1996 and 2015? Efficiency apartments in Fort Collins/Loveland saw the largest increase in rent between 1996 and 2015. During this 19 year period, rent rose 226.5% from $239.26 to$781.18. ### How have rents decreased between 1996 and 1996? 3 Bed apartments in Aspen saw the largest decrease in rent between 1996 and 2015. During this 19 year period, rent fell -40.31% from $1600 to$751.89. ### As of 2015, what are rent prices? The most expensive apartment to rent is a 1 bed in Boulder/Broomfield which goes for $2008.62 per month. On the opposite end of the sepectrum, you can rent a very affordable efficiency in Grand Junction for$258.78 per month! ##", "this reasonable price? Please advice. Thanks in advance. Answers:The average price is between $100/sq. ft to$200/sq.ft depending in the type of construction & the amenities. \\$28,000 is about ballpark", "gross at all. (This is not really a practical issue — the worst movie in our data set had$1.1 million in box office gross — but we do need an intercept to define our line.) 4.3331 ( ) is the incremental increase in video units sold that we expect to see in response to every $1 million in box office gross. (e) Predict the average video unit sales for a movie that had a box office gross of$20 million. (f) What other factors in addition to box office gross might be useful in predicting video unit sales? Some possibilities might include: the time of year the movie was released, which famous actors starred in the movie, what other movies were showing at the same time as this movie, unemployment levels at the time of the movie's release, etc. Example : 9. An agent for a residential real estate company in a large city would like to be able to predict the monthly rental costs for apartments based on the size of apartment as defined by square footage. A sample of 25 apartments in a particular residential neighborhood was selected and the information gathered revealed the following: APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) APARTMENT MONTHLY RENT ($) SIZE (SQUARE FEET) 1 950 850 14 1,800 1,369 2 1,600", "that we would expect to see for every unit of increase in the square foot variable. (e) Predict the average monthly rental cost for an apartment that has 1,000 square feet. (f) Why would it not be appropriate to use the model to predict the monthly rental for apartments that have 500 square feet? We don't have data for apartments in that size range. Therefore, our model may not produce reliable results for those apartments. (g) Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet for a monthly rent of $1,250 and the other with 1,200 square feet for a monthly rent of$1,425. What would you recommend to them? Why? We can use our model to see whether these apartments are a relatively good deal for the money. 1,000 square foot apartment: ($1,250 is a little more expensive than we would expect.) 1,200 square foot apartment: ($1,425 seems like a pretty good deal.) Of course, there may be other important factors to consider; this analysis only considers square feet. If the 1,000 square foot apartment is on Central Park West and the 1,200 square foot apartment is in Yonkers, Jim and Jennifer might ignore the results of", "your own research on this, but I concluded that saving $500 a month and dealing with a few inconveniences was worth it for me and I am happy with my decision. ***In full disclosure - a lot of Boothies are dissatisfied with MPP because of the construction going on (they are turning the bottom two floors into retail). I definitely understand the frustration, but when I looked into moving somewhere else in the loop the options were much more expensive. Do your own research and be aware that while MPP is very cheap for the Loop, there are some trade-offs.*** Some people also live at MDA City Apartments which is less than a block away from MPP. MDA is a nicer and more expensive than MPP and has washer/dryer in-unit. The units tend to be quite a bit smaller or if you want a unit comparable to MPP size, it's a lot more expensive. Park Millennium is a condo building that rents some units and there are a handful of Boothies that live there. It is nicer than MPP. The downside is that it's harder to plan ahead for this building, the condos don't come on the market until a month or two before they are available. Lakeshore East: This is a much nicer area just east of", "Jennifer are considering signing a lease for an apartment in this residential neighborhoOd. They are trying to dec.ide between two apartments, one with 1,000 square feet for a monthly rent of 1,275 and the other with 1,200 square feet for a monthly rem of 1,425. Based on (a) through (d), which apartment do you think is a better deal? 13.10 A company that holds the DVD distribution rights to movies previously released only in theaters wants to esti· mate sales of DVDs based on box office success. The file ~:.lists the box office gross (in millions) for each of30 movies and the number of DVDs sold (in thousands). For these data, a. construct a scatter plot. b. assuming a linear relationship. use the least-squares method to determine the regression coefficients bQ and hi' c. interpret the meaning of the slope, bj, in this problem. d. predict the sales for a movie DVD that had a box office gross of75 million. Table for Problem 13.9 Apartment 1 2 3 4 5 6 7 8 9 10 Il 12 13 Monthly Rent () Size (Sq. Feet) 950 1,600 1,200 1,500 950 1,700 1,650 935 875 1,150 1,400 1,650 2,300 850 1,450 1,085 1,232 718 1,485 1,136 726 700 956 1,100 1,285 1,985 Apartment 14 15 16 17 18 19 20", "feet of apartment space costs a ridiculous $2,575 per month in Brooklyn Heights and a much-more-ridiculous$3,650 per month in the Upper West Side. There are a lot of examples like this, where a Manhattan neighborhood isn’t obviously superior to a Brooklyn or Queens neighborhood but is considerably pricier; a comparison between Harlem (Manhattan) and Fort Greene (Brooklyn) follows this pattern, for instance. * Housing costs in New York appear to increase somewhat exponentially with increasing neighborhood desirability, whereas in the other city I’ve lived in as an adult (Chicago), they’re fairly linear. In New York, the five most expensive neighborhoods are about 5x as expensive as the median neighborhood on a price-per-square-foot basis; in Chicago, the ratio is just 2:1. There might be any number of reasons for this, but I suspect that a lot of it has to do with the fact that in cities like Chicago that are even slightly less dense, if you want to pour more money into your home you can usually do so by getting a bigger home. In New York, the constrains on space are such that to a large extent, the only real variable you can permute is where you live, and the most desirable neighborhoods may be priced in at a super-premium. Neighborhoods like the West Village, Tribeca and", "player on sight even if they didn't do anything to aggravate the townsfolk AI. Tomorrow morning, I have to fire someone. It's been a tough three months with a lot of big life changes. My girlfriend and I were unable to pay our apartment rent in downtown Seattle, for the months of April and May. So, we were strongly encouraged to move out. We were paying $2461 per month in rent, plus$200 a month for parking, plus utilities, all for a 940 square foot two bedroom apartment with no air conditioning. Then they raised the rent. So, we moved out and found a house to rent in Edmonds, a small sleepy town about 12 miles north of Seattle. We doubled the square footage and only pay $2000 a month in rent. It's amazing. It's so peaceful and quiet. It is far superior to living in an apartment. The Seattle apartment was two blocks away from a fire station, so you would often have fire engines roaring down the street with sirens blaring at 3am. Or, there'd be someone unloading product all night for the business next door, operating a hydraulic lift. Or, maybe there'd be homeless or drunk people having an argument outside my window. I don't miss it one bit. The only thing I miss is my", "$700 ans$900 a month respectively. last month they had six vacant apartments and reported \\$4600 in lost rent. how many of each type of apartment were vacant?", "389.59 4.98 < 0.01 Shaugnessey 2369.72 517.99 4.57 < 0.01 Bedrooms and square feet have positive coefficients. This means that on average, holding everything else constant, the more bedrooms and the more square feet an apartment has, the more expensive it will be. The most expensive neighborhoods are Shaugnessey, Coal Harbour, and Yaletwon. The p-values for these neighborhoods are statistically significant. SqFtbedrooms, and neighborhood explain around 50% of the variation in rents. This means that the other 50% is due to other factors and remains unexplained in our model. The F-statistic is again statistically significant, which means that the fitted model with the covariates is better than the intercept-only model. If you are interested in the full table with all neighborhoods then you can look at this github document. ## What Are the Market Rates for Rent Prices in the Greater Vancouver Area? The market rate for renting the same 1 bedroom apartment with 800 square feet in Yaletown would cost$3284 dollars. The equation would look like this: So, whenever there is an apartment with a renting price below the one you’ll get when plugging in the information into the equation above, then this apartment’s rent is likely below market value. Likewise, is the renting price is above the price you’ll get from the equation, then the rent", "Suppose that the demand and supply schedules for rental apartments in the city of Gotham are as given in the table below. a. What is the market equilibrium rental price per month and the market equilibrium number of apartments demanded and supplied? b. If the local government can enforce a rent-control law that sets the maximum monthly rent at $1500, will there be a surplus or a shortage? Of how many units? And how many units will actually be rented each month? c. Suppose that a new government is elected that wants to keep out the poor. It declares that the minimum rent that can be charged is$2500 per month. If the government can enforce that price floor, will there be a surplus or a shortage? Of how many units? And how many units will actually be rented each month? d. Suppose that the government wishes to decrease the market equilibrium monthly rent by increasing the supply of housing. Assuming that demand remains unchanged, by how many units of housing would the government have to increase the supply of housing in order to get the market equilibrium rental price to fall to $1500 per month? To$1000 per month? To $500 per month? ## Answers (1) • (a) 12,500 apartments at a rent of$2000 per month; (b) A shortage", "of both apartments. b. Which apartment has closets with more square footage? Justify your thinking. The suburban apartment has greater square footage in the closet floors. c. Which apartment has the largest bathroom? Justify your thinking. The city apartment has the largest bathroom. d. A one-year lease for the suburban apartment costs $750 per month. A one-year lease for the city apartment costs$925. Which apartment offers the greater value in terms of the cost per square foot? The suburban cost per square foot is $$\\frac{750}{720}$$, or approximately $1.04 per square foot. The city cost per square foot is $$\\frac{925}{768}$$, or approximately$1.20 per square foot. The suburban apartment offers a greater value (cheaper cost per square foot), $1.04 versus$1.20. ### Eureka Math Grade 7 Module 1 Lesson 19 Exit Ticket Answer Key A 1-inch length in the scale drawing below corresponds to a length of 12 feet in the actual room. Question 1. Describe how the scale or the scale factor can be used to determine the area of the actual dining room. Scale drawing area of dining room: (1$$\\frac{1}{2}$$ in.×$$\\frac{3}{4}$$ in.) + ($$\\frac{3}{4}$$ in. × $$\\frac{1}{2}$$in.) = $$\\frac{12}{8}$$ in2 or 1$$\\frac{1}{2}$$ in2 Actual area of the dining room: $$\\frac{12}{8}$$ ft.× 144 ft. = 216 ft2 The narrower section of the dining room measures $$\\frac{3}{4}$$ by $$\\frac{1}{2}$$ in the", "# Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential.. Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet with a monthly rent of $1,275 and the other with 1,200 square feet for a monthly rent of$1,425. What would you recommend to them? Why? Your friends Jim and Jennifer are considering signing a lease for an apartment in this residential neighborhood. They are trying to decide between two apartments, one with 1,000 square feet with a..." ]

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[ "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "3 + 4 = \\boxed{21}$", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910}$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "5. $7^{24}=7^{12}\\times7^{12}$ 6. $7^{48}=7^{24}\\times7^{24}$ 7. $7^{51}=7^{48}\\times7^3$ 8. $7^{102}=7^{51}\\times7^{51}$ 9. $7^{204}=7^{102}\\times7^{102}$ 10. $7^{408}=7^{204}\\times7^{204}$ 11. $7^{816}=7^{408}\\times7^{408}$ 12. $7^{818}=7^{816}\\times7^2$ It takes at least 12 multiplications to calculate $7^{818}$ - Using the binary representation of 818, yields: $$818_{10} = 1100110010_2$$ There are ones in the $2^{1}, 2^{4}, 2^{5}, 2^{8}, 2^{9}$ positions. So we can write: $$7^{818} \\pmod {1637} = 7^{(2+16+32+256+512)} \\pmod {1637}$$ $$7^{2} \\equiv 49 \\pmod {1637}$$ $$7^{16} \\equiv 899 \\pmod {1637}$$ $$7^{32} \\equiv 1160 \\pmod {1637}$$ $$7^{256} \\equiv 765 \\pmod {1637}$$ $$7^{512} \\equiv 816 \\pmod {1637}$$ So, we just multiply these, arriving at $$7^{818} \\pmod {1637} \\equiv 49 \\cdot 899 \\cdot \\ 1160 \\cdot 765 \\cdot 816 \\pmod {1637} \\equiv 1636 \\pmod {1637}$$ Regards - Excellent demonstration! +1 – amWhy May 4 '13 at 0:28 Code in Pari/GP {powmod(base,expon,modul)=local(a,vb,result,fk,j); vb=binary(expon); \\\\ a vector containing the binary \\\\ representation of the exponent result=1; fk=base; j=#vb+1; \\\\ supplement index which goes from the tail of for(k=1,#vb,j--; if(vb[j]==1 ,result = (result * fk) % modul); fk = (fk*fk) % modul ); return(result);} \\\\ test it powmod(7,818,1637) %931 = 1636 powmod(7,818818818818,1637) %934 = 1636 powmod(7,100000000000,1637) %935 = 908 powmod(7,10^32 ,1637) %936 = 379 -", "'14 at 18:09 • Yes: $1,000,001 = 1,000,000 + 1$ so $999,999·1,000001 = 999,999,000,000 + 999,999$ wich \"fit\" preciselly into $999,999,999,999$ – Darth Geek Aug 16 '14 at 18:11 For starters, it's $999,999\\times1,000,001$, so if you can factor each of those, you're done. Obviously $999,999=3\\times3\\times111,111$, and that last is obviously divisible by $3$, getting $999,999=3\\times3\\times3\\times37,037$. Then you've obviously got divisibility by $37$, so $999,999=3\\times3\\times3\\times37\\times1001$. Factoring $1001$ isn't hard since if you start checking small prime factors you see it's divisbly by $7$, so $1001=7\\times143$, and $143=11\\times13$. Now deal with $1,000,001$. Since you're handed the fact that it's divisible by $101$, you've got $1,000,001=101\\times 9901$. If I'm not mistaken, $9901$ is prime, but I don't know how to show that except by brute force: divide by every prime number $\\le \\sqrt{9901}\\approx99.5$. With $1000001=101\\cdot 9901$ and $999999=3^3\\cdot 7 \\cdot 11 \\cdot 13 \\cdot 37$ we obtain all prime factors of $999999999999=999999\\cdot 1000001$.", "before addition, so $2\\times 3^k+3^k$ actually means $(2\\times 3^k)+3^k$. This is $(2\\times 3^k)+(1\\times 3^k)$, which is clearly just $3\\times 3^k$, or $3^{k+1}$. - It looks like you are confusing $2\\cdot 3^k+3^k$ with $2 \\cdot (3^k+3^k)$. The first gives $3 \\cdot 3^k=3^{k+1}$, while the second gives $2 \\cdot (2 \\cdot 3^k)=4\\cdot 3^k$. Parentheses are important. - $2\\times 3^k+3^k=(2+1)\\times3^k=3\\times3^k$ -", "$0.451$ $0.569$ $0.574$ $0.577$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.04\\times10^{-2}$ $1.52\\times10^{-2}$ $3.88\\times10^{-3}$ $4.48\\times10^{-4}$ $\\|M(w_h, \\mathbf{q}_h)\\|_{L^2(Q_T)}$ $1.13\\times10^5$ $4.45\\times10^4$ $1.48\\times10^4$ $2.86\\times10^3$ $\\sharp\\ \\text{triangles}$ $110$ $1197$ $2880$ $8636$ $\\|\\mathbf{q}_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.46$ $0.57$ $0.574$ $0.577$ $\\|v-\\mathbf{q}_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.24\\times10^{-2}$ $1.55\\times10^{-2}$ $3.72\\times10^{-3}$ $5.18\\times10^{-4}$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.451$ $0.569$ $0.574$ $0.577$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.04\\times10^{-2}$ $1.52\\times10^{-2}$ $3.88\\times10^{-3}$ $4.48\\times10^{-4}$ $\\|M(w_h, \\mathbf{q}_h)\\|_{L^2(Q_T)}$ $1.13\\times10^5$ $4.45\\times10^4$ $1.48\\times10^4$ $2.86\\times10^3$ Non uniform mesh - Conjugate gradient method - Number of iterates for $r = 1$ (top), $r = 10^{-2}$ and $r = h$ (bottom) $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $31$ $41$ $54$ $77$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.469$ $0.576$ $0.589$ $0.586$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $3.21\\times 10^{-1}$ $1.72\\times 10^{-1}$ $1.43\\times 10^{-1}$ $1.25\\times 10^{-1}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $46$ $103$ $125$ $133$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.55$ $0.566$ $0.569$ $0.571$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $2.05\\times 10^{-1}$ $1.47\\times 10^{-1}$ $1.12\\times 10^{-1}$ $8.71\\times 10^{-2}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $36$ $43$ $56$ $80$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.523$ $0.566$ $0.574$ $0.573$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $2.39\\times 10^{-1}$ $1.46\\times 10^{-1}$ $1.19\\times 10^{-1}$ $9.54\\times 10^{-2}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $31$ $41$ $54$ $77$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.469$ $0.576$ $0.589$ $0.586$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $3.21\\times 10^{-1}$ $1.72\\times 10^{-1}$ $1.43\\times 10^{-1}$ $1.25\\times 10^{-1}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$", "+ 31 = \\boxed{058}$.", "373248 = 72^3 = (3\\times 24)^3$ $T_{644} + T_{654} = 421875 = 75^3 = (3\\times 25)^3$ $T_{450} + T_{800} = 421875 = 75^3 = (3\\times 25)^3$", "420^2} = 29 \\cdot 20 = \\boxed{580}$.", "• $23068673=2^{21}\\times11+1$ • $69206017=2^{21}\\times3\\times11+1$ • $81788929=2^{21}\\times3\\times13+1$ • $104857601=2^{22}\\times5^{2}+1$ • $111149057=2^{21}\\times53+1$ • $113246209=2^{22}\\times3^{3}+1$ • $132120577=2^{21}\\times3^{2}\\times7+1$ • $136314881=2^{21}\\times5\\times13+1$ • $138412033=2^{22}\\times3\\times11+1$ • $155189249=2^{22}\\times37+1$ • $163577857=2^{22}\\times3\\times13+1$ • $167772161=2^{25}\\times5+1$ • $169869313=2^{21}\\times3^{4}+1$ • $186646529=2^{21}\\times89+1$ • $199229441=2^{21}\\times5\\times19+1$ • $211812353=2^{21}\\times101+1$ • $230686721=2^{22}\\times5\\times11+1$ • $249561089=2^{21}\\times7\\times17+1$ • $257949697=2^{21}\\times3\\times41+1$ • $270532609=2^{21}\\times3\\times43+1$ • $274726913=2^{21}\\times131+1$ • $377487361=2^{23}\\times3^{2}\\times5+1$ • $383778817=2^{21}\\times3\\times61+1$ • $387973121=2^{21}\\times5\\times37+1$ • $415236097=2^{22}\\times3^{2}\\times11+1$ • $459276289=2^{21}\\times3\\times73+1$ • $463470593=2^{21}\\times13\\times17+1$ • $469762049=2^{26}\\times7+1$ • $576716801=2^{21}\\times5^{2}\\times11+1$ • $595591169=2^{23}\\times71+1$ • $597688321=2^{21}\\times3\\times5\\times19+1$ • $635437057=2^{21}\\times3\\times101+1$ • $639631361=2^{21}\\times5\\times61+1$ • $645922817=2^{23}\\times7\\times11+1$ • $648019969=2^{21}\\times3\\times103+1$ • $666894337=2^{22}\\times3\\times53+1$ • $683671553=2^{22}\\times163+1$ • $710934529=2^{21}\\times3\\times113+1$ • $715128833=2^{21}\\times11\\times31+1$ • $740294657=2^{21}\\times353+1$ • $754974721=2^{24}\\times3^{2}\\times5+1$ • $786432001=2^{21}\\times3\\times5^{3}+1$ • $799014913=2^{21}\\times3\\times127+1$ • $824180737=2^{21}\\times3\\times131+1$ • $880803841=2^{23}\\times3\\times5\\times7+1$ • $897581057=2^{23}\\times107+1$ • $899678209=2^{21}\\times3\\times11\\times13+1$ • $918552577=2^{22}\\times3\\times73+1$ • $924844033=2^{21}\\times3^{2}\\times7^{2}+1$ • $935329793=2^{22}\\times223+1$ • $943718401=2^{22}\\times3^{2}\\times5^{2}+1$ • $950009857=2^{21}\\times3\\times151+1$ • $962592769=2^{21}\\times3^{3}\\times17+1$ • $975175681=2^{21}\\times3\\times5\\times31+1$ • $985661441=2^{22}\\times5\\times47+1$ • $998244353=2^{23}\\times7\\times17+1$ • $1004535809=2^{21}\\times479+1$ • $1012924417=2^{21}\\times3\\times7\\times23+1$ Kerry Su", "# Make numbers from 1-10 with 5 3s In this puzzle book, I came across a question that goes like this: Make the numbers from 1-10 using 5 3s. Rules: You are only allowed to use the four basic arithmetic operators ($$+, -, \\times, \\div$$). You are allowed to use brackets. Here is what I have so far: $$3+3-3+3-3=3$$ $$3+3+3+3-3=9$$ $$(3\\times3)+(3\\div3)-3=7$$ $$3\\div3+3+3+3=10$$ $$3+3\\div3-3+3=4$$ Can someone help me figure out the rest? Complete answer(thanks to Ross Millikan and BAWS): $$3-3\\div3-3\\div3=1$$ $$3-3\\div3\\times3\\div3=2$$ $$3+3-3+3-3=3$$ $$3+3\\div3-3+3=4$$ $$((3\\times3)+(3+3))\\div3=5$$ $$(3\\times3)-(3-3)-3 = 6$$ $$(3\\times3)+(3\\div3)-3=7$$ $$3+3+((3+3)\\div3)=8$$ $$3+3+3+3-3=9$$ $$3\\div3+3+3+3=10$$ $$((3\\times3)+(3+3))\\div3 = 5$$ $$(3\\times3)-(3-3)-3 = 6$$ I'll leave the rest with this new way of looking at it. • Hi, on this site, for answers, you need to hide them in spoiler blocks(>!) most of the times. Thank you. Dec 13 '20 at 5:37 Here are $$1,2$$ and $$8$$ $$3-\\frac 33 - \\frac 33 =1$$ $$3-\\frac 33 \\cdot \\frac 33=2$$ $$3+3+\\frac{3+3}3=8$$ We obtain the missing numbers from the following. $$(3-3)\\times3+\\frac{3}{3}=1$$ $$3-3+\\frac{3+3}{3}=2$$ $$3+\\frac{3}{3}+\\frac{3}{3}=5$$ $$3+3+3\\times (3-3)=6$$", "x = 49 mod 127. Note that $51$ is the inverse of $5 \\pmod {127}$ thus $$5x \\equiv 118 \\pmod {127}\\iff51\\cdot5x\\equiv x\\equiv 51\\cdot 118 \\pmod {127}\\iff x\\equiv 49 \\pmod {127}$$ $305\\cdot 5 - 12\\cdot 127 = 1$ $118\\cdot 305\\cdot 5 - 118\\cdot 12\\cdot 127 = 118$ $5\\cdot 35990 \\equiv 118 \\pmod{127}$ Then subtract the biggest multiple of 127 less than 35990 which is $283\\cdot 127 = 35941$ Therefore $x = 49 + 127a$ where $a$ is integer. • Best to use $\\cdot :\\;$ \\cdot instead of $*$. Also, using \\pmod{127} yields $\\pmod{127}$ – Namaste Jan 9 '18 at 0:03 To solve $5x \\equiv 118 \\mod 127$, first find the $\\gcd$ of $5$ and $127$ using the Euclidean method. $$127 = 5 \\times 25 + 2, 5 = 2 \\times 2 + 1$$ this tells us the $\\gcd$ is $1$. Furthermore: $$1 = 5 - 2 \\times 2 = 5 - 2 \\times (127 - 25 \\times 5) = 5 - 2 \\times 127 + 50 \\times 5 = 51 \\times 5 - 2 \\times 127$$ therefore: $$118 = (51 \\times 118) \\times 5 - (2 \\times 118) \\times 127 \\implies (51 \\times 118) \\times 5 \\equiv 118 \\mod 127$$ Therefore, $x = 51 \\times 118$ does the job. Upon a little more simplification:$$51 \\times 118 \\equiv 51 \\times", "add all three cases: $1+17+243=\\boxed{261}$", "+ 1\\cdot3 + 2\\cdot3 + 2\\cdot3 + 3\\cdot3 = \\boxed{29}$.", "our answer is $m+n=61+4=\\boxed{065}$.", "2 ) + 1This gives the sequence 1,1,2,3,4,6,9,14... Suppose m box can give out any two amounts summing to at most f(m), then we havef(m) = f(m-1) + floor( f(m-1)/2 ) + 1and f(2)=2, f(3)=4, f(4)=7, f(5)=11, f(6)=17, f(7)=26, f(8)=40,...A good estimate for f(m) would be f(m) ~= 1.5^(m+1), and 1.5^(m+1)=2005 gives us m=17.75. So we'd expect to use something like 18 boxes. Last edited by cuinuc on December 4th, 2006, 11:00 pm, edited 1 time in total." ]

[ "\\times \\cdots \\times 100 .$ This product is divisible by some $2^m ,$ where $m$ is an integer. What is the largest possible value of $m?$ ×", "'14 at 18:09 • Yes: $1,000,001 = 1,000,000 + 1$ so $999,999·1,000001 = 999,999,000,000 + 999,999$ wich \"fit\" preciselly into $999,999,999,999$ – Darth Geek Aug 16 '14 at 18:11 For starters, it's $999,999\\times1,000,001$, so if you can factor each of those, you're done. Obviously $999,999=3\\times3\\times111,111$, and that last is obviously divisible by $3$, getting $999,999=3\\times3\\times3\\times37,037$. Then you've obviously got divisibility by $37$, so $999,999=3\\times3\\times3\\times37\\times1001$. Factoring $1001$ isn't hard since if you start checking small prime factors you see it's divisbly by $7$, so $1001=7\\times143$, and $143=11\\times13$. Now deal with $1,000,001$. Since you're handed the fact that it's divisible by $101$, you've got $1,000,001=101\\times 9901$. If I'm not mistaken, $9901$ is prime, but I don't know how to show that except by brute force: divide by every prime number $\\le \\sqrt{9901}\\approx99.5$. With $1000001=101\\cdot 9901$ and $999999=3^3\\cdot 7 \\cdot 11 \\cdot 13 \\cdot 37$ we obtain all prime factors of $999999999999=999999\\cdot 1000001$.", "5. $7^{24}=7^{12}\\times7^{12}$ 6. $7^{48}=7^{24}\\times7^{24}$ 7. $7^{51}=7^{48}\\times7^3$ 8. $7^{102}=7^{51}\\times7^{51}$ 9. $7^{204}=7^{102}\\times7^{102}$ 10. $7^{408}=7^{204}\\times7^{204}$ 11. $7^{816}=7^{408}\\times7^{408}$ 12. $7^{818}=7^{816}\\times7^2$ It takes at least 12 multiplications to calculate $7^{818}$ - Using the binary representation of 818, yields: $$818_{10} = 1100110010_2$$ There are ones in the $2^{1}, 2^{4}, 2^{5}, 2^{8}, 2^{9}$ positions. So we can write: $$7^{818} \\pmod {1637} = 7^{(2+16+32+256+512)} \\pmod {1637}$$ $$7^{2} \\equiv 49 \\pmod {1637}$$ $$7^{16} \\equiv 899 \\pmod {1637}$$ $$7^{32} \\equiv 1160 \\pmod {1637}$$ $$7^{256} \\equiv 765 \\pmod {1637}$$ $$7^{512} \\equiv 816 \\pmod {1637}$$ So, we just multiply these, arriving at $$7^{818} \\pmod {1637} \\equiv 49 \\cdot 899 \\cdot \\ 1160 \\cdot 765 \\cdot 816 \\pmod {1637} \\equiv 1636 \\pmod {1637}$$ Regards - Excellent demonstration! +1 – amWhy May 4 '13 at 0:28 Code in Pari/GP {powmod(base,expon,modul)=local(a,vb,result,fk,j); vb=binary(expon); \\\\ a vector containing the binary \\\\ representation of the exponent result=1; fk=base; j=#vb+1; \\\\ supplement index which goes from the tail of for(k=1,#vb,j--; if(vb[j]==1 ,result = (result * fk) % modul); fk = (fk*fk) % modul ); return(result);} \\\\ test it powmod(7,818,1637) %931 = 1636 powmod(7,818818818818,1637) %934 = 1636 powmod(7,100000000000,1637) %935 = 908 powmod(7,10^32 ,1637) %936 = 379 -", "• $23068673=2^{21}\\times11+1$ • $69206017=2^{21}\\times3\\times11+1$ • $81788929=2^{21}\\times3\\times13+1$ • $104857601=2^{22}\\times5^{2}+1$ • $111149057=2^{21}\\times53+1$ • $113246209=2^{22}\\times3^{3}+1$ • $132120577=2^{21}\\times3^{2}\\times7+1$ • $136314881=2^{21}\\times5\\times13+1$ • $138412033=2^{22}\\times3\\times11+1$ • $155189249=2^{22}\\times37+1$ • $163577857=2^{22}\\times3\\times13+1$ • $167772161=2^{25}\\times5+1$ • $169869313=2^{21}\\times3^{4}+1$ • $186646529=2^{21}\\times89+1$ • $199229441=2^{21}\\times5\\times19+1$ • $211812353=2^{21}\\times101+1$ • $230686721=2^{22}\\times5\\times11+1$ • $249561089=2^{21}\\times7\\times17+1$ • $257949697=2^{21}\\times3\\times41+1$ • $270532609=2^{21}\\times3\\times43+1$ • $274726913=2^{21}\\times131+1$ • $377487361=2^{23}\\times3^{2}\\times5+1$ • $383778817=2^{21}\\times3\\times61+1$ • $387973121=2^{21}\\times5\\times37+1$ • $415236097=2^{22}\\times3^{2}\\times11+1$ • $459276289=2^{21}\\times3\\times73+1$ • $463470593=2^{21}\\times13\\times17+1$ • $469762049=2^{26}\\times7+1$ • $576716801=2^{21}\\times5^{2}\\times11+1$ • $595591169=2^{23}\\times71+1$ • $597688321=2^{21}\\times3\\times5\\times19+1$ • $635437057=2^{21}\\times3\\times101+1$ • $639631361=2^{21}\\times5\\times61+1$ • $645922817=2^{23}\\times7\\times11+1$ • $648019969=2^{21}\\times3\\times103+1$ • $666894337=2^{22}\\times3\\times53+1$ • $683671553=2^{22}\\times163+1$ • $710934529=2^{21}\\times3\\times113+1$ • $715128833=2^{21}\\times11\\times31+1$ • $740294657=2^{21}\\times353+1$ • $754974721=2^{24}\\times3^{2}\\times5+1$ • $786432001=2^{21}\\times3\\times5^{3}+1$ • $799014913=2^{21}\\times3\\times127+1$ • $824180737=2^{21}\\times3\\times131+1$ • $880803841=2^{23}\\times3\\times5\\times7+1$ • $897581057=2^{23}\\times107+1$ • $899678209=2^{21}\\times3\\times11\\times13+1$ • $918552577=2^{22}\\times3\\times73+1$ • $924844033=2^{21}\\times3^{2}\\times7^{2}+1$ • $935329793=2^{22}\\times223+1$ • $943718401=2^{22}\\times3^{2}\\times5^{2}+1$ • $950009857=2^{21}\\times3\\times151+1$ • $962592769=2^{21}\\times3^{3}\\times17+1$ • $975175681=2^{21}\\times3\\times5\\times31+1$ • $985661441=2^{22}\\times5\\times47+1$ • $998244353=2^{23}\\times7\\times17+1$ • $1004535809=2^{21}\\times479+1$ • $1012924417=2^{21}\\times3\\times7\\times23+1$ Kerry Su", "x = 49 mod 127. Note that $51$ is the inverse of $5 \\pmod {127}$ thus $$5x \\equiv 118 \\pmod {127}\\iff51\\cdot5x\\equiv x\\equiv 51\\cdot 118 \\pmod {127}\\iff x\\equiv 49 \\pmod {127}$$ $305\\cdot 5 - 12\\cdot 127 = 1$ $118\\cdot 305\\cdot 5 - 118\\cdot 12\\cdot 127 = 118$ $5\\cdot 35990 \\equiv 118 \\pmod{127}$ Then subtract the biggest multiple of 127 less than 35990 which is $283\\cdot 127 = 35941$ Therefore $x = 49 + 127a$ where $a$ is integer. • Best to use $\\cdot :\\;$ \\cdot instead of $*$. Also, using \\pmod{127} yields $\\pmod{127}$ – Namaste Jan 9 '18 at 0:03 To solve $5x \\equiv 118 \\mod 127$, first find the $\\gcd$ of $5$ and $127$ using the Euclidean method. $$127 = 5 \\times 25 + 2, 5 = 2 \\times 2 + 1$$ this tells us the $\\gcd$ is $1$. Furthermore: $$1 = 5 - 2 \\times 2 = 5 - 2 \\times (127 - 25 \\times 5) = 5 - 2 \\times 127 + 50 \\times 5 = 51 \\times 5 - 2 \\times 127$$ therefore: $$118 = (51 \\times 118) \\times 5 - (2 \\times 118) \\times 127 \\implies (51 \\times 118) \\times 5 \\equiv 118 \\mod 127$$ Therefore, $x = 51 \\times 118$ does the job. Upon a little more simplification:$$51 \\times 118 \\equiv 51 \\times", "before addition, so $2\\times 3^k+3^k$ actually means $(2\\times 3^k)+3^k$. This is $(2\\times 3^k)+(1\\times 3^k)$, which is clearly just $3\\times 3^k$, or $3^{k+1}$. - It looks like you are confusing $2\\cdot 3^k+3^k$ with $2 \\cdot (3^k+3^k)$. The first gives $3 \\cdot 3^k=3^{k+1}$, while the second gives $2 \\cdot (2 \\cdot 3^k)=4\\cdot 3^k$. Parentheses are important. - $2\\times 3^k+3^k=(2+1)\\times3^k=3\\times3^k$ -", "# Fifty factorial 2 Find the integer value of $$n$$ such that $$2^n$$ gives the best estimate of $$50!$$. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "$0.451$ $0.569$ $0.574$ $0.577$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.04\\times10^{-2}$ $1.52\\times10^{-2}$ $3.88\\times10^{-3}$ $4.48\\times10^{-4}$ $\\|M(w_h, \\mathbf{q}_h)\\|_{L^2(Q_T)}$ $1.13\\times10^5$ $4.45\\times10^4$ $1.48\\times10^4$ $2.86\\times10^3$ $\\sharp\\ \\text{triangles}$ $110$ $1197$ $2880$ $8636$ $\\|\\mathbf{q}_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.46$ $0.57$ $0.574$ $0.577$ $\\|v-\\mathbf{q}_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.24\\times10^{-2}$ $1.55\\times10^{-2}$ $3.72\\times10^{-3}$ $5.18\\times10^{-4}$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.451$ $0.569$ $0.574$ $0.577$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $8.04\\times10^{-2}$ $1.52\\times10^{-2}$ $3.88\\times10^{-3}$ $4.48\\times10^{-4}$ $\\|M(w_h, \\mathbf{q}_h)\\|_{L^2(Q_T)}$ $1.13\\times10^5$ $4.45\\times10^4$ $1.48\\times10^4$ $2.86\\times10^3$ Non uniform mesh - Conjugate gradient method - Number of iterates for $r = 1$ (top), $r = 10^{-2}$ and $r = h$ (bottom) $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $31$ $41$ $54$ $77$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.469$ $0.576$ $0.589$ $0.586$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $3.21\\times 10^{-1}$ $1.72\\times 10^{-1}$ $1.43\\times 10^{-1}$ $1.25\\times 10^{-1}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $46$ $103$ $125$ $133$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.55$ $0.566$ $0.569$ $0.571$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $2.05\\times 10^{-1}$ $1.47\\times 10^{-1}$ $1.12\\times 10^{-1}$ $8.71\\times 10^{-2}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $36$ $43$ $56$ $80$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.523$ $0.566$ $0.574$ $0.573$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $2.39\\times 10^{-1}$ $1.46\\times 10^{-1}$ $1.19\\times 10^{-1}$ $9.54\\times 10^{-2}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$ $3.80\\times 10^{-2}$ $1.90\\times 10^{-2}$ $\\sharp$ iterates $31$ $41$ $54$ $77$ $\\|\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $0.469$ $0.576$ $0.589$ $0.586$ $\\|v-\\lambda_h(1, \\cdot)\\|_{L^2(0, T)}$ $3.21\\times 10^{-1}$ $1.72\\times 10^{-1}$ $1.43\\times 10^{-1}$ $1.25\\times 10^{-1}$ $h$ $1.52\\times 10^{-1}$ $7.60\\times 10^{-2}$", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "each addition/subtraction before the next multiplication. On our calculator, it would be: . . $3\\;\\boxed{\\times}\\;2\\;\\boxed{-}\\;7\\;\\boxed{=}\\;\\;\\boxed{\\times}\\;2\\;\\boxed{+}\\;4 \\;\\boxed{=}\\;\\;\\boxed{\\times}\\;2\\;\\boxed{-}\\;9\\;\\boxed{=}\\;\\;\\boxed{\\times}\\;2\\;\\boxed{+}\\;8 \\;\\boxed{=}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Back to base-five . . . Example: Convert $3024_5$ to base-ten. The value is: . $3\\cdot5^3 + 0\\cdot5^2 + 2\\cdot5 + 4$ .and we have a \"polynomial in 5\". So we have: . $3\\cdot5 + 0\\cdot 5 + 2\\cdot 5 + 4\\quad\\Rightarrow\\quad\\begin{array}{cccccc}\\text{3 times 5} \\\\ \\text{plus 0}\\\\ \\text{times 5}\\\\ \\text{plus 2}\\\\ \\text{times 5}\\\\ \\text{plus 4}\\end{array}$ $\\begin{array}{cccccc}15\\\\15\\\\75\\\\77\\\\385\\\\\\boxed{3 89}\\end{array}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ This procedure is even more spectacular in base-two. Example: Write $101011_2$ in base-ten. We have: . $1\\times2 + 0\\times2 + 1\\times 2 +0\\times2 + 1\\times2 + 1$ . . . . . . . . . . . $2\\quad\\,2\\quad\\;4\\quad5\\;\\;\\;10\\;\\;\\;10\\;\\;\\;20\\; \\;\\;21\\;\\;42\\;\\;\\:\\boxed{43} $ You have the answer while others are discovering that it begins with $2^5.$ • Jul 29th 2006, 03:46 PM JakeD Quote: Originally Posted by Soroban But has anyone ever seen this method? It is Horner's Rule. And reading that I see \"The Horner", "(\\bmod 103) \\equiv 2(100)! \\times 1 (\\bmod 103)$ But we know that $(102)! \\equiv 102 (\\bmod 103)$ (1) Therefore $2(100)! \\equiv 102 (\\bmod 103)$ $\\boxed{x=102}$", "2 ) + 1This gives the sequence 1,1,2,3,4,6,9,14... Suppose m box can give out any two amounts summing to at most f(m), then we havef(m) = f(m-1) + floor( f(m-1)/2 ) + 1and f(2)=2, f(3)=4, f(4)=7, f(5)=11, f(6)=17, f(7)=26, f(8)=40,...A good estimate for f(m) would be f(m) ~= 1.5^(m+1), and 1.5^(m+1)=2005 gives us m=17.75. So we'd expect to use something like 18 boxes. Last edited by cuinuc on December 4th, 2006, 11:00 pm, edited 1 time in total.", "373248 = 72^3 = (3\\times 24)^3$ $T_{644} + T_{654} = 421875 = 75^3 = (3\\times 25)^3$ $T_{450} + T_{800} = 421875 = 75^3 = (3\\times 25)^3$", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "3 + 4 = \\boxed{21}$", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "$1.31244559310830 \\times 10^{-31}$ $0$ $1.312 \\times 10^{-31}$ -", "+ 31 = \\boxed{058}$.", "still get a value nonetheless. So for the warmup: Since $7$ and $4$ are relatively prime, we get: $28-7-4=17$. So we can definitely measure all times over $17$ min. Note that as others have computed already, you can measure all values over $6$ min. Easy: Since $5$ and $8$ are relatively prime, we get: $40-5-8 = 27$. So we can definitely measure all times over $27$ min. Medium: Since $6$ and $11$ are relatively prime, we get: $66-6-11 = 49$ So we can definitely measure all times over $49$ min. Hard: If $m$ and $n$ are relatively prime, then we know all numbers above $mn-m-n$ can be measured by the two timers. In general, you can measure all combinations of the following times: 1. $n$ 2. $m$ 3. $2m - n$ 4. $4n - m$ 5. $3m - 2n$ 6. $6n - 2m$ And times $t(x)=min(A \\cdot m,B \\cdot n) + (x \\cdot \\delta)$ where $max(A \\cdot m,B \\cdot n)-min(A \\cdot m,B \\cdot n)=\\delta$ Item 1 is found by starting timer $n$ and noting when expired. Item 2 is found by starting timer $m$ and noting when expired. Item 3 is found by starting timers $m$ and $n$ at the same time. When timer $n$ has expired reset $n$. When timer m has expired flip $n$. This gives", "cancelled by the numerator of the second term and so on… So, the final value = 1/40 $5^{5} \\div (5^{3} \\times 2^{2})$=$5^{2}/2^{2}$ =25/4=6.25 $2^{4} \\times 3^{4} \\times 8^{4} \\div (2 \\times 3 \\times 8)^{3}$ =$\\frac{2^{16} \\times 3^{4}}{2^{12} \\times 3^{3}}=2^{4} \\times 3=48$ 9321+5406+1001= 15728 498+929+660= 2087 15728$\\div$2087= 7.536 So option D is the right answer. $561204\\times58 = ? \\times 55555$ implies ? = $561204\\times58 \\div 55555$ = 586 7857 + 3596 + 4123 = 15576 15576 / 96 = 162.25 $(-216 \\times 1728)^{\\frac{1}{3}}$ =$(-6^{3} \\times 12^{3})^{\\frac{1}{3}}$" ]

[ "information, and, as a result, will facilitate focusing on long-term trends rather than the noise. In [6]: # Creating a column with smoothed values of EUR-USD exchange rates euro_to_dollar['rolling_mean'] = euro_to_dollar['US_dollar'].rolling(30).mean() Now, we'll plot 2 graphs: EUR-USD exchange rate evolution with and without smoothing, to see the difference: In [7]: columns = ['US_dollar', 'rolling_mean'] titles = ['EUR-USD Exchange Rate Evolution\\nNo Rolling Mean', 'EUR-USD Exchange Rate Evolution\\nWith Rolling Mean'] def create_line_plot(df, title, xlabel='Month', column='rolling_mean', title_font=25, label_font=20, tick_font=14, x_min='1999-01-01', x_max='2021-01-08', y_min=None, y_max=None): plt.plot(df['Time'], df[column], color='slateblue') plt.title(title, fontsize=title_font) plt.xlabel(xlabel, fontsize=label_font) plt.ylabel('Exchange rate', fontsize=label_font) plt.xticks(fontsize=tick_font) plt.yticks(fontsize=tick_font) plt.xlim(x_min, x_max) plt.ylim(y_min, y_max) sns.despine() plt.figure(figsize=(12,5)) for i in range(1,3): plt.subplot(1, 2, i) create_line_plot(df=euro_to_dollar, column=columns[i-1], title=titles[i-1], xlabel='Year', title_font=17, label_font=14, tick_font=None, y_min=0.8, y_max=1.62) As we expected, the second graph is much easier to interpret. We can distinguish various features on it: initial decrease lasted up until 2003 and followed by a rapid increase. After a series of going up and down, we observe a clear sharp drop at the end of 2015, followed by some fluctuations at lower levels. Obviously, over the last 22 years, a lot of events happened in the USA both at national and international scales that influenced the EUR-USD exchange rate variation. The graph above can tell us numerous stories, but the one we want to visualize and explore in this project is", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "on your browser. Formulas like and should be aligned with the surrounding text. 3. The first matching end delimiter is always interpreted as the end of the expression. Thus, $\\$$ produces an error but $$\\$$ produces . 4. To produce a dollar sign type [=$=]. This prevents the Wiki engine from interpreting it as part of an inline formula. 5. You must explicitly change the font of inline to match changes in the surrounding text font. For example, if you enter [+++'''Let $A\\in 2^\\omega$ be 1-random.'''+++] then you get Let be 1-random. In order to generate Let be 1-random. you should enter [+++'''Let \\@\\boldmath \\LARGE $A\\in 2^\\omega$\\@ be 1-random.'''+++]. Note that math font information is generally set before entering math mode, so the \\@...\\@ markup is usefull here. Although not perfect, there is a fairly reasonable correspondence between font sizes in and in the text markup on this site. The following tables suggests the code to use in various situations. Matching Font Sizes PmWiki markup Corresponding [+++...+++] \\LARGE [++...++] \\Large [+...+] \\large [-...-] \\footnotesize [--...--] \\scriptsize PmWiki markup Corresponding ! \\boldmath\\LARGE or \\bf\\LARGE !! \\boldmath\\Large or \\bf\\Large !!! \\boldmath\\large or \\bf\\large", "(c) 258.33 (d) 151.33 (e) None of these Explanations- \\begin{align} & \\text{131}.\\text{289} \\\\ & \\,\\,\\,\\text{16}.0\\text{42} \\\\ & \\,\\,\\,\\,\\,\\,\\text{2}.\\text{468} \\\\ & +\\,\\,\\,\\text{8}.\\text{234} \\\\ & \\overline{\\underline{\\text{158}.0\\text{33}}} \\\\ \\end{align} Find the value of A. If A = 331.000 ? 96.042 (a) 344.345 (b) 453.786 (c) 234.958 (d) 234.876 (e) None of these Explanation \\begin{align} & \\,\\,\\,\\text{331}.000 \\\\ & -\\text{ 96}.0\\text{42} \\\\ & \\,\\,\\,\\overline{\\underline{\\text{234}.\\text{958}}} \\\\ \\end{align} $''0.02002\\times {{10}^{4}}=200.2''$In the given expression 0.02002 is multiplied by 104 and the product is 200.2. Jack replaces${{10}^{4}}$P thus he finds the product gets changed and it becomes 2.002. Find the value of P (a) ${{10}^{1}}$ (b) ${{10}^{2}}$ (c)${{10}^{5}}$ (d)${{10}^{3}}$ (e) None of these By what number 4562.2 should be divided to obtain the decimal 4.5622? (a) 10 (b) 100 (c) 1000 (d) 10000 (e) None of these Which one of the following is true for$0.\\text{45}\\times$$\\text{1}00\\times \\text{56}$? (a) 2205 (b) 1825 (c) 2025 (d) 2520 (e) None of these", "chicken but the price was still an integral number in dollar. In the afternoon she could sell all the chickens, and she got totally $\\textdollar 132$ for the whole day. How many chickens were sold in the morning? ## Problem I11 A rectangle $ABCD$ is made up of five small congruent rectangles as shown in the given figure. Find the perimeter, in cm, of $ABCD$ if its area is $6750\\text{ cm}^2$. $[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(3.45cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.75, xmax = 39.09, ymin = -10.43, ymax = 20.84; /* image dimensions */ /* draw figures */ draw((0,3.45)--(0,0)); draw((0,0)--(4.29,0)); draw((0,3.45)--(4.29,3.45)); draw((4.29,3.45)--(4.29,0)); draw((0,1.32)--(4.29,1.32)); draw((2.14,0)--(2.14,1.32)); draw((1.43,1.32)--(1.43,3.45)); draw((2.86,1.32)--(2.86,3.45)); /* dots and labels */ label(\"B\", (-0.2,0), SW * labelscalefactor); label(\"A\", (-0.2,3.45), NW * labelscalefactor); label(\"C\", (4.29,0), SE * labelscalefactor); label(\"D\", (4.29,3.45), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ //Credit to dasobson for the diagram[/asy]$ ## Problem I12 In a die, $1$ and $6$, $2$ and $5$, $3$ and $4$ appear on opposite faces. When $2$ dice are thrown, product of numbers appearing on the", "in Number1 box; specify Constant2 (for example: c. or copy); enter specific numbers separated by commas (for example: 1,2,3,4) or range of numbers (for example: 1-4) in Number2 box. To produce labels for a single item that contains multiple volumes or parts: specify Constant1 (for example: v., vol, part., or pt.); enter specific range of part numbers within quotes (for example: \"2-4\") in Number1 box. To produce labels for multiple volumes published annually: leave Constant1 box blank; enter specific years separated by commas (for example: 1990, 1993, 1995) or a range of years (for example: 1990-1995) or a specific range (for example: 1990,\"1991-1993\",1994-1995). To produce multiple copies of the same label: type the number of copies you want. The default number is 1. ATTENTION: You can not leave Number1 box blank. If you do not need to type any numbers in the Number1 box, enter a space (press the Space key) 9.5.8 Changing font type, size, and style You can change font type, size, and style with the command buttons located under the menus: 1. Select the label data you want to format. 2. Click on the down arrow to the right of the font list to display the list of fonts, and select a font; or click on the down arrow to the right of the size", "(9.5, 2.05){\\rotatebox{-90}{$Government$}}; \\node at (0, 2.2) [align = left, right] {Wages \\ and \\\\ profits}; \\node at (6.3, 2.2) [align = left, left] {Spending \\\\ on \\ goods \\\\ and \\\\ services}; \\draw [very thick] (1.8, 0) -- (4.7, 0) -- (4.7, 1) -- (1.8, 1) -- (1.8, 0); \\draw [very thick] (1.8, 3.4) -- (4.7, 3.4) -- (4.7, 4.4) -- (1.8, 4.4) -- (1.8, 3.4); \\draw [very thick, ->] [loosely dashed] (4.6, 4.4) to [out=100, in=0] (3.2, 5.6) to [out=180, in=90] (1.9, 4.4); \\draw [very thick] (7.3, 0.9) -- (8, 0.9) -- (8, 3.2) -- (7.3, 3.2) -- (7.3, 0.9); \\draw [very thick] (9.1, 0.9) -- (9.8, 0.9) -- (9.8, 3.2) -- (9.1, 3.2) -- (9.1, 0.9); \\draw [very thick, ->] (1.8, 3.9) to [out=-180, in=90] (0, 2.2) to [out=-90, in=180] (1.8, 0.5); \\draw [very thick, ->] (4.7, 0.5) to [out=10, in=-90] (6.5, 2.2) to [out=90, in=-10] (4.7, 3.7); \\draw [very thick, ->] [dashed] (4.7, 0.4) ..controls (6.2, 0.3) and (7.2, 0.4) ..(7.6, 0.9) node [below left] {$S$}; \\draw [very thick, ->] [dashed] (4.7, 0.3) ..controls (7.4, 0.1) and (8.7, 0.3) ..(9.4, 0.9) node [below left] {$T$}; \\draw [very thick, <-] [dashed] (4.7, 3.9) ..controls (6.2, 3.7) and (7.2, 3.5) ..(7.6, 3.2); \\draw [very thick, <-] [dashed] (4.7, 4.1) ..controls (7.4, 3.9) and (8.7, 3.7) ..(9.4, 3.2);", "\\qquad \\textbf{(C)}\\ \\frac 35 \\qquad \\textbf{(D)}\\ \\frac 23 \\qquad \\textbf{(E)}\\ \\frac 34$ ## Problem 18 A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$. And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircle. $[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label(\"A\",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label(\"B\",(2.61,-0.26),NE*lsf); dot((0,4),ds); label(\"D\",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label(\"C\",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}\\ 2:3 \\qquad\\textbf{(B)}\\ 3:2\\qquad\\textbf{(C)}\\ 6:\\pi \\qquad\\textbf{(D)}\\ 9: \\pi \\qquad\\textbf{(E)}\\ 30 : \\pi$ ## Problem 19 The two circles pictured have the same center $C$. Chord $\\overline{AD}$ is tangent to the inner circle at $B$, $AC$ is $10$, and chord $\\overline{AD}$ has length $16$. What is the area between the two circles? $[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label(\"A\", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label(\"B\", (1.97,-0.31),NE*lsf); dot((2,1),ds); label(\"C\", (1.96,1.09),NE*lsf); dot((4,0),ds); label(\"D\", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]$ $\\textbf{(A)}\\ 36 \\pi \\qquad\\textbf{(B)}\\ 49 \\pi\\qquad\\textbf{(C)}\\ 64 \\pi\\qquad\\textbf{(D)}\\ 81 \\pi\\qquad\\textbf{(E)}\\ 100 \\pi$ ## Problem 20 In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is", "Ribbon and change the font style to Arial, change the font size to 10 points, and select the bold and italics commands. 23. Type Dollars in Millions. This tells the audience that the numbers have been truncated and represent denominations in millions. This means you would add six zeros to the end of each number on the chart. Therefore, the Out of Pocket value for 1969 is shown as $22,617 but is actually$22,617,000,000, or 22.6 billion. Figure 4.43 Lines Added to the Stacked Column Chart 24. Repeat steps 19–22 to add a second text box to the chart. Begin drawing this text box below the first box approximately one inch in from the left edge of the chart (see Figure 4.43 \"Lines Added to the Stacked Column Chart\"). Complete the formatting changes in step 22 and select the Align Text Right command. 25. Type 100% = in the second text box. 26. Repeat steps 19–22 to add a third text box to the chart. Center this text box over the 1969 stack. In addition to the formatting commands in step 22, select the Center align command and the Underline command. 27. Type66,172 in the third text box. 28. Repeat steps 19–22 to add a fourth text box to the chart. Center this text box over the 2009 stack.", "30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 % MATLAB 2020b & newer clear OP = py.importlib.import_module('openpyxl'); styles = py.importlib.import_module('openpyxl.styles'); Font = styles.Font; Alignment = styles.Alignment; PatternFill = styles.PatternFill; book = OP.Workbook(); sheet_data = book.create_sheet(\"Profits 2022\", int64(0)); sheet_data.sheet_properties.tabColor = \"15AA15\"; Region = {'N America', 'S America', 'Europe', 'Asia', 'Australia', 'Antarctica'}; nR = int64(length(Region)); profits = 10000 * (0.75 - rand(nR,2)); dollars = '\"$\"#,##0.00;[Red](\"$\"#,##0.00)'; pct = '0.0%;[Red]0.0%'; sheet_data.merge_cells(\"B2:E3\"); B2 = sheet_data.cell(int64(2),int64(2)); B2.value = 'Worldwide Sales'; B2.font = Font(pyargs(\"name\",\"Arial\",\"size\",int64(16),... \"color\",\"4444FF\",\"bold\",py.True)); B2.alignment = Alignment(pyargs(\"horizontal\",\"center\",\"vertical\",\"center\")); row = int64(5); col = int64(2); header_font = Font(pyargs(\"name\",\"Arial\",\"size\",int64(12),\"bold\",py.True)); region_cell = sheet_data.cell(row, int64(2)); % B5 region_cell.value = \"Region\"; region_cell.font = header_font; last_year = sheet_data.cell(row, int64(3)); % C5 last_year.value = \"Last Year\"; last_year.font = header_font; this_year = sheet_data.cell(row, int64(4)); % D5 this_year.value = \"This Year\"; this_year.font = header_font; growth = sheet_data.cell(row, int64(5)); % E5 growth.value = \"Growth\"; growth.font = header_font; rank = sheet_data.cell(row,", "cycle); /* manual arrowhead? avoid resizing */ /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype(\"4 4\")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < n+1; ++j) filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i)); /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label(\"n\",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label(\"1\",shiftR+(n,-n),S,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is $n(n+1)/2$. $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; real r = 0.35, h = 3/4; /* radius size and horizontal spacing */ pair shiftR = (h*(n+1)+r, 0); pen colors(int i){ /* shading */ if(i == n) return red; return rgb(5/n,0.4+5/(2n),1-5/n); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } pair getCenter(int i, int j){ return ((2*j-i)*h,-i);} /* triangle */ for(int i = 0; i < n+1; ++i){ draw((-i*h,-i)--(i*h,-i)); /* horizontal lining */ for(int j = 0; j <= i; ++j) filldraw(circle(getCenter(i,j),r), colors(i)); } /* fill in circle in row 4, column 3 */ filldraw(circle(getCenter(3,2),r),blue); draw(getCenter(n,2)-- getCenter(3,2)-- getCenter(n,n+2-3)); makeshiftarrow(getCenter(n,2),pi/4,0.5); makeshiftarrow(getCenter(n,n+2-3),3*pi/4,0.5); htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),E,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is ${n+1 \\choose 2}$.[1] $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35;", "12 A lucky year is one in which at least one date, when written in the form month/day/year, has the following property: The product of the month times the day equals the last two digits of the year. For example, 1956 is a lucky year because it has the date 7/8/56 and $7\\times 8 = 56$. Which of the following is NOT a lucky year? $\\text{(A)}\\ 1990 \\qquad \\text{(B)}\\ 1991 \\qquad \\text{(C)}\\ 1992 \\qquad \\text{(D)}\\ 1993 \\qquad \\text{(E)}\\ 1994$ ## Problem 13 In the figure, $\\angle A$, $\\angle B$, and $\\angle C$ are right angles. If $\\angle AEB = 40^\\circ$ and $\\angle BED = \\angle BDE$, then $\\angle CDE =$ $[asy] dot((0,0)); label(\"E\",(0,0),SW); dot(dir(85)); label(\"A\",dir(85),NW); dot((4,0)); label(\"D\",(4,0),SE); dot((4.05677,0.648898)); label(\"C\",(4.05677,0.648898),NE); draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle); dot((2,2)); label(\"B\",(2,2),N); draw((0,0)--(2,2)--(4,0)); pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898)); dot(x[0]); dot(x[1]); label(\"F\",x[0],SE); label(\"G\",x[1],SW); [/asy]$ $\\text{(A)}\\ 75^\\circ \\qquad \\text{(B)}\\ 80^\\circ \\qquad \\text{(C)}\\ 85^\\circ \\qquad \\text{(D)}\\ 90^\\circ \\qquad \\text{(E)}\\ 95^\\circ$ ## Problem 14 A team won 40 of its first 50 games. How many of the remaining 40 games must this team win so it will have won exactly 70% of its games for the season? $\\text{(A)}\\ 20 \\qquad \\text{(B)}\\ 23 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 35$ ## Problem 15 What is the $100^\\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$?", "any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad \\text{(D)}\\frac{21}{32}\\qquad \\text{(E)}\\frac{3}{4}$ Problem 26 If the base $8$ representation of a perfect square is $ab3c$, where $a\\ne 0$, then $c$ equals $\\text{(A)} 0\\qquad \\text{(B)}1 \\qquad \\text{(C)} 3\\qquad \\text{(D)} 4\\qquad \\text{(E)} \\text{not uniquely determined}$ Problem 27 Suppose $z=a+bi$ is a solution of the polynomial equation $c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0$, where $c_0, c_1, c_2, c_3, a$, and $b$ are real constants and $i^2=-1$. Which of the following must also be a solution? $\\text{(A)} -a-bi\\qquad \\text{(B)} a-bi\\qquad \\text{(C)} -a+bi\\qquad \\text{(D)}b+ai \\qquad \\text{(E)} \\text{none of these}$ Problem 28 A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\\frac{7}{17}$. What number was erased? $\\text{(A)} 6\\qquad \\text{(B)}7 \\qquad \\text{(C)}8 \\qquad \\text{(D)} 9\\qquad \\text{(E)}\\text{cannot be determined}$ Problem 29 Let $x,y$, and $z$ be three positive real numbers whose sum is $1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$", "\\varsigma ς \\approx \\div ÷ \\tau τ \\Re \\neq \\equiv \\oplus \\aleph \\Im \\cup \\wp \\otimes \\subseteq \\oslash \\cap \\in \\supseteq \\supset \\lceil \\subset \\int \\cdot · \\o ο \\rfloor \\neg ¬ \\nabla \\lfloor \\times x \\ldots ... \\perp \\surd \\prime ´ \\wedge \\varpi ϖ \\0 \\rceil \\rangle \\mid | \\vee \\langle \\copyright #### LaTeX Markup To use LaTeX markup, set the TickLabelInterpreter property to 'latex'. Use dollar symbols around the labels, for example, use '$\\int_1^{20} x^2 dx$' for inline mode or '$$\\int_1^{20} x^2 dx$$' for display mode. The displayed text uses the default LaTeX font style. The FontName, FontWeight, and FontAngle properties do not have an effect. To change the font style, use LaTeX markup within the text. The maximum size of the text that you can use with the LaTeX interpreter is 1200 characters. For multiline text, the maximum size of the text reduces by about 10 characters per line. Tick label format, specified as a character vector or string. The default format is based on the data. Example: ax.XAxis.TickLabelFormat = 'yyyy-MM-dd'; displays a date and time such as 2014-04-19. Example: ax.XAxis.TickLabelFormat = 'eeee, MMMM d, yyyy HH:mm:ss'; displays a date and time such as Saturday, April 19, 2014 21:41:06. Example: ax.XAxis.TickLabelFormat = 'MMMM d, yyyy HH:mm:ss Z'; displays a date and time such as April", "x = 0.65; y = 0.24; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 493; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.24; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents how many dollars you have.\"; //--- ToDo: Localize; }; { idc = 494; text = \"Cost:\"; //--- ToDo: Localize; x = 0.65; y = 0.28; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 495; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.28; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents the cost of the item selected in dollars.\"; //--- ToDo: Localize; }; { idc = 496; text = \"Total:\"; //--- ToDo: Localize; x = 0.65; y = 0.32; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 497; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.32; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents much money you", "rate for an online store of CDs. Let’s further study the distribution of total repeat times of the customers. # get the matrix of customer ID ~ the customer’s total number of transactions > TimesByID <-as.data.frame(table(df$ID)) #get the matrix of total number of transactions ~ number of customers who have the total number > GroupByTimes <- as.data.frame(table(TimesByID$Freq)) > head(GroupByTimes,12) Times Customers 1 1 1205 2 2 406 3 3 208 4 4 150 5 5 98 6 6 56 7 7 65 8 8 35 9 9 23 10 10 21 11 11 8 12 12 10 >plot(GroupByTimes,xlab=”Total Number of Purchases”,ylab=”Number of Customers”,pch=16,col=”blue”,type=”o”) > text(2,1220,”1205″) > text(3,425,”406″) > text(4,220,”208″) > text(5,170,”150″) > text(6,120,”98″) > text(12,50,”10″) > text(30,50,”1″) Figure – 1 As we can see from Figure – 1 above, the number of customers decreases very quickly while the total number of purchases increases from 1 to 6. Almost of half of the customers only made one purchase during the 1.5 year period! Let’s examine the percentage of customers making (x) purchases more closely. > percentages<-round(GroupByTimes$Customers / 2357 , 3) > percentages [1] 0.511 0.172 0.088 0.064 0.042 0.024 0.028 0.015 0.010 0.009 [11] 0.003 0.004 0.006 0.003 0.003 0.003 0.004 0.001 0.001 0.000 [21] 0.001 0.002 0.002 0.000 0.000 0.000 0.000 0.000 0.000 0.000 [31] 0.000 0.000 0.001", "\\times \\text{12})$$. In this case, $$n$$ represents the number of years; the product $$(n \\times \\text{12})$$ represents the number of times the bank pays interest into the account. \\begin{align*} \\text{619,09} & = \\text{450} \\left( 1 + \\frac{\\text{0,0711}}{\\text{12}} \\right) ^ {(n \\times \\text{12})} \\end{align*} \\begin{align*} \\text{619,09} & = \\text{450}(\\text{1,00592} \\ldots)^{\\text{12}n} \\\\ \\frac{\\text{619,09}}{\\text{450}} & = (\\text{1,00592} \\ldots)^{\\text{12}n} \\\\ \\text{1,37575} \\ldots & = (\\text{1,00592} \\ldots )^{\\text{12}n} \\end{align*} At this point we must change the equation to logarithmic form: \\begin{align*} \\text{12}n & = \\log_{\\text{1,005925}} (\\text{1,37575} \\ldots ) \\\\ \\text{12}n & = \\text{54} \\end{align*} To find the number of years, we solve for $$n$$: \\begin{align*} \\text{12}n & = \\text{54} \\\\ n & = \\frac{\\text{54}}{\\text{12}} \\\\ n & = \\text{4,5} \\end{align*} The money has been in the Simosethu's account for $$\\text{4,5}$$ years.", "# Get the data data = data.get_data_yahoo(ticker, start_date, end_date) Out[2]: High Low Open Close Volume Adj Close Date 1997-05-15 2.500000 1.927083 2.437500 1.958333 72156000.0 1.958333 1997-05-16 1.979167 1.708333 1.968750 1.729167 14700000.0 1.729167 1997-05-19 1.770833 1.625000 1.760417 1.708333 6106800.0 1.708333 1997-05-20 1.750000 1.635417 1.729167 1.635417 5467200.0 1.635417 1997-05-21 1.645833 1.375000 1.635417 1.427083 18853200.0 1.427083 In [3]: import matplotlib.pyplot as plt %matplotlib inline plt.show() In [4]: # Plot the adjusted close price # Define the label for the title of the figure plt.title(\"Adjusted Close Price of %s\" % ticker, fontsize=16) # Define the labels for x-axis and y-axis plt.ylabel('Price', fontsize=14) plt.xlabel('Year', fontsize=14) # Plot the grid lines plt.grid(which=\"major\", color='k', linestyle='-.', linewidth=0.5) # Show the plot plt.show() In [5]: # Plot the adjusted close price # Define the label for the title of the figure plt.title(\"Adjusted Close Price of %s\" % ticker, fontsize=16) # Define the labels for x-axis and y-axis plt.ylabel('Price', fontsize=14) plt.xlabel('Year', fontsize=14) # Plot the grid lines plt.grid(which=\"major\", color='k', linestyle='-.', linewidth=0.5) # Show the plot plt.show() # Ride Santa Barbara 100, a climb too far Well, I actually tried the RIDESB100 this weekend. There where four routes: 39 miles, 100 km, 100 km with Gibraltar, and the century. I did not finish the ride. At around 43 miles and two miles to the top of Gibraltar, my legs", "Modified commas an periods by forester2015 */ /* Import and Set variables */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */ /* draw figures */ draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1)); draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1)); draw((10,4)--(14,4), linewidth(1)); draw((4,6)--(8,6), linewidth(1)); draw((4,2)--(8,2), linewidth(1)); draw((8,2)--(8,6), linewidth(1)); draw((4,6)--(4,2), linewidth(1)); draw((6,6)--(6,2), linewidth(1)); draw((-6,6)--(-6,2), linewidth(1)); draw((-6,6)--(2,6), linewidth(1)); draw((2,6)--(2,2), linewidth(1)); draw((2,2)--(-6,2), linewidth(1)); draw((-4,2)--(-4,6), linewidth(1)); draw((-2,6)--(-2,2), linewidth(1)); draw((0,2)--(0,6), linewidth(1)); draw((50,6)--(50,2), linewidth(1)); draw((50,2)--(58,2), linewidth(1)); draw((58,2)--(58,6), linewidth(1)); draw((58,6)--(50,6), linewidth(1)); draw((52,6)--(52,2), linewidth(1)); draw((54,6)--(54,2), linewidth(1)); draw((56,6)--(56,2), linewidth(1)); draw((32,6)--(32,2), linewidth(1)); draw((46,2)--(46,6), linewidth(1)); draw((34,6)--(34,2), linewidth(1)); draw((36,2)--(36,6), linewidth(1)); draw((38,6)--(38,2), linewidth(1)); draw((40,2)--(40,6), linewidth(1)); draw((42,6)--(42,2), linewidth(1)); draw((44,2)--(44,6), linewidth(1)); draw((16,6)--(16,2), linewidth(1)); draw((28,2)--(28,6), linewidth(1)); draw((18,6)--(18,2), linewidth(1)); draw((20,6)--(20,2), linewidth(1)); draw((22,6)--(22,2), linewidth(1)); draw((24,6)--(24,2), linewidth(1)); draw((26,6)--(26,2), linewidth(1)); draw((16,6)--(22,6), linewidth(1)); draw((24,6)--(28,6), linewidth(1)); draw((16,2)--(22,2), linewidth(1)); draw((24,2)--(28,2), linewidth(1)); draw((32,6)--(36,6), linewidth(1)); draw((32,2)--(36,2), linewidth(1)); draw((38,6)--(40,6), linewidth(1)); draw((38,2)--(40,2), linewidth(1)); draw((42,6)--(46,6), linewidth(1)); draw((42,2)--(46,2), linewidth(1)); /* dots and labels */ label(\",\",(59,2)); label(\".\",(60,2)); label(\".\",(61,2)); label(\".\",(62,2)); label(\",\",(29,2)); label(\",\",(47,2)); [/asy]$ Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth? $\\textbf{(A)} ~(6, 1, 1) \\qquad\\textbf{(B)} ~(6, 2, 1) \\qquad\\textbf{(C)} ~(6, 2, 2) \\qquad\\textbf{(D)} ~(6, 3, 1) \\qquad\\textbf{(E)} ~(6, 3, 2)$", "level p','color','k','fontsize',18); text(1,0.05,'Interest rate i','color','b','fontsize',18); print -depsc2 sticky_inflation.eps; print -dpng sticky_inflation.png; % price stickiness and policy [xt pit vt qt it rnt pt] = solveit(sig,kap, bet, thpi, thx, rho, omeg,T); figure; hold on; plot(tim,it,'-b','linewidth',2); plot(tim, pit, '-rv','linewidth',2); plot(tim, pt, '--k','linewidth',2); plot(tim,0*tim,'-k'); xlabel('Time'); ylabel('Percent'); axis([0 7 -0.1 1.1]) text(2,0.3,'Inflation \\pi','color','r','fontsize',18); text(3.5,0.75,'Price level p','color','k','fontsize',18); text(5,1,['p(\\infty) = ' num2str(pt(end),'%4.2f')],'color','k','fontsize',18); text(1,0.15,'Interest rate i','color','b','fontsize',18); print -depsc2 policy_inflation.eps; print -dpng policy_inflation.png; disp('total inflation'); disp(pt(end)); %****** function [xt pit vt qt it rnt pt] = solveit(sig,kap, bet, thpi, thx, rho, omeg,T); show_results = 1; N = 4; A = [... 1 sig 0 0 ; 0 bet 0 0 ; 0 1 rho -omeg ; 0 0 0 omeg ]; % x pi q ui us B = [ ... 1+sig*thx sig*thpi 0 0 ; -kap 1 0 0 ; 0 0 1 -1 ; thx thpi 0 1 ]; C = [... 0 ; 0 ; 1 ; 0 ]; D = [... 1 sig 0 ; 0 bet 0 ; 0 0 0 ; 0 0 omeg ]; % Solve by eigenvalues taken from ftmp_model_final A1 = inv(A); F = A1*B; [Q L] = eig(F); Q1 = inv(Q); if show_results; disp('Eigenvalues'); disp(abs(diag(L)')); end % produce Ef, Eb, that select forward and backward % eigvenvalues. If L>=1 in position 1, 3, % produce %" ]

[ "the patient’s temperature showed an upward trend. ## Chapter 15 Ex.15.1 Question 2 The following line graph shows the yearly sales figure for a manufacturing company. (a) What were the sales in (i) $$2002$$ (ii) $$2006$$? (b) What were the sales in (i) $$2003$$ (ii) $$2005$$? (c) Compute the difference between the sales in $$2002$$ and $$2006.$$ (d) In which year was there the greatest difference between the sales as compared to its previous year? ### Solution Reasoning: Horizontal axis represents ‘Years’ and vertical represents ‘Sales’ in Rs crore. We can find the value of ‘Sales’ on vertical axis according to the given value of ‘Years’ on the horizontal axis value or vice versa. Steps: (a) (i) In $$2002,$$ the sales were $$\\rm{Rs}\\; 4$$ crores. (ii) In $$2006,$$ the sales were $$\\rm{Rs}\\; 8$$ crores. (b) (i) In $$2003,$$ the sales were $$\\rm{Rs}\\; 7$$ crores. (ii) In $$2005,$$ the sales were $$\\rm{Rs}\\; 10$$ crores. (c) (i) In $$2002,$$ the sales were $$\\rm{Rs}\\; 4$$ crores and in $$2006,$$ the sales were $$\\rm{Rs}\\; 8$$ crores. Difference between the sales in $$2002$$ and $$2006$$ $$= \\rm{Rs} \\;(8 − 4)$$ crores $$=$$ $$\\rm{Rs}\\; 4$$ crores (d) Difference between the sales of the year $$2006$$ and $$2005$$ $$= \\rm{Rs}\\; (10 − 8)$$ crores $$=\\rm{ Rs}\\; 2$$ crores Difference between the sales of the year", "charts is 2 boxes and there are 2 examples where the difference between the two sales charts is 3 boxes. Since you are looking for the largest percent increase, you only need to compare the largest percent increase from each of these two sets. The amazing thing here is that you can know which chart has the largest percent increase in each of those two sets without actually doing any computation since the percent increase will be largest when the starting point is the smallest. So, just eyeballing the charts now we can see that February’s percent increase (an increase of about 66%) is the one that we need to compare to March’s (an increase of 50%) and we see that February had the largest percent increase. Again, we can see that February has the largest percent increase without calculating what the percent increase was in January, April, or May. An amazing idea when you are seeing it for the first time – or at least that’s the impression I got this morning! Finally an old V.I. Arnold problem that I saw on Tanya Kovanova’s blog a while ago. Unfortunately her blog has been damaged by a spam attack so the problem isn’t up right now. Hopefully she’s able to save the blog. Now, this problem shows basically", "the dollar sales level necessary to provide the 10 percent increase in profits, assuming that the maximum price increase is implemented c. If the volume of sales were to remain at 87,000 units, what price change would be required to attain the 10 percent increase in profits? Calculate the new price.", "Chart represents annual sales in 2016, broken down by four product lines. The right-hand chart has the same breakdown, but for 2017. This provides some context to our discussions. Suppose what is of interest is how the sales for each product line in the 2016 chart compare to their counterparts in the right-hand one; e.g. A and A’, B and B’ and so on. Well for the As, we have the helpful fact that they both start from a vertical line and then swing down and round, initially rightwards. This can be used to gauge that A’ is a bit bigger than A. What about B and B’? Well they start in different places and end in different places, looking carefully, we can see that B’ is bigger than B. C and C’ are pretty easy, C is a lot bigger. Then we come to D and D’, I find this one a bit tricky, but we can eventually hazard a guess that they are pretty much the same. So we can compare Pie Charts and talk about how sales change between two years, what’s the problem? The issue is that it takes some time and effort to reach even these basic conclusions. How about instead of working out which is bigger, A or A’, I ask the", "by both the companies is made in 2009 and the least profit in 2004. Consider statement A: M made the highest profit in 2009. This statement is true. Consider statement B: L made the least profit in 2005. This statement is false. So, the answer is option a) (Since option e) is not true anymore).", "# Q.2 The following line graph shows the yearly sales figure for a manufacturing company. (c) Compute the difference between sales in 2002 and 2006 (c) The difference between sales in 2002 and 2006 $=8-4 =4$.", "## Thinking Mathematically (6th Edition) a. The point on the axis representing the percentage in 2001 looks to be a bit after the middle between $5\\%$ and $10\\%$, so we can estimate that it's $\\approx 8\\%$. b. To see between which two years the percentage increased at the greatest rate, we don't need to estimate the value for each point and calculate the difference between them. We simply need to see between which two years the slope is the steepest and heading upwards. We can see that the steepest slope in the graph is between the years 2005 and 2006; therefore, the percentage decreased the most between 2005 and 2006. c. Looking at the points on the graph for each year and their distance from the dotted line for $10\\%$, we can see that the only point that could be at $12\\%$ is 2004.", "is calculated in million dollars. Solution: The double line graph of above data is given below: By comparing both the lines, we can easily say that overall profit of company A is greater than that of company B and profits of both the companies were equal in year 2004-05. ## Interpreting Line Graphs Line graph gives a lot of information about the data. Line graph can be used for interpreting various results and outcomes based on which different strategies can be generated. However, interpretation of a line graph depends upon the presentation of the data. Few interpretations are listed below: • Determination of dependent and independent variable. • Determination of item measured and its unit. • Determination of rate of change of an item. • Comparison of two or more data sets in case of double line and multiple line graphs. • Estimating peaks or ups and valleys or downs. • Forecasting future results. ## Line Graph Examples Few examples based on line graphs are given below: ### Solved Examples Question 1: Construct a line graph for the information given below and answer the following questions by the graph: Months January February March April May June Number of cars sold 20 35 25 40 45 30 • Which month were the highest number of cars sold? • Which", "The price of products may increase due to inflation and decrease due to depreciation. Derek is studying the change in the price of two products, A and B, over time. The price f(x), in dollars, of product A after x years is represented by the function below: f(x) =... Using the data shown on the graph, which statements are correct? $Using the data shown on the graph, which statements are correct?$...", "# Easy Bar Charts Solved QuestionData Interpretation Discussion Common Information Study the bar chart and answer the questions. Q. Common Information Question: 4/4 The percentage increases in sales from 2001 to 2002 was? ✖ A. 115% ✖ B. 128% ✔ C. 122% ✖ D. 118% Solution: Option(C) is correct The percentage increase exhibited is: $=\\left[\\dfrac{40 - 18}{18} × 100\\right]\\%$ = 122% approximately", "any reason for the year being squared? #### skeeter ##### Elite Member to make sure we get this correct, please verify what both axes of the graph represent with regard to units ... x axis has what units? y axis has what units? #### casherr ##### New member x is years y is the income per person in dollars #### skeeter ##### Elite Member x is years y is the income per person in dollars then the slope would be (income per person) per year, $$\\displaystyle m = \\dfrac{\\text{income/person}}{\\text{year}}$$", "the number of sales in the following years: (i) 2012 (ii) 2014 (b) Give the number of sales in the following years: (i) 2011 (ii) 2015 What is the difference between the number of sales done in the year 2012 and 2015? (d) In which year was the difference greatest in the numbers of sales compared to its previous year? (2011- 5, 2012-8, 2013-7, 2014-11, 2015-9) Ans.) (a) The numbers of sales: (i) The sales were Rs. 8 crores, in 2012. (ii) The sales were Rs. 11 crores, in 2014. (b) The numbers of sales: (i) The sales were Rs. 5 crores, in 2011. (ii) The sales were Rs. 9 crores, in 2015. I The sales were Rs. 8 crores, in 2012 and the sales were Rs. 9 crores, in 2015. Therefore, the difference between the sales in 2012 and 2015 = Rs. (9 – 8) crores = 1 crore (d) The difference between the sales in 2015 and 2014 = Rs. (11 – 9) crores = Rs. 2 crores The difference between the sales in 2014 and 2013 = Rs. (11 – 7) crores = Rs. 4 crores The difference between the sales in 2013 and 2012 = Rs. (8 – 7) crores = Rs. 1 crore The difference between the sales in 2012 and 2011 =", "## me123456789 2 years ago if the sales of a company increased exponentially by 50% from 1985 to 1990, by what percent will the sales increase from 1990 to 1996? 1. mubzz 2. me123456789 no 3. Kainui Another business came around in 1993 and wiped out their profits so they went bankrupt and had no sales increase. 4. Kainui So let's consider the exponential equation. What does it look like? $F(t)=Ae^{rt}$ So that tells us that A is our initial amount and F(t) is our final amount. Since we don't know the amounts but we do know the percents, you can divide out A to get a fraction, which translates to your percentage! 5. me123456789 can you show it? i cant figure this one out 6. PaxPolaris assuming they continue the same trend... you can PREDICT how much the sales will increase It increases by 1.5 times ... every 5 years .... IF you have a general formula for exponential function... you can plug these values in 7. PaxPolaris Using the formula above: ( i substituted $$\\large c=e^r$$) $\\large F(t)= A \\cdot c^t$$\\implies \\large 1.5 \\cancel A= \\cancel A \\cdot c^5$$\\implies \\Large c=\\sqrt[5]{1.5}$ 8. PaxPolaris So in 6 years ... it should increase by c^6 times : $\\Large = \\left( \\sqrt[5]{1.5} \\right)^6=1.5^\\left( 6/5 \\right)$ 9. PaxPolaris 62.67%", "if previously they were counting charity toys as sold and accounting for its sale, then changing that policy will increase there revenues and will actively resolve the paradox. b)Toys4Them sold a higher proportion of more expensive toys last year than the previous year. --- This resolves the paradox if last year they hold more expensive toys than previous year, the increase revenue could be explained given the fact the total number of toys sold did not increased. c)Last year, the number of consumers shopping for toys increased over the previous year. --- hmnnn. Classic example of what i mentioned above.If an answer supports or proves only one side of the paradox, that answer will be incorrect. The correct answer must show how both sides coexist. This only explains what may have caused the increase in revenue but does not address the fact that number of toys sold did not increased significantly. Correct Answer d)Last year, Toys4Them experienced an unprecedented boom in its divisions that do not sell toys. --- This again holds both side of the conversation. If this is true then the increased revenue could be from this division and not from the sale of the toys. e)Because of an economic downturn, Toys4Them heavily discounted its toys during the holiday season two years ago--- This hold both", "that DOL is higher at lower level of sales and it declines as sales increase.", "228 views A line graph on a graph sheet shows the revenue for each year from $1990$ through $1998$ by points and joints the successive points by straight line segments. The point for revenue of $1990$ is labeled $\\text{A}$, that for $1991$ as $\\text{B}$, and that for $1992$ as $\\text{C}$. What is the ratio of growth in revenue between $1991-92$ and $1990-91?$ Statement I: The angle between $\\text{AB}$ and $x$- axis when measured with a protractor is $40$ degrees, and the angle between $\\text{CB}$ and $x$-axis is $80$ degrees. Statement II: The scale of $y$-axis is $1$ cm = $1000$. 1", "revenue, not a net increase. The correct answer is: \"The company's revenue generally decreased from 1994 to 1996.\" Notice the data points matching the years 1994, 1995, and 1996 are linked by a line that has a downward slope. This indicates a general decrease in revenue. Example Question #2 : Recognize Possible Associations And Trends In Data An individual's income over 10 years is represented by the following line chart. Approximate individual's highest annual income from 1994 to 1998. $500,000$460,000 $210,000$500 $460 Correct answer:$460,000 Explanation: Notice that the question narrows the time period you are examining to 1994–1998. While there are years with higher incomes than in this time period (namely, 1992 and 1999), we can only consider the years from 1994 to 1998. The highest point within this period is in the year 1998. Matching this data point with the values on the y-axis (the income values), you will see that this data point lies between $400,000 and$500,000. Looking more closely, you can even say that it is above halfway between the values, so it appears to be greater than $450,000. Thus, the value that approximates the income in 1998 is$460,000. Be careful to take into account the units on each axis. The y-axis measures income in thousands of dollars, so the answer is $460,000 and not$460.", "### Home > INT1 > Chapter A > Lesson A.1.3 > ProblemA-35 A-35. The two lines at right represent the growing profits of Companies A and B. 1. Sketch this graph on your paper. If Company A started out with more profit than Company B, determine which line represents A and which represents B. Label the lines appropriately. Company A's line should have a greater y-intercept, since they started with more profit than Company B. 1. Are the profits of Company A a proportional situation? Why or why not? 2. In how many years will both companies have the same profit? Find where the two lines intersect. Determine the x-coordinate for this point. $3$ years 3. Approximately what will that profit be? The profit is the y-coordinate of the intercept. Each tick mark represents $10{,}000$. 4. Which company's profits are growing more quickly? How can you tell? Growth is related to how steep a line is. Which line is steeper?", "here and it relates to what we said Pie Charts show earlier in this piece. What we have in our two Pie Charts above is the make-up of a whole number (in the example we have been working through, this is total annual sales) by categories (product lines). These are percentages and what we have been doing above is to compare the fact that A made up 30% of the total sales in 2016 and 33% in 2017. What we cannot say based on just the above exhibits is how actual sales changed. The total sales may have gone up or down, the Pie Chat does not tell us this, it just deals in how the make-up of total sales has shifted. Some people try to address this shortcoming, which can result in exhibits such as: Here some attempt has been made to show the growth in the absolute value of sales year on year. The left-hand Pie Chart is smaller and so we assume that annual sales have increased between 2016 and 2017. The most logical thing to do would be to have the change in total area of the two Pie Charts to be in proportion to the change in sales between the two years (in this case – based on the underlying data – 2017", "pennies made in 2015 than there are people in the world! Or we could use powers of 10 to write these numbers: $$\\displaystyle 7.4 \\boldcdot 10^9$$ for people in the world and $$\\displaystyle 8.9 \\boldcdot 10^9$$ for the number of pennies. For a visual representation, we could plot these two numbers on a number line. We need to carefully choose our end points to make sure that the numbers can both be plotted. Since they both lie between $$10^9$$ and $$10^{10}$$, if we make a number line with tick marks that increase by one billion, or $$10^9$$, we start the number line with 0 and end it with $$10 \\boldcdot 10^{9}$$, or $$10^{10}$$. Here is a number line with the number of pennies and world population plotted:" ]

[ "information, and, as a result, will facilitate focusing on long-term trends rather than the noise. In [6]: # Creating a column with smoothed values of EUR-USD exchange rates euro_to_dollar['rolling_mean'] = euro_to_dollar['US_dollar'].rolling(30).mean() Now, we'll plot 2 graphs: EUR-USD exchange rate evolution with and without smoothing, to see the difference: In [7]: columns = ['US_dollar', 'rolling_mean'] titles = ['EUR-USD Exchange Rate Evolution\\nNo Rolling Mean', 'EUR-USD Exchange Rate Evolution\\nWith Rolling Mean'] def create_line_plot(df, title, xlabel='Month', column='rolling_mean', title_font=25, label_font=20, tick_font=14, x_min='1999-01-01', x_max='2021-01-08', y_min=None, y_max=None): plt.plot(df['Time'], df[column], color='slateblue') plt.title(title, fontsize=title_font) plt.xlabel(xlabel, fontsize=label_font) plt.ylabel('Exchange rate', fontsize=label_font) plt.xticks(fontsize=tick_font) plt.yticks(fontsize=tick_font) plt.xlim(x_min, x_max) plt.ylim(y_min, y_max) sns.despine() plt.figure(figsize=(12,5)) for i in range(1,3): plt.subplot(1, 2, i) create_line_plot(df=euro_to_dollar, column=columns[i-1], title=titles[i-1], xlabel='Year', title_font=17, label_font=14, tick_font=None, y_min=0.8, y_max=1.62) As we expected, the second graph is much easier to interpret. We can distinguish various features on it: initial decrease lasted up until 2003 and followed by a rapid increase. After a series of going up and down, we observe a clear sharp drop at the end of 2015, followed by some fluctuations at lower levels. Obviously, over the last 22 years, a lot of events happened in the USA both at national and international scales that influenced the EUR-USD exchange rate variation. The graph above can tell us numerous stories, but the one we want to visualize and explore in this project is", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "on your browser. Formulas like and should be aligned with the surrounding text. 3. The first matching end delimiter is always interpreted as the end of the expression. Thus, $\\$$ produces an error but $$\\$$ produces . 4. To produce a dollar sign type [=$=]. This prevents the Wiki engine from interpreting it as part of an inline formula. 5. You must explicitly change the font of inline to match changes in the surrounding text font. For example, if you enter [+++'''Let $A\\in 2^\\omega$ be 1-random.'''+++] then you get Let be 1-random. In order to generate Let be 1-random. you should enter [+++'''Let \\@\\boldmath \\LARGE $A\\in 2^\\omega$\\@ be 1-random.'''+++]. Note that math font information is generally set before entering math mode, so the \\@...\\@ markup is usefull here. Although not perfect, there is a fairly reasonable correspondence between font sizes in and in the text markup on this site. The following tables suggests the code to use in various situations. Matching Font Sizes PmWiki markup Corresponding [+++...+++] \\LARGE [++...++] \\Large [+...+] \\large [-...-] \\footnotesize [--...--] \\scriptsize PmWiki markup Corresponding ! \\boldmath\\LARGE or \\bf\\LARGE !! \\boldmath\\Large or \\bf\\Large !!! \\boldmath\\large or \\bf\\large", "(c) 258.33 (d) 151.33 (e) None of these Explanations- \\begin{align} & \\text{131}.\\text{289} \\\\ & \\,\\,\\,\\text{16}.0\\text{42} \\\\ & \\,\\,\\,\\,\\,\\,\\text{2}.\\text{468} \\\\ & +\\,\\,\\,\\text{8}.\\text{234} \\\\ & \\overline{\\underline{\\text{158}.0\\text{33}}} \\\\ \\end{align} Find the value of A. If A = 331.000 ? 96.042 (a) 344.345 (b) 453.786 (c) 234.958 (d) 234.876 (e) None of these Explanation \\begin{align} & \\,\\,\\,\\text{331}.000 \\\\ & -\\text{ 96}.0\\text{42} \\\\ & \\,\\,\\,\\overline{\\underline{\\text{234}.\\text{958}}} \\\\ \\end{align} $''0.02002\\times {{10}^{4}}=200.2''$In the given expression 0.02002 is multiplied by 104 and the product is 200.2. Jack replaces${{10}^{4}}$P thus he finds the product gets changed and it becomes 2.002. Find the value of P (a) ${{10}^{1}}$ (b) ${{10}^{2}}$ (c)${{10}^{5}}$ (d)${{10}^{3}}$ (e) None of these By what number 4562.2 should be divided to obtain the decimal 4.5622? (a) 10 (b) 100 (c) 1000 (d) 10000 (e) None of these Which one of the following is true for$0.\\text{45}\\times$$\\text{1}00\\times \\text{56}$? (a) 2205 (b) 1825 (c) 2025 (d) 2520 (e) None of these", "chicken but the price was still an integral number in dollar. In the afternoon she could sell all the chickens, and she got totally $\\textdollar 132$ for the whole day. How many chickens were sold in the morning? ## Problem I11 A rectangle $ABCD$ is made up of five small congruent rectangles as shown in the given figure. Find the perimeter, in cm, of $ABCD$ if its area is $6750\\text{ cm}^2$. $[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(3.45cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.75, xmax = 39.09, ymin = -10.43, ymax = 20.84; /* image dimensions */ /* draw figures */ draw((0,3.45)--(0,0)); draw((0,0)--(4.29,0)); draw((0,3.45)--(4.29,3.45)); draw((4.29,3.45)--(4.29,0)); draw((0,1.32)--(4.29,1.32)); draw((2.14,0)--(2.14,1.32)); draw((1.43,1.32)--(1.43,3.45)); draw((2.86,1.32)--(2.86,3.45)); /* dots and labels */ label(\"B\", (-0.2,0), SW * labelscalefactor); label(\"A\", (-0.2,3.45), NW * labelscalefactor); label(\"C\", (4.29,0), SE * labelscalefactor); label(\"D\", (4.29,3.45), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ //Credit to dasobson for the diagram[/asy]$ ## Problem I12 In a die, $1$ and $6$, $2$ and $5$, $3$ and $4$ appear on opposite faces. When $2$ dice are thrown, product of numbers appearing on the", "in Number1 box; specify Constant2 (for example: c. or copy); enter specific numbers separated by commas (for example: 1,2,3,4) or range of numbers (for example: 1-4) in Number2 box. To produce labels for a single item that contains multiple volumes or parts: specify Constant1 (for example: v., vol, part., or pt.); enter specific range of part numbers within quotes (for example: \"2-4\") in Number1 box. To produce labels for multiple volumes published annually: leave Constant1 box blank; enter specific years separated by commas (for example: 1990, 1993, 1995) or a range of years (for example: 1990-1995) or a specific range (for example: 1990,\"1991-1993\",1994-1995). To produce multiple copies of the same label: type the number of copies you want. The default number is 1. ATTENTION: You can not leave Number1 box blank. If you do not need to type any numbers in the Number1 box, enter a space (press the Space key) 9.5.8 Changing font type, size, and style You can change font type, size, and style with the command buttons located under the menus: 1. Select the label data you want to format. 2. Click on the down arrow to the right of the font list to display the list of fonts, and select a font; or click on the down arrow to the right of the size", "(9.5, 2.05){\\rotatebox{-90}{$Government$}}; \\node at (0, 2.2) [align = left, right] {Wages \\ and \\\\ profits}; \\node at (6.3, 2.2) [align = left, left] {Spending \\\\ on \\ goods \\\\ and \\\\ services}; \\draw [very thick] (1.8, 0) -- (4.7, 0) -- (4.7, 1) -- (1.8, 1) -- (1.8, 0); \\draw [very thick] (1.8, 3.4) -- (4.7, 3.4) -- (4.7, 4.4) -- (1.8, 4.4) -- (1.8, 3.4); \\draw [very thick, ->] [loosely dashed] (4.6, 4.4) to [out=100, in=0] (3.2, 5.6) to [out=180, in=90] (1.9, 4.4); \\draw [very thick] (7.3, 0.9) -- (8, 0.9) -- (8, 3.2) -- (7.3, 3.2) -- (7.3, 0.9); \\draw [very thick] (9.1, 0.9) -- (9.8, 0.9) -- (9.8, 3.2) -- (9.1, 3.2) -- (9.1, 0.9); \\draw [very thick, ->] (1.8, 3.9) to [out=-180, in=90] (0, 2.2) to [out=-90, in=180] (1.8, 0.5); \\draw [very thick, ->] (4.7, 0.5) to [out=10, in=-90] (6.5, 2.2) to [out=90, in=-10] (4.7, 3.7); \\draw [very thick, ->] [dashed] (4.7, 0.4) ..controls (6.2, 0.3) and (7.2, 0.4) ..(7.6, 0.9) node [below left] {$S$}; \\draw [very thick, ->] [dashed] (4.7, 0.3) ..controls (7.4, 0.1) and (8.7, 0.3) ..(9.4, 0.9) node [below left] {$T$}; \\draw [very thick, <-] [dashed] (4.7, 3.9) ..controls (6.2, 3.7) and (7.2, 3.5) ..(7.6, 3.2); \\draw [very thick, <-] [dashed] (4.7, 4.1) ..controls (7.4, 3.9) and (8.7, 3.7) ..(9.4, 3.2);", "\\qquad \\textbf{(C)}\\ \\frac 35 \\qquad \\textbf{(D)}\\ \\frac 23 \\qquad \\textbf{(E)}\\ \\frac 34$ ## Problem 18 A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$. And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircle. $[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label(\"A\",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label(\"B\",(2.61,-0.26),NE*lsf); dot((0,4),ds); label(\"D\",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label(\"C\",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$ $\\textbf{(A)}\\ 2:3 \\qquad\\textbf{(B)}\\ 3:2\\qquad\\textbf{(C)}\\ 6:\\pi \\qquad\\textbf{(D)}\\ 9: \\pi \\qquad\\textbf{(E)}\\ 30 : \\pi$ ## Problem 19 The two circles pictured have the same center $C$. Chord $\\overline{AD}$ is tangent to the inner circle at $B$, $AC$ is $10$, and chord $\\overline{AD}$ has length $16$. What is the area between the two circles? $[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label(\"A\", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label(\"B\", (1.97,-0.31),NE*lsf); dot((2,1),ds); label(\"C\", (1.96,1.09),NE*lsf); dot((4,0),ds); label(\"D\", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]$ $\\textbf{(A)}\\ 36 \\pi \\qquad\\textbf{(B)}\\ 49 \\pi\\qquad\\textbf{(C)}\\ 64 \\pi\\qquad\\textbf{(D)}\\ 81 \\pi\\qquad\\textbf{(E)}\\ 100 \\pi$ ## Problem 20 In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is", "Ribbon and change the font style to Arial, change the font size to 10 points, and select the bold and italics commands. 23. Type Dollars in Millions. This tells the audience that the numbers have been truncated and represent denominations in millions. This means you would add six zeros to the end of each number on the chart. Therefore, the Out of Pocket value for 1969 is shown as $22,617 but is actually$22,617,000,000, or 22.6 billion. Figure 4.43 Lines Added to the Stacked Column Chart 24. Repeat steps 19–22 to add a second text box to the chart. Begin drawing this text box below the first box approximately one inch in from the left edge of the chart (see Figure 4.43 \"Lines Added to the Stacked Column Chart\"). Complete the formatting changes in step 22 and select the Align Text Right command. 25. Type 100% = in the second text box. 26. Repeat steps 19–22 to add a third text box to the chart. Center this text box over the 1969 stack. In addition to the formatting commands in step 22, select the Center align command and the Underline command. 27. Type66,172 in the third text box. 28. Repeat steps 19–22 to add a fourth text box to the chart. Center this text box over the 2009 stack.", "30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 % MATLAB 2020b & newer clear OP = py.importlib.import_module('openpyxl'); styles = py.importlib.import_module('openpyxl.styles'); Font = styles.Font; Alignment = styles.Alignment; PatternFill = styles.PatternFill; book = OP.Workbook(); sheet_data = book.create_sheet(\"Profits 2022\", int64(0)); sheet_data.sheet_properties.tabColor = \"15AA15\"; Region = {'N America', 'S America', 'Europe', 'Asia', 'Australia', 'Antarctica'}; nR = int64(length(Region)); profits = 10000 * (0.75 - rand(nR,2)); dollars = '\"$\"#,##0.00;[Red](\"$\"#,##0.00)'; pct = '0.0%;[Red]0.0%'; sheet_data.merge_cells(\"B2:E3\"); B2 = sheet_data.cell(int64(2),int64(2)); B2.value = 'Worldwide Sales'; B2.font = Font(pyargs(\"name\",\"Arial\",\"size\",int64(16),... \"color\",\"4444FF\",\"bold\",py.True)); B2.alignment = Alignment(pyargs(\"horizontal\",\"center\",\"vertical\",\"center\")); row = int64(5); col = int64(2); header_font = Font(pyargs(\"name\",\"Arial\",\"size\",int64(12),\"bold\",py.True)); region_cell = sheet_data.cell(row, int64(2)); % B5 region_cell.value = \"Region\"; region_cell.font = header_font; last_year = sheet_data.cell(row, int64(3)); % C5 last_year.value = \"Last Year\"; last_year.font = header_font; this_year = sheet_data.cell(row, int64(4)); % D5 this_year.value = \"This Year\"; this_year.font = header_font; growth = sheet_data.cell(row, int64(5)); % E5 growth.value = \"Growth\"; growth.font = header_font; rank = sheet_data.cell(row,", "cycle); /* manual arrowhead? avoid resizing */ /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype(\"4 4\")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < n+1; ++j) filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i)); /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label(\"n\",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label(\"1\",shiftR+(n,-n),S,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is $n(n+1)/2$. $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; real r = 0.35, h = 3/4; /* radius size and horizontal spacing */ pair shiftR = (h*(n+1)+r, 0); pen colors(int i){ /* shading */ if(i == n) return red; return rgb(5/n,0.4+5/(2n),1-5/n); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } pair getCenter(int i, int j){ return ((2*j-i)*h,-i);} /* triangle */ for(int i = 0; i < n+1; ++i){ draw((-i*h,-i)--(i*h,-i)); /* horizontal lining */ for(int j = 0; j <= i; ++j) filldraw(circle(getCenter(i,j),r), colors(i)); } /* fill in circle in row 4, column 3 */ filldraw(circle(getCenter(3,2),r),blue); draw(getCenter(n,2)-- getCenter(3,2)-- getCenter(n,n+2-3)); makeshiftarrow(getCenter(n,2),pi/4,0.5); makeshiftarrow(getCenter(n,n+2-3),3*pi/4,0.5); htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),E,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is ${n+1 \\choose 2}$.[1] $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35;", "12 A lucky year is one in which at least one date, when written in the form month/day/year, has the following property: The product of the month times the day equals the last two digits of the year. For example, 1956 is a lucky year because it has the date 7/8/56 and $7\\times 8 = 56$. Which of the following is NOT a lucky year? $\\text{(A)}\\ 1990 \\qquad \\text{(B)}\\ 1991 \\qquad \\text{(C)}\\ 1992 \\qquad \\text{(D)}\\ 1993 \\qquad \\text{(E)}\\ 1994$ ## Problem 13 In the figure, $\\angle A$, $\\angle B$, and $\\angle C$ are right angles. If $\\angle AEB = 40^\\circ$ and $\\angle BED = \\angle BDE$, then $\\angle CDE =$ $[asy] dot((0,0)); label(\"E\",(0,0),SW); dot(dir(85)); label(\"A\",dir(85),NW); dot((4,0)); label(\"D\",(4,0),SE); dot((4.05677,0.648898)); label(\"C\",(4.05677,0.648898),NE); draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle); dot((2,2)); label(\"B\",(2,2),N); draw((0,0)--(2,2)--(4,0)); pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898)); dot(x[0]); dot(x[1]); label(\"F\",x[0],SE); label(\"G\",x[1],SW); [/asy]$ $\\text{(A)}\\ 75^\\circ \\qquad \\text{(B)}\\ 80^\\circ \\qquad \\text{(C)}\\ 85^\\circ \\qquad \\text{(D)}\\ 90^\\circ \\qquad \\text{(E)}\\ 95^\\circ$ ## Problem 14 A team won 40 of its first 50 games. How many of the remaining 40 games must this team win so it will have won exactly 70% of its games for the season? $\\text{(A)}\\ 20 \\qquad \\text{(B)}\\ 23 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 35$ ## Problem 15 What is the $100^\\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$?", "any given morning, he goes through $C$. $[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label(\"A\", (6,19), SE); label(\"B\", (23,4), NW); label(\"C\", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label(\"N\", (28,13.5), N); label(\"W\", (26,11.5), W); label(\"E\", (30,11.5), E); label(\"S\", (28,9.5), S);[/asy]$ $\\text{(A)}\\frac{11}{32}\\qquad \\text{(B)}\\frac{1}{2}\\qquad \\text{(C)}\\frac{4}{7}\\qquad \\text{(D)}\\frac{21}{32}\\qquad \\text{(E)}\\frac{3}{4}$ Problem 26 If the base $8$ representation of a perfect square is $ab3c$, where $a\\ne 0$, then $c$ equals $\\text{(A)} 0\\qquad \\text{(B)}1 \\qquad \\text{(C)} 3\\qquad \\text{(D)} 4\\qquad \\text{(E)} \\text{not uniquely determined}$ Problem 27 Suppose $z=a+bi$ is a solution of the polynomial equation $c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0$, where $c_0, c_1, c_2, c_3, a$, and $b$ are real constants and $i^2=-1$. Which of the following must also be a solution? $\\text{(A)} -a-bi\\qquad \\text{(B)} a-bi\\qquad \\text{(C)} -a+bi\\qquad \\text{(D)}b+ai \\qquad \\text{(E)} \\text{none of these}$ Problem 28 A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\\frac{7}{17}$. What number was erased? $\\text{(A)} 6\\qquad \\text{(B)}7 \\qquad \\text{(C)}8 \\qquad \\text{(D)} 9\\qquad \\text{(E)}\\text{cannot be determined}$ Problem 29 Let $x,y$, and $z$ be three positive real numbers whose sum is $1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$", "\\varsigma ς \\approx \\div ÷ \\tau τ \\Re \\neq \\equiv \\oplus \\aleph \\Im \\cup \\wp \\otimes \\subseteq \\oslash \\cap \\in \\supseteq \\supset \\lceil \\subset \\int \\cdot · \\o ο \\rfloor \\neg ¬ \\nabla \\lfloor \\times x \\ldots ... \\perp \\surd \\prime ´ \\wedge \\varpi ϖ \\0 \\rceil \\rangle \\mid | \\vee \\langle \\copyright #### LaTeX Markup To use LaTeX markup, set the TickLabelInterpreter property to 'latex'. Use dollar symbols around the labels, for example, use '$\\int_1^{20} x^2 dx$' for inline mode or '$$\\int_1^{20} x^2 dx$$' for display mode. The displayed text uses the default LaTeX font style. The FontName, FontWeight, and FontAngle properties do not have an effect. To change the font style, use LaTeX markup within the text. The maximum size of the text that you can use with the LaTeX interpreter is 1200 characters. For multiline text, the maximum size of the text reduces by about 10 characters per line. Tick label format, specified as a character vector or string. The default format is based on the data. Example: ax.XAxis.TickLabelFormat = 'yyyy-MM-dd'; displays a date and time such as 2014-04-19. Example: ax.XAxis.TickLabelFormat = 'eeee, MMMM d, yyyy HH:mm:ss'; displays a date and time such as Saturday, April 19, 2014 21:41:06. Example: ax.XAxis.TickLabelFormat = 'MMMM d, yyyy HH:mm:ss Z'; displays a date and time such as April", "x = 0.65; y = 0.24; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 493; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.24; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents how many dollars you have.\"; //--- ToDo: Localize; }; { idc = 494; text = \"Cost:\"; //--- ToDo: Localize; x = 0.65; y = 0.28; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 495; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.28; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents the cost of the item selected in dollars.\"; //--- ToDo: Localize; }; { idc = 496; text = \"Total:\"; //--- ToDo: Localize; x = 0.65; y = 0.32; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; }; { idc = 497; text = \"\"; //--- ToDo: Localize; x = 0.75; y = 0.32; w = 0.0875; h = 0.04; colorText[] = {1,1,1,1}; colorBackground[] = {0.33,0.55,0.25,0}; colorActive[] = {0.33,0.55,0.25,0}; tooltip = \"This number represents much money you", "rate for an online store of CDs. Let’s further study the distribution of total repeat times of the customers. # get the matrix of customer ID ~ the customer’s total number of transactions > TimesByID <-as.data.frame(table(df$ID)) #get the matrix of total number of transactions ~ number of customers who have the total number > GroupByTimes <- as.data.frame(table(TimesByID$Freq)) > head(GroupByTimes,12) Times Customers 1 1 1205 2 2 406 3 3 208 4 4 150 5 5 98 6 6 56 7 7 65 8 8 35 9 9 23 10 10 21 11 11 8 12 12 10 >plot(GroupByTimes,xlab=”Total Number of Purchases”,ylab=”Number of Customers”,pch=16,col=”blue”,type=”o”) > text(2,1220,”1205″) > text(3,425,”406″) > text(4,220,”208″) > text(5,170,”150″) > text(6,120,”98″) > text(12,50,”10″) > text(30,50,”1″) Figure – 1 As we can see from Figure – 1 above, the number of customers decreases very quickly while the total number of purchases increases from 1 to 6. Almost of half of the customers only made one purchase during the 1.5 year period! Let’s examine the percentage of customers making (x) purchases more closely. > percentages<-round(GroupByTimes$Customers / 2357 , 3) > percentages [1] 0.511 0.172 0.088 0.064 0.042 0.024 0.028 0.015 0.010 0.009 [11] 0.003 0.004 0.006 0.003 0.003 0.003 0.004 0.001 0.001 0.000 [21] 0.001 0.002 0.002 0.000 0.000 0.000 0.000 0.000 0.000 0.000 [31] 0.000 0.000 0.001", "\\times \\text{12})$$. In this case, $$n$$ represents the number of years; the product $$(n \\times \\text{12})$$ represents the number of times the bank pays interest into the account. \\begin{align*} \\text{619,09} & = \\text{450} \\left( 1 + \\frac{\\text{0,0711}}{\\text{12}} \\right) ^ {(n \\times \\text{12})} \\end{align*} \\begin{align*} \\text{619,09} & = \\text{450}(\\text{1,00592} \\ldots)^{\\text{12}n} \\\\ \\frac{\\text{619,09}}{\\text{450}} & = (\\text{1,00592} \\ldots)^{\\text{12}n} \\\\ \\text{1,37575} \\ldots & = (\\text{1,00592} \\ldots )^{\\text{12}n} \\end{align*} At this point we must change the equation to logarithmic form: \\begin{align*} \\text{12}n & = \\log_{\\text{1,005925}} (\\text{1,37575} \\ldots ) \\\\ \\text{12}n & = \\text{54} \\end{align*} To find the number of years, we solve for $$n$$: \\begin{align*} \\text{12}n & = \\text{54} \\\\ n & = \\frac{\\text{54}}{\\text{12}} \\\\ n & = \\text{4,5} \\end{align*} The money has been in the Simosethu's account for $$\\text{4,5}$$ years.", "# Get the data data = data.get_data_yahoo(ticker, start_date, end_date) Out[2]: High Low Open Close Volume Adj Close Date 1997-05-15 2.500000 1.927083 2.437500 1.958333 72156000.0 1.958333 1997-05-16 1.979167 1.708333 1.968750 1.729167 14700000.0 1.729167 1997-05-19 1.770833 1.625000 1.760417 1.708333 6106800.0 1.708333 1997-05-20 1.750000 1.635417 1.729167 1.635417 5467200.0 1.635417 1997-05-21 1.645833 1.375000 1.635417 1.427083 18853200.0 1.427083 In [3]: import matplotlib.pyplot as plt %matplotlib inline plt.show() In [4]: # Plot the adjusted close price # Define the label for the title of the figure plt.title(\"Adjusted Close Price of %s\" % ticker, fontsize=16) # Define the labels for x-axis and y-axis plt.ylabel('Price', fontsize=14) plt.xlabel('Year', fontsize=14) # Plot the grid lines plt.grid(which=\"major\", color='k', linestyle='-.', linewidth=0.5) # Show the plot plt.show() In [5]: # Plot the adjusted close price # Define the label for the title of the figure plt.title(\"Adjusted Close Price of %s\" % ticker, fontsize=16) # Define the labels for x-axis and y-axis plt.ylabel('Price', fontsize=14) plt.xlabel('Year', fontsize=14) # Plot the grid lines plt.grid(which=\"major\", color='k', linestyle='-.', linewidth=0.5) # Show the plot plt.show() # Ride Santa Barbara 100, a climb too far Well, I actually tried the RIDESB100 this weekend. There where four routes: 39 miles, 100 km, 100 km with Gibraltar, and the century. I did not finish the ride. At around 43 miles and two miles to the top of Gibraltar, my legs", "Modified commas an periods by forester2015 */ /* Import and Set variables */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */ /* draw figures */ draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1)); draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1)); draw((10,4)--(14,4), linewidth(1)); draw((4,6)--(8,6), linewidth(1)); draw((4,2)--(8,2), linewidth(1)); draw((8,2)--(8,6), linewidth(1)); draw((4,6)--(4,2), linewidth(1)); draw((6,6)--(6,2), linewidth(1)); draw((-6,6)--(-6,2), linewidth(1)); draw((-6,6)--(2,6), linewidth(1)); draw((2,6)--(2,2), linewidth(1)); draw((2,2)--(-6,2), linewidth(1)); draw((-4,2)--(-4,6), linewidth(1)); draw((-2,6)--(-2,2), linewidth(1)); draw((0,2)--(0,6), linewidth(1)); draw((50,6)--(50,2), linewidth(1)); draw((50,2)--(58,2), linewidth(1)); draw((58,2)--(58,6), linewidth(1)); draw((58,6)--(50,6), linewidth(1)); draw((52,6)--(52,2), linewidth(1)); draw((54,6)--(54,2), linewidth(1)); draw((56,6)--(56,2), linewidth(1)); draw((32,6)--(32,2), linewidth(1)); draw((46,2)--(46,6), linewidth(1)); draw((34,6)--(34,2), linewidth(1)); draw((36,2)--(36,6), linewidth(1)); draw((38,6)--(38,2), linewidth(1)); draw((40,2)--(40,6), linewidth(1)); draw((42,6)--(42,2), linewidth(1)); draw((44,2)--(44,6), linewidth(1)); draw((16,6)--(16,2), linewidth(1)); draw((28,2)--(28,6), linewidth(1)); draw((18,6)--(18,2), linewidth(1)); draw((20,6)--(20,2), linewidth(1)); draw((22,6)--(22,2), linewidth(1)); draw((24,6)--(24,2), linewidth(1)); draw((26,6)--(26,2), linewidth(1)); draw((16,6)--(22,6), linewidth(1)); draw((24,6)--(28,6), linewidth(1)); draw((16,2)--(22,2), linewidth(1)); draw((24,2)--(28,2), linewidth(1)); draw((32,6)--(36,6), linewidth(1)); draw((32,2)--(36,2), linewidth(1)); draw((38,6)--(40,6), linewidth(1)); draw((38,2)--(40,2), linewidth(1)); draw((42,6)--(46,6), linewidth(1)); draw((42,2)--(46,2), linewidth(1)); /* dots and labels */ label(\",\",(59,2)); label(\".\",(60,2)); label(\".\",(61,2)); label(\".\",(62,2)); label(\",\",(29,2)); label(\",\",(47,2)); [/asy]$ Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth? $\\textbf{(A)} ~(6, 1, 1) \\qquad\\textbf{(B)} ~(6, 2, 1) \\qquad\\textbf{(C)} ~(6, 2, 2) \\qquad\\textbf{(D)} ~(6, 3, 1) \\qquad\\textbf{(E)} ~(6, 3, 2)$", "level p','color','k','fontsize',18); text(1,0.05,'Interest rate i','color','b','fontsize',18); print -depsc2 sticky_inflation.eps; print -dpng sticky_inflation.png; % price stickiness and policy [xt pit vt qt it rnt pt] = solveit(sig,kap, bet, thpi, thx, rho, omeg,T); figure; hold on; plot(tim,it,'-b','linewidth',2); plot(tim, pit, '-rv','linewidth',2); plot(tim, pt, '--k','linewidth',2); plot(tim,0*tim,'-k'); xlabel('Time'); ylabel('Percent'); axis([0 7 -0.1 1.1]) text(2,0.3,'Inflation \\pi','color','r','fontsize',18); text(3.5,0.75,'Price level p','color','k','fontsize',18); text(5,1,['p(\\infty) = ' num2str(pt(end),'%4.2f')],'color','k','fontsize',18); text(1,0.15,'Interest rate i','color','b','fontsize',18); print -depsc2 policy_inflation.eps; print -dpng policy_inflation.png; disp('total inflation'); disp(pt(end)); %****** function [xt pit vt qt it rnt pt] = solveit(sig,kap, bet, thpi, thx, rho, omeg,T); show_results = 1; N = 4; A = [... 1 sig 0 0 ; 0 bet 0 0 ; 0 1 rho -omeg ; 0 0 0 omeg ]; % x pi q ui us B = [ ... 1+sig*thx sig*thpi 0 0 ; -kap 1 0 0 ; 0 0 1 -1 ; thx thpi 0 1 ]; C = [... 0 ; 0 ; 1 ; 0 ]; D = [... 1 sig 0 ; 0 bet 0 ; 0 0 0 ; 0 0 omeg ]; % Solve by eigenvalues taken from ftmp_model_final A1 = inv(A); F = A1*B; [Q L] = eig(F); Q1 = inv(Q); if show_results; disp('Eigenvalues'); disp(abs(diag(L)')); end % produce Ef, Eb, that select forward and backward % eigvenvalues. If L>=1 in position 1, 3, % produce %" ]

[ "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "q); label(\"P\", P, dir(250), q); label(\"Q\", Q, 0.25*(Q-P), q); label(\"\\sqrt{3}\",B--Q, dir(60), p); label(\"\\sqrt{3}\",P--B, dir(210), p); label(\"\\sqrt{3}\",P--Q, dir(135), p); label(\"2\",P--C, dir(120), p); label(\"2\",A--Q, dir(-20), p); label(\"1\",A--P, dir(210), p); [/asy]$ Note that $\\triangle QAP$ is a $30^\\circ$-$60^\\circ$-$90^\\circ$ triangle, hence $\\angle PQA=30^\\circ$, and $\\angle BQA=90^\\circ$, so $$AB^2=PQ^2+AQ^2=2^2+3=7,$$ and the answer is $\\boxed{\\textbf{(B) } \\sqrt{7}}$. ## Solution 2 (Intuition) $[asy] unitsize(1inch); pen p = fontsize(10pt); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label(\"A\", (0,0), SW, p); label(\"C\", (2.646,0), SE, p); label(\"B\", (1.323,2.291), N, p); label(\"P'\", (0.756,0.655), NW, p); label(\"1\", (2.174,0.164), N, p); label(\"1\", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]$ Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments). Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to", "C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label(\"A\",A,(1,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0));label(\"P\",P,(0,-1));label(\"Q\",Q,(1,0));label(\"R\",R,(-1,1)); [/asy]$", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "of $n$ and three times the $n-1$th triangular number. If $P_n$ denotes the $n$th pentagonal number, then $P_n = 3T_{n-1}+n$. $[asy]defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); int n = 5, fib = 1, fib2 = 1, xsum = 1, ysum = 0; real h = 0.15; void fillsq(pair A = (0,0), real s, pen p = invisible, pen l = linewidth(1)){ filldraw(shift(A)*xscale(s)*yscale(s)*unitsquare, p, l); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < n; ++i) { fillsq((0,h*ysum),h*fib2,rgb(0.9,1,0.9)); fillsq((h*xsum,0),h*fib,rgb(1,0.9,0.9)); if(i == n-1){ label(\"F_{n}^2\",h*(xsum+fib/2,fib/2),sm); label(\"F_{n-1}^2\",h*(fib2/2,ysum+fib2/2),sm); } else if(i == n-2){ label(\"F_{n-2}^2\",h*(xsum+fib/2,fib/2),sm); label(\"F_{n-3}^2\",h*(fib2/2,ysum+fib2/2),sm); } fib = fib + fib2; fib2 = fib - fib2; xsum = fib; ysum = fib2; fib = fib + fib2; fib2 = fib - fib2; } htick(h*(xsum,0)+(1,0),h*(xsum,ysum)+(1,0)); label(\"F_n\",h*(xsum,ysum/2)+(1,0), E, sm); htick(h*(0,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label(\"F_{n-1}\",h*((xsum-fib+fib2)/2,ysum)+(0,1), N, sm); htick(h*(xsum,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label(\"F_{n}\",h*((2*xsum-fib+fib2)/2,ysum)+(0,1), N, sm); [/asy]$ The identity $F_1^2 + F_2^2 + \\cdots + F_n^2 = F_{n} \\cdot F_{n+1}$, where $F_i$ is the $i$th Fibonacci number. ### Geometric Series $[asy]defaultpen(linewidth(0.7)); unitsize(15); int n = 10; /* # of iterations */ real s = 6; /* square size */ pair shiftR = (s+2,0); pen sm = fontsize(10); void fillrect(pair A, pair B = (0,0), pen p = invisible, pen l = linewidth(1)){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, l); } void htick(pair A, pair B,", "+0 0 92 1 Let $X$, $Y$, and $Z$ be points on a circle. Let $\\overline{XY}$ and the tangent to the circle at $Z$ intersect at $W$. If $WX = 4$, $WZ = 8$, and $\\overline{WY} \\perp \\overline{WZ}$, then find $YZ$. [asy] unitsize(2 cm); pair A, B, C, D; D = (0,1); B = dir(150); C = dir(210); A = (-sqrt(3)/2,1); draw(Circle((0,0),1)); draw(D--A--C); label(\"$W$\", A, NW); label(\"$X$\", B, SE); label(\"$Y$\", C, SW); label(\"$Z$\", D, N); label(\"$8$\", (A + D)/2, N); label(\"$4$\", (A + B)/2, W); [/asy] Feb 14, 2020", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "QR but not on the actual segment or in the rectangle. $[asy] size(200); pair A,B,C,Q,R,S; real r = (pi/2 - asin(1/5))/2; A = (0,0); B = 40*dir(r*180/pi); C = 31*dir(90-r*180/pi); draw(A--B--C--cycle); Q = (40*cos(r),0); R = (40*cos(r),31*cos(r)); S = (0, 31*cos(r)); draw(A--Q--R--S--cycle); label(\"A\",A,SW); label(\"B\",B,NE); label(\"C\",C,N); label(\"Q\",Q,SE); label(\"R\",R,E); label(\"S\",S,NW); [/asy]$ The actual answer is a radical near $728$ (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer $744$ despite the invalid problem statement.", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "# 1974 USAMO Problems/Problem 5 ## Problem Consider the two triangles $\\triangle ABC$ and $\\triangle PQR$ shown in Figure 1. In $\\triangle ABC$, $\\angle ADB = \\angle BDC = \\angle CDA = 120^\\circ$. Prove that $x=u+v+w$. $[asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label(\"A\",A,N); label(\"B\",B,ESE); label(\"C\",C,SW); label(\"D\",D,NE); label(\"a\",(B+C)/2,S); label(\"b\",(C+A)/2,WNW); label(\"c\",(A+B)/2,NE); label(\"u\",(A+D)/2,E); label(\"v\",(B+D)/2,N); label(\"w\",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NE); label(\"x\",(P+Q)/2,S); label(\"x\",(Q+R)/2,E); label(\"x\",(R+P)/2,W); label(\"a\",(P+M)/2,SE); label(\"b\",(Q+M)/2,N); label(\"c\",(R+M)/2,E); label(\"Figure 1\",P-(0,1),S); [/asy]$ ## Solutions ### Solution 1 We rotate figure $PRQM$ by a clockwise angle of $\\pi/3$ about $Q$ to obtain figure $RR'QM'$: $[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NW); label(\"R'\",RR,NE); label(\"M'\",MM,ESE); [/asy]$ Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then $$[ABC] + \\frac{b^2 \\sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .$$ Then by symmetry, $$3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2\\bigl( [PMQ] + [QMR] + [RMP] \\bigr) = 2 [PRQ] .$$ But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}$. Therefore, the area of $ABC$ is $$\\frac{(wv+vu+uw)\\sqrt{3}}{4} .$$ Also, by the Law", "draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\\frac{1^2\\sqrt{3}}{4}\\implies\\frac{1\\sqrt{3}}{4}$, and $\\frac{2^2\\sqrt{3}}{4}\\implies\\frac{4\\sqrt{3}}{4}\\implies\\sqrt{3}$ and the area of the large equilateral triangle is $\\frac{7^2\\sqrt{3}}{4}\\implies\\frac{49\\sqrt{3}}{4}$ so the area of the hexagon would be $\\frac{49\\sqrt{3}}{4}-\\frac{\\sqrt {3}}{4}-\\frac{\\sqrt{3}}{4}-\\sqrt{3}\\implies\\boxed{\\frac{43\\sqrt{3}}{4}}$ Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can", "1996 AJHSME Problems/Problem 5 Problem 5 The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown. $[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label(\"-3\",(-3,-0.1),S); label(\"-2\",(-2,-0.1),S); label(\"-1\",(-1,-0.1),S); label(\"0\",(0,-0.1),S); label(\"1\",(1,-0.1),S); label(\"2\",(2,-0.1),S); label(\"3\",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label(\"P\",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label(\"Q\",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label(\"R\",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label(\"S\",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label(\"T\",(1.5,0.25),N); [/asy]$ Which of the following expressions represents a negative number? $\\text{(A)}\\ P-Q \\qquad \\text{(B)}\\ P\\cdot Q \\qquad \\text{(C)}\\ \\dfrac{S}{Q}\\cdot P \\qquad \\text{(D)}\\ \\dfrac{R}{P\\cdot Q} \\qquad \\text{(E)}\\ \\dfrac{S+T}{R}$ Solution 1 First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\\boxed{A}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in", "+0 I need help asap! 0 557 1 Points $A$, $B$, and $C$ are on sides $\\overline{WX}$, $\\overline{YZ}$, and $\\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA = 4$, $YB = 18$, $\\angle ACB= 90^\\circ$, and $YC = 2XC$. Find $AB$. [asy] pair A,B,C,D,P,Q,R; A = (0,0); P = (0,0.7); B = (0,1); R = (0.4,1); C = (1,1); Q = (1,0.2); D = (1,0); draw(P--B--C--D--A--P--R--Q--P); draw(rightanglemark(P,R,Q,2)); label(\"$W$\", A, SW); label(\"$Z$\",D,SE); label(\"$X$\",B,NW); label(\"$Y$\",C,NE); label(\"$C$\",R,N); label(\"$A$\",P,W); label(\"$B$\",Q,E); label(\"$4$\", (B+P)/2,W); label(\"$18$\", (C+Q)/2,E); [/asy] Sorry if the latex is hard to read Feb 16, 2020", "Find the remainder when $N$ is divided by 144. (Note: $\\binom{a}{b} = 0$ if $a, and $\\binom{0}{0} = 1$.) ### Geometry As in the following diagram, square $ABCD$ and square $CEFG$ are placed side by side (i.e. $C$ is between $B$ and $E$ and $G$ is between $C$ and $D$). If $CE = 14$, $AB > 14$, compute the minimal area of $\\triangle AEG$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(real x, real y) { pair P = (x,y); dot(P,linewidth(3)); return P; } int big = 30, small = 14; filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5)); draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare); label(\"A\",D2(0,big),NW); label(\"B\",D2(0,0),SW); label(\"C\",D2(big,0),SW); label(\"D\",D2(big,big),N); label(\"E\",D2(big+small,0),SE); label(\"F\",D2(big+small,small),NE); label(\"G\",D2(big,small),NE); [/asy]$ In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\\sqrt{c}$. Find $\\frac{abc}{4}$. Triangle $ABC$ has $AB = 4$, $AC = 5$, and $BC = 6$. An angle bisector is drawn from angle $A$, and meets $BC$ at $M$. What is the nearest integer to $100 \\frac{AM}{CM}$? In regular", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "$MN$ down so that $N$ coincides with $B$, and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$, the longer leg is the same as $MN=x$, and the hypotenuse is $AB=12$. We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$, and solve), you find that $x=\\sqrt{130}$, so $x^2 = \\boxed{130}$. Solution by Kinglogic ### Full Proof that R, A, B are collinear $[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); label(\"T\", t , NW); [/asy]$ Let $M$ and $N$ be the feet of the perpendicular from $A$ to $PQ$ and $B$ to $PR$ respectively. It is well known that a perpendicular from the center of a circle to a chord of that circle bisects the chord, so $QM = MP = PN = NR$, since the problem told us $QP = PR$. We will show that $R$ lies on $AB$. Let $T$ be the intersection of circle centered at $B$", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The" ]

[ "45-45-90 triangle.", "# Question 1 of 52 The figure below is not drawn to scale. ABCD is a square. XYZ is a right-angled isosceles triangle of area 180cm$^2$. Find the area of Square ABCD. A 46cm B 48cm C 50cm D 52cm E None of the above", "\\cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \\boxed{660}$.", "= 90^\\circ$, when the isosceles triangle will be a 45-45-90 triangle. However, when $\\theta > 90^\\circ$, a different picture will be needed. I’ll consider this in tomorrow’s post.", "− xyz^2, if x = − 2, y = 1, and z = 3? 9 11 Apr 2017, 22:59 34 XYZ is a right angled triangle with YZ = 6cm and XY = 8cm. XZ is the h 8 05 Mar 2017, 06:42 7 If w=(1+x*y/z^2)(x+y), x=z/2, and y=3z/4, what is the value 10 16 Dec 2015, 19:13 Display posts from previous: Sort by", "# 45-45-90 Triangle Side Ratios Showing the ratios of the sides of a 45-45-90 triangle are 1:1:sqrt(2)", "Squares $$AXYZ$$ and $$A_{1}X_{1}Y_{1}Z_{1}$$ are inscribed in the two congruent right triangle $$ABC$$ and $$PQR$$, as shown below. Find $$AB+AC$$ if $$[AXYZ]=441$$ and $$[A_{1}X_{1}Y_{1}Z_{1}]=440$$. (第五届AIME1987 第15题)", "292 views In a triangle $XYZ$, $P$ and $Q$ are points on ${XY,XZ}$ respectively such that $XP=2PY$, $XQ=2QZ$, then the ratio, of area of $\\triangle XPQ$ and area of $\\triangle XYZ$ is: 1. $4:9$ 2. $2:3$ 3. $3:2$ 4. $9:4$ Correct option is A) Given, XP=2PY, XQ=2QZ and the included angle is equal. Hence the triangles XPQ and XYZ are similar with their sides in the ratio of 2:3. Thus the ratio of areas of the triangles is 4:9 378 points 1 259 views", "right triangle, and $a^2+b^2 = 64$. So $(a+b)^2 = 64+2(24) = 112$, so $2(a+b) = \\sqrt{\\boxed{448}}$.", "triangle is $\\frac{1}{2}$ of the parallelogram, so the answer is $\\boxed{(A) \\frac{4}{9}}.$ By babyzombievillager", "+0 # help geometry 0 109 1 Triangle WYZ is equilateral and triangle WXY is isosceles with a vertex angle of angle XWY. If the perimeter of the entire figure is 28 units and the area of the entire figure is 9 \\sqrt{3} + 4 \\sqrt{7}, what are the side lengths of each side of each triangle? Feb 14, 2022", "of the triangle $XYZ$ is $28cm$. Use this method to calculate the perimeter of any triangle in mathematical system. Save (or) Share", "is the ratios between the sides of a 30-60-90 triangle. Contact Supplier ## Request for Quotation You can get the price list and we will contact you within one business day.", "# Q. In triangle XYZ, $\\angle$XZY= 90$°$, XY = z units, YZ = x units, ZX = y units. ZD is perpendicular to XY and ZD = p units. Prove that : $\\frac{1}{{p}^{2}}=\\frac{1}{{x}^{2}}+\\frac{1}{{y}^{2}}$ This question has not been answered yet! What are you looking for?", "$\\triangle PQR$ in the y direction is $2$ and $MP=MQ=\\sqrt{2}$, so the answer is $\\boxed{\\textbf{(C)}\\ 2 \\sqrt{2}}$ ~Geometry285", "# Triangles in 3D! $O,X,Y,Z$ are four points on a 3D Cartesian space, where $O$ is the origin, $X$ a point on the $x$-axis, $Y$ a point on the $y$-axis, and $Z$ a point on the $z$-axis. Now, the area of $\\triangle OXY$ is given by $a$, the area of $\\triangle OYZ$ by $b$, and the area of $\\triangle OXZ$ by $c$. Find the area of $\\triangle XYZ$ in terms of $a,b,c$. ×", "of Triangle XYZ? [#permalink] Show Tags 22 Apr 2018, 17:05 Bunuel wrote: What is the area of Triangle XYZ? A. 4√3 B. 8 C. 8√3 D. 16 E. 16√3 Attachment: TriangleXYZ.png Watch for special triangles, especially if you see $$\\sqrt{3}$$ or $$\\sqrt{2}$$ in answer choices. Rule: A right triangle whose hypotenuse is twice the length of one of its legs is a 30-60-90 triangle 30-60-90 triangles have corresponding sides opposite those angles in ratio $$x : x\\sqrt{3}: 2x$$ $$2x$$ , opposite the 90° angle = $$8$$ $$x= 4$$ So the other leg, $$x\\sqrt{3}=4\\sqrt{3}$$ Area, $$A=\\frac{b*h}{2}$$ $$A= \\frac{4*4\\sqrt{3}}{2}=8\\sqrt{3}$$ _________________ Never look down on anybody unless you're helping them up. --Jesse Jackson What is the area of Triangle XYZ? &nbs [#permalink] 22 Apr 2018, 17:05 Display posts from previous: Sort by What is the area of Triangle XYZ? Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.", "xz, xy, y ^ 2-z ^ 3, x ^ 2-z ^ 3)$$ but here I don't know if there are more (there probably are).", "# Question 1 of 12 What is the 87$^{th}$ shape in the following pattern? A Box B Hart C Square D Circle E None of the above", "is $$148$$. · 2 years, 5 months ago Siddhartha, how did you figure out that the side lengths were 4,5 and 6? · 2 years, 5 months ago Pretty much a lucky guess. We see that its the only solution to $$xyz = 120$$ and $$(x-2)(y-2)(z-2) = 24$$. The first equation is obvious. The second one comes is from the fact that the total volume of the cuboid not colored is $$(x-2)(y-2)(z-2)$$, since we have to discount the cubes which appear on either ends. · 2 years, 5 months ago" ]

[ "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "q); label(\"P\", P, dir(250), q); label(\"Q\", Q, 0.25*(Q-P), q); label(\"\\sqrt{3}\",B--Q, dir(60), p); label(\"\\sqrt{3}\",P--B, dir(210), p); label(\"\\sqrt{3}\",P--Q, dir(135), p); label(\"2\",P--C, dir(120), p); label(\"2\",A--Q, dir(-20), p); label(\"1\",A--P, dir(210), p); [/asy]$ Note that $\\triangle QAP$ is a $30^\\circ$-$60^\\circ$-$90^\\circ$ triangle, hence $\\angle PQA=30^\\circ$, and $\\angle BQA=90^\\circ$, so $$AB^2=PQ^2+AQ^2=2^2+3=7,$$ and the answer is $\\boxed{\\textbf{(B) } \\sqrt{7}}$. ## Solution 2 (Intuition) $[asy] unitsize(1inch); pen p = fontsize(10pt); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label(\"A\", (0,0), SW, p); label(\"C\", (2.646,0), SE, p); label(\"B\", (1.323,2.291), N, p); label(\"P'\", (0.756,0.655), NW, p); label(\"1\", (2.174,0.164), N, p); label(\"1\", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]$ Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments). Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to", "$W$, $X$, $Y$, and $Z$ divide the sides into lengths of one unit. Find the ratio of the area of the shaded quadrilateral $UWXY$ to the area of the triangle $PQR$. $[asy] draw((1/2,0)--(-1/2,0)--(0,sqrt(3)/2)--cycle); dot((1/6,sqrt(3)/3)); dot((-1/6,sqrt(3)/3)); dot((1/3,sqrt(3)/6)); dot((-1/3,sqrt(3)/6)); dot((1/6,0)); dot((1/6,0)); dot((-1/6,0)); filldraw((-1/6,sqrt(3)/3)--(1/3,sqrt(3)/6)--(1/6,0)--(-1/6,0)--cycle); label(\"P\",(0,sqrt(3)/2),N); label(\"Z\",(1/6,sqrt(3)/3),NE); label(\"Y\",(1/3,sqrt(3)/6),NE); label(\"R\",(1/2,0),E); label(\"X\",(1/6,0),S); label(\"W\",(-1/6,0),S); label(\"Q\",(-1/2,0),W); label(\"V\",(-1/3,sqrt(3)/6),NW); label(\"U\",(-1/6,sqrt(3)/3),NW); //Credit to chezbgone2 for the diagram[/asy]$ ## Problem T2 Evaluate $\\begin{eqnarray*} && 1 \\left(\\dfrac{1}{1}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right) \\\\ &+& 3\\left(\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&5\\left(\\dfrac{1}{3}+\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&7\\left(\\dfrac{1}{4}+\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&9\\left(\\dfrac{1}{5}+\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+11\\left(\\dfrac{1}{6}+\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&13\\left(\\dfrac{1}{7}+\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+15\\left(\\dfrac{1}{8}+\\dfrac{1}{9}+\\dfrac{1}{10}\\right)\\\\ &+&17\\left(\\dfrac{1}{9}+\\dfrac{1}{10}\\right)+19\\left(\\dfrac{1}{10}\\right)\\end{eqnarray*}$ ## Problem T3 To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed? ## Problem T4 In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$. The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey? ## Problem T5 During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued: 'A': It was 'B' or 'C' 'B': Neither 'E'", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "QR but not on the actual segment or in the rectangle. $[asy] size(200); pair A,B,C,Q,R,S; real r = (pi/2 - asin(1/5))/2; A = (0,0); B = 40*dir(r*180/pi); C = 31*dir(90-r*180/pi); draw(A--B--C--cycle); Q = (40*cos(r),0); R = (40*cos(r),31*cos(r)); S = (0, 31*cos(r)); draw(A--Q--R--S--cycle); label(\"A\",A,SW); label(\"B\",B,NE); label(\"C\",C,N); label(\"Q\",Q,SE); label(\"R\",R,E); label(\"S\",S,NW); [/asy]$ The actual answer is a radical near $728$ (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer $744$ despite the invalid problem statement.", "(1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label(\"y\", (q+R)/2, NW); label(\"A'\", A, SW); label(\"B'\", B, SE); label(\"C'\", C, N); label(\"Q\", (q-(0,0.3))); label(\"R\", R, NE); label(\"S\", S, NE); label(\"T\", T, W); [/asy]$ Similary, $\\triangle A'B'C'$ and $\\triangle RB'Q$ are similar, so $RB' = \\frac{4}{3}y$, and $C'S = \\frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \\frac{4}{3}y + y + \\frac{3}{4}y = 5$. Solving for $y$, we get $y = \\frac{60}{37}$. Thus, $\\frac{x}{y} = \\frac{37}{35}$.", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label(\"A\",A,(1,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0));label(\"P\",P,(0,-1));label(\"Q\",Q,(1,0));label(\"R\",R,(-1,1)); [/asy]$", "# Difference between revisions of \"2010 USAMO Problems/Problem 1\" ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "+0 0 92 1 Let $X$, $Y$, and $Z$ be points on a circle. Let $\\overline{XY}$ and the tangent to the circle at $Z$ intersect at $W$. If $WX = 4$, $WZ = 8$, and $\\overline{WY} \\perp \\overline{WZ}$, then find $YZ$. [asy] unitsize(2 cm); pair A, B, C, D; D = (0,1); B = dir(150); C = dir(210); A = (-sqrt(3)/2,1); draw(Circle((0,0),1)); draw(D--A--C); label(\"$W$\", A, NW); label(\"$X$\", B, SE); label(\"$Y$\", C, SW); label(\"$Z$\", D, N); label(\"$8$\", (A + D)/2, N); label(\"$4$\", (A + B)/2, W); [/asy] Feb 14, 2020", "# Difference between revisions of \"2010 USAMO Problems/Problem 1\" ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution 1 Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars", "draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\\frac{1^2\\sqrt{3}}{4}\\implies\\frac{1\\sqrt{3}}{4}$, and $\\frac{2^2\\sqrt{3}}{4}\\implies\\frac{4\\sqrt{3}}{4}\\implies\\sqrt{3}$ and the area of the large equilateral triangle is $\\frac{7^2\\sqrt{3}}{4}\\implies\\frac{49\\sqrt{3}}{4}$ so the area of the hexagon would be $\\frac{49\\sqrt{3}}{4}-\\frac{\\sqrt {3}}{4}-\\frac{\\sqrt{3}}{4}-\\sqrt{3}\\implies\\boxed{\\frac{43\\sqrt{3}}{4}}$ Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "# 2010 USAMO Problems/Problem 1 ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution 1 Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from Y pair P", "# 1974 USAMO Problems/Problem 5 ## Problem Consider the two triangles $\\triangle ABC$ and $\\triangle PQR$ shown in Figure 1. In $\\triangle ABC$, $\\angle ADB = \\angle BDC = \\angle CDA = 120^\\circ$. Prove that $x=u+v+w$. $[asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label(\"A\",A,N); label(\"B\",B,ESE); label(\"C\",C,SW); label(\"D\",D,NE); label(\"a\",(B+C)/2,S); label(\"b\",(C+A)/2,WNW); label(\"c\",(A+B)/2,NE); label(\"u\",(A+D)/2,E); label(\"v\",(B+D)/2,N); label(\"w\",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NE); label(\"x\",(P+Q)/2,S); label(\"x\",(Q+R)/2,E); label(\"x\",(R+P)/2,W); label(\"a\",(P+M)/2,SE); label(\"b\",(Q+M)/2,N); label(\"c\",(R+M)/2,E); label(\"Figure 1\",P-(0,1),S); [/asy]$ ## Solutions ### Solution 1 We rotate figure $PRQM$ by a clockwise angle of $\\pi/3$ about $Q$ to obtain figure $RR'QM'$: $[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"P\",P,W); label(\"Q\",Q,E); label(\"R\",R,W); label(\"M\",M,NW); label(\"R'\",RR,NE); label(\"M'\",MM,ESE); [/asy]$ Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then $$[ABC] + \\frac{b^2 \\sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .$$ Then by symmetry, $$3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2\\bigl( [PMQ] + [QMR] + [RMP] \\bigr) = 2 [PRQ] .$$ But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}$. Therefore, the area of $ABC$ is $$\\frac{(wv+vu+uw)\\sqrt{3}}{4} .$$ Also, by the Law", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ ### Solution 2 (Easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. ### Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and", "be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then $\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.$ Doing routine algebra on the above equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = \\boxed{130}.$ Solution 2 (easiest) $[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"A\",a,S); label(\"B\",b,S); label(\"M\",m,NE); label(\"N\",n,NE); label(\"P\",p,N); label(\"Q\",q,NW); label(\"R\",r,E); label(\"12\",(14,0),SW); label(\"6\",(23,0),S); [/asy]$ Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=\\boxed{130}$. Solution 3 Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}(-11/24)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}(x/16)$ and $\\cos^{-1}(x/12)$. So we have $\\cos^{-1}(x/16)+\\cos^{-1}(-11/24)=180-\\cos^{-1}(x/12).$ Taking the $\\cos$ of both sides and simplifying using the cosine addition identity gives $x^2=130$. Solution 4 (quickest) Let $QP = PR = x.$ Extend the line containing the centers of the two circles to meet R and the other side of the circle the large circle. The line segment consisting of R and the first intersection of the larger circle has length 10. The" ]

[ "# Inscribed Rectangle Geometry Level 2 A rectangle is inscribed within a circle of radius $$7\\text{ cm}$$, and this rectangle has a perimeter $$32\\text{ cm}$$. What is the area of this rectangle? ×", "# Rectangles Geometry Level 2 Given perimeter of a rectangle is 26 and area of the rectangle is 42. If the angle between diagonals is $$P$$, find $$\\cos P$$. ×", "# Relationship Between Perimeter And Area Geometry Level 3 A circle with perimeter $$x \\text{ cm}$$ has an area of $$x \\text{ cm}^2.$$ A square with perimeter $$y \\text{ cm}$$ has an area of $$y \\text{ cm}^2.$$ Which of these shapes has a larger area, the circle or the square? (That is, $$x>y$$ or $$x<y?$$) ×", "# Prime Square Geometry Level 2 A red rectangle is reshaped into a blue square, as shown below, such that both quadrilaterals have the same perimeter. The area of the square is $4\\text{ cm}^2$ larger than that of the rectangle. If all the sides of both quadrilaterals have prime number lengths $($in $\\text{cm}),$ then what is the perimeter of the square? ×", "a square with an area of $100$ and still have a different perimeter form the first square (it would be like trying to get another combination of four numbers that have the same value, but that when you multiply two of them together they give you $100$). In conclusion, that is why the second rectangle isn't (and can't be) a square. *A square can be a rectangle, but a rectangle can't be a square so, the first rectangle was originally a square.", "# rectangle inscribed in a circle • November 7th 2009, 07:15 AM sri340 rectangle inscribed in a circle The area of the largest rectangle inscribed in a circle of radius 5 cm is • November 7th 2009, 07:27 AM ... when it's a square. And the area is $4\\times\\tfrac12\\times5\\times5=50 \\text{ cm}^2$.", "# Perimeter from Area Geometry Level 1 The length of a rectangle is three times of its width. If the area of the rectangle is $$300 \\text{ m}^2$$, what is the perimeter of the rectangle (in $$\\text{m}$$)? ×", "# The least possible perimeter of a rectangle of area $100m^{2}$ is $\\begin{array}{1 1}(1)10&(2)20\\\\(3)40&(4)60\\end{array}$", "# Largest Rectangle with Given Perimeter is Square/Historical Note It is a more specific (and accessible) version of Proposition $27$: Similar Parallelogram on Half a Straight Line.", "## Basic College Mathematics (9th Edition) A square consists of 4 sides of the same length. The length of a side of a square can be denoted $s$. The area of a square can be defined as $s$$^{2}$. The perimeter of a square can be defined as 4$s$. Each corner of a square is a 90$^{\\circ}$ angle. A rectangle consists of two pairs of sides of the same length. These can be labeled $l$ and $w$. The perimeter of a rectangle can be defined as 2$l$ * 2$w$. The area of a rectangle can be defined as $l$*$w$. Each corner of a rectangle is a 90$^{\\circ}$ angle.", "is the perimeter of the rectangle.", "the greatest area given a fixed perimeter of $202$. As a square has the maximal area among rectangles given a fixed perimeter, the solution should occur when all sides have equal length, that is, when $x = 101 - x$.", "area of the tile is 1 square inch. ## Example Each of two square tiles is $$1$$ square inch. Two tiles are shown together. What is the perimeter of the figure? What is the area? ### Solution The perimeter is the distance around the figure. The perimeter is $$6$$ inches. The area is the surface covered by the figure. There are $$2$$ square inch tiles so the area is $$2$$ square inches. ### Optional Video: Perimeter and Area Formulas Do you want to suggest a correction or an addition to this content? Leave Contribution", "square (D) equal to the square root of the perimeter of the square (E) none of the above _________________ Intern Joined: 29 Aug 2017 Posts: 35 Re: If the area of a rectangle is equal to the area of a square, then the [#permalink] Show Tags 22 Oct 2017, 06:22 1 Bunuel wrote: If the area of a rectangle is equal to the area of a square, then the perimeter of the rectangle must be (A) one half of the perimeter of the square (B) equal to the perimeter of the square (C) equal to twice the perimeter of the square (D) equal to the square root of the perimeter of the square (E) none of the above Area of a rectangle = Area of a square lb = s*s Perimeter of the square = 4s = $$4\\sqrt{lb}$$ Perimeter of the rectangle = 2l + 2b which can't be expressed just in terms of the perimeter of the square Manager Joined: 09 Aug 2017 Posts: 62 Location: United States Concentration: Technology GMAT 1: 640 Q44 V33 GMAT 2: 630 Q47 V29 WE: Research (Investment Banking) Re: If the area of a rectangle is equal to the area of a square, then the [#permalink] Show Tags 25 Oct 2017, 12:57 Bunuel, besides you who can I follow to get", "# rectangle inscribed in a circle • Nov 7th 2009, 07:15 AM sri340 rectangle inscribed in a circle The area of the largest rectangle inscribed in a circle of radius 5 cm is • Nov 7th 2009, 07:27 AM ... when it's a square. And the area is $4\\times\\tfrac12\\times5\\times5=50 \\text{ cm}^2$.", "is the area of the triangle? a) 1856 $cm^2$ b) 1658 $cm^2$ c) 1558 $cm^2$ d) 1586 $cm^2$ e) None of these Question 5: Area of the circle is 616 $cm^{2}$. What is the area of the rectangle? (One side of the rectangle passes through the center of the circle) a) 786 $cm^{2}$ b) 196 $cm^{2}$ c) 392 $cm^{2}$ d) Cannot be determined e) None of these Question 6: The area of a square is thrice the area of a rectangle If the area of the square is 225 sq cm and the length of the rectangle is 15 cm what is the difference between the breadth of the rectangle and the side of the square? a) 8 cm b) 10 cm c) 12 cm d) 6 cm e) None of these Question 7: The respective ratio of curved surface area and total surface area of a cylinder is 4:5. If the curved surface area of the cylinder is 1232 cm2, what is the height? (in cm) a) 392 cm b) 394 cm c) 396 cm d) 300 cm e) None of these Question 8: The perimeter of a rectangle whose length is 6rn more than its breadth is 84 m. What would be the area of a triangle whose base is equal to the diagonal of the", "the rectangle? ×", "}$$ and $$21\\text{ in }$$. ## Vocabulary Term Definition area The amount of space inside a figure. Area is measured in square units. perimeter The distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write units. Area of a Rectangle To find the area '$$A$$' of a rectangle, calculate $$A = bh$$, where $$b$$ is the base (width) and h is the height (length). Perimeter of a Rectangle The perimeter '$$P$$' of a rectangle is equal to twice the base added to twice the height: $$P = 2b + 2h$$.", "# area-and-perimeter-of-circle-problem-solving-02 ## No 01 The area and perimeter of the shaded region respectively are ... (A) $$1232 \\text{ cm}^2 \\text{ and } 352\\text{ cm}$$ (B) $$1232 \\text{ cm}^2 \\text{ and } 354\\text{ cm}$$ (C) $$1233 \\text{ cm}^2 \\text{ and } 355\\text{ cm}$$ (D) $$1234\\text{ cm}^2 \\text{ and } 357\\text{ cm}$$", "the Top Brands at Competitive Prices. In other words, the area of a rectangle is the product of its length and width. 4-H is the nation’s largest youth development organization, involving young people in urban, suburban and rural communities. In 1930, there were 300,000 deer species in the United States. 14 x 20 2 = 3. It is now open for burials. A rectangular park, with dimensions of 1500 feet by 2000 feet, has a diagonal walking path that goes from the top northeast corner to the bottom southwest corner. If you prefer to order by phone, call us at (877) 588-2530. Then, mark the following statements as. The average annual runoff decreased more than 20% during the last 50 years (Qinghai Provincial Government 2005). find the dimensions of the rectangle that will enclose the most area, and show 1 Educator Answer. Then we can express the above algebraically: 300 (perimeter) = 2L + 2W 3600 (area) = L x W Now let's find one variable in. Our staff can't provide legal advice, interpret the law or conduct research. Responsibly Sourced, Shelf-Stable, Nutritious Food. A local grocery store has plans to construct a rectangular parking lot on land that is bordered on one side by a highway. The largest area rectangle that can be enclosed by 550 feet" ]

[ "to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\\boxed{260}$.", "# Perimeter of regular hexagon Geometry Level 1 A regular hexagon has side length $$8$$ cm. What is the perimeter (in cm) of the hexagon? ×", "400 Radius = (√400 – 10) ÷ 2 Radius = (20 – 10) ÷ 2", "Therefore, perimeter of a rhombus (P) = 4 × side = 4 × 5 cm = 20 cm", "with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) The perimeter is $2(61)+2(69)=\\boxed{260}$.", "# The perimeter of a square is 20 cm. How do you find the length of a diagonal? Let $a$ be the side of the square .Its perimeter equals with $4 a$ hence $4 a = 20 \\implies a = 5 c m$ .Now the diagonal $d$ is the hypotenese of the right triangle with sides those of the square so ${d}^{2} = {a}^{2} + {a}^{2} \\implies d = \\sqrt{2} \\cdot a \\implies d = 5 \\cdot \\sqrt{2}$", "+0 # help me with perimeter and area 0 76 2 Find the perimeter and area. Aug 5, 2020 #1 0 The answer is $$4 \\times (\\frac{25 \\pi}{2} - 25) = \\boxed{50 \\pi - 100}$$ So, each petal shaped figure is area of 2 quarter circles of length 5 - the area of a square with side 5 Aug 5, 2020 #2 +1421 +1 Perimeter P = 2*10pi Aug 5, 2020", "Ans: We know, Perimeter of Triangle=$10+14+15$ =$39$cm So, Perimeter of Triangle=$39$cm 15. Find the perimeter of a regular hexagon with each side measuring $8$ cm. Ans: We know, Perimeter of Hexagon=$6\\times 8$ =$48$m So, Perimeter of Hexagon=$48$m 16. Find the side of the square whose perimeter is $20$ m. Ans: We know, Perimeter of square=$4\\times side$ $20=4\\times side$ Side=$\\frac{20}{4}$ =$5$ m So, Perimeter of Square=$5$m 17. The perimeter of a regular pentagon is $100$ cm. How long is its each side? Ans: We know, Perimeter of regular pentagon=$100$cm $100=5\\times side$ Side=$\\frac{100}{5}$ =$20$ cm So, Perimeter of Square=$20$cm 18. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a square Ans: We know, Perimeter of square=$4\\times side$ $30=4\\times side$ Side=$\\frac{30}{4}$ =$7.5$ cm So, Perimeter of Square=$7.5$cm 19. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: an equilateral triangle Ans: We know, Perimeter of Equilateral Triangle=$3\\times side$ $30=3\\times side$ Side=$\\frac{30}{3}$ cm =$10$ cm So, Perimeter of Equilateral Triangle=$10$cm 20. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a regular hexagon Ans: We know, Perimeter of Hexagon=$30$cm", "# 2021 JMPSC Accuracy Problems/Problem 3 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem In a regular octagon, the sum of any three consecutive sides is $90.$ A square is constructed using one of the sides of this octagon. What is the area of the square? ## Solution We are given that the sum of any three sides of the octagon is 90, and since the octagon is regular, we know that all sides are equal. Thus, each side of the octagon must equal to $\\frac{90}{3} = 30$. Since one side of the square shares a side with the octagon, we know the side lengths of the square are also 30. Thus, our answer is $30^2 = \\boxed{900}$ ## Solution 2 We have each side of the octagon is $s$, so $3s=90$, or $s=30$. This means the area of the square is $30^2=\\boxed{900}$ ~Geometry285", "triangle = Sum of the lengths of all sides of the triangle Perimeter = 15 + 12 + 13 = 40 cm Q8. A regular hexagon of each side measuring 6 m is given. Find its perimeter. Ans: Perimeter of regular hexagon = 6 x Side of regular hexagon Perimeter of the regular hexagon = 6 x 6 = 36 m Q9. Find the side of the square whose perimeter is 16 m. Ans: Perimeter of square = 4 x Side 16 = 4 x Side $$Side = \\frac{16}{4} = 4\\ cm$$ Side = 4 Q10. Find the side of a regular pentagon if its perimeter is 150 cm. Ans: Perimeter of regular pentagon = 5 x Length of side 150 = S x Side $$Side = \\frac{150}{5} = 30\\ cm$$ Therefore, Side = 30 cm Q11. A piece of string is 120 cm long. What will be the length of each side if the string is used to form (a) A regular hexagon? (b) A square? (c) An equilateral triangle? Ans: (a) Perimeter = 4 x Side 120 = 4 x Side $$Side = \\frac{120}{4} = 30\\ cm$$ (b) Perimeter = 3 x Side 120 = 3 x Side $$Side = \\frac{120}{3} = 40\\ cm$$ (c) Perimeter = 6 x Side 120 = 6 x Side $$Side", "## Thinking Mathematically (6th Edition) perimeter = $1000$ inches RECALL: (1) The perimeter of a polygon is equal to the sum of the lengths of all its sides. (2) A square has four equal sides. This means that the three unknown side lengths are all 250 inches. Add the lengths of all the sides to find the perimeter: perimeter = $250+250+250+250= 1000$ inches", "of each side of the triangle is $10\\ cm$. (c) The perimeter of the regular hexagon form of the string $=30\\ cm$ We know that, The perimeter of a regular hexagon is six times its side $6\\times$ each side of the regular hexagon $=30\\ cm$ Each side of the regular hexagon $=\\frac{30\\ cm}{6}$ $=5\\ cm$ The length of each side of the regular hexagon is $5\\ cm$. Updated on 10-Oct-2022 13:36:27 Advertisements", "+0 # help me with perimeter and area 0 175 2 Find the perimeter and area. Aug 5, 2020 ### 2+0 Answers #1 0 The answer is $$4 \\times (\\frac{25 \\pi}{2} - 25) = \\boxed{50 \\pi - 100}$$ So, each petal shaped figure is area of 2 quarter circles of length 5 - the area of a square with side 5 Aug 5, 2020 #2 +1468 +1 Perimeter P = 2*10pi Aug 5, 2020", "Geometry Puzzle – Challenge 73 Time for another great math challenge for those who enjoy solving mathematics and critical thinking challenges. Challenge: The perimeter of a square is 13 cm and the perimeter of an equilateral triangle is 7 cm. What is the difference of a side of the square and a side of a triangle? A- $$\\frac{7}{13}$$ cm B- $$\\frac{11}{12}$$ cm C- $$1\\frac{1}{4}$$ cm D- $$2\\frac{1}{4}$$ cm E- $$\\frac{13}{7}$$ cm $11.99 Satisfied 123 Students The correct answer is B. The perimeter of a square is 13cm. Therefore, one side of the square is $$\\frac{13}{4}$$ cm. The perimeter of the equilateral triangle is 7cm. Therefore, one side of the triangle is $$\\frac{7}{3}$$ cm. The difference of a side of the square and a side of a triangle is: $$\\frac{13}{4} – \\frac{7}{3} = \\frac{39-28}{12} = \\frac{11}{12}$$ cm More math articles What people say about \"Geometry Puzzle - Challenge 73\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE$5 It was $16.99 now it is$11.99", "$2$). That's enough to confirm that the answer has to be $\\boxed{\\textbf{(A) }3}$. ## Solution 5 (When You're Running Out of Time) Since triangles $T'$ and $T$ have the same area and the same perimeter, $2a+b=18$ and $9*(9-a)^2(9-b) = 9*4^2*1$ By trying each answer choice, it is clear that the answer is $\\boxed{\\textbf{(A) }3}$.", "5. $$\\frac{3}{10}$$ + $$\\frac{1}{10}$$ = ___________ $$\\frac{3}{10}$$ is 3/10 $$\\frac{1}{10}$$ is 1/10 we need to add these values: 3/10+1/10 =4+1/10 =5/10 =1/2. Therefore, $$\\frac{3}{10}$$ + $$\\frac{1}{10}$$ is 1/2. Question 6. 20 × 3 – 99 ÷ 3 = ___________ 20 × 3 =60 99 ÷ 3 = 33 now subtract the values. 60-33=27 Therefore, 20 × 3 – 99 ÷ 3 is 27. Question 7. 63 ÷ 9 = 56 ÷ 63 ÷ 9 = 56 ÷ why 8? if we divide 63 by 9 then we get 7 as quotient and 0 as remainder. If we divide 56 by 7 then we get 7 as quotient and 0 as remainder. we got equal quotients and equal remainders. so, 8 is the correct answer. Question 8. Calculate the perimeter of a pentagon with 3-cm sides. the perimeter of pentagon formula: Using the perimeter of a pentagon formula, you can find the perimeter of a regular pentagon with relative ease. To find the perimeter of a regular pentagon with sides of length, s, you use this formula: p=5×s In our formula, 5 is the number of sides, and S is the length of the side that we know. Just like with the perimeter of a square, or the perimeter of a polygon in general, you find the perimeter of", "# What is the perimeter of a parallelogram with sides of 5 and 7? $P = 24$ Perimeter is the sum of all sides, so The given are $5$ and $7$ which means there are $2 \\left(5\\right)$ and $2 \\left(7\\right)$. $= 10 + 14$ $= 24$", "from the 360 Blog: Problem 5.2.4: A box has three possible perimeters. Suppose box A has perimeters 12, 16, and 20, while box B has perimeters 12, 16, and 24. Which box has the greater volume? If you're looking for good sources of Math videos check out this article at Free Technology for Teachers. I'll leave you today with this fantastic cartoon from xkcd.", "for each triangle. If $h$ must be an integer, then the perimeter must be an integer, and the answer of $17.5$ is not possible.", "## Elementary Geometry for College Students (7th Edition) Perimeter of $GHJKL=75$ Given pentagon $ABCDE\\sim GHJKL$, side $AB$ corresponds to side $GH$. $\\frac{AB}{GH}=\\frac{6}{9}=\\frac{2}{3}$ The ratio of the sides of the pentagons are 2:3. The ratio of the perimeters of the pentagons will also be 2:3. So... $\\frac{2}{3}=\\frac{50}{x}$ Cross Multiply... $2x=150$ Divide by 2... $x=75$ or The perimeter of $GHJKL=75$" ]

[ "# Inscribed Rectangle Geometry Level 2 A rectangle is inscribed within a circle of radius $$7\\text{ cm}$$, and this rectangle has a perimeter $$32\\text{ cm}$$. What is the area of this rectangle? ×", "## Basic College Mathematics (9th Edition) A square consists of 4 sides of the same length. The length of a side of a square can be denoted $s$. The area of a square can be defined as $s$$^{2}$. The perimeter of a square can be defined as 4$s$. Each corner of a square is a 90$^{\\circ}$ angle. A rectangle consists of two pairs of sides of the same length. These can be labeled $l$ and $w$. The perimeter of a rectangle can be defined as 2$l$ * 2$w$. The area of a rectangle can be defined as $l$*$w$. Each corner of a rectangle is a 90$^{\\circ}$ angle.", "# Rectangles Geometry Level 2 Given perimeter of a rectangle is 26 and area of the rectangle is 42. If the angle between diagonals is $$P$$, find $$\\cos P$$. ×", "# Prime Square Geometry Level 2 A red rectangle is reshaped into a blue square, as shown below, such that both quadrilaterals have the same perimeter. The area of the square is $4\\text{ cm}^2$ larger than that of the rectangle. If all the sides of both quadrilaterals have prime number lengths $($in $\\text{cm}),$ then what is the perimeter of the square? ×", "# Relationship Between Perimeter And Area Geometry Level 3 A circle with perimeter $$x \\text{ cm}$$ has an area of $$x \\text{ cm}^2.$$ A square with perimeter $$y \\text{ cm}$$ has an area of $$y \\text{ cm}^2.$$ Which of these shapes has a larger area, the circle or the square? (That is, $$x>y$$ or $$x<y?$$) ×", "us back at risk. Another discussion. Every square has a perimeter, and an area. If the perimeter is 12, the area is 9 (check that). But one special square has its area equal to its perimeter. (Answer is in a few lines). You could just guess at it, and get it. You could do some algebra: $(side)^2 = 4(side)$ Call the side s, and $s^2 = 4s$ and $s^2 - 4s = 0$ and $s(s-4) = 0$ and now we have 4 or 0, but 0 makes no sense. And yeah, the area of a square with side 4 is 16, and so is the perimeter. ## Rectangles (here’s the puzzle) Now, rectangles come in more varieties than squares. That can make them more interesting, or more complicated. Can a rectangle have its area equal its perimeter? Yes, 4×4 works (remember, a square is a special rectangle). But there is at least one more. Can you find another rectangle with its area equal its perimeter? Can you find all of them? How do you know when you have them all? I’m teaching a Number Theory elective this Spring, and trying to have some fun. I chose a text, and I knew we would finish it in April, which we did. And I gave students some choices of what", "a square with an area of $100$ and still have a different perimeter form the first square (it would be like trying to get another combination of four numbers that have the same value, but that when you multiply two of them together they give you $100$). In conclusion, that is why the second rectangle isn't (and can't be) a square. *A square can be a rectangle, but a rectangle can't be a square so, the first rectangle was originally a square.", "two trapezoids. The dimensions of the square is $13$ by $13$, which means that the area is $169$ square units. The same polygons above are used to create the rectangle below. The dimensions of the rectangle is $8$ by $21$, which means that the area is $168$. Continue reading", "# Perimeter-Area Inequality For all rectangles with perimeter $P$ and area $A$, what is the minimum value of $\\frac{P^2}{A}?$ ×", "# Just a rectangle Geometry Level 3 A rectangle is inscribed in a triangle with sides equal to $$10cm,17cm$$ and $$21cm$$ so that two of its vertices are situated on one side of the triangle , and two others on other two sides of the triangle .Find the sides of the rectangle if it is known that its perimeter is $$22.5cm$$. If your answer come as $$a$$ cm and $$b$$ cm , submit it as descending order of their digits (excluding decimal) in a single linear order. ×", "area of the tile is 1 square inch. ## Example Each of two square tiles is $$1$$ square inch. Two tiles are shown together. What is the perimeter of the figure? What is the area? ### Solution The perimeter is the distance around the figure. The perimeter is $$6$$ inches. The area is the surface covered by the figure. There are $$2$$ square inch tiles so the area is $$2$$ square inches. ### Optional Video: Perimeter and Area Formulas Do you want to suggest a correction or an addition to this content? Leave Contribution", "# area-and-perimeter-of-circle-problem-solving-02 ## No 01 The area and perimeter of the shaded region respectively are ... (A) $$1232 \\text{ cm}^2 \\text{ and } 352\\text{ cm}$$ (B) $$1232 \\text{ cm}^2 \\text{ and } 354\\text{ cm}$$ (C) $$1233 \\text{ cm}^2 \\text{ and } 355\\text{ cm}$$ (D) $$1234\\text{ cm}^2 \\text{ and } 357\\text{ cm}$$", "square (D) equal to the square root of the perimeter of the square (E) none of the above _________________ Intern Joined: 29 Aug 2017 Posts: 35 Re: If the area of a rectangle is equal to the area of a square, then the [#permalink] Show Tags 22 Oct 2017, 06:22 1 Bunuel wrote: If the area of a rectangle is equal to the area of a square, then the perimeter of the rectangle must be (A) one half of the perimeter of the square (B) equal to the perimeter of the square (C) equal to twice the perimeter of the square (D) equal to the square root of the perimeter of the square (E) none of the above Area of a rectangle = Area of a square lb = s*s Perimeter of the square = 4s = $$4\\sqrt{lb}$$ Perimeter of the rectangle = 2l + 2b which can't be expressed just in terms of the perimeter of the square Manager Joined: 09 Aug 2017 Posts: 62 Location: United States Concentration: Technology GMAT 1: 640 Q44 V33 GMAT 2: 630 Q47 V29 WE: Research (Investment Banking) Re: If the area of a rectangle is equal to the area of a square, then the [#permalink] Show Tags 25 Oct 2017, 12:57 Bunuel, besides you who can I follow to get", "Area of rectangle=20 cm x 10 cm =20 cm x 10 cm=200 sq cm Therefore, Square has more area. X. Find the perimeter of a square whose area is 400 m². Solution: Given that, Area of square=400 m² We know that area=side2 400 m²=side2 side=$$\\sqrt{ 400 m² }$$ =20 m we know that perimeter of a square=4side p=4 x 20 m =80 m Therefore, the perimeter of a square is 80 m.", "# The least possible perimeter of a rectangle of area $100m^{2}$ is $\\begin{array}{1 1}(1)10&(2)20\\\\(3)40&(4)60\\end{array}$", "our reasoning above, we have $AB=100x$. Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\\boxed{40400 \\text{ (E)}}$.", "• # question_answer If the radius of the circle is 4 cm, what is the perimeter of the square? If the radius of the circle is 4 cm, what is the perimeter of the square? A) 28 cm B) 24 cm C) 32 cm D) 36 cm The side of the square is twice the n of the circle in it $=\\text{ 2 }\\times \\text{ 4 cm }=\\text{ 8 cm}$ So, the perimeter of the square $=\\text{4 }\\times \\text{ 8 cm }=\\text{32 cm}$", "20 meter-by-20 meter square pen to get maximal area. 109 External dimensions of a closed wooden box are in the ratio 5:4:3. For example, a garden shaped as a rectangle with a length of 10 yards and width of 3 yards has an area of 10 x 3 = 30 square yards. a Find the area of the. Show transcribed image text Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible. For the best answers, search on this site https://shorturl. Label the side lengths of Rectangle A on the grid below. If x = 10, y = 10, so the minimum perimeter comes from a rectangle that is a square, 10m by 10m, with a 40m perimeter. The greatest common factor is the largest possible integer by which both numbers named in the fraction are. Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible. An exact expression of the perimeter P of an ellipse was first published in 1742 by the Scottish. Area = 1, perimeter = 5. Find the cost of fertilizing the field at$0. If the area of the rectangle is 15 in2, find the unknown number. is possible, find the length of the side of the square and the", "+ 2) ∙ 14 cm = $$\\frac{36}{7}$$ × 14 cm = 36 × 2 cm = 72 cm. 2. The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Find its area. (Use π = $$\\frac{22}{7}$$.) Solution: Then, perimeter = ($$\\frac{π}{2}$$ + 2)r ⟹ 75 cm = ($$\\frac{1}{2}$$ ∙ π + 2)r ⟹ 75 cm = ($$\\frac{ 1 }{2}$$ ∙ $$\\frac{22}{7}$$ + 2)r ⟹ 75 cm = ($$\\frac{11}{7}$$ + 2)r ⟹ 75 cm = $$\\frac{25}{7}$$r ⟹ $$\\frac{25}{7}$$r = 75 cm ⟹ r = 75 × $$\\frac{7}{25}$$ cm ⟹ r = 3 × 7 cm ⟹ r = 21 cm. Therefore, the radius of the circle is 21 cm. Now, area = $$\\frac{1}{4}$$πr^2 = $$\\frac{1}{4}$$ ∙ $$\\frac{22}{7}$$ ∙ 21^2 cm^2 = $$\\frac{1}{4}$$ ∙ $$\\frac{22}{7}$$ ∙ 21 ∙ 21 cm^2 = $$\\frac{693}{2}$$ cm^2 = 346.5 cm^2. Therefore, area of the sheet of paper is 346.5 cm^2. ## You might like these • ### Area of a Rectangle | Area of Rectangle = Length × Breadth | Examples Area of a rectangle is discussed here. We know, that a rectangle has length and breadth. Let us look at the rectangle given below. Each rectangle is made of squares. The side of each square is 1 cm long. The area of each square", "}$$ and $$21\\text{ in }$$. ## Vocabulary Term Definition area The amount of space inside a figure. Area is measured in square units. perimeter The distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write units. Area of a Rectangle To find the area '$$A$$' of a rectangle, calculate $$A = bh$$, where $$b$$ is the base (width) and h is the height (length). Perimeter of a Rectangle The perimeter '$$P$$' of a rectangle is equal to twice the base added to twice the height: $$P = 2b + 2h$$." ]

[ "REQUIREMENT THAT VESSELS BE NUMBERED, SO AS TO REVISE THE CIRCUMSTANCES WHEN A VESSEL IS NOT REQUIRED TO BE NUMBERED; TO AMEND SECTION 50-23-345, RELATING TO TEMPORARY CERTIFICATES OF BOAT NUMBER, SO AS TO FURTHER PROVIDE FOR WHEN THE DEPARTMENT MAY ISSUE TEMPORARY CERTIFICATES; TO AMEND SECTION 50-23-370, RELATING TO TERMS AND RENEWAL OF CERTIFICATES OF BOAT NUMBER ISSUED BY THE DEPARTMENT, SO AS TO FURTHER PROVIDE FOR THEIR EXPIRATION AND RENEWAL AND THE CIRCUMSTANCES WHEN THESE CERTIFICATES MAY BE ISSUED; BY ADDING SECTION 50-23-375 SO AS TO PROVIDE THAT IT IS UNLAWFUL TO DISPLAY A REGISTRATION NUMBER OR VALIDATION DECAL OR AN OUTBOARD MOTOR TITLE DECAL ON ANY WATERCRAFT OR OUTBOARD MOTOR EXCEPT ON THOSE FOR WHICH IT WAS ISSUED; TO AMEND SECTION 50-23-380, RELATING TO TRANSFER OF REGISTRATION UPON CHANGE OF OWNERSHIP, SO AS TO REVISE A REFERENCE; TO AMEND SECTION 50-23-400, RELATING TO NOTICE OF CHANGE OF ADDRESS OF A HOLDER OF A CERTIFICATE OF BOAT NUMBER, SO AS TO REVISE THE TIME WITHIN WHICH THIS NOTICE MUST BE PROVIDED; TO REPEAL SECTION 50-21-35 RELATING TO THE USE OF DEALER DEMONSTRATION NUMBERS FOR WATERCRAFT, SECTION 50-21-60 RELATING TO PERSONNEL, EXPENSES AND SALARIES OF DEPARTMENT EMPLOYEES, SECTIONS 50-23-15, 50-23-40, 50-23-50, 50-23-65, 50-23-100, AND 50-23-160 ALL RELATING TO CERTIFICATES OF TITLE OR MARINE DEALER PERMITS, AND SECTION", "Manual offers information on vessel number, title laws Author: Publish date: The third edition of the National Vessel Numbering and Titling Manual was released by the National Association of State Boating Law Administrators. The manual will help numbering and titling professionals improve their state programs by providing information about state numbering and titling laws, regulations and other areas of interest. This manual summarizes information gathered in a survey of states' numbering and titling practices and requirements conducted by the association's vessel identification, numbering and titling committee. The manual features information about which states title vessels and which vessels require a title in each state, as well as contact information for state agencies, marine theft units, registration agencies and titling agencies. \"By sharing this information among the states, the Numbering and Titling Manual will advance NASBLA's goals of uniformity and reciprocity across the nation, as well as assist in reducing boat fraud and theft,\" NASBLA executive director John Johnson said in a statement. Click here to order a copy of the manual.", "By admin / August 27, 2019 It is essential to obtain a boating license if you want to operate a boat under the age of 18. If you do not get your consent under this age, it will be illegal to operate a vessel or Jet Ski. To get your license, you will need to enroll in the concerned course first, take the class, and have to pass the test. Here are the three simple steps mentioned below, which you will have to follow to acquire your license and enjoy boating legally. ## Step 1: Find A Course ### Look on a State Level The first step to get your boating license is to pick a course. These courses are done on state-level. Do your research and learn about your state’s requirements because they vary from one state to another. You should complete an approved boating license course to operate legally on state waters. ### Choose Classes Now it’s up to you whether you want to pick online or in-person classes. Most states offer you both facilities and your license will be equally valued either way. It is significant to know that your state will have their website for in-person classes. If you are interested in taking online courses, then they will conduct your tests and lectures properly.", "employment,; vehicle make, model, color, and license tag number, including work vehicles, and the permanent or frequent location where all vehicles are kept;, fingerprints and palm prints;, Internet identifiers; passport and immigration documents; and a photograph; (b) the name, address, and county of each institution of higher learning, including the specific campus location, if the person is enrolled, employed, volunteers at, interns at, or carries on a vocation there; (c) the vehicle identification number, license tag number, registration number, and a description, including the color scheme, if the person lives in a motor vehicle, trailer, mobile home, or manufactured home and the permanent or frequent location where all vehicles, trailers, mobile homes and manufactured homes are kept; and (d) the hull identification number, the manufacturer's serial number, the name of the vessel, live-aboard vessel, or houseboat, the registration number, and a description of the color scheme, if the person lives in a vessel, live-aboard vessel, or houseboat; and (e) the tail number, manufacturer's serial number, and model of any aircraft, and a description of the aircraft, including the color scheme, and the permanent or frequent location where all aircraft are kept, if the person owns or operates an aircraft.\" SECTION 9. This act takes effect upon approval by the Governor./ Renumber sections to conform. Amend title to conform.", "license to operate a vessel issued by the United States Coast Guard in the person's name, regardless of the expiration date on the license; (3) is in possession of a merchant mariner credential issued by the United States Coast Guard in the person's name, regardless of the expiration date on the credential; (4) is a nonresident in possession of a boater education certificate, or an equivalency, issued by another state in the nonresident's name; or (5) is accompanied by a person at least 18 years old who: (a) is in possession of a South Carolina boating safety certificate issued by the department in the person's name; or (b) meets one of the criteria in items (1) through (5) of this paragraph. (C) The department shall partner with motorboat rental businesses to ensure that: (1) a motorboat, including a personal watercraft, is not rented or leased to any person for operation on the waters of the state unless the renter or lessee has a valid watercraft operator's permit or is an exempt operator and is 18 years of age or older; (2) the motorboat rental business lists on each motorboat rental or lease agreement the name and age of each operator who is authorized to operate the motorboat or personal watercraft. The renter or lessee of the motorboat must", "moped must contain: (1) the name, bona fide residence and mailing address of the owner or business address of the owner if a firm, association or corporation; (2) a description of the moped including, insofar as this exists with respect to a given moped, the make, model, type of body, serial number or other identifying number, whether the vehicle is new or used, and the date of sale by the manufacturer or seller to the person intending to operate the moped; (3) other information that reasonably may be required to enable the department to determine whether the vehicle is lawfully entitled to registration and licensing. (B) The application must be accompanied by a bill of sale and a vehicle registration certificate, Manufacturer's Certificate of Origin, or an affidavit from the applicant certifying that he is the legal and rightful owner of the moped. The documentation provided must list the vehicle specifications, including the total cubic centimeters of the engine or wattage of the engine, as applicable. Section 56-2-3050. The department, at the request of the owner, may issue a title for the moped in conjunction with the moped registration, provided that the owner makes application for title on the appropriate form and provides the department with a Manufacturer's Statement of Origin or a prior title. If an owner", "obtain a lifetime combination license which grants the same privileges as a statewide combination license from the department at its designated licensing locations. The licensing fees are: (1) for a Type 3 if at the time of application the individual is under two years of age: three hundred dollars; (2) for a Type 4 if at the time of application the individual is at least two years of age but under sixteen years of age: four hundred dollars; (3) for a Type 5 if at the time of application the individual is at least sixteen years of age but under sixty-four years of age: five hundred dollars. (B) A resident holder of a lifetime combination license may add the privilege of statewide saltwater recreational fishing for the following fees: (1) for a Type 13 if at the time of application the individual is under two years of age: one hundred twenty dollars; (2) for a Type 14 if at the time of application the individual is at least two years of age but under sixteen years of age: one hundred sixty dollars; (3) for a Type 15 if at the time of application the individual is at least sixteen years of age but under sixty-four years of age: two hundred dollars. (C) A resident age sixty-three who holds", "boats despite the limited amount of information collected for each boat-time unit. The presentation and code may be found on Nathan Danneman's web site. The audio for the presentation is also available.", "least what age? 12 with a valid boating safety certificate Which statement is true concerning visual distress signal? Flares are rated for day, night or combined day/night use Which of the following indicates an emergency situation aboard Orange smoke billowing from a boat On Maryland waters all children under what age must wear a u.s coastguard approved PFD while underway in a recreational vessel under 21 feet in length 13 Which type of power boat engine must have a backfire flame arrester Gasoline inboard Which of the following is a feature of a type IV PFD It is designed to be thrown Where is the best place to store PFDs on a boat A location that makes them readily accessible When operating on Maryland waters at what depth must a PWC operate at minimum wake speed 18 inches ### Sets with similar terms Boating Safety Course- Texas 10 terms MIA_LUTON PA Boating Safety Test 47 terms cocodorey Safe Boating⛵️ 61 terms Kayleigh_Moses va boaters course quiz 2 10 terms ali357 ### Sets found in the same folder Page 1 chap 4 quiz 7 terms shel449 Page 3 8 terms shel449 Page 4 7 terms shel449 Boating Test Part 2 20 terms tony_pham20 ### Other sets by this creator Vocab 15 20 terms shel449 ap lang vocab 14 24", "permits, we will also briefly look at their $E_{\\infty}$-version. Please email vb8 at illinois dot edu for the zoom details.", "of the USCG Form 1270 for vessels of 5 net tons or greater, or a current copy of a state registration form for vessels under 5 net tons. (B) For a request to change a vessel registration and/or change in permit ownership or permit holder for sablefish-endorsed permits with a tier assignment for which a corporation or partnership is listed as permit owner and/or holder, an Identification of Ownership Interest Form must be completed and included with the application form. (C) For a request to change permit ownership for an MS permit or for a request to change a vessel registration and/or change in permit ownership or permit holder for an MS/CV-endorsed limited entry trawl permit, an Identification of Ownership Interest Form must be completed and included with the application form. (D) For a request to change the vessel registration to a permit, the permit owner must submit to SFD a current marine survey conducted by a certified marine surveyor in accordance with USCG regulations to authenticate the length overall of the vessel being newly registered with the permit. Marine surveys older than 3 years at the time of the request for change in vessel registration will not be considered current'' marine surveys for purposes of this requirement. (E) For a request to change a permit's ownership where", "ensure that only listed authorized operators operate the motorboat or personal watercraft; (3) the motorboat rental business provides each authorized operator (a) a summary of this provision together with any other relevant laws, regulations, and rules governing operation of motorboats and personal watercraft in the state and (b) instructions for safe operation. Each authorized operator must review the summary provided under this paragraph and must either successfully complete a boating education course administered or approved by the department; provide satisfactory proof to the motorboat rental business that the person was issued a boater education certificate, or an equivalency, by another state; or provide satisfactory proof to the motorboat rental business that the person was issued a license to operate a vessel by the United States Coast Guard or was issued a merchant mariner credential by the United States Coast Guard before the motorboat or personal watercraft leaves the motorboat rental business premises; and (4) that the motorboat rental business provides a United States Coast Guard (USCG) approved wearable personal flotation device with a USCG label indicating it either is approved for or does not prohibit use with personal watercraft or water-skiing and any other required safety equipment to all persons who rent a personal watercraft at no additional cost. (D) A person who violates this provision shall be", "number 49 appears • Boats 2 Three quarters of the boats in the marina are white 4/7 of the remaining boats are blue, rest are red if there are 9 red boats. How many boats are in the marina?", "the limited time of the charter agreement. (ii) Qualifying criteria for MS permit. To qualify for initial issuance of an MS permit, a person must own, or operate under a bareboat charter, a vessel on which at least 1,000 mt of Pacific whiting was processed in the mothership sector in each year for at least two years between 1997 and 2003 inclusive. (iii) MS permit application. Persons may apply for initial issuance of an MS permit in one of two ways: complete and submit a prequalified application received from NMFS, or complete and submit an application package. The completed application must be either postmarked or hand- delivered within normal business hours no later than November 1, 2010. If an applicant fails to submit a completed application by the deadline date, they forgo the opportunity to receive consideration for initial issuance of an MS permit. (A) Prequalified application. A prequalified application'' is a partially pre-filled application where NMFS has preliminarily determined the processing history that may qualify the applicant for an initial issuance of an MS permit. NMFS will mail prequalified application packages to the owners or bareboat charterer of vessels which NMFS determines may qualify for an MS permit. NMFS will mail the application by certified mail to the current address of record in the NMFS permit database.", "VDS unless it's an emergency. Breaking the law can come with serious penalties! All recreational boats operating in U.S. Coastal Waters or the Great Lakes, or bodies of water directly connected to U.S. Coastal Waters or the Great Lakes - up to a point where those waters are less than 2 miles wide---are required by law to be equipped with visual distress signals. U.S. owned boats must also carry visual distress signals when operating in international waters. There are some exceptions. During daytime hours the following boats are not required to carry visual distress signals: Boats less than 16 feet in length; Boats participating in organized events, such as regattas; Open sailboats that are less than 26 feet in length and not equipped with an engine; And manually propelled boats, such as canoes. These boats are only required to carry visual distress signals approved for nighttime use when operating at night in the above listed waters. ## Visual Distress Signals: Pyrotechnic One of the most common types of visual distress signals are pyrotechnic signals such as flares. Federal regulations require that all pyrotechnic distress signals be Coast Guard approved, in good condition, unexpired and readily accessible in case of an emergency. Launchers for visual distress signals that were produced before 1981 do not need to be Coast Guard", "lifetime statewide combination license from the department at its designated licensing locations, which grants the same privileges as an annual combination license. The license fee is based on the age of the applicant. If at the time of application the individual is: (1) under two years of age, the fee is three hundred dollars; (2) at least two years of age, but less than sixteen years of age, the fee is four hundred dollars; (3) at least sixteen years of age but less than sixty-four years of age, the fee is five hundred dollars. (B) A resident who holds a lifetime combination license may obtain the privilege of statewide saltwater recreational fishing from the department at its designated licensing locations. The license fee is based on the age of the applicant. If at the time of application the individual is: (1) under two years of age, the fee is one hundred twenty dollars; (2) at least two years of age but less than sixteen years of age, the fee is one hundred sixty dollars; (3) at least sixteen years of age but less than sixty-four years of age, the fee is two hundred dollars. (C) A resident who holds a lifetime combination license may obtain the privilege of hunting migratory waterfowl from the department at its designated licensing", "The Norwegian Registers of Fishing Boats are available online by clicking on the link. These cover approximatley every second year from 1922 to 2002. The Norwegian Port Letters and Numbers are made up of three rather than the usual two parts. The first/leftmost letter denotes the Fishery District, the right hand letter/s denote the Home Port within that Fishery District and the number between the two sets of letters is that particular boat’s number within her Home Port. For example: In 1922 F-4-H, the Lyngen was from the District of Finnmark and No 4 in the port of Hammerfest in that District.", "transfer the title and registration for any vehicles brought with you during the move. • Any vehicle used upon the highways of Louisiana must be insured. When you apply for a license plate, you must have proof of the required liability insurance or other allowable substitute • Every automobile, truck, trailer, boat trailer and motorcycle operated on the highways of this state must have a current motor vehicle inspection sticker. 5 out of 56 Which class of persons must have a driver's license in Louisiana? 6 out of 56 7 out of 56 Louisiana law requires which of the following to operate a motor vehicle on public roads and highways: 8 out of 56 ### Traffic Laws and Regulations • You are required to obey all traffic laws when driving a vehicle upon a street or highway. • You must operate your vehicle as indicated by traffic signs and signals and pavement markings. • It is unlawful to negligently fail to maintain reasonable and proper control of your vehicle. • You must obey all lawful orders and directions of a police officer. • You must not drive a vehicle that is overloaded with passengers or any other thing that will obstruct your view in any direction or interfere with your control of the vehicle. 9 out of 56", "TO FURTHER PROVIDE FOR THIS APPLICABILITY AND REVISE WHEN CERTAIN PENALTY PROVISIONS APPLY; TO AMEND SECTION 50-23-280, RELATING TO PENALTIES FOR VIOLATING CERTAIN WATERCRAFT PROVISIONS, SO AS TO REVISE A SPECIFIC PENALTY PROVISION; TO AMEND SECTION 50-23-290, RELATING TO CONDITIONAL TITLES, SO AS TO FURTHER PROVIDE FOR THE CIRCUMSTANCES UNDER WHICH THE DEPARTMENT MAY ISSUE A CONDITIONAL TITLE; TO AMEND SECTION 50-23-320, RELATING TO EXCEPTIONS TO THE REQUIREMENT THAT VESSELS BE NUMBERED, SO AS TO REVISE THE CIRCUMSTANCES WHEN A VESSEL IS NOT REQUIRED TO BE NUMBERED; TO AMEND SECTION 50-23-345, RELATING TO TEMPORARY CERTIFICATES OF BOAT NUMBER, SO AS TO FURTHER PROVIDE FOR WHEN THE DEPARTMENT MAY ISSUE TEMPORARY CERTIFICATES; TO AMEND SECTION 50-23-370, RELATING TO TERMS AND RENEWAL OF CERTIFICATES OF BOAT NUMBER ISSUED BY THE DEPARTMENT, SO AS TO FURTHER PROVIDE FOR THEIR EXPIRATION AND RENEWAL AND THE CIRCUMSTANCES WHEN THESE CERTIFICATES MAY BE ISSUED; BY ADDING SECTION 50-23-375 SO AS TO PROVIDE THAT IT IS UNLAWFUL TO DISPLAY A REGISTRATION NUMBER OR VALIDATION DECAL OR AN OUTBOARD MOTOR TITLE DECAL ON ANY WATERCRAFT OR OUTBOARD MOTOR EXCEPT ON THOSE FOR WHICH IT WAS ISSUED; TO AMEND SECTION 50-23-380, RELATING TO TRANSFER OF REGISTRATION UPON CHANGE OF OWNERSHIP, SO AS TO REVISE A REFERENCE; TO AMEND SECTION 50-23-400, RELATING TO NOTICE OF CHANGE OF ADDRESS", "of Natural Resources, there is established a pilot program within the department named the Public/Private Partnerships - Boating Safety. The pilot program shall administer a boating education course and increase overall boater safety and boater safety education. A list of approved courses must be provided on the department's website. The following persons must be provided a South Carolina boating safety certificate in both physical and electronic forms by the department: (1) a person who successfully completes a boating education course administered or approved by the department; (2) a person who provides satisfactory proof to the department that the person was issued a boater education certificate, or an equivalency, by another state; and (3) a person who provides satisfactory proof to the department that the person was issued a license to operate a vessel by the United States Coast Guard or was issued a merchant mariner credential by the United States Coast Guard. (B) A person may not operate, upon the waters of this State, a vessel powered by an engine of ten horsepower or greater, a personal watercraft, or a specialty propcraft without having possession of a South Carolina boating safety certificate issued by the department in the person's name, unless the person: (1) was born on or before July 1, 2006; (2) is in possession of a" ]

[ "the reasoning. Each of those $10^2$ ways to post the first two letters can be combined with any of $10$ different ways to post the third letter, so there are $10^2\\cdot10=10^3$ ways to post the first three letters. Two more arguments of the same kind lead to the conclusion that there are $10^5$ different ways to post the $5$ letters. Note that I’m assuming that you’re allowed to post more than one letter in the same letter box. If that’s not the case, the reasoning is similar, but the answer is quite different. If you can post at most one letter in a box, there are still $10$ different ways to post the first letter. After you’ve done that, however, there are only $9$ ways to post the second letter, since you can’t use the first letter box again. Thus, you get only $10\\cdot9$ different combinations of letter boxes for the first two letters. Similarly, after they’ve been posted you must pick one of the $8$ remaining letter boxes for the third letter, so each of the $10\\cdot9$ ways of posting the first two letters gives you only $8$ ways of posting the first three. As a result, there are $10\\cdot9\\cdot8$ ways of posting the first three letters. Two more arguments of the same kind lead this time", "to the conclusion that there are $10\\cdot9\\cdot8\\cdot7\\cdot6=30,240$ different ways to post the letters if you can use each letter box at most once.", "This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5* 5 = \\boxed{\\textbf{(E) }25}$", "for its position. . . There are $3\\cdot1 \\,=\\3$ ways. In the fifth row, place a letter. There are 2 choices for the letter, 3 choices for its position. . . There are $2\\cdot3 \\,=\\,6$ ways. So, there are: . $18\\cdot5\\cdot12\\cdot3\\cdot6 \\:=\\:19,\\!400$ ways . . to place one letter in each row. For the last letter, there 6 choices of boxes. Hence, there are: . $19,\\!400\\cdot6 \\:=\\:116,\\!640$ ways . . to place the six letters in the boxes. But there is duplication caused by . . the two identical L's and two identical E's. Therefore, there are: . $\\frac{116,\\!640}{2!\\,2!} \\:=\\:29,\\!160$ ways . . to place the letters of the word LETTER in the boxes.", "# Find number of ways.. The number of ways in which we can post 5 letters in 10 letter boxes is..... I already have the answer given to me but I am not able to get there(Only answer is given not solution). Please answer with appropriate solution for understanding. - If you had only one letter, there would be $10$ ways to post it. With two letters you’d have $10$ ways to post the first letter, and no matter which letter box you used, you could still post the second letter in any of the $10$ letter boxes. Thus, you’d have $10^2$ possible choices. If this isn’t clear, imagine that the letter boxes are labelled A through J. If you post the first letter in box A, you can post the second in any of the $10$ boxes; that’s a total of $10$ different ways to post the two letters. If you post the first letter in box B, you can again post the second in any of the $10$ boxes; that’s another $10$ different ways to post the two letters. Thus, there are $10$ different ways to post the two letters for each of the $10$ ways to post the first letter, for a grand total of $10\\cdot10=10^2$ ways to post the two letters. Now just extent", "there are $2$ slots. This means we have \"$6$ choose $2$\", which I calculated to be $15$. They are saying the answer is $B$. 1. Four seniors and two juniors: ${5 \\choose 4}{5 \\choose 2} = 5 \\cdot 10 = 50$ possibilities 2. Five seniors and one junior: ${5 \\choose 5}{5 \\choose 1} = 1 \\cdot 5 = 5$ possibilities In total, we find $50 + 5 = 55$ possible arrangements.", "both Ts and both Ms, there are no more letters left to choose, so we get $$1$$ more choice. In total, we have $$2(1) + 5(3) + 1 = 18$$ distinct collections of consonants. Therefore, the number of distinct collections of letters is $$4 \\cdot 18 = \\boxed{72}.$$", "select the duplicated letter, ${4 \\choose 2}=6$ ways to choose the positions for that letter, and $2$ ways to place the other two. So we have $24+3\\cdot 6 \\cdot 2=60$ successes.", "refer to the symbols \"0\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\", and \"9\". There are a total of ten digits. The problem told us that we cannot have the first digit of the code be zero -- this means there are 9 total choices for the first digit of the lock box code. For the second and third digits of the lock box code, we are not told a restriction, and so there are 10 choices for each. For the fourth digit of the code, we may pick any odd digit: there are 5 total choices here. This means there are a total of $$9 \\cdot 10 \\cdot 10 \\cdot 5=4500$$ choices for a code.", "be $0 \\pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \\cdot 2 \\cdot 2 = \\boxed{040}$ numbers.", "are going to consider the second slot--how many choices do we have to fill that slot?0208 Well, we used to have 5 objects; we used to have a, b, c, d, and e.0213 But we used one of them in this first slot--we used one object on the first slot, so now we only have four objects for the second one.0218 Thus, there are four choices; now, notice: by this logic, we never actually mention which object was picked--just that we used some object.0227 We don't have to care what object is picked; let's really go ahead and make this clear.0234 Consider if we had a, b, c, d, and e; if we had chosen a to go into that first one, well, then, we would have b, c, d, and e for this box here.0238 So, we would have four possibilities for that box.0249 But what if we had chosen a different letter to go into the first box?0252 If instead we had chosen letter d to go into it, well, once again, we are still going to have 1, 2, 3, 4 choices: a, b, c, e.0256 So, we are going to have the same number of choices to go into that slot.0264 So, it doesn't matter which choice went into the first slot; we are", "$2$ ways to serve their meals. Thus, there are $(3\\cdot2)^3=\\boxed{216}$ ways to serve their meals. Note: This solution gets the correct answer through coincidence and should not be used. ~ dolphin7", "Dec 14 '15 at 17:45 The first letter can be posted in any of the $2$ post boxes. Therefore, it has $2$ choices. Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the $2$ post boxes. Therefore, the total number of ways the $5$ letters can be posted in $2$ boxes is $\\color{red}{ 2\\times 2\\times 2\\times 2\\times 2=32}.$", "$D$ together? Duct-tape $C$ and $A$ together. There are 2 ways: . $\\boxed{CA}$ or $\\boxed{AC}$ Duct-tape $N$ and $D$ together. There are 2 ways: . $\\boxed{ND}$ or $\\boxed{DN}$ We have six \"letters\" to arrange: . $\\{A,E,L,R,\\boxed{CA},\\,\\boxed{ND}\\, \\}$ . . They can be arranged in: ${\\color{blue}6!}$ ways. Hence, there are: . $2\\times 2 \\times 6! \\:=\\:2880$ ways to have $C,A$ and $N,D$ together. Therefore, there are: . $10,080 - 2,880 \\;=\\;\\boxed{7,200}$ ways . . in which $C$ and $A$ are together, but $N$ and $D$ are not.", "pairs of letters you could choose from the remaining six letters C, D, E, F, G and H. You now have $15$ possible sets of four letters. Each set of four letters can be arranged in $4! = 24$ different ways, giving you $15 \\cdot 24 = 360$ possible strings.", "first card, $4$ available choices for the second card, $3$ for the third card, $5$ for the fourth card, and $4$ choices for the last card. This leads to a total of $5\\cdot 4\\cdot 3\\cdot 5\\cdot 4=1200$ configurations for this case. There are $4$ cases of this type. Adding the cases up gives $2\\cdot 120+4\\cdot 600+4\\cdot 1200=7440$ \"happy\" configurations in total. This means that the probability that Kathy is happy will be $\\dfrac{7440}{10\\cdot 9\\cdot 8\\cdot 7\\cdot 6},$ which simplifies to $\\dfrac{31}{126}.$ So the answer is $31+126=\\boxed{157.}$", "$$2{6\\choose 2}+6{5\\choose2}=90$$ admissible placements for these three letters. For each such placement the remaining $5$ letters $RRGED$ can then be placed arbitrarily in the $5$ remaining cells in ${5!\\over2!}=60$ ways. It follows that there are $90\\cdot60=5400$ admissible arrangements in all.", "The total number of combinations are $7735+70+15+6=\\boxed{7826}.$", "Precalculus (10th Edition) $1600000.$ We know that the number of arrangements of $n$ objects in $r$ slots (where in a slot only $1$ of the $n$ elements can be put) is: $n^r$. If there are $k$ experiments, the first one can be done in $a_1$ ways, the second one in $a_2$ ways... the last one in $a_k$, then there are $a_1\\cdot a_2\\cdot...\\cdot a_k$ ways of doing the experiments together. Hence here, since there are $8$ digits for the first choice, $2$ digits for the last choice, $10$ digits for the remaining five, the number of possibilities: $8\\cdot10^5\\cdot2=1600000.$", "problem. In problems like this, we should use stars and bars. There are $\\binom{5}{2}=10$ ways to assign three 2s to three distinct letters, and there are $\\binom{3}{2}=3$ ways to assign one 251 to three distinct letters. Multiplying, we get $3\\cdot10=\\boxed{30}$." ]

[ "obtain a lifetime combination license which grants the same privileges as a statewide combination license from the department at its designated licensing locations. The licensing fees are: (1) for a Type 3 if at the time of application the individual is under two years of age: three hundred dollars; (2) for a Type 4 if at the time of application the individual is at least two years of age but under sixteen years of age: four hundred dollars; (3) for a Type 5 if at the time of application the individual is at least sixteen years of age but under sixty-four years of age: five hundred dollars. (B) A resident holder of a lifetime combination license may add the privilege of statewide saltwater recreational fishing for the following fees: (1) for a Type 13 if at the time of application the individual is under two years of age: one hundred twenty dollars; (2) for a Type 14 if at the time of application the individual is at least two years of age but under sixteen years of age: one hundred sixty dollars; (3) for a Type 15 if at the time of application the individual is at least sixteen years of age but under sixty-four years of age: two hundred dollars. (C) A resident age sixty-three who holds", "lifetime statewide combination license from the department at its designated licensing locations, which grants the same privileges as an annual combination license. The license fee is based on the age of the applicant. If at the time of application the individual is: (1) under two years of age, the fee is three hundred dollars; (2) at least two years of age, but less than sixteen years of age, the fee is four hundred dollars; (3) at least sixteen years of age but less than sixty-four years of age, the fee is five hundred dollars. (B) A resident who holds a lifetime combination license may obtain the privilege of statewide saltwater recreational fishing from the department at its designated licensing locations. The license fee is based on the age of the applicant. If at the time of application the individual is: (1) under two years of age, the fee is one hundred twenty dollars; (2) at least two years of age but less than sixteen years of age, the fee is one hundred sixty dollars; (3) at least sixteen years of age but less than sixty-four years of age, the fee is two hundred dollars. (C) A resident who holds a lifetime combination license may obtain the privilege of hunting migratory waterfowl from the department at its designated licensing", "By admin / August 27, 2019 It is essential to obtain a boating license if you want to operate a boat under the age of 18. If you do not get your consent under this age, it will be illegal to operate a vessel or Jet Ski. To get your license, you will need to enroll in the concerned course first, take the class, and have to pass the test. Here are the three simple steps mentioned below, which you will have to follow to acquire your license and enjoy boating legally. ## Step 1: Find A Course ### Look on a State Level The first step to get your boating license is to pick a course. These courses are done on state-level. Do your research and learn about your state’s requirements because they vary from one state to another. You should complete an approved boating license course to operate legally on state waters. ### Choose Classes Now it’s up to you whether you want to pick online or in-person classes. Most states offer you both facilities and your license will be equally valued either way. It is significant to know that your state will have their website for in-person classes. If you are interested in taking online courses, then they will conduct your tests and lectures properly.", "permits, we will also briefly look at their $E_{\\infty}$-version. Please email vb8 at illinois dot edu for the zoom details.", "# How Many License Numbers Consist of $4$ digits and $4$ letters, where one letter or digit is repeated? If license numbers consist of four letters and four numbers how many different licenses can be created having at least one letter or digit repeated? $$26 \\cdot 26 \\cdot 26 \\cdot 26 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10 = 4569760000$$ Is this right? - You should consider permutations of letters (L) and numbers (N). For example, LLLLNNNN, LNLNLNLN, NNNNLLLL, ... - are different. – Oleg567 Aug 6 '13 at 3:23 @Oleg567 so you are saying the answer is? – suffix Aug 6 '13 at 3:25 @Oleg567 Thank you for your comment. I didn't realize that. – Alraxite Aug 6 '13 at 3:32 @Alraxite so my answer is not right? – suffix Aug 6 '13 at 3:39 @suffix No, what you've calculated is the total number of licences which are of the form $\\text{LLLLNNNN}$ where $\\text{L}$ and $\\text{N}$ denotes a letter and a number respectively. – Alraxite Aug 6 '13 at 3:47 Assuming the licences are of the form $\\text{LLLLNNNN}$ where $\\text{L}$ and $\\text{N}$ denote letters and numbers respectively: Number of licences with no letters or numbers repeated: $26\\cdot 25\\cdot 24\\cdot 23\\cdot 10\\cdot 9\\cdot 8\\cdot 7$. So, number of licences with at least one letter or number repeated:", "must have a number $M$ of box states available, and infront of each part of the sum in $Z$ there must also be $M \\choose N-n$. – s.harp Jan 5 '16 at 10:03", "license to operate a vessel issued by the United States Coast Guard in the person's name, regardless of the expiration date on the license; (3) is in possession of a merchant mariner credential issued by the United States Coast Guard in the person's name, regardless of the expiration date on the credential; (4) is a nonresident in possession of a boater education certificate, or an equivalency, issued by another state in the nonresident's name; or (5) is accompanied by a person at least 18 years old who: (a) is in possession of a South Carolina boating safety certificate issued by the department in the person's name; or (b) meets one of the criteria in items (1) through (5) of this paragraph. (C) The department shall partner with motorboat rental businesses to ensure that: (1) a motorboat, including a personal watercraft, is not rented or leased to any person for operation on the waters of the state unless the renter or lessee has a valid watercraft operator's permit or is an exempt operator and is 18 years of age or older; (2) the motorboat rental business lists on each motorboat rental or lease agreement the name and age of each operator who is authorized to operate the motorboat or personal watercraft. The renter or lessee of the motorboat must", "employment,; vehicle make, model, color, and license tag number, including work vehicles, and the permanent or frequent location where all vehicles are kept;, fingerprints and palm prints;, Internet identifiers; passport and immigration documents; and a photograph; (b) the name, address, and county of each institution of higher learning, including the specific campus location, if the person is enrolled, employed, volunteers at, interns at, or carries on a vocation there; (c) the vehicle identification number, license tag number, registration number, and a description, including the color scheme, if the person lives in a motor vehicle, trailer, mobile home, or manufactured home and the permanent or frequent location where all vehicles, trailers, mobile homes and manufactured homes are kept; and (d) the hull identification number, the manufacturer's serial number, the name of the vessel, live-aboard vessel, or houseboat, the registration number, and a description of the color scheme, if the person lives in a vessel, live-aboard vessel, or houseboat; and (e) the tail number, manufacturer's serial number, and model of any aircraft, and a description of the aircraft, including the color scheme, and the permanent or frequent location where all aircraft are kept, if the person owns or operates an aircraft.\" SECTION 9. This act takes effect upon approval by the Governor./ Renumber sections to conform. Amend title to conform.", "least what age? 12 with a valid boating safety certificate Which statement is true concerning visual distress signal? Flares are rated for day, night or combined day/night use Which of the following indicates an emergency situation aboard Orange smoke billowing from a boat On Maryland waters all children under what age must wear a u.s coastguard approved PFD while underway in a recreational vessel under 21 feet in length 13 Which type of power boat engine must have a backfire flame arrester Gasoline inboard Which of the following is a feature of a type IV PFD It is designed to be thrown Where is the best place to store PFDs on a boat A location that makes them readily accessible When operating on Maryland waters at what depth must a PWC operate at minimum wake speed 18 inches ### Sets with similar terms Boating Safety Course- Texas 10 terms MIA_LUTON PA Boating Safety Test 47 terms cocodorey Safe Boating⛵️ 61 terms Kayleigh_Moses va boaters course quiz 2 10 terms ali357 ### Sets found in the same folder Page 1 chap 4 quiz 7 terms shel449 Page 3 8 terms shel449 Page 4 7 terms shel449 Boating Test Part 2 20 terms tony_pham20 ### Other sets by this creator Vocab 15 20 terms shel449 ap lang vocab 14 24", "$\\text{Total number of licences} - 26\\cdot 25\\cdot 24\\cdot 23\\cdot 10\\cdot 9\\cdot 8\\cdot 7$. Where the total number of licences is $26^4\\cdot 10^4$ which is what you calculated. Assuming the letters and numbers can appear in any order: You can obtain this answer by multiplying the answer obtained above by $\\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic. Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$: Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$): The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values. Now, number of licences with no letters or numbers repeated: $36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$. So, number of licences with at least one letter or number repeated: $\\text{Total number of licences} - 36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$ The total number is given by $36^8$. - If the licence plate is of the form $LLLLNNNN$ where L stands for letter and N for integer between 0 and 9 inclusive then no. Number", "# A bicycle license plate requires 3 digits, each ranging from 0-9. How many different combinations are possible? Mar 20, 2018 $10 \\cdot 10 \\cdot 10$ to find the number of combinations. $10 \\cdot 10 \\cdot 10 = 1000$ possible combinations", "1 T. $${\\small 10. \\enspace (\\textrm{a}).\\hspace{0.7em}}$$ A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats. $${\\small \\hspace{1.6em} (\\textrm{b}).\\hspace{0.6em}}$$ Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases. $${\\small \\hspace{2.8em} (\\textrm{i}).\\hspace{0.7em}}$$ The odd digits are together and the even digits are together. $${\\small \\hspace{2.8em} (\\textrm{ii}).\\hspace{0.6em}}$$ The 2s are not together. As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) . 1. I have a question about the answer to 5a problem 6; the envelope problem. If the total number of possible ways to fill the five addressed envelopes is 5! = 120, and the question asked “how many ways can the letters be placed in each envelope without getting every letter in the right envelope” then shouldn’t that one possible way of getting them placed right be subtracted from that total, leaving the answer 120 – 1 = 119 ? Please help me understand the why the answer remains", "of Natural Resources, there is established a pilot program within the department named the Public/Private Partnerships - Boating Safety. The pilot program shall administer a boating education course and increase overall boater safety and boater safety education. A list of approved courses must be provided on the department's website. The following persons must be provided a South Carolina boating safety certificate in both physical and electronic forms by the department: (1) a person who successfully completes a boating education course administered or approved by the department; (2) a person who provides satisfactory proof to the department that the person was issued a boater education certificate, or an equivalency, by another state; and (3) a person who provides satisfactory proof to the department that the person was issued a license to operate a vessel by the United States Coast Guard or was issued a merchant mariner credential by the United States Coast Guard. (B) A person may not operate, upon the waters of this State, a vessel powered by an engine of ten horsepower or greater, a personal watercraft, or a specialty propcraft without having possession of a South Carolina boating safety certificate issued by the department in the person's name, unless the person: (1) was born on or before July 1, 2006; (2) is in possession of a", "the numbers. If repetitions are allowed in the letters but not in the digits _____ ---- How many different computer passwords are possible? a. So to find the total number of letter combinations we multiply the number of choices for the first digit by the number of choices for the next digit, in this case 26*26, for a total of 676. For example, if you want to use the alphabets a,b,c and three. To get the number of permutations here, you need to multiply the different values each of the characters can have: $26\\cdot10\\cdot10\\cdot10=26,000$ 2) A letter followed by four numbers. The Most Beautiful Equation in Math - Duration: 3:50. In this case you first are choosing one letter out of 26, followed by one lett. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. a 7-character password consists of two numbers (from 1 to 9) followed by three letters (from a to z) followed by two more numbers (from 1 to 9). An encyclopedia has eight volumes. For example, John Doe 123-45-6789 would be jdo56789. 4 million passwords each containing four digits, the DataGenetics Web site reported that the most commonly used four-digit sequence (representing 11 percent of. 5", "The Norwegian Registers of Fishing Boats are available online by clicking on the link. These cover approximatley every second year from 1922 to 2002. The Norwegian Port Letters and Numbers are made up of three rather than the usual two parts. The first/leftmost letter denotes the Fishery District, the right hand letter/s denote the Home Port within that Fishery District and the number between the two sets of letters is that particular boat’s number within her Home Port. For example: In 1922 F-4-H, the Lyngen was from the District of Finnmark and No 4 in the port of Hammerfest in that District.", "}shoes|=8$ $|Outfit|=|Shirts|\\cdot|Pairs\\text{ } of\\text{ }pants|\\cdot |Pairs\\text{ } of\\text{ }shoes|=8\\cdot 4\\cdot5=160$ • Two dice are rolled, one blue and one red. How many outcomes are possible? $Outcomes=\\{(color,position)|color\\in Colors,position\\in Positions\\}$ $|Colors|=2$, $|Positions|=6$ $|Outcomes|=2\\cdot6=12$ • The Braille system of representing characters was developed early in the nineteenth century by Louis Braille. The characters, used by the blind, consist of raised dots. The positions for the dots are selected from two vertical columns of three dots each. At least one raised dot must be present. How many distinct Braille characters are possible? $|Dot|=|Raised \\text{ } dot|+|No \\text{ } raised \\text{ } dot|=2$ $|Dot|^{n\\text{ }dots={columns\\times rows}}=2^{6}=64$ are Braille characters possibles. • The options available on a particular model of a car are five interior colors, six exterior colors, two types of seats, three types of engines, and three types of radios. How many different possibilities are available to the consumer? $|Possibilities|=|Interior\\text{ } colors|\\cdot |Exterior\\text{ } colors|\\cdot|Seats|\\cdot|Engines|\\cdot|Radios|=5 \\cdot 6 \\cdot 2 \\cdot 3 \\cdot 3=540$ • How many different car license plates can be constructed if the licenses contain three letters followed by two digits if repetitions are allowed? if repetitions are not allowed? $\\text{Car license plates}=\\{(letter,letter,letter,digit,digit)|letter \\in \\text{English alphabet},digit \\in \\text{Decimal numeral system}\\}$ $|\\text{English alphabet}|=26$ $|\\text{Decimal numeral system}|=10$ If repetitions are allowed, then $\\text{Car license plates}=26\\cdot26\\cdot26\\cdot10\\cdot10=1757600$ If repetitions are not allowed, then $\\text{Car", "there are a total of 22 counselors on staff, then how many of the counselors have a first aid certification? A) 7 B) 9 C) 11 D) 13 E) 15 Given data: 7 counselors have a sailing certification | P(Sailing C) = 7 Formula used: P(Total having C) = P(Sailing C) + P(First Aid C) - P(Both) If there are x counselors at the summer camp, x - 4 have either certification. For 22 counselors at the summer camp, only 18 have either of the 2 certifications. Since twice as many counselors have neither certifications as have both certifications, P(Both) = $$\\frac{4}{2} = 2$$ (as 4 have neither certification) Substituting values, we get $$18 = 7 + F - 2$$ -> $$18 = 5 + F$$ -> $$F = 13$$ Therefore, the total number of counselors that have a first aid certification is 13(Option D) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 12 Mar 2018 Posts: 85 Re: All but 4 of the counselors at a certain summer camp have a sailing ce [#permalink] ### Show Tags 02 Dec 2018, 05:53 Could anyone explain \"All but 4 of the counselors at a certain summer camp have a sailing certification, first aid certification, or both.\" this statement means? I cannot solve", "figure out the number of ways to choose each and then multiply them. For the initial sign-on your Username and Password will be the same. A password consists of three digits, 0 through 9, followed by three letters from an alphabet having 26 letters. The password has 3 letters followed by 3 digits. How many different possible passwords are there?. Glad we were able to help! fn+F11 fixed mine- thank you! my keyboard instead of typing 'E' it 'pastes information,, my 'o' changes the scgreen to full screen and 'u' adds a h so it becomes 'hu' and now each time I type 'p' this sign. A license plate is to consist of three letters followed by two digits. Repeated characters are allowed. You have to multiply the possibilities - you need two letters and five numbers, so for each one you figure out the number of ways to choose each and then multiply them. In the first space, you can place any of the digits from 0 to 9. How many different words can you make from the one word \" mississippi\" , with out using two letters twice , and with out using the actual word? W hat do the numbers in my wi driver's license number mean? the letter is the 1st letter of my", "# There are 10 bicyclists entered in a race. In how many different orders could these 10 bicyclist finish? Apr 13, 2017 #### Answer: 10! is the answer. #### Explanation: this is just like you are given 10 lines on a paper and you have to arrange 10 names on those 10 lines in all different ways . so , starting with the lowermost line, you can put one of 10 names on that line and then on the line above that you can put 1 of 9 names and so on . so the total ways to arrange all the names on those lines in all ways will be : $10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ = 10!", "# Counting different kinds of permutations • What's the number of strings of digits and letters of length $7$ without repetition? • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\\binom{10}{2}\\binom{21}{4} \\cdot 5 \\cdot 7!$ • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\\binom72\\binom54 \\cdot 10 \\cdot 9 \\cdot 21 \\cdot 20 \\cdot 19 \\cdot 18 \\cdot 5.$ I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below: • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there? We choose places first and for each of those we choose symbols: $\\binom63\\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$ What makes these two problems different? Why does choosing places first in the second problem results in wrong answer? For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\\times9\\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third" ]

[ "units, the hypotenuse will be 5 units if you have a true 90-degree angle between the sides. {\"smallUrl\":\"https:\\/\\/www.wikihow.com\\/images\\/thumb\\/5\\/58\\/Construct-a-90-Degrees-Angle-Using-Compass-and-Ruler-Step-1.jpg\\/v4-460px-Construct-a-90-Degrees-Angle-Using-Compass-and-Ruler-Step-1.jpg\",\"bigUrl\":\"\\/images\\/thumb\\/5\\/58\\/Construct-a-90-Degrees-Angle-Using-Compass-and-Ruler-Step-1.jpg\\/aid3172782-v4-728px-Construct-a-90-Degrees-Angle-Using-Compass-and-Ruler-Step-1.jpg\",\"smallWidth\":460,\"smallHeight\":345,\"bigWidth\":728,\"bigHeight\":546,\"licensing\":\"", "Let length a=3b=4, and hypotenuse c=5. Construct a square using leg aa as the right side of the square. It will be 9 square units (${a}^{2}$). Construct a square using leg b as the top side of its square, so it is 16 square units (${b}^{2}$). Cut out another 5 x 5 square and line it up with hypotenuse, c, so the square is ${c}^{2}$. Think: what is 9 square units + 16 square units? It is 25 square units, the area of ${c}^{2}$. ## Right triangle altitude theorem The right triangle altitude theorem tells us that the altitude of a right triangle drawn to the hypotenuse c forms two similar right triangles that are also similar to the original right triangle. Construct △ABC so that hypotenuse c is horizontal and opposite right angle C, meaning legs aa and bb are intersecting above c to form the right angle C. This puts ∠A to the bottom left, and ∠B to the bottom right. Construct an altitude (or height) h from the interior right angle C to hypotenuse c (so it is perpendicular to c). This altitude h creates two smaller triangles inside our original triangle. The altitude divided ∠C, and also created two right angles where it intersected hypotenuse c. Call the point where the altitude h touches", "# The figure below shows a right triangle: A right triangle is shown with hypotenuse equal to q units ###### Question: The figure below shows a right triangle: A right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units. The angle b What is r ÷ q equal to? (6 points) Select one: a. tan y° b. sin x° c. sin y° d. tan x° $The figure below shows a right triangle: A right triangle is shown with hypotenuse equal to q units$ ### Atank was filled with 4 and half litres of water. 1.175 litres of water was poured out from the tank. Atank was filled with 4 and half litres of water. 1.175 litres of water was poured out from the tank. how much water was left in the tank?... ### Which organizational style would best fit a piece written in response to thisprompt?Prompt: Air pollution is a major problem in cities Which organizational style would best fit a piece written in response to this prompt? Prompt: Air pollution is a major problem in cities such as Los Angeles, Mexico City, and Beijing. Write an essay that offers several different ways cities can reduce air pollution. When you finish, you will enter ...", "is 180 - 90 - 70 = 20 degrees. Theorem: If the altitude (dotted) is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to each other and the original B ~ ~ ADB BDC ABC. 1 Similar Right Triangles. Example 1: Use Figure 3 to write three proportions involving geometric means. The altitude divides the original triangle into two smaller, similar triangles that are also similar to the original triangle. Example 2 : In the diagram given below, prove that ΔABC ≅ ΔFGH. You are calculating the hypotenuse before you're actually acceptin height and width as parameters. So if we can find the length of one side, then we can multiply that length by $3$ to get the perimeter of the triangle. When working with right triangles your hypotenuse is always opposite the 90 degree angle. If you draw out your triangle you will see that RT is opposite the right angle so that is your hypotenuse. Explain why ABC ∼ ACD. A right triangle is a type of triangle that has one angle that measures 90°. The two theorems above, states that in the triangle below Let and be the legs of the triangle with hypotenuse. Find the two sides and the hypotenuse of the triangle. Altitude CD is drawn", "He essentially manipulated right triangles to produce isosceles triangles, scalene triangles, rectangles, isosceles trapezoids, isosceles trapezoids with three equal sides, and In a right triangle, the altitude with the hypotenuse \"c\" as base divides the hypotenuse into two lengths \"p\" and \"q\". Not completely sure why, but I know not everything I type in there could possibly be a right triangle. Important Terms Associated to Triangles H. The side opposite to the right angle is called as the hypotenuse while the other two sides are legs of the right angled triangle. Altitude - A line segment joining a vertex of a triangle with the opposite side such that the segment is perpendicular to the opposite side. In a right angle triangle a line drawn form the right angle vertex to the mid point of the hypotenuse will create two isoscolese triangles. For each mathematical calculation, your program should prompt the user for each. A garden is designed in the shape of a rhombus formed from 4 identical 30°-60°-90° triangles. So you can set up a proportion:. Intellectual Property Rights. 2, what is the length of ST, to the nearest tenth? 1) 6. The altitude drawn to the leg of a right triangle forms two triangles that are similar to each other. Learn how to make over 43 Triangle", "60 degrees. The hypotenuse is x√2 where x=side length. Area of a triangle is equal to 1/2 times the base of the triangle times the height of the triangle. an equilateral triangle is described on the diagonal of a square. The height of an equilateral triangle is x√3/2 where x=side length. $\\triangle ABC$ is an equilateral triangle with area 24. Therefore the hypotenuse is 6√2. IF the diagonal of the square is 20 cm. Right triangles are perhaps the easiest triangle to make, because of the fact that two of their sides are straight lines, going either vertical or horizontal. The following grids show how to make a right triangle with a 4 block base and height, and one with a 9 block base and height. Right Triangle Equations. Calculations at an equilateral triangle or regular trigon. Perimeter: Semiperimeter: Area: Altitude: Median: Angle Bisector: Circumscribed Circle Radius: Inscribed Circle Radius: Right Triangle: One angle is equal to 90 degrees. Needed the height of an equilateral triangle for a sewing project so I could compute what size square to cut for making 1.5 inch wide bias strips when using a diagonal cut of … That's just basic geometry. A rectangle has two diagonals, and each is the same length. It is interesting to note that the altitude of an", "where you have 60 degrees and 30 degrees. The special triangles always come in handy. So what this means is that if we have a right triangle in our actual question with legs of length <i>l</i> over two, and so we’re going to be multiplying each leg here by <i>l</i> over two, which is five centimeters over two. In which case the hypotenuse is going to be five over two multiplied by root two, which works out to five over root two. That’s the way I think about it, but maybe you have a preference for different method, which would be to say well, I know that sine of 45 degrees is the opposite which is <i>l</i> over two, divided by the hypotenuse which is <i>r</i>. This five over two is <i>l</i> over two. And I want to solve for <i>r</i>, in which case <i>r</i> is going to be <i>l</i> over two divided by <i>sin 45</i>, but <i>sin 45</i> is one over root two, or you can get its decimal form, plug that in into your calculator. Either way, you’ll get the same answer. So it’s <i>l</i> over two times one over, or divided by one over root two, so multiply top and bottom by root two, and these cancel, leaving us with <i>l</i> root two over two,", "triangles evaluation pi answer key unit 3 lesson 1 c3 lesson 1 classifying. NCERT curriculum (for CBSE/ICSE)Class 9 - Triangle. Lesson Guide. Solution: The hypotenuse is 2 times the length of either leg, so. For example, 3 is a triangular number and can be drawn like this. Math Tutor Math Skills Math Lessons Teaching Math Math Education Math Word Walls Math Charts Maths Solutions Math Notes. Possible answer: AC#and FD# 16. 8 - Key C) TRANSFORMATIONS L 1, 4 1) If triangle JKL in the xy-plane shown above is shifted 7 units to the right and 4 units up, what would be the coordinates of point L after the shift? e) 8,0 1 2) Triangle DEF in the xy-plane above will be translated 3 units to the right and then 2 units down. UE9 use common present simple forms [positive, negative, question] and contractions to talk about what you want and like and habits and facts (also is practiced. lessons 1-2 lessons 3-4 lessons 5-6 lesson 7 lesson 8 lesson 9 Consolidation Lessons 12-13 Test yourself. Draw a triangle with angle measures of 45°, 45°, and 90°. Explain how you could find the total area of the figure. Other students may duplicate the triangle to create a square and figure out the area of the square is double", "hypotenuse] of the given right triangle [long leg 8 units, angle 30 degrees between long leg and hypotenuse] 6. Stonehenge in Salisbury Plains, England, was constructed using solid stone blocks weighing over 97000 pounds each. Lifting a single stone required 550 people, who pulled the stone up a ramp inclined at an angle of 8 degrees. To the nearest tenth of a foot, approximate the distance that a stone was moved along the ramp in order to raise it to a height of 34 feet above the level ground. 7. Approximate sec(78 degrees 18') to four decimal places. 8. csc(x)/cot^2(x) is equivalent to which of the following? csc(x)cot(x), sec(x)cot(x), csc(x)tan(x), cos(x)sin(x), sec(x)tan(x) 9. Find the exact value of tan(theta) if theta is in standard position and the terminal side of theta is in quadrant II and is parallel to the line 3x+5y = 9. 10. If sec theta = 8 and cot theta < 0, find the exact value of sin theta. 11. Let P(t) = (-7/25, -24/25) be the point on the unit circle that corresponds to t. Find the exact value of P(-t+pi). 12. Complete the statement: As x -> pi/2+, tan(x) -> ?. 13. Approximate, to the nearest 0.01 radians, all angles theta in the interval [0, 2pi) that satisfy the equation cot(theta) = -2.3412.", "practice to continue. These circles can be used if the website does not work. Skip practice slide Let’s consider our special right triangles and their trigonometric values: A B C A B C A B C √𝟑 𝟐 1 1 √𝟐 𝟐 1 𝟏 𝟐 30° 45° 60° √𝟑 𝟐 √𝟐 𝟐 𝟏 𝟐 Now, let’s practice! If we let the hypotenuse be 1, then sinΘ=y and cosΘ=x Since the hypotenuse is the radius of a unit circle, we consider it to be radius = r = 1. sinΘ cosΘ tanΘ 30° √3 45° 1 60° √𝟑 𝟐 𝟏 𝟐 √𝟐 𝟐 √𝟐 𝟐 √𝟑 𝟐 √𝟑 𝟑 𝟏 𝟐 Let’s consider our special right triangles and their trigonometric values: B C A B C A B C √𝟑 𝟐 1 1 √𝟐 𝟐 1 𝟏 𝟐 30° 45° 60° √𝟑 𝟐 √𝟐 𝟐 𝟏 𝟐 Next Slide If we let the hypotenuse be 1, then sinΘ=y and cosΘ=x Since the hypotenuse is the radius of a unit circle, we consider it to be radius = r = 1. or Answers Now, you practice… sinΘ cosΘ tanΘ 30° 45° 60° Use the following finger tricks! Hold your hand to remind you of the special angles in the first quadrant. Think of cosine on top and sine on the", "can find the area of the triangle by Heron's Formula: Let have circumcenter and incenter .Then. This 3, 4, 5 right triangle has an inradius of 1. 9. Triangle ABC has circumcenter O. This line containing the opposite side is called the extended base of the altitude. A line parallel to hypotenuse AB of a right triangle ABC passes through the incenter I. And this formula comes from the area of Heron and . Also, because they both subtend arc . and then simplifying to get If has inradius and semi-perimeter, then the area of is .This formula holds true for other polygons if the incircle exists. The incenter of a right triangle is equidistant from the midpoint of the hy-potenuse and the vertex of the right angle. Divide 60 cubic inches by 4 to get 15 cubic inches. We let , , , , and . JavaScript is not enabled. Hence, since triangle AED Author has 26.1K answers and 13.3M answer views The circumradius of a right angled triangle is half the hypotenuse and the centre is the midpoint of the hypotenuse. The intersection of the extended base and the altitude is called the foot of the altitude. Examples: Input: r = 2, R = 5 Output: 2.24 What are the advantages and disadvantages of individual sports", "triangle, with a=H=height, and angles as 90, 45 and 45 degrees, then b becomes the hypoteneus=28, with M=1 kg, F=3 Kg, the equation goes negative. i guess i am wrong somewhere... Last edited: Jan 9, 2009 4. Jan 10, 2009 ### Thaakisfox The unit for force is Newtons. So I guess under a force of 3kg, you mean weight of 3kg, this means that: F=3kg*10m/s^2 = 30N And your equation wont go negative. But if you say you want a right angled triangle with 45 45 90, and you know the height, then you already determine all of the sides... :D", "Emotionally Unavailable Woman Traits, Bi-fold Window Shutters, What Is Not A Polynomial, Goochland County Schools, Ach Medical Term, Gavita Pro 1000e Review, Silicone Caulk For Showers, \" /> # segments in triangles practice answers #### segments in triangles practice answers Perpendicular Bisector Angle Bisector Median Circumcenter Incenter Centroid Orthocenter Obtuse Triangle two of the altitudes are outside the triangle. Melen Math . An altitude of a triangle is a segment from any vertex perpendicular to the line containing the opposite side. To start, identify what kind of segment LK is. 7-4 Practice Form K Similarity in Right Triangles Identify the following in right kXYZ. Neither; the distance is the same because BC O AX and AB O XC. Shed the societal and cultural narratives holding you back and let step-by-step SpringBoard Geometry textbook solutions reorient your old paradigms. 6. Practice questions. (Caution: Make sure the three sides satisfy the Triangle Inequality Theorem.) Share practice link. 1. the hypotenuse 2. the segments of the hypotenuse 3. the altitude to the hypotenuse 4. the segment of the hypotenuse adjacent to leg ZY Write a similarity statement relating the three triangles in each diagram. 5-1 Practice (continued) Form G Midsegments of Triangles 13 mi 2.9 mi 3.5 km 70 73 46 41.5 BC is shorter because BC is half of 5", "hypotenuse: 15 m. PDF: Practice-Area 2 squares: 11: WS PDF: Practice-Area 3 rectangles: 13: WS PDF: Practice-Area 4 parallelograms: 5: WS PDF: Practice-Area 5 rhombuses: 5: WS PDF: Practice-Area 6 trapezoids: 10: WS PDF: Practice-Area 7 n>4: 8: WS PDF: Practice-Area 8 compositions: 12: WS PDF: Practice-Geometric Probability: 8: WS PDF: Practice-Compositions of. The focus is on calculating the area of triangles and different quadrilaterals. Right Triangle: In this example, the children know that the area should be 4. (x +5)° 90° 155° 60; right scalene Classify each dashed triangle by its angles and by its. Explain why the two shaded triangles have the same area. The most important thing is that the base and height are at right angles. Area Addition Postulate The area of a region is the sum of the areas of its nonoverlapping parts. EXTRA PRACTICE Epigraphics, Inc. 4 \\text{ inches squared} . 75 km² 3) 1. Practice: Area challenge. Recommended Videos Detailed Description for All Area and Perimeter Worksheets. The height of the triangle is the perpendicular drawn on the base of the triangle. Determine the area of the triangle to the nearest tenth. Properties of Triangles. Area and Perimeter Worksheets. Area of #ABC =1bc(sin A) 2 B c a b AC Practice and Problem Solving EXERCISES For more exercises, see Extra Skill,", "left. Summer 2013 layout as of the end of Monday 24 June on the right. Summer 2013 the center was set at (0,0). On Tuesday I will introduce the transform attribute on a graphics element g to move the triangle: <g transform=\"translate(100,100)\"> and </g> The notes below are not yet edited to reflect this shift. Homework is to find the coordinates of the red circles based on the center being at (0,0) Homework: Using adjacent = r cos θ and opposite = r sin θ, try to work out the coordinates of the circle at the vertices of the equilateral triangle. Note that r = 100, the center of the triangle is located at (0, 0), and the angle θ is 60°. The grid we are using is on the right. #### Tuesday The angle θ is 60°. Your text book notes that a 30-60-90 triangle is another \"special\" triangle. They are special because they build equilaterial triangles and hexagons, among other reasons. The hypotenuse r is 100 pixels long. The angle θ is 60°. The adjacent side uses the cosine function: $\\mathrm{cosine}\\left(\\theta \\right)=\\mathrm{cos}\\left(\\theta \\right)=\\frac{\\mathrm{adjacent}}{\\mathrm{hypotenuse}}$ adjacent = r cos(θ). r = 100 and θ = 60° Thus adjacent = 100*cos(60°) To calculate this use WolframAlpha: 100*cos(60 degrees) The opposite side still uses the sine function: $\\mathrm{sine}\\left(\\theta \\right)=\\mathrm{sin}\\left(\\theta \\right)=\\frac{\\mathrm{opposite}}{\\mathrm{hypotenuse}}$", "altitude shown is. Our tool can help work out the circumscribed circle radius of any triangle to equiangular... You agree to our Cookie Policy you to use h is h b or, the altitude of.... Triangle step-by-step their properties including its perimeter, area of a right isosceles triangle with area 24 b,. Or, the altitude shown h is h b or, the altitude of b click you know! Of all parameters of the hypotenuse formula your questionnaire.Sending completion, area altitude or perpendicular height area = a² √3... Unique features: ( 1 ) all three sides are equal, so internal angles are 60° each linked. Triangle whose area S is 1 length of 8 cm, equilateral triangle area calculator degrees/3! Customize '' button above to learn more of any triangle, based on one known value, agree... Or, the within a fraction of seconds side given: area = a² * equilateral triangle area calculator 4... And difficult math problems or 180 degrees/3 - calculate area, perimeter of the hypotenuse formula the perimeter an... Are of equal length hb or, the solution for given inputs within a fraction of seconds to. Polygon formed by three line segments linked end-to-end just three formulas 's formula ) area of equilateral. Substitute the height in this triangle area calculator is designed to give the", "A is 60°, then BD is half of AB, because AB is equal to cos 60° =.. We are looking at two 30-60-90 triangles are similar ; that is, look the... Decide which of those angles is the proof that in a 30°-60°-90° triangle the are! By combining two other constructions: a 30 60 90 triangles is that the ratio 1 \\sqrt3:2\\. You won ’ t have to use triangle properties like the Pythagorean.. Cotangent is the University of Central Florida, where she majored in.... Triangles and also 30-60-90 triangles expert admissions guidance — for free which will become the hypotenuse --,. Major, a Guide to the opposite base side length 2 and with point D as cotangent. … the altitude of an equilateral triangle ABC is an equilateral triangle inscribed a. Is called the hypotenuse of course, when it ’ s exactly 45 degrees, the longer leg is same! Of cotangent function is the value of x constructions: a 30 60 90 triangle have... 3 inches Divorced parents with side length 2 and with point D as the cotangent of x what. Chancing engine takes into consideration your SAT Score for angle b and side c. example 3 the a... Colleges with an equilateral triangle each side is s, and the hypotenuse at b is its,. 3", "you are viewing a single comment's thread. [–]Theory of Computing 1 point2 points (2 children) sorry, this has been archived and can no longer be voted on Oh, now that I work it out, I see that you additionally need to memorize that tan(60 degrees) = sqrt(3). But that's all! 1. The 45-45-90 triangle is easy. Using the distance formula you can find out that the side lengths on a unit circle are 1-1-sqrt(2). So draw the triangle and use the soh-cah-toa definition to find the sin,cos,tan of either 45 degree angle. 2. The 30-60-90 triangle is a little harder. Let the shortest side be 1 unit long. Then by the tan(60 degrees) rule you memorized, the other leg of the triangle is sqrt(3). Now by the distance formula you can determine that the hypotenuse is 2. Then just apply soh-cah-toa. 3. At this point you can also get the sin,cos,tan of 0,90,180,270 degrees by applying the soh-cah-toa definition to values very close to 0,90,180,270. If you're suspicious that the limit will always exist, you're right; the tangent doesn't exist at 90 or 270 degrees. 4. Signature in each quadrant is the easiest of all. Just draw the 45 degree angle in that quadrant. Make it a triangle by going down (or up) to the x axis. Label", "side 1 unit long, then the hypotenuse must be 2 units long since the triangle can be thought of as half of an equilateral triangle. Call the length of the altitude $$a$$. By the Pythagorean Theorem, we can say $$a^2+1^2=2^2$$ so $$a=\\sqrt3$$. Now, consider another right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side $$y$$ units long. By the Angle-Angle Triangle Similarity Theorem, it must be similar to the right triangle with angles 30, 60, and 90 degrees and with sides 1, $$\\sqrt3$$, and 2 units long. The scale factor is $$y$$, so a triangle with angles 30, 60, and 90 degrees has side lengths $$y, y\\sqrt3,$$ and $$2y$$ units long. In triangle $$ABC, 2y=5$$ so $$y=\\frac 52$$. That means $$DB$$ is $$\\frac 52$$ units and $$DC$$ is $$\\frac 52 \\sqrt 3$$ units. In triangle $$EGH, y\\sqrt 3=4$$ so $$y=\\frac{4}{\\sqrt 3}$$. That means $$FG$$ is $$\\frac{4}{\\sqrt 3}$$ units and $$EG$$ is $$2 \\frac{4}{\\sqrt 3}$$ or $$\\frac{8}{\\sqrt 3}$$ units.", "175 -- 3 from PQR PQ^2 + QR^2 = 625 --- 4 Solve eqn 3 qnd 4 PQ^2 = 400 . Sub in 1 and find QS^2 = 144 so QS = 12 so area of triangle PQR = 1/2* 25* 12 = 150. Senior Manager Joined: 30 Nov 2008 Posts: 483 Schools: Fuqua ### Show Tags 13 Feb 2009, 07:38 2 KUDOS There is one formula that is worth remembering. In a right triangle, when a perpendicular(altitude) is drawn from the vertex opposite to hypotenuse on the hypotenuse, then (length of the altitude)^2 = product of the segments that it divides on the hypotenuse. Based on the given figure, applying the above formula, $$(QS)^2 = PS * SR$$ ==> $$QS = sqrt(16 * 9) = 12.$$ Now height is known, base is known. Applying the formula we get area of the trainagle to be 150. SVP Joined: 07 Nov 2007 Posts: 1789 Location: New York ### Show Tags 13 Feb 2009, 13:27 mrsmarthi wrote: There is one formula that is worth remembering. In a right triangle, when a perpendicular(altitude) is drawn from the vertex opposite to hypotenuse on the hypotenuse, then (length of the altitude)^2 = product of the segments that it divides on the hypotenuse. Based on the given figure, applying the above formula, $$(QS)^2 =" ]

[ "betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"?\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr This is a 30-60-90 triangle. Thus, the shorter leg is half as long as the hypotenuse. BC = \\frac12 \\cdot AB = \\frac12 \\cdot (( 2 * BC ) + BCrs) = ( BC ) + BCrs 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ABs. How long is AC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", \"?\", ABs ); AC * AC * ACr This is a 30-60-90 triangle. Thus, the longer leg is \\frac{\\sqrt{3}}{2} times as long as the hypotenuse. AC = \\frac{\\sqrt{3}}{2} \\cdot AB = \\frac{\\sqrt{3}}{2} \\cdot (ABs) = AC + ACrs", "$AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$. $[asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*\"24\", A--T, N); label(rotate(degrees(C-A))*\"24\", A--C, 2*NW); label(\"12\\sqrt 3\", C--D, E); label(\"12\\sqrt 3\", D--B, S); label(\"12\", A--D, S); pen p = fontsize(8)+red; MA(\"45^\\circ\", C,B,A,2); MA(\"30^\\circ\", B,A,T,2.5); MA(\"30^\\circ\", T,A,C,3.5); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, N); dot(\"T\", T, NE); dot(\"D\", D, S); [/asy]$ By 30-60-90 triangles, $AD=12$ and $CD=12\\sqrt{3}$. We also notice that $\\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\\sqrt{3}$. The base $AB$ is $12+12\\sqrt{3}$, and the altitude $CD=12\\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=\\boxed{291}$. -youyanli ## See also 2001 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "# 2006 AIME I Problems/Problem 14 ## Problem A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\\frac m{\\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\\lfloor m+\\sqrt{n}\\rfloor.$ (The notation $\\lfloor x\\rfloor$ denotes the greatest integer that is less than or equal to $x.$) ## Solution 1 $[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"T\",T,N);label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,NE);label(\"S\",S,NW);label(\"O\",O,SW);label(\"M\",M,NE); label(\"4\",(S+T)/2,NW);label(\"1\",(S+A)/2,NW);label(\"5\",(B+T)/2,NE);label(\"4\",(O+T)/2,W); dot(S);dot(O); [/asy]$ We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters). Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let", "(opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using cosine: betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", AC + ACrs, \"x\" ); arc([0, 5*sqrt(3)/2], 0.8, 270, 300); label([-0.1, (5*sqrt(3)/2)-1], \"{30}^{\\\\circ}\", \"below right\"); Cosine is adjacent over hypotenuse (SOH CAH TOA), so \\cos {30}^{\\circ} = \\dfrac{ACdisp}{x}. We also know that \\cos{30}^{\\circ} = \\dfrac{\\sqrt{3}}{2}. Solving for x, we get \\qquad x \\cdot \\cos{30}^{\\circ} = ACdisp \\qquad x \\cdot \\dfrac{\\sqrt{3}}{2} = ACdisp \\qquad x = ACdisp \\cdot \\dfrac{2}{\\sqrt{3}} \\qquad x = ACdisp \\cdot \\dfrac{2\\cdot\\sqrt{3}}{3} So, x = ABs. randRange( 2, 6 ) randFromArray([ [1, \"\"], [3, \"\\\\sqrt{3}\"] ]) 2*BC + BCrs randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"x\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr We know the length of the hypotenuse of this triangle, and want to find the length of a leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "endpoints of a side of the larger hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy]", "\"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and BC = BC + BCrs. How long is AB? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", BC + BCrs, \"\", \"?\" ); 4 * BC * BC * BCr This is a 30-60-90 triangle. Thus, the hypotenuse is twice as long as the shorter leg. AB = 2 \\cdot BC = 2 \\cdot (BC + BCrs) = ( 2 * BC ) + BCrs 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AC = AC + ACrs. How long is AB? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", AC + ACrs, \"?\" ); AB This is a 30-60-90 triangle. Thus, the hypotenuse is \\frac{2}{\\sqrt{3}} times as long as the longer leg. AB = \\frac{2}{\\sqrt{3}} \\cdot AC = \\frac{2}{\\sqrt{3}} \\cdot (AC + ACrs) = ABs randRange( 2, 6 ) randFromArray([ [1, \"\"], [3, \"\\\\sqrt{3}\"] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC?", "yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$: $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); label(\"P\",P,dir(235)); [/asy]$ This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \\boxed{\\textbf{(E) } 500}$ ## Solution 6 (Ptolemy's Theorem) $[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D(\"A\", dir(50), dir(50)); pair B = D(\"B\", dir(90), dir(90)); pair C = D(\"C\", dir(130), dir(130)); pair D = D(\"D\", dir(170), dir(170)); pair O = D(\"O\", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]$ Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\\triangle OBC$. Since $\\triangle OBC$ is isosceles we can compute its area to be $s^2 \\sqrt7/4$, hence $CA = 2 \\tfrac{2 \\cdot s^2\\sqrt7/4}{s\\sqrt2} = s\\sqrt{7/2}$. Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \\cdot s \\implies AD = (7/2-1)s.$ This gives us: $$\\boxed{\\textbf{(E) } 500.}$$ ## Solution 7 (Trigonometry) Since all three sides equal $200$, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\\sqrt{7},200\\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\\frac{100}{200\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$. Similarly, the cosine is $\\frac{100\\sqrt{7}}{200\\sqrt{2}}=\\frac{\\sqrt{14}}{4}$. Since there are three sides, and since", "$H$. $CH=4/3\\div 68/15\\cdot 8/5=8/17$. So $HG=8-8/17=128/17$. By Power of a Point, $CH\\cdot HG=AH\\cdot HE$. $AH=16/5\\cdot 5\\sqrt{13}/17=16\\sqrt{13}/17.$ So $HE=1024/289\\cdot 17/(16\\sqrt{13})=64/(17\\sqrt{13})$. The height from $E$ to $CG$ is $17/(5\\sqrt{13})\\cdot 64/(17\\sqrt{13})=64/65$. Thus, $[\\triangle CEG]=64/65\\cdot 8\\div 2=256/65$. The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$. Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$. So $64+65=\\boxed{129}$. ## Solution 5 (HARD computation) $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); path circle2 = Circle((4,0),4); N = (-8/3,0); pair X =rotate(180,O_2)*E; pair Y = (8,0); draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); dot(\"X\", X, NE);dot(\"Y\", Y, NE); [/asy]$ Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \\parallel XY$, $XYE$ is right angle. First, $\\frac{NO_1}{NO_1+5} = \\frac{1}{4}$, so $NO_1=\\frac{5}{3}$. And, $$\\cos{\\angle{AO_2C}}=\\cos{2\\angle{AYC}} = \\frac{O_2C}{NO_1+5} = \\frac{3}{5}$$ $$\\sin{\\angle{AYC}} = \\sqrt{\\dfrac{1-\\cos{2\\angle{AYC}}}{2}}=\\frac{1}{\\sqrt{5}}$$ $$\\cos{\\angle{AYC}} = \\frac{2}{\\sqrt{5}}$$ Then, $$[AEC] = \\frac{1}{2}AE*CE*\\sin{\\angle{ACE}}=\\frac{1}{2}AE*8\\sin{\\angle{CYE}}*\\frac{1}{\\sqrt{5}}$$ $$[CXY] = \\frac{1}{2}CX*XY*\\sin{\\angle{CXY}}=\\frac{1}{2}*8\\sin{\\angle{XYC}}*XY*\\sin{\\angle{CAY}}$$ Since $\\angle{CAY} = 90 - \\angle{AYC}$, $\\angle{XYC} = 90 - \\angle{CYE}$, $XY = AE$, we have $$[CXY] = \\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\cos{\\angle{AYC}}=\\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\frac{2}{\\sqrt{5}}$$ Since $\\triangle{CXY}$ is four times in scale to $\\triangle{AEC}$, their area ratio is 16. Divide the two equations for the two areas, we have $$\\tan{\\angle{CYE}} = \\frac{1}{8}$$ With this", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "to be $\\frac{11}{2}$ and $\\frac{5}{2}$, so you add them up: $\\frac{11}{2}+\\frac{5}{2}=\\boxed{\\textbf{(A) 8}}$. ~kevinmathz ### Solution 4 (Geometric Interpretation) Draw a right triangle $ABC$ with a hypotenuse $AC$ of length $7$ and leg $AB$ of length $x$. Draw $D$ on $BC$ such that $AD=5$. Note that $BC=\\sqrt{49-x^2}$ and $BD=\\sqrt{25-x^2}$. Thus, from the given equation, $BC-BD=DC=3$. Using Law of Cosines on triangle $ADC$, we see that $\\angle{ADC}=120^{\\circ}$ so $\\angle{ADB}=60^{\\circ}$. Since $ADB$ is a $30-60-90$ triangle, $\\sqrt{25-x^2}=BD=\\frac{5}{2}$ and $\\sqrt{49-x^2}=\\frac{5}{2}+3=\\frac{11}{2}$. Finally, $\\sqrt{49-x^2}+\\sqrt{25-x^2}=\\frac{5}{2}+\\frac{11}{2}=\\boxed{\\textbf{(A)~8}}$. $[asy] var s = sqrt(3); pair A = (-5*s/2, 0); pair B = (0,0); pair C = (0,5.5); pair D = (0,2.5); draw(A--B--C--A--D); rightanglemark(A, B, D); label(\"A\", A, SW); label(\"B\", B, SE); label(\"C\", C, NE); label(\"D\", D, E); label(\"7\", (-5*s/4, 5.5/2), NW); label(\"120^\\circ\", D, NW); label(\"60^\\circ\", (0,2), SW); label(\"x\", 0.5*A, S); draw(rightanglemark(A, B, C)); draw(anglemark(A, D, B)); markscalefactor = 0.04; draw(anglemark(C, D, A)); label(\"\\frac{5}{2}\", (0,1.25), E); label(\"3\", (0,4), E); label(\"5\", (-5*s/4, 5/4), N); [/asy]$ ### Solution 5 (No Square Roots, Fastest) We notice that the two expressions are conjugates, and therefore we can write them in a \"difference-of-squares\" format. Namely, we can write it as $((49-x^2) - (25 - x^2)) = (49 - x^2 - 25 + x^2) = 24$. Given the $3$ in the problem, we can divide $24 / 3 = \\boxed{\\textbf{(A) } 8}$. -aze.10 ### Solution 6" ]

[ "$AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$. $[asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*\"24\", A--T, N); label(rotate(degrees(C-A))*\"24\", A--C, 2*NW); label(\"12\\sqrt 3\", C--D, E); label(\"12\\sqrt 3\", D--B, S); label(\"12\", A--D, S); pen p = fontsize(8)+red; MA(\"45^\\circ\", C,B,A,2); MA(\"30^\\circ\", B,A,T,2.5); MA(\"30^\\circ\", T,A,C,3.5); dot(\"A\", A, SW); dot(\"B\", B, SE); dot(\"C\", C, N); dot(\"T\", T, NE); dot(\"D\", D, S); [/asy]$ By 30-60-90 triangles, $AD=12$ and $CD=12\\sqrt{3}$. We also notice that $\\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\\sqrt{3}$. The base $AB$ is $12+12\\sqrt{3}$, and the altitude $CD=12\\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=\\boxed{291}$. -youyanli ## See also 2001 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS", "# Difference between revisions of \"1973 AHSME Problems/Problem 4\" ## Problem Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $\\textbf{(A)}\\ 6\\sqrt3\\qquad\\textbf{(B)}\\ 8\\sqrt3\\qquad\\textbf{(C)}\\ 9\\sqrt3\\qquad\\textbf{(D)}\\ 12\\sqrt3\\qquad\\textbf{(E)}\\ 24$ ## Solution $[asy] fill((0,3.464)--(-6,0)--(6,0)--cycle,yellow); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-6,0)--(6,0)--(3,5.196)--(-6,0)); draw((0,3.464)--(0,0)); label(\"6\",(-3,0),S); label(\"6\",(3,0),S); [/asy]$ Note that the altitude of the shared region bisects the hypotenuse of the original two right triangles (this can be confirmed by using AAS on the two right triangles with that altitude of the shared region). By using 30-60-90 triangles, the altitude’s length is $2\\sqrt{3}$. The area of the shared region is $\\tfrac12 \\cdot 12 \\cdot 2\\sqrt{3} = \\boxed{\\textbf{(D)}\\ 12\\sqrt3}$.", "(opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using cosine: betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", AC + ACrs, \"x\" ); arc([0, 5*sqrt(3)/2], 0.8, 270, 300); label([-0.1, (5*sqrt(3)/2)-1], \"{30}^{\\\\circ}\", \"below right\"); Cosine is adjacent over hypotenuse (SOH CAH TOA), so \\cos {30}^{\\circ} = \\dfrac{ACdisp}{x}. We also know that \\cos{30}^{\\circ} = \\dfrac{\\sqrt{3}}{2}. Solving for x, we get \\qquad x \\cdot \\cos{30}^{\\circ} = ACdisp \\qquad x \\cdot \\dfrac{\\sqrt{3}}{2} = ACdisp \\qquad x = ACdisp \\cdot \\dfrac{2}{\\sqrt{3}} \\qquad x = ACdisp \\cdot \\dfrac{2\\cdot\\sqrt{3}}{3} So, x = ABs. randRange( 2, 6 ) randFromArray([ [1, \"\"], [3, \"\\\\sqrt{3}\"] ]) 2*BC + BCrs randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"x\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr We know the length of the hypotenuse of this triangle, and want to find the length of a leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine", "# Difference between revisions of \"2009 AMC 10A Problems/Problem 10\" ## Problem Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\\triangle ABC$? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C)); dot(ps); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"3\",midpoint(A--D),NE); label(\"4\",midpoint(D--C),NE); [/asy]$ $\\mathrm{(A)}\\ 4\\sqrt3 \\qquad \\mathrm{(B)}\\ 7\\sqrt3 \\qquad \\mathrm{(C)}\\ 21 \\qquad \\mathrm{(D)}\\ 14\\sqrt3 \\qquad \\mathrm{(E)}\\ 42$ ## Solution It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\\cdot CD$. (See below for a proof.) Then $BD = \\sqrt{ 3\\cdot 4 } = 2\\sqrt 3$, and the area of the triangle $ABC$ is $\\frac{AC\\cdot BD}2 = 7\\sqrt3\\Rightarrow\\boxed{\\text{(B)}}$. Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get: 1. $AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \\cdot AD \\cdot DC$. 2. $AB^2 = AD^2 + BD^2$ 3. $BC^2 = BD^2 + CD^2$ Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD \\cdot DC$. Alternatively, note that $\\triangle ABD \\sim \\triangle BCD \\Longrightarrow \\frac{AD}{BD} = \\frac{BD}{CD}$.", "# Difference between revisions of \"2009 AMC 10A Problems/Problem 10\" ## Problem Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\\triangle ABC$? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C)); dot(ps); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"3\",midpoint(A--D),NE); label(\"4\",midpoint(D--C),NE); [/asy]$ $\\mathrm{(A)}\\ 4\\sqrt3 \\qquad \\mathrm{(B)}\\ 7\\sqrt3 \\qquad \\mathrm{(C)}\\ 21 \\qquad \\mathrm{(D)}\\ 14\\sqrt3 \\qquad \\mathrm{(E)}\\ 42$ ## Solution 1 It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\\cdot CD$. (See below for a proof.) Then $BD = \\sqrt{ 3\\cdot 4 } = 2\\sqrt 3$, and the area of the triangle $ABC$ is $\\frac{AC\\cdot BD}2 = 7\\sqrt3\\Rightarrow\\boxed{\\text{(B)}}$. Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get: 1. $AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \\cdot AD \\cdot DC$. 2. $AB^2 = AD^2 + BD^2$ 3. $BC^2 = BD^2 + CD^2$ Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD \\cdot DC$. Alternatively, note that $\\triangle ABD \\sim \\triangle BCD \\Longrightarrow \\frac{AD}{BD} =", "endpoints of a side of the larger hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy]", "betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"?\", \"\", ( 2 * BC ) + BCrs ); BC * BC * BCr This is a 30-60-90 triangle. Thus, the shorter leg is half as long as the hypotenuse. BC = \\frac12 \\cdot AB = \\frac12 \\cdot (( 2 * BC ) + BCrs) = ( BC ) + BCrs 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ABs. How long is AC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", \"?\", ABs ); AC * AC * ACr This is a 30-60-90 triangle. Thus, the longer leg is \\frac{\\sqrt{3}}{2} times as long as the hypotenuse. AC = \\frac{\\sqrt{3}}{2} \\cdot AB = \\frac{\\sqrt{3}}{2} \\cdot (ABs) = AC + ACrs", "1. ## Trig! Exam tomorrow! Hello, I am here reviewing for my exam tmw.. this question I can't figure out - pease help. Two triangular building lots, ABC and ACD, have side AC in common. (in attachment) a) Determine an exact expression for the side length of side AC b) Determine an exact expression for the length of side AB I get confused from the beginning because I don't know which trig ratio to solve with? (Sin/Tan or Cos?) Hello, I am here reviewing for my exam tmw.. this question I can't figure out - pease help. Two triangular building lots, ABC and ACD, have side AC in common. (in attachment) a) Determine an exact expression for the side length of side AC b) Determine an exact expression for the length of side AB I get confused from the beginning because I don't know which trig ratio to solve with? (Sin/Tan or Cos?) These angles are special angles. $\\displaystyle \\frac{\\pi}{4}=45^{\\circ}\\:\\:and\\:\\:\\frac{\\pi}{6}=3 0^{\\circ}$ Use your rules for a 45-45-90 triangle and then use your rules for a 30-60-90 triangle. First, $\\displaystyle 60=AC\\sqrt{2}$ $\\displaystyle AC=\\frac{60}{\\sqrt{2}}$ Rationalizing the denominator, we get $\\displaystyle AC=\\frac{60}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{60\\sqrt{2}}{2}=30\\ sqrt{2}$ This value is the hypotenuse of the 30-60-90 triangle, so we have CB as the short leg and it's equal to one-half the hypotenuse... $\\displaystyle CB=15\\sqrt{2}$ And...AB", "\"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and BC = BC + BCrs. How long is AB? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", BC + BCrs, \"\", \"?\" ); 4 * BC * BC * BCr This is a 30-60-90 triangle. Thus, the hypotenuse is twice as long as the shorter leg. AB = 2 \\cdot BC = 2 \\cdot (BC + BCrs) = ( 2 * BC ) + BCrs 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AC = AC + ACrs. How long is AB? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", AC + ACrs, \"?\" ); AB This is a 30-60-90 triangle. Thus, the hypotenuse is \\frac{2}{\\sqrt{3}} times as long as the longer leg. AB = \\frac{2}{\\sqrt{3}} \\cdot AC = \\frac{2}{\\sqrt{3}} \\cdot (AC + ACrs) = ABs randRange( 2, 6 ) randFromArray([ [1, \"\"], [3, \"\\\\sqrt{3}\"] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ( 2 * BC ) + BCrs. How long is BC?", "the answer is $\\boxed{\\textbf{(D) }\\frac{50 - 25\\sqrt{3}}{2}}$ ## Solution 2 The area of $\\triangle ABC$ is $12.5$, and so the leg length of $45 - 45 - 90$ $\\triangle ABC$ is $5.$ Thus, the altitude to hypotenuse $AB$, $CF$, has length $\\dfrac{5}{\\sqrt{2}}$ by $45 - 45 - 90$ right triangles. Now, it is clear that $\\angle{ACD} = \\angle{BCE} = 30^\\circ$, and so by the Exterior Angle Theorem, $\\triangle{CDE}$ is an isosceles $30 - 75 - 75$ triangle. Thus, $DF = CF \\tan 15^\\circ = \\dfrac{5}{\\sqrt{2}} (2 - \\sqrt{3})$ by the Half-Angle formula, and so the area of $\\triangle CDE$ is $DF \\cdot CF = \\dfrac{25}{2} (2 - \\sqrt{3})$. The answer is thus $\\boxed{\\textbf{(D) } \\frac{50 - 25\\sqrt{3}}{2}}$ ## Solution 3 Because the area of triangle $ABC$ is $12.5$, and the triangle is right and isosceles, we can quickly see that the leg length of the triangle $ABC$ is 5. If we put the triangle on the coordinate plane, with vertex $C$ at the origin, and the hypotenuse in the first quadrant, we can use slope-intercept form and tangents to get three lines that intersect at the origin, $D$, and $E$. Then, you can use the distance formula to get the length of $DE$. The height is just $\\frac{5}{\\sqrt{2}}$, so the area is just $DE*\\frac{5}{\\sqrt{2}}*\\frac{1}{2}=\\boxed{\\textbf{(D) } \\frac{50 - 25\\sqrt{3}}{2}}$", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "# 1983 AIME Problems/Problem 11 ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label(\"A\",A,W); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); [/asy]$ ## Solutions ### Solution 1 First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); label(\"12\\sqrt{2}\",(E+F)/2,N); label(\"6\\sqrt{2}\",(A+B)/2,S); label(\"6\",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\\cdot \\left(", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] 1. 👍 2. 👎 6. 4/3 is the answer. 1. 👍 2. 👎 7. It is 4/3. 1. 👍 2. 👎 ## Similar Questions 1. ### Math please helppp 1 Find the missing angle measure in the polygon A ] 77 B ] 87 C ] 97 D ] 107 i think its b 2 Find the sum of the interior angles in 10 - sided polygon A ] 1,260 B ] 1,440 c ] 1,620 d ] 1,800 i think its c or b 3 Find the measure 2. ### geometry please please please help What is the value of x in", "# Math Help - Forming Functions from Verbal Descriptions 1. ## Forming Functions from Verbal Descriptions Hey, noob here. I was wondering if any of you smarter guys out there could help me out. Coming from the textbook \"Advanced Mathematics\" by Richard Brown... PROBLEM: Express the area A of a 30*-60*-90* triangle as a function of the length h of the hypotenuse. TIA 2. Originally Posted by guywhohatesmath Hey, noob here. I was wondering if any of you smarter guys out there could help me out. Coming from the textbook \"Advanced Mathematics\" by Richard Brown... PROBLEM: Express the area A of a 30*-60*-90* triangle as a function of the length h of the hypotenuse. TIA 1. Draw the triangle including the height h. 2. The height divide the right angle into 60° and 30° and the hypotenuse c into 2 parts: q under the 60°-angle, p under the 30°-angle. 3. The area of the right triangel is calculated by: $A=\\dfrac12\\cdot c \\cdot h$ 4. $q = h \\cdot \\tan(60^\\circ) = h \\cdot \\sqrt{3}$ 5. $p = h \\cdot \\tan(30^\\circ) = h \\cdot \\sqrt{\\frac13}$ Plug in these values into the equation of the area: $A= \\dfrac12 \\left(h \\cdot \\sqrt{3}+h \\cdot \\sqrt{\\frac13}\\right)\\cdot h$ After factor out the common factor you'll get: $A(h)=\\dfrac23\\sqrt{3}\\cdot h^2$ 3. Thanks for your time and help.", "wrote: ConnectTheDots wrote: Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon? (A) 24% (B) 30% (C) 36% (D) 42% (E) 48% Let's look for a shortcut way. Dear ConnectTheDots, I'm happy to help. The sides of the regular hexagon inscribed in a circle are equal to the radius of the circle. You can easily see this if you divide the hexagon into six congruent equilateral triangles --- then each triangle length equals a hexagon side, and two of those triangle lengths are the diameter of the circle. Look at triangle ABC. Sides AB = BC = r. The angle at B is 120 degrees, so the smaller angles are 30 degrees. Consider the midpoint of AC --- call that point P. Then segment BP divides triangle ABC into two congruent 30-60-90 triangles, each with hypotenuse = r. In one of those 30-60-90 triangles: hypotenuse = r short leg = r/2 long leg = (r*sqrt(3))/2 We could put two of those triangles together to make a rectangle with sides of (short leg) and (long leg), so the area of this rectangle would equal the area of triangle ABC. That area is A = [(r^2)*sqrt(3)]/4 The entire diagram has three triangles that side in the shaded region, so the", "# Difference between revisions of \"1983 AIME Problems/Problem 11\" ## Problem The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\\sqrt{2}$, what is the volume of the solid? $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label(\"A\",A); label(\"B\",B); label(\"C\",C); label(\"D\",D); label(\"E\",E,N); label(\"F\",F,N); [/asy]$ ## Solution 1 First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to find that the height is equal to $6$. $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A); label(\"B\",B); label(\"C\",C); label(\"D\",D); label(\"E\",E,N); label(\"F\",F,N); label(\"12\\sqrt{2}\",(E+F)/2,N); label(\"6\\sqrt{2}\",(A+B)/2); label(\"6\",(3*s/2,s/2,3),ENE); [/asy]$ Next, we complete the figure into a triangular prism, and find the area, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined area is", "(adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using cosine: betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"x\", \"\", ( 2 * BC ) + BCrs ); arc([5/2,0], 0.5, 120, 180); label([5/2-0.2, 0], \"{60}^{\\\\circ}\", \"above left\"); Cosine is adjacent over hypotenuse (SOH CAH TOA), so \\cos {60}^{\\circ} = \\dfrac{x}{ABdisp}. We also know that \\cos{60}^{\\circ} = \\dfrac{1}{2}. Solving for x, we get \\qquad x = ABdisp \\cdot \\cos{60}^{\\circ} \\qquad x = ABdisp \\cdot \\dfrac{1}{2} So, x = BC + BCrs. 3 * randRange( 2, 6 ) randFromArray([ [1, \"\", (AC * 2 / 3) + \"\\\\sqrt{3}\", AC * AC * 4 / 3], [3, \"\\\\sqrt{3}\", (AC * 2), AC * AC * 4] ]) randFromArray([ \"\\\\angle A = 30^\\\\circ\", \"\\\\angle B = 60^\\\\circ\" ]) In the right triangle shown, mAB and AB = ABs. How long is AC? betterTriangle( 1, sqrt(3), \"A\", \"B\", \"C\", \"\", \"x\", ABs ); AC * AC * ACr We know the length of the hypotenuse of this triangle, and want to find the length of a leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse).", "equilateral triangle is altitude of the larger triangle which is \\sqrt{3}/2 * 1. So half of this altitude will be \\sqrt{3}/4 AB^2 = (1/2)^2 -(\\sqrt{3}/4)^2 = 1/16 So AB = 1/4 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 15 Jul 2012 Posts: 33 Followers: 0 Kudos [?]: 1 [0], given: 5 Re: Geometry [#permalink] 22 Oct 2013, 22:31 VeritasPrepKarishma wrote: anu1706 wrote: See the idea what I got is for 60degree = square root 3 times 30Degree & same way for 90 degree = 2 times 30 degree. i.e. 1:root3:2.... In this case i have understood root3/2 on a larger altitude. Now to find a side opposit to 60degree in smaller altitude, we have been given only 90degree side root3/2.. so how you calculated please tell. What you wrote is hard to understand. I am assuming you are talking about the small 30-60-90 triangle which includes side AB as the side opposite to the 30 degree angle. You know the hypotenuse is 1/2 (since it is half of side 1 of the equilateral triangle. ) The side opposite 60 degree angle is half of the side is the smaller", "and multiplying these together proves the formula for isogonal lines. Hence, we have $$\\frac{BE}{15-BE}\\cdot \\frac{9}{6}=\\frac{169}{196}\\implies BE=\\frac{2535}{463}$$ so our desired answer is $\\boxed{463}.$ ## Solution 6 (Tangent subtraction formulas) Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle. $[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label(\"G\",G,N); label(\"F\",F,N); [/asy]$ Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\\angle DAG = \\alpha$. Now we know that $\\overline{DG} = \\frac{24}{5}$ and $\\overline{GC} = \\frac{18}{5}$. Therefore, $\\overline{AG} = \\frac{52}{5}$, so $\\tan{(\\alpha)} = \\frac{6}{13}$. Our goal now is to use tangent $\\angle EAG$ in triangle $AEG$. We set $\\overline{BE}$ to $x$, so $\\overline{ED} = 9 - x$ and $\\overline{EC} = 15 - x$, so $\\overline{EG} = \\frac{4}{5}(15-x)$ and $\\overline{GC} = \\frac{3}{5}(15-x)$ so $\\overline{AG} = \\frac{3x+25}{5}$. Now we just need tangent of $\\angle EAG$. We find this using $\\tan{(EAG)} =" ]

[ "# Allergy Allergies, also known as allergic diseases, are a number of conditions caused by hypersensitivity of the immune system to typically harmless substances in the environment.", "immune • question_answer200) It is now well known that stress 191 up a 192 of body chemicals that eventually 193 your health. Now scientists 194 further findings, It 195 that every time one goes through 196 stress, the bodys stress thermostat gets 197 to a greater 198 Researchers compare this to an allergy once the body is 199 to a substance. It reacts strongly even at a mere 200 of the stuff. A) disturbance B) whiff", "more, they’re sending students to schools in … b The test does assume an identically shaped and scaled distribution for each group, except for any difference in medians. are the ranks of the Percentage of the people in the U.S. who have a food allergy : 4% of adults and 5% of children. {\\displaystyle a_{ij}=-a_{ji}} to different observations of a particular variable.", "50% of Americans) claim that they won't change their mind.", "food allergy (24) 10.21% Yes - an autoimmune disorder (i.e coeliac, colitis) (7) 2.98% Yes - I have an intolerance and allergy (8) 3.4% No (170) 72.34%", "of people in the US", "that you may not be allergic to a specific allergen. Keep in mind that skin tests aren't always accurate. Sometimes it indicates an allergy when there isn't a single allergy (false positive), or a skin test might not trigger a reaction when you're exposed to something you're allergic to (false negative). You may react differently to the same test done on different occasions. Or, you may react positively to a substance during the test but not interact with it in everyday life. Your allergy treatment plan may include medications, immunotherapy, changes to your work environment or home, or dietary changes. Ask your doctor to explain anything you do not understand about your diagnosis or treatment. With test results that identify your allergens and a treatment plan to help you gain control, you'll be able to reduce or eliminate allergy signs and symptoms. Are there any risks involved? Since a very small amount of the allergen is used, the reaction is also very small and clears up within a few hours. Most of the time, the reaction can appear as a few itchy bumps. Asthma patients tend to develop a severe allergic reaction or asthma attack within the next 24 hours. Therefore, they should always keep the physician informed of their condition and always carry inhalers and medications while", "# Sinus Congestion Studies ### Higher Sinus Congestion Predicts Slightly Higher Guiltiness This individual’s Guiltiness is generally lowest after an average of 1.9 out of 5 of Sinus Congestion over the previous 7 days.", "and 45% of all adults…", "allergies account for more than thirty thousand Wed Nov 22, 2017 2:20 pm Food allergies account for more than thirty thousand emergency department visits each year. Often, victims of these episodes are completely unaware of their allergies until they experience a major reaction. Studies show that ninety percent of food allergy reactions are caused by only eight distinct foods. For this reason, individuals should sample a minuscule portion of each of these foods to determine whether a particular food allergy is present. Which of the following must be studied in order to evaluate the recommendation made in the argument? (A) The percentage of allergy victims who were not aware of the allergy before a major episode (B) The percentage of the population that is at risk for allergic reactions (C) Whether some of the eight foods are common ingredients used in cooking (D) Whether an allergy to one type of food makes someone more likely to be allergic to other types of food (E) Whether ingesting a very small amount of an allergen is sufficient to provoke an allergic reaction in a susceptible individual Which of the Options is most fit in the argument? OA E ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 14169 messages Followed by: 1818 members 13060 GMAT Score: 790 Thu", "being about \\$17 billion a year. Scope of the Problem According to a common estimate, at any one time 10 percent of the American population has mental health problems sufficiently serious to warrant care; recent evidence suggests that this figure may be closer to 15 percent. Not all the people who need help receive it, however; in 1975 only 3 percent of the American population received mental health service. One major reason for this is that people still fear the stigma attached to mental illness and hence often fail to report it or to seek help. Analysis of the figures on mental illness shows that schizophrenia afflicts an estimated 2 million Americans, another 2 million suffer from profound depressive disorders, and 1 million have organic psychoses or other permanently disabling mental conditions. As much as 25 percent of the population is estimated to suffer from mild or moderate depression, anxiety, and other types of emotional problems. Some 10 million Americans have problems related to alcohol abuse, and millions more are thought to abuse drugs. Some 5 to 15 percent of children between the ages of 3 and 15 are the victims of persistent mental health problems, and at least 2 million are thought to have severe learning disabilities that can seriously impair their mental health. In addition, according", "# Question A recent Health of Boston report suggests that 14% of female residents suffer from asthma as opposed to 6% of males (Boston Globe, August 16, 2010). Suppose 250 females and 200 males responded to the study. a. Develop the appropriate null and alternative hypotheses to test whether the proportion of females suffering from asthma is greater than the proportion of males. b. Calculate the value of the test statistic and the p-value. c. At the 5% significance level, what is the conclusion? Do the data suggest that females suffer more from asthma than males? Sales1 Views64", "under the supervision of an allergist, as it can cause very serious reactions.", "interesting - and very telling! 6 of 29 hold licenses (20%). 80% don't. I think that explains a lot of what I've read on this forum! 2", "1 (604) 999-1884 - U.S & Canada|[email protected] Do you have asthma? Take the asthma quiz 2020-04-21T16:01:05-07:00 # Do You Have Asthma ## Take The 30 Second Quiz It is estimated that there are 22 million people diagnosed with asthma in North America; nearly nine million of them are children. More than 5000 North Americans will die from an asthma attack this year. It is believed a large percentage of those deaths could be prevented through understanding and preventative actions. WHO IS AT RISK • Asthma is common, affecting 5 – 10% of the population. • Asthma is one of the most common conditions of childhood. • Asthma is closely linked to allergies. • Asthma can run in families. • Most but not all people with asthma have allergies. • Children with a family history of allergy and asthma are more likely to have asthma. • Although asthma affects people of all ages, it most often starts in childhood. • More boys have asthma than girls. • In adulthood, more women have asthma than men. • Although asthma affects people of all races, African American are more likely than Caucasians to be hospitalized for asthma attacks. ### AND GET A FREE EBOOK! I wheeze (breathing out with a noisy whistling sound) I cough due to stuff (mucus) in my", "# Poop Quantity Rating Studies ### Higher Poop Quantity Rating Predicts Slightly Lower Overall Mood This individual’s Overall Mood is generally highest after an average of 2.3 out of 5 of Poop Quantity Rating over the previous 7 days.", "diagnosis of allergic rhinitis (OR: 2.1; CI: 1.1, 3.7) were also predictors for asthma. The population-attributable risk of having 1 or more residential exposures associated with doctor-diagnosed asthma was 44.4% (95% CI: 29–60), or an estimated 2 million excess cases. The attributable cost of asthma resulting from residential exposures was $405 million (95% CI:$264–\\$547 million) annually. Conclusions. The elimination of identified residential exposures, if causally associated with asthma, would result in a 44% decline in doctor-diagnosed asthma among older children and adolescents in the United States.", "percent chance that exactly 3 of the children attending the party prefer milk?", "Contents Allergies 1 to 5 Rating Symptoms Factors Effects Allergies Allergy Causes of Allergies User reported causes of Allergies based on their intuition. Cause Agree 86% 77% 65% 65% 65% 58% 58% 58% 55% 54% 51% 50% 47% 38% 36% 35% 34% 34% 34% 34% 33% 31% 30% 28% 28% 27% 27% 26% 26% 25% 24% 22% 20% 19% 19% 19% 19% 18% 18% 18% 18% 17% 16% 16% 16% 15% 15% 14% 14% 13% 13% 13% 12% 12% 12% 11% 11% 11% 11% 10% 10% 10% 9% 9% 9% 9% 9% 9% 9% 8% 8% 8% 8% 8% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 6% 6% 6% 6% 6% 6% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 4% 4% 4% 4% 4% 3% 3% 3% 3% 3% 3% 3% 3% 2% 2% 2% 1% 1% 1% 1% Conditions Resulting from Allergies User-reported conditions resulting from Allergies based on their intuition. Resulting Condition Agree 90% 74% 64% 49% 48% 42% 25% 23% 11% Treatments User-reported effectiveness in treating Allergies Treatment Major Improvement Moderate Improvement Much Worse No Effect Worse Responses 12% 31% 3% 49% 5% 221 9% 15% 4% 72% 0% 53 7% 27% 2% 62% 3% 116 9% 24% 1% 65% 1% 101 31% 50% 0% 19%", "Dec 21, 2017 5:56 am lheiannie07 wrote: Food allergies account for more than thirty thousand emergency department visits each year. Often, victims of these episodes are completely unaware of their allergies until they experience a major reaction. Studies show that ninety percent of food allergy reactions are caused by only eight distinct foods. For this reason, individuals should sample a minuscule portion of each of these foods to determine whether a particular food allergy is present. Which of the following must be studied in order to evaluate the recommendation made in the argument? (A) The percentage of allergy victims who were not aware of the allergy before a major episode (B) The percentage of the population that is at risk for allergic reactions (C) Whether some of the eight foods are common ingredients used in cooking (D) Whether an allergy to one type of food makes someone more likely to be allergic to other types of food (E) Whether ingesting a very small amount of an allergen is sufficient to provoke an allergic reaction in a susceptible individual Plan: Individuals should sample a minuscule portion of the eight foods known to cause 90% of allergic reactions. Conclusion: An individual will be able to determine whether a particular food allergy is present. Rephrase the answer choices as STATEMENTS. The" ]

[ "more, they’re sending students to schools in … b The test does assume an identically shaped and scaled distribution for each group, except for any difference in medians. are the ranks of the Percentage of the people in the U.S. who have a food allergy : 4% of adults and 5% of children. {\\displaystyle a_{ij}=-a_{ji}} to different observations of a particular variable.", "250? that is your answer. (25)", "allergy\" with a variable $$A$$. One crucial thing to note is that there is a notable difference between children and adults. The analysis is going to be different depending on whether you're desperately searching the grocery store non-nut butter at 10 pm because your kid needs a sandwich in the morning, or searching for a gluten-free restaurant for a night out with friends. The article gives the probability of a food allergy in children as being between 4%-8% and for adults 1%-2%. For our analysis, we'll assume the middle values, but keep in mind that we could be more conservative in our estimates. Now we can write these down as two probabilities: $$P(A_{\\text{child}})=0.06$$ $$P(A_{\\text{adult}})=0.015$$ The only other piece of data we need is the probability that someone will claim they have a food allergy. We'll use the variable $$C$$ for \"claims they have an allergy\" (or that their kids do). Luckily the article states that both adults and parents of children claim that either they or their children have a food allergy 30% of the time. For the probability that someone will claim they or their child has a food allergy we get: $$P(C) = 0.3$$ The article says \"up to 30%\", so it's important to realize that this is an upper bound. It is critical to keep", "food allergy (24) 10.21% Yes - an autoimmune disorder (i.e coeliac, colitis) (7) 2.98% Yes - I have an intolerance and allergy (8) 3.4% No (170) 72.34%", "## Statistics Question 23. In a 2006 poll of 1000 Americans, the proportion who were satisfied with the way things were going in the U.S. was only 0.27.Which of these is denoted by X in this situation? A) 0.27 B) 270 C) 1000 D) none of the above • B • d • B 270 • .27 *1000 = 270 hence b) • D) none of the above n=1000 p=0.27 but both are not x so answer is none of above • .27 *1000 = 270 hence b) Get homework help", "# Math Help - [SOLVED] Statistics and Probability: 1. ## [SOLVED] Statistics and Probability: Out of 20 questions I have managed to complete all but these four, so any assistance would be awesome. Not sure if I am in brain freeze mode or what: 6.22: According to the national Marine manufacturers Association, 50% of the population of Vermont were boating participants during the most recent year. For the randomly selected sample of 20 Vermont residents, with the discrete random variable x= the number in the sample who were boating participants that year, determine the following: a) E (x) b) P (x ≤ 8) c) P (x=10) d) P (x=12) e) P (7 ≤ x ≤ 13) 6.30: It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home. Assuming that Hargrove students are typical in terms of susceptibility to allergies and that the college presidents home happens to contain just about every common allergin to which afflicted persons react, what is the probability that at least 8 of the students will be able to stay for the duration of the event? 7.33: It has been reported that households in the west spend an", "Contents Allergies 1 to 5 Rating Symptoms Factors Effects Allergies Allergy Causes of Allergies User reported causes of Allergies based on their intuition. Cause Agree 86% 77% 65% 65% 65% 58% 58% 58% 55% 54% 51% 50% 47% 38% 36% 35% 34% 34% 34% 34% 33% 31% 30% 28% 28% 27% 27% 26% 26% 25% 24% 22% 20% 19% 19% 19% 19% 18% 18% 18% 18% 17% 16% 16% 16% 15% 15% 14% 14% 13% 13% 13% 12% 12% 12% 11% 11% 11% 11% 10% 10% 10% 9% 9% 9% 9% 9% 9% 9% 8% 8% 8% 8% 8% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 6% 6% 6% 6% 6% 6% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 4% 4% 4% 4% 4% 3% 3% 3% 3% 3% 3% 3% 3% 2% 2% 2% 1% 1% 1% 1% Conditions Resulting from Allergies User-reported conditions resulting from Allergies based on their intuition. Resulting Condition Agree 90% 74% 64% 49% 48% 42% 25% 23% 11% Treatments User-reported effectiveness in treating Allergies Treatment Major Improvement Moderate Improvement Much Worse No Effect Worse Responses 12% 31% 3% 49% 5% 221 9% 15% 4% 72% 0% 53 7% 27% 2% 62% 3% 116 9% 24% 1% 65% 1% 101 31% 50% 0% 19%", "expect the same fraction of $B$'s to have $A$ as the whole population. Of the $260 B$'s, $100$ are also $A$'s, about $38\\%$. But of the whole population, $A$'s are rare-a little less than $4\\%$. Similarly the other way around. They are far from independent.", "So let's just multiply that times 120. That gets us to 25 point-- I'll just round it-- 25.3 people should have gotten sick. So the expected-- so let me write it over here, I'll do expected in yellow-- so the expected right over here. If you assume that 21% of each group should have gotten sick is that you would have expected 25.3 people to get sick in group one, in herb one group. And then the remainder will not get sick. So let's just subtract or I could actually multiply 79% times 120, either one of those would be good. But let me just take 120 minus 25.3, and then I get 94.7. So you would have expected 94.7 to not get sick. So this is expected again. 94.7 to not get sick. And now let's do that for each of these groups. So once again, group two, you would've expected 21% to get sick. 21% of the total people in that group, so that's 140, so that's 29.4. And then the remainder-- let's see, 140 minus 29.4-- should not have gotten sick. So that gets us this right here. We have 29.4 should have gotten sick if the herbs did nothing. And then, over here, we would have 110.6 should not have gotten sick. And these", "allergies account for more than thirty thousand Wed Nov 22, 2017 2:20 pm Food allergies account for more than thirty thousand emergency department visits each year. Often, victims of these episodes are completely unaware of their allergies until they experience a major reaction. Studies show that ninety percent of food allergy reactions are caused by only eight distinct foods. For this reason, individuals should sample a minuscule portion of each of these foods to determine whether a particular food allergy is present. Which of the following must be studied in order to evaluate the recommendation made in the argument? (A) The percentage of allergy victims who were not aware of the allergy before a major episode (B) The percentage of the population that is at risk for allergic reactions (C) Whether some of the eight foods are common ingredients used in cooking (D) Whether an allergy to one type of food makes someone more likely to be allergic to other types of food (E) Whether ingesting a very small amount of an allergen is sufficient to provoke an allergic reaction in a susceptible individual Which of the Options is most fit in the argument? OA E ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 14169 messages Followed by: 1818 members 13060 GMAT Score: 790 Thu", "is provided below. Allergic to nuts Not allergic to nuts Allergic to dairy $10$10 $6$6 Not allergic to dairy $6$6 $14$14 1. How many students are allergic to nuts? 2. How many students are allergic to nuts or dairy, or both? 3. How many students are allergic to at most one of the two things? ##### QUESTION 5 In a biology study, the appearance of an animal when carrying a certain gene is recorded. Assume that all animals must cary Gene $A$A or Gene $B$B but not both. Gene $A$A Gene $B$B Light fur $36$36 $27$27 Dark fur $28$28 $x$x 1. If, out of all the animals, the proportion carrying Gene A and having light fur is $\\frac{4}{19}$419, what is the value of $x$x? ### Outcomes #### S-ID.5' Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. '[Linear focus; discuss general principle.]", "| 250 | 500)", "1 to 20 of 269 total", "> > > > > > > > > > > > > > > > > > > > > > > Are 50 parents (one classroom) responsible for the > > > safety > > > > of > > > > > > > someone > > > > > > > > > else's child? > > > > > > > > > > > > > > > > > > Are 500 parents (entire school) responsible for the > > > safety > > > > > of > > > > > > > someone > > > > > > > > > else's child? > > > > > > > > > > > > > > > > > As the parent of a child with not only celiac but > also > > a > > > > > > > complicated > > > > > > > > > list of allergies and intolerances, I do not expect > > > other > > > > > > parents > > > > > > > to > > > > > > > > > be responsible for my child's safety. To date, he > has > > > not > >", "ensemble, which is less than 1/3 of the original 300.", "It has been reported that households in the west spend an annual average of$6050.00 for groceries. Assume a normal distribution with a standard deviation of $1500. a) what is the probability that randomly selected western household spends more than$6350 for groceries? b) How much money would a western household have to spend on groceries per year in order to be at the 99th percentile (ie only 1% of western households would spend more on groceries)? [snip] a) $z = \\frac{6350 - 6050}{1500} = 0.2$. Calculate Pr(Z > 0.2) = 1 - Pr(Z < 0.2). Using tables, I assume ....? b) Use tables (in reverse) to get the value of z* such that Pr(Z < z*) = 0.99. Substitute this value into $z^* = \\frac{x - 6050}{1500}$ and solve for $x$. 4. Originally Posted by loriemomof2 [snip] 6.30: It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home. Assuming that Hargrove students are typical in terms of susceptibility to allergies and that the college presidents home happens to contain just about every common allergin to which afflicted persons react, what is the probability that at least 8 of the students will be able", "50 | 100 | 250 | 500)", "inspected four times. . . But there are only 2,500 boxes. 2) Make a list of the numbers of all the boxes which are checked by 3 inspectors. $LCM(5,7,13) = 455$ Every 455th box would be inspected for its sticker, cardboard and logo. These would be: #455, 910, 1365, 1820. 2275 $LCM(5,7,23) = 805$ Every 805th box would be inspected for its sticker, cardboard and packaging. These would be: #805, 1610, 2415 $LCM(5,13,23) = 1495$ Every 1495th box would be inspected for its sticker, logo and packaging. This would be: #1495 $LCM(7,13,23) = 2093$ Every 2093rd box would be inspected for its cardboard, logo and packaging. This would be: #2093", "determine which food type the patient is allergic to with the smallest number of steps. Hopefully this is clear. I have a feeling someone may mention a Travelling Salesman type solution but is this correct and how would I incorporate probabilities into this? I am comfortable with R if a solution is provided in this format. With $6$ foods, we have $2^6 = 64$ possible states the patient can be in, corresponding to all subsets of that set of foods: a patient's state is the set of foods that patient is allergic to. Each state $i$ has a certain probability $p_i$, the sum of which is $1$. Now our knowledge of the patient's state at any time corresponds to a set of states: these are the states that are possible for this patient, given the tests that have been done. At the beginning, all $64$ states are possible, and at the end you want to narrow it down to only one. You say you want to do this as quickly as possible, by which I interpret you want to minimize the expected number of tests performed. A test will consist of administering a certain set $T$ of foods and seeing if there is an allergic response. If there is, we know the patient is allergic to at least", "so, the age at which they were diagnosed. Further detail regarding information on allergies included asking the participant to indicate the number of individual allergies (0, 1, 2, 3+) within each of the following types/groups of allergy: seasonal (pollens, molds, etc.), pets, medications, foods (wheat, peanuts, soy, tree nuts, eggs, milk, fish/ shellfish, other), and other allergies. For example, a person allergic to cats, dogs, ragweed, and aspirin would have reported 2 pet allergies, 1 seasonal allergy, 1 medication allergy, and no food or other allergies. The overall number of types of allergies was calculated by summing whether a person positively reported any medically diagnosed allergies to seasonal triggers, pets, medications, foods, or other allergens for a maximum potential score of 5. For the example given above, the number of types of allergies would be 3 (pets, seasonal, and medications). Years since diagnosis of allergies were estimated by calculating the difference between the age at brain tumor diagnosis or interview and the age at allergy diagnosis. In addition, participants were asked about medications they may have used on a “regular basis for at least 1 month” before 2 years ago. Detailed information on antihistamine use was collected through a number of questions which probed for specific details on the specific brands of prescription and over-the-counter antihistamines used for" ]

[ "more, they’re sending students to schools in … b The test does assume an identically shaped and scaled distribution for each group, except for any difference in medians. are the ranks of the Percentage of the people in the U.S. who have a food allergy : 4% of adults and 5% of children. {\\displaystyle a_{ij}=-a_{ji}} to different observations of a particular variable.", "food allergy (24) 10.21% Yes - an autoimmune disorder (i.e coeliac, colitis) (7) 2.98% Yes - I have an intolerance and allergy (8) 3.4% No (170) 72.34%", "allergies account for more than thirty thousand Wed Nov 22, 2017 2:20 pm Food allergies account for more than thirty thousand emergency department visits each year. Often, victims of these episodes are completely unaware of their allergies until they experience a major reaction. Studies show that ninety percent of food allergy reactions are caused by only eight distinct foods. For this reason, individuals should sample a minuscule portion of each of these foods to determine whether a particular food allergy is present. Which of the following must be studied in order to evaluate the recommendation made in the argument? (A) The percentage of allergy victims who were not aware of the allergy before a major episode (B) The percentage of the population that is at risk for allergic reactions (C) Whether some of the eight foods are common ingredients used in cooking (D) Whether an allergy to one type of food makes someone more likely to be allergic to other types of food (E) Whether ingesting a very small amount of an allergen is sufficient to provoke an allergic reaction in a susceptible individual Which of the Options is most fit in the argument? OA E ### GMAT/MBA Expert GMATGuruNY GMAT Instructor Joined 25 May 2010 Posted: 14169 messages Followed by: 1818 members 13060 GMAT Score: 790 Thu", "# Question A recent Health of Boston report suggests that 14% of female residents suffer from asthma as opposed to 6% of males (Boston Globe, August 16, 2010). Suppose 250 females and 200 males responded to the study. a. Develop the appropriate null and alternative hypotheses to test whether the proportion of females suffering from asthma is greater than the proportion of males. b. Calculate the value of the test statistic and the p-value. c. At the 5% significance level, what is the conclusion? Do the data suggest that females suffer more from asthma than males? Sales1 Views64", "So let's just multiply that times 120. That gets us to 25 point-- I'll just round it-- 25.3 people should have gotten sick. So the expected-- so let me write it over here, I'll do expected in yellow-- so the expected right over here. If you assume that 21% of each group should have gotten sick is that you would have expected 25.3 people to get sick in group one, in herb one group. And then the remainder will not get sick. So let's just subtract or I could actually multiply 79% times 120, either one of those would be good. But let me just take 120 minus 25.3, and then I get 94.7. So you would have expected 94.7 to not get sick. So this is expected again. 94.7 to not get sick. And now let's do that for each of these groups. So once again, group two, you would've expected 21% to get sick. 21% of the total people in that group, so that's 140, so that's 29.4. And then the remainder-- let's see, 140 minus 29.4-- should not have gotten sick. So that gets us this right here. We have 29.4 should have gotten sick if the herbs did nothing. And then, over here, we would have 110.6 should not have gotten sick. And these", "being about \\$17 billion a year. Scope of the Problem According to a common estimate, at any one time 10 percent of the American population has mental health problems sufficiently serious to warrant care; recent evidence suggests that this figure may be closer to 15 percent. Not all the people who need help receive it, however; in 1975 only 3 percent of the American population received mental health service. One major reason for this is that people still fear the stigma attached to mental illness and hence often fail to report it or to seek help. Analysis of the figures on mental illness shows that schizophrenia afflicts an estimated 2 million Americans, another 2 million suffer from profound depressive disorders, and 1 million have organic psychoses or other permanently disabling mental conditions. As much as 25 percent of the population is estimated to suffer from mild or moderate depression, anxiety, and other types of emotional problems. Some 10 million Americans have problems related to alcohol abuse, and millions more are thought to abuse drugs. Some 5 to 15 percent of children between the ages of 3 and 15 are the victims of persistent mental health problems, and at least 2 million are thought to have severe learning disabilities that can seriously impair their mental health. In addition, according", "Contents Allergies 1 to 5 Rating Symptoms Factors Effects Allergies Allergy Causes of Allergies User reported causes of Allergies based on their intuition. Cause Agree 86% 77% 65% 65% 65% 58% 58% 58% 55% 54% 51% 50% 47% 38% 36% 35% 34% 34% 34% 34% 33% 31% 30% 28% 28% 27% 27% 26% 26% 25% 24% 22% 20% 19% 19% 19% 19% 18% 18% 18% 18% 17% 16% 16% 16% 15% 15% 14% 14% 13% 13% 13% 12% 12% 12% 11% 11% 11% 11% 10% 10% 10% 9% 9% 9% 9% 9% 9% 9% 8% 8% 8% 8% 8% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 7% 6% 6% 6% 6% 6% 6% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 5% 4% 4% 4% 4% 4% 3% 3% 3% 3% 3% 3% 3% 3% 2% 2% 2% 1% 1% 1% 1% Conditions Resulting from Allergies User-reported conditions resulting from Allergies based on their intuition. Resulting Condition Agree 90% 74% 64% 49% 48% 42% 25% 23% 11% Treatments User-reported effectiveness in treating Allergies Treatment Major Improvement Moderate Improvement Much Worse No Effect Worse Responses 12% 31% 3% 49% 5% 221 9% 15% 4% 72% 0% 53 7% 27% 2% 62% 3% 116 9% 24% 1% 65% 1% 101 31% 50% 0% 19%", "diagnosis of allergic rhinitis (OR: 2.1; CI: 1.1, 3.7) were also predictors for asthma. The population-attributable risk of having 1 or more residential exposures associated with doctor-diagnosed asthma was 44.4% (95% CI: 29–60), or an estimated 2 million excess cases. The attributable cost of asthma resulting from residential exposures was $405 million (95% CI:$264–\\$547 million) annually. Conclusions. The elimination of identified residential exposures, if causally associated with asthma, would result in a 44% decline in doctor-diagnosed asthma among older children and adolescents in the United States.", "# Sinus Congestion Studies ### Higher Sinus Congestion Predicts Slightly Higher Guiltiness This individual’s Guiltiness is generally lowest after an average of 1.9 out of 5 of Sinus Congestion over the previous 7 days.", "1 (604) 999-1884 - U.S & Canada|[email protected] Do you have asthma? Take the asthma quiz 2020-04-21T16:01:05-07:00 # Do You Have Asthma ## Take The 30 Second Quiz It is estimated that there are 22 million people diagnosed with asthma in North America; nearly nine million of them are children. More than 5000 North Americans will die from an asthma attack this year. It is believed a large percentage of those deaths could be prevented through understanding and preventative actions. WHO IS AT RISK • Asthma is common, affecting 5 – 10% of the population. • Asthma is one of the most common conditions of childhood. • Asthma is closely linked to allergies. • Asthma can run in families. • Most but not all people with asthma have allergies. • Children with a family history of allergy and asthma are more likely to have asthma. • Although asthma affects people of all ages, it most often starts in childhood. • More boys have asthma than girls. • In adulthood, more women have asthma than men. • Although asthma affects people of all races, African American are more likely than Caucasians to be hospitalized for asthma attacks. ### AND GET A FREE EBOOK! I wheeze (breathing out with a noisy whistling sound) I cough due to stuff (mucus) in my", "that you may not be allergic to a specific allergen. Keep in mind that skin tests aren't always accurate. Sometimes it indicates an allergy when there isn't a single allergy (false positive), or a skin test might not trigger a reaction when you're exposed to something you're allergic to (false negative). You may react differently to the same test done on different occasions. Or, you may react positively to a substance during the test but not interact with it in everyday life. Your allergy treatment plan may include medications, immunotherapy, changes to your work environment or home, or dietary changes. Ask your doctor to explain anything you do not understand about your diagnosis or treatment. With test results that identify your allergens and a treatment plan to help you gain control, you'll be able to reduce or eliminate allergy signs and symptoms. Are there any risks involved? Since a very small amount of the allergen is used, the reaction is also very small and clears up within a few hours. Most of the time, the reaction can appear as a few itchy bumps. Asthma patients tend to develop a severe allergic reaction or asthma attack within the next 24 hours. Therefore, they should always keep the physician informed of their condition and always carry inhalers and medications while", "Dec 21, 2017 5:56 am lheiannie07 wrote: Food allergies account for more than thirty thousand emergency department visits each year. Often, victims of these episodes are completely unaware of their allergies until they experience a major reaction. Studies show that ninety percent of food allergy reactions are caused by only eight distinct foods. For this reason, individuals should sample a minuscule portion of each of these foods to determine whether a particular food allergy is present. Which of the following must be studied in order to evaluate the recommendation made in the argument? (A) The percentage of allergy victims who were not aware of the allergy before a major episode (B) The percentage of the population that is at risk for allergic reactions (C) Whether some of the eight foods are common ingredients used in cooking (D) Whether an allergy to one type of food makes someone more likely to be allergic to other types of food (E) Whether ingesting a very small amount of an allergen is sufficient to provoke an allergic reaction in a susceptible individual Plan: Individuals should sample a minuscule portion of the eight foods known to cause 90% of allergic reactions. Conclusion: An individual will be able to determine whether a particular food allergy is present. Rephrase the answer choices as STATEMENTS. The", "2 of them. How many people suffered from exactly one symptom? Math Expert Joined: 02 Sep 2009 Posts: 56307 Re: Of the 300 patients who suffered from at least one symptom of A, B, an [#permalink] ### Show Tags 20 Aug 2010, 08:43 Of the 300 patients who suffered from at least one symptom of A, B, and C, 35 % suffered from A, 45% suffered from B, 40 % suffered from C, 10% suffered from exactly 2 of them. How many people suffered from exactly one symptom? Total=A+B+C - {People who suffered from exactly 2 symptoms} - 2*{People who suffered from exactly 3 symptoms} + {People who suffered from neither of symptoms}; 100=35+45+40-10-2*{People who suffered from exactly 3 symptoms}+0 --> {People who suffered from exactly 3 symptoms}=5%; So as {People who suffered from exactly 2 symptoms}=10% and {People who suffered from exactly 3 symptoms}=5% then {People who suffered from exactly one symptom} is 100-10-5=85% --> in numbers 300*85%=255. Check for more on this formula at: formulae-for-3-overlapping-sets-69014.html#p729340 Hope it helps. _________________ ##### General Discussion Manager Joined: 28 Aug 2004 Posts: 194 Re: Of the 300 patients who suffered from at least one symptom of A, B, an [#permalink] ### Show Tags 21 May 2005, 00:10 1 I deferred solution till OA is out, but here it goes.. T", "> > > > > > > > > > > > > > > > > > > > > > > Are 50 parents (one classroom) responsible for the > > > safety > > > > of > > > > > > > someone > > > > > > > > > else's child? > > > > > > > > > > > > > > > > > > Are 500 parents (entire school) responsible for the > > > safety > > > > > of > > > > > > > someone > > > > > > > > > else's child? > > > > > > > > > > > > > > > > > As the parent of a child with not only celiac but > also > > a > > > > > > > complicated > > > > > > > > > list of allergies and intolerances, I do not expect > > > other > > > > > > parents > > > > > > > to > > > > > > > > > be responsible for my child's safety. To date, he > has > > > not > >", "# Math Help - [SOLVED] Statistics and Probability: 1. ## [SOLVED] Statistics and Probability: Out of 20 questions I have managed to complete all but these four, so any assistance would be awesome. Not sure if I am in brain freeze mode or what: 6.22: According to the national Marine manufacturers Association, 50% of the population of Vermont were boating participants during the most recent year. For the randomly selected sample of 20 Vermont residents, with the discrete random variable x= the number in the sample who were boating participants that year, determine the following: a) E (x) b) P (x ≤ 8) c) P (x=10) d) P (x=12) e) P (7 ≤ x ≤ 13) 6.30: It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home. Assuming that Hargrove students are typical in terms of susceptibility to allergies and that the college presidents home happens to contain just about every common allergin to which afflicted persons react, what is the probability that at least 8 of the students will be able to stay for the duration of the event? 7.33: It has been reported that households in the west spend an", "It is possible that you have bronchitis and your cough just happened to go away by chance during this particular hour, but that is unlikely (certainly less likely than 1 in 10 for a long lasting, consistent cough). On the other hand, allergy medication tends to be fairly effective against coughs caused by allergies, so we should expect there to be at least a 1 in 3 chance that your cough would go away after taking the medication if you did in fact have allergies. Hence, in this case we can produce a conservatively low estimate for the Bayes factor with just a couple minutes of thought. How likely the cough was to have stopped given A we put at least at 1/3. How likely the cough was to stop given B we put at most at 1/10. Hence, the Bayes Factor, which is how likely the cough was to stop given A compared to how likely it was to stop given B, is greater than (1/3) / (1/10) = 3.3. This should be interpreted as saying that we should now believe A is at least 3.3 times more likely compared to B than we used to think it was. In other words, the Bayes Factor tells us how much our new evidence should cause us to update", "to further investigate 1. ### Statistics The monthly incomes from a random sample of workers in a factory are shown below. Monthly Income (In \\$1,000) 4.0 5.0 7.0 4.0 6.0 6.0 7.0 9.0 a. Compute the standard error of the mean (in dollars). b. Compute the margin of error 2. ### Statistics #3 The estimate of the population proportion is to be within plus or minus .10, with a 99 percent level of confidence. The best estimate of the population proportion is .45. How large a sample is required? 3. ### Math (Statistics) The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary 4. ### Statistics (1) Gluten sensitivity, which is also known as wheat intolerance, affects approximately 15% of people in U.S. Let p be the proportion in a random sample of 800 individuals who have gluten sensitivity. Find the probability that the", "exactly how flawed it is because the authors, in breach of all guidelines on such matters, have refused, for over a year, to make available the raw data that would allow a correct analysis to be done. Annals of Allergy, Asthma, & Immunology ©2000;84:639-640 To the Editor: We should like to comment on the paper by Forsyth et al. 1 This paper claims, on the basis of a randomized double-blind clinical trial, that NADH is effective in the treatment of chronic fatigue syndrome. It seems that there are several problems with the work described in this paper. • On page 187, the ‘raw symptom score’ is defined as a score from a questionnaire normalized to lie between 0 and 1 (the questionnaire had 50 questions, each scored 1, 2, 3 or 4, so raw scores are in the range 50 to 200). This was normalized to lie between 0 and 1 by first subtracting 50 and then dividing by 150. However, patients with chronic fatigue syndrome are said to have scores ranging from 10.9 to 55.8, so the normalized score has presumably been expressed as a percentage, whereas in the case of normal patients the values of 0.01 to 0.13 suggest that proportions have been used. Furthermore, even using percentages rather than proportions, we cannot reproduce the ranges", "many people will test positive for it 22. anonymous of the 20 people that have it, the test is 98% accurate what is 98% of 20? 23. anonymous btw i hope it is clear that you do $$.98\\times 20$$ 24. anonymous 19.6? 25. anonymous right 26. anonymous yeah I'm good with the percentage....most of the time 27. anonymous i guess we have to work with the decimal, should have chosen a population of 10,000 then it would be whole numbers but too late now 28. anonymous now we know $$980$$ so NOT have the allergy, lets see how many of those will also test positive 29. anonymous for them the test is 96% accurate so 96% of them will test negative, but that means 6% of them will test positive what is 6% of 980? 30. anonymous 31. anonymous 4% will test positive what is 4% of 980 32. anonymous 39.2? 33. anonymous yeah i get that too 34. anonymous ok so now how many people total (it is a decimal) will test positive? 35. anonymous 0.04 people? I'm confused 36. anonymous we computed two numbers of people that test positive right? the ones with the allergy, 19.6 and the ones without the allergy, 39.2 what is the total? 37. anonymous oh, 58.8? 38. anonymous right, that is", "About 29% of ..." ]

[ "+0 Decimals 0 49 1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ May 15, 2022 #1 +13731 +1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ Hello Guest! $$x=0.\\overline{81}\\\\ 100x=81.\\overline{81}\\\\ 99x=81\\\\ x=\\dfrac{9}{11}$$ $$\\color{blue}0.\\overline{81}=\\dfrac{9}{11}$$ ! May 15, 2022", "+0 # Fractions 0 82 1 Express as a fraction in lowest terms: $0.\\overline{12} + 0.\\overline{001}$ May 25, 2022", "+0 # Express $4.\\overline{054}$ as a common fraction in lowest terms. 0 126 1 Express $4.\\overline{054}$ as a common fraction in lowest terms. Aug 31, 2020 #1 0 If we expand it out, we get 4.054054054..... Setting this equal to x, we have: 1000x = 4054.054054054... x = 4.054054054... Subtracting the equations, we get: 999x = 4050 x = 4050/999 x = 450/111 x = 150/37 Jan 14, 2021", "+0 # Pls help 0 92 1 Compute:$$0.\\overline{7}-0.\\overline{4}+0.\\overline{2}$$ . Express your answer as a common fraction. Sep 22, 2020", "# Express it in fraction Express $$0.461538461538...$$ as a fraction. (In other words, 461538 repeats forever which is called 0 point 461538 recurring.) ×", "a fraction! #### Examples ##### Question 1 Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. ##### Question 2 Starting with $x=0.\\overline{216}$x=0.216, express $x$x as a fraction in simplest form. ##### Question 3 Starting with $x=0.6\\overline{87}$x=0.687, express $x$x as a fraction in simplest form.", "c) $7.\\overline{09}$. Discussion", "+0 # Decimals 0 35 5 Express as a common fraction the reciprocal of $0.\\overline{72}$. Jun 28, 2022 #1 +1 Express as a common fraction the reciprocal of $0.\\overline{72}$. I can give you the answer. The decimal equals 8 / 11 so its reciprocal would be 11 / 8. I can't tell you how to calculate it though. I guessed and got lucky. I guessed 8/9 ... that was too big, so I guessed 8/11 and that was it. . Jun 28, 2022 #2 +2275 +1 Let $$x = 0. \\overline{72}$$ This means $$100x = 72.\\overline{72}$$ Subtracting the 2 gives $$99x = 72$$, meaning $$x = {72 \\over 99} ={ 8 \\over 11}$$. The reciprocal of this, as Guest found, is $$\\color{brown}\\boxed{11 \\over 8}$$ Jun 28, 2022 #3 +69 +2 Just a fun fact, if it repeats every one digit, then the fraction version is [digit] / 9. If it repeats every two digits, then it is [digit] / 99. You can see the pattern here. ⚡⚡⚡ ShockFish Jun 28, 2022 edited by ShockFish Jun 28, 2022 #4 +2275 +1 Hi Shockfish, nice to see you here BuilderBoi Jun 28, 2022 #5 +69 +2 Heyyy just wondering if you were here too ShockFish Jun 28, 2022", "# Convert recurring decimals to fractions ## Interactive practice questions Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. Easy Approx a minute Starting with $x=2.\\overline{3}$x=2.3, express $x$x as a fraction in simplest form. Starting with $x=1.\\overline{7}$x=1.7, express $x$x as a fraction in simplest form. Starting with $x=3.\\overline{6}$x=3.6, express $x$x as a fraction in simplest form.", "downloads about What Is Rewrite 4 As A Fraction In Simplest Form. Find answers researching ebooks, papers or essays. ... Express as a fraction in simplest form: a) 0.806 b) 0.0256 c) 11.375 6. Rewrite as a decimal: a) 9 16 b) 17 40 c) 52 464 25i d) 7 12 e) 3 7 f) 24 11 ...", "Express $\\dfrac{14}{-5}$ as a rational number with numerator $56$. • A $\\dfrac{56}{-20}$ • B $\\dfrac{56}{-5}$ • C $\\dfrac{56}{-10}$ • D $\\dfrac{56}{-15}$", "Q # Express the following numbers in standard form. (ii) 0.00000000000942 1. Express the following numbers in standard form. (ii) 0.00000000000942 The standard form is $9.42\\times 10^{-12}$", "Math Help - Fraction 1. Fraction Express as a simple fraction: [(1/2(x+h))-(2/3x)] / h 2. Is this what you're trying to show? $[1/2(x+h) - 2/3x]/h$ 3. Yeah, now I need to figure out where to start though.", "Browse Questions # Express the following in the standard form $a + ib$: $\\frac{2(i-3)}{(1+i)^2}$ This is part 1 of a multipart question, answered separately on Clay6.com", "$v$ and $\\rho$ and then express $\\rho$ in terms of $z$.)", "\"common fraction\" • fraction", "21:13 The problems says \"express in terms of $f_1$ and $f_1 + f_2$.\" I defined $f_2' = f_1 + f_2$ for convenience. – Muphrid May 30 '13 at 21:17 Thank you for your answer. – user4167 May 30 '13 at 21:20", "Q # Express the following numbers in standard form. (i) 0.0000000000085 1. Express the following numbers in standard form. (i) 0.0000000000085 The standard form is $8.5\\times 10^{-12}$", "#### Problem 54E Express the number as a ratio of integers. $10.1 \\overline{35}=10.135353535 \\ldots$", "so i can under. please. • How can you express 18.7? • If you want it as a fraction it would be 18 7/10 • how do u change 5.03 as a fraction please? • They should fix the last question because it says express 5.03 as a fraction and some people will think oh that's easy it's 503/100 but it says it's wrong and the right answer is 5 and 3/100 which is right" ]

[ "# Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \\overline{6}$<br/> <br/> (ii) $0.4 \\overline{7}$ <br/> <br/>(iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$", "Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. (i) $0 . \\overline{6}$ (ii) $0.4 \\overline{7}$ (iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$ Editor", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$", "# Why is ⅓ = 0.3333....? • ## Why is it true that $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ ? I wish I had a very simple and clever way to show this, but all the simple and clever ways usually use some idea that is already a step more advanced. For example, you could say that $$0.\\overline{999}\\ldots = 1$$ and then divide both sides by $$3$$ (but knowing the fact that $$0.\\overline{999}\\ldots = 1$$ is even harder to prove!). Or you could start from the fact that $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ and multiply both sides by $$3,$$ but wait, isn't that just the reverse of what we did to get to this place? We were originally trying to show $$\\frac{1}{9} = 0.\\overline{111}\\ldots$$ in the first place, assuming that we knew the decimal representation of $$\\frac{1}{3}!$$ (Round and round we go!) I think the best way to show why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is just to show the long division. Then you'll find that $$1$$ divided by $$3$$ equals $$0.\\overline{333}\\ldots.$$ . This isn't hard, but there are some basic concepts to lay down first. : place value and the idea of division as a reverse-multiplication operation. ## Place value I have some affectionate names that I personally like to use when referring to place value digits. You can make up your own names,", "first by $10^{s}$, where $s$ is the length of the period of the pure recurring decimal $x$. Hence, $$10^{s}x = q_{1}q_{2}\\ldots q_{s-1}q_{s}.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + 0.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + x.$$ Therefore, $$x = \\frac{q_{1}q_{2}\\ldots q_{s}}{10^{s} - 1} = \\frac{q_{1}q_{2}\\ldots q_{s}}{999....9},$$ where the denominator in the last expression is consisted only of $9$'s and it contains the same number of $9$'s as that of the length $s$ of the period of the pure recurring decimal. If $d = g.c.d((q_{1}q_{2}\\ldots q_{s}), (999....9))$, then $$x = \\frac{\\frac{q_{1}q_{2}\\ldots q_{s}}{d}}{\\frac{999....9}{d}},$$ gives the equivalent irreducible proper fraction form of the recurring decimal $x$. Let us illustrate this through an example. Consider $x = 0.\\overline{285714}$, then $x = \\frac{285714}{10^{6} - 1} = \\frac{285714}{999999} = \\frac{2}{7}.$ Case II (Mixed Recurring Decimal) : Let, $0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}}$ be the mixed recurring decimal and we need to find out its equivalent irreducible proper fraction form. $x = 0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}},$ then we will multiply $x$ first by $10^{t},$ $t$ is the length of the starting non-repeating part of the decimal representation. Hence, $$10^{t}x = (r_{1}r_{2}\\ldots r_{t}) + 0.\\overline{q_{1}q_{2}\\ldots q_{s}}.$$ The above expression will then be multiplied by $10^{s},$ where $s$ is the length of the period of the mixed recurring decimal $x$. We, therefore, obtain $$10^{s + t}x = 10^{s}(r_{1}r_{2}\\ldots r_{t}) + q_{1}q_{2}\\ldots q_{s}", "Thus, the number of time 9 to be repeated in the denominator becomes three. $$\\begin{array}{l}0.\\overline{125} = \\frac{125}{999}\\end{array}$$ ### Case 2: Fraction of the type $$\\begin{array}{l}0.ab..\\overline{cd}\\end{array}$$ The formula to convert this type of repeating decimal to the fraction is given by: $$\\begin{array}{l}0.ab..\\overline{cd} =\\frac{(ab….cd…..) – ab……}{Number \\; of \\; time \\; 9’s \\; the \\; repeating \\; term \\; followed \\; by \\; the \\; number \\; of \\; times \\; 0’s \\; for \\; the \\; non-repeated \\; terms }\\end{array}$$ Example 3: Convert $$\\begin{array}{l}0.12\\overline{34}\\end{array}$$ to the fractional form. Solution: In the given decimal number, 12 is a non-repeated decimal value, and 34 is in the repeating form. Thus denominator becomes 9900. $$\\begin{array}{l}0.12\\overline{34} = \\frac{1234 – 12 }{9900} = \\frac{1222}{9900}\\end{array}$$ Example 4: Convert $$\\begin{array}{l}0.00\\overline{69}\\end{array}$$ in the fraction form. Solution: In the given decimal number, the number 00 is a non-repeated decimal value, and 69 is in the repeating form. Thus, the denominator becomes 9900. $$\\begin{array}{l}0.00\\overline{69} = \\frac{0069}{9900} = \\frac{69}{9900}\\end{array}$$ It is all about the conversion of repeating decimal to the fraction form. To learn more interesting topics in Maths, download BYJU’S – The Learning App and learn with ease. Test your Knowledge on Repeating Decimal to Fraction", "$3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Algebraic Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$ ### Limits $0.999\\ldots = \\lim_{n\\to\\infty}0.\\underbrace{ 99\\ldots9 }_{n} = \\lim_{n\\to\\infty}\\sum_{k = 1}^n\\frac{9}{10^k} = \\lim_{n\\to\\infty}\\left(1-\\frac{1}{10^n}\\right) = 1-\\lim_{n\\to\\infty}\\frac{1}{10^n} = 1$", "$1+4+7+ \\ldots +295$ \\item $4+2+0-2- \\ldots - 146$ \\item $1+3+9+ \\ldots + 2187$ \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\begin{multicols}{3} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots + \\frac{1}{256}\\vphantom{\\displaystyle \\sum_{n = 1}^{10} -2n + \\left(\\frac{5}{3}\\right)^{n}}$ \\item $3 - \\frac{3}{2} + \\frac{3}{4} - \\frac{3}{8}+- \\dots +\\frac{3}{256} \\vphantom{\\displaystyle \\sum_{n = 1}^{10} -2n + \\left(\\frac{5}{3}\\right)^{n}}$ \\item $\\displaystyle \\sum_{n = 1}^{10} -2n + \\left(\\frac{5}{3}\\right)^{n}$ \\label{findsumformulalast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} In Exercises \\ref{dectofracfirst} - \\ref{dectofraclast}, use Theorem \\ref{geoseries} to express each repeating decimal as a fraction of integers. \\begin{multicols}{4} \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item $0.\\overline{7}$ \\label{dectofracfirst} \\item $0.\\overline{13}$ \\item $10.\\overline{159}$ \\item $-5.8\\overline{67}$ \\label{dectofraclast} \\setcounter{HW}{\\value{enumi}} \\end{enumerate} \\end{multicols} \\pagebreak In Exercises \\ref{annuityfirst} - \\ref{annuitylast}, use Equation \\ref{fvannuity} to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly. \\begin{enumerate} \\setcounter{enumi}{\\value{HW}} \\item payments are \\$300, interest rate is 2.5\\%, term is 17 years. \\label{annuityfirst} \\item payments are \\$50, interest rate is 1.0\\%, term is 30 years. \\item payments are \\$100, interest rate is 2.0\\%, term is 20 years \\item payments are \\$100, interest rate is 2.0\\%, term is 25 years \\item payments are \\$100, interest rate is 2.0\\%, term is 30 years \\item payments are \\$100, interest rate is 2.0\\%, term is 35 years \\label{annuitylast} \\item", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "# What is 0.8959595... as a fraction ? Jan 9, 2017 $0.8 \\overline{95} = \\frac{887}{990}$ #### Explanation: Given: $0.8 \\overline{95}$ First multiply by $10 \\left(100 - 1\\right) = 1000 - 10$ to get an integer. The first multiplier $10$ is to shift the repeating pattern to just after the decimal point. Then $\\left(100 - 1\\right)$ shifts the number $2$ places to the left and subtracts the original value to cancel out the repeating tail: $\\left(1000 - 10\\right) \\cdot 0.8 \\overline{95} = 895. \\overline{95} - 8. \\overline{95} = 887$ Then divide both ends by $\\left(1000 - 10\\right)$ to find: $0.8 \\overline{95} = \\frac{887}{1000 - 10} = \\frac{887}{990}$ $887$ and $990$ have no common factor larger than $1$, so this fraction is in simplest form.", "one tenth, the three hundredths would turn into a one hundredth, the three thousandths would turn into a one thousandth, etc. Dividing the equivalent fraction, $$\\frac{1}{3},$$ by $$3,$$ gives us the fraction $$\\frac{1}{9}.$$ This will give us the decimal that we want, $$0.\\overline{111}\\ldots,$$ which is then equal to $$\\frac{1}{9}!$$ The same cool pattern applies to the repeating decimal representations of similar fractions whose denominator is $$9,$$ like the ones below: $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ $$0.\\overline{222}\\ldots = \\frac{2}{9}$$ $$0.\\overline{444}\\ldots = \\frac{4}{9}$$ $$0.\\overline{555}\\ldots = \\frac{5}{9}$$ $$0.\\overline{777}\\ldots = \\frac{7}{9}$$ $$0.\\overline{888}\\ldots = \\frac{8}{9}$$ • could you please further elucidate for the students like myself i understood this part that we divide .333 by 3 to arrive at 0.111 which turns 3 tenths into 1 tenths and the rest accordingly but how did we derive 1 / 3 ? did we derive from the fact that 3 divides .333 by .111 so it becomes 1/3 i understood this part - for dividing the 1/3 by 3 is actually multiplying with its reciprocal 1/3 which gives us 1/9 and if i am correct in my above computation then here's my other queries - For the other numbers - 0.222 if we divide 2 into 0.222 - this would give us .111 same as before, but when we are going to divide 2 by 1/2 which", "≠ 0. (p, q ϵ Z, q ≠ 0). (i) $$0 . \\overline{6}$$ (ii) $$0 . \\overline{47}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.6666………. Let x = 0.6666…………. ∴ x = 0.6666……….. Multiplying both sides by 10, 10x = 6.666….. 10x = 6 + 0.666 10x = 6 + x 10x – x = 6 9x = 6 $$x=\\frac{6}{9}=\\text { form of } \\frac{\\mathrm{p}}{\\mathrm{q}}$$ ∴ $$0 . \\overline{6}=\\frac{2}{3}$$ (ii) $$0 . \\overline{47}$$ = 0.4777….. Let x = $$0 . \\overline{47}$$ then x = 0.4777…….. Here digit 4 is not repeating but 7 is repeated. Let x = 0.4777 Multiplying both sides by 10. 10x = 4.3 + x 10x -x = 4.3 9x = 4.3 Multiplying both sides by 10. 90x = 43 ∴ $$0.4 \\overline{7}=\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ x = 0.001001………. Multiplying both sides by 1000 1000x = 1.001001 1000x = 1 + x 1000x – x = 1 999x = 1 $$x=\\frac{1}{999}$$ ∴ $$0 . \\overline{001}=\\frac{1}{999}$$ Question 4. Express 0.99999…….. in the form $$\\frac{p}{q}$$ , Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense. Express 0.99999 in the form of $$\\frac{p}{q}$$, Let x = 0.99999 … Multiplying by 10. 10x = 9.9999…… 10x = 9 + 0.9999……. 10x = 9 + x 10x", "us: $0. \\overline{3} = \\frac{3}{{10}^{1} - 1} = \\frac{3}{9} = \\frac{1}{3}$. We can also modify the technique to handle cases where the repeating portion does not begin until after a finite number of non-repeating digits, such as numbers like $0.1232323 \\ldots = 0.1 \\overline{23}$. The idea is still to multiply by an appropriate power of $10$ and then subtracting away the repeating portion. As an exercise, a student can try finding the fractional form of such a number.", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "# Are numbers with repeating patterns in their decimal expansion (e.g. $0.123123123\\ldots$) rational? There's a question that I've been thinking about for quite some time now. We all know that numbers with infinite decimal expansion such as $0.\\overline{3}$ or $0.\\overline{1}$ are not necessarily irrational. $$0.\\overline{3}=0.333333\\dots=\\frac{1}{3}$$ $$0.\\overline{1}=0.111111\\ldots=\\frac{1}{9}$$ Therefore we can say that for instance $0.\\overline{4}$ is rational because we have: $$0.\\overline{4}=4\\cdot 0.\\overline{1}=\\frac{4}{9}$$ My question is: can we generalize these results to any number that has a repeating pattern in its decimal expansion? Can we claim for instance that $0.\\overline{123}=0.123123123\\ldots$ is rational? This question boils down to asking whether or not expressions like $0.\\overline{01}$ and $0.\\overline{001}$ etcetera are rational. If so, in the case of my example we could say: $$0.\\overline{123}=123\\cdot 0.\\overline{001}$$ A naive thought of mine was that $0.\\overline{01}$ is simply $\\frac{1}{10}\\cdot 0.\\overline{1}$ but this is of course not true. Then I thought that maybe we can represent $0.\\overline{01}$ as $\\sum\\limits_{i=1}^{\\infty}\\frac{1}{10^{2i}}$ which converges by the ratio test. However this does not tell us if it is rational or not. This is not for homework or anything, just something I've been thinking about and I was wondering if any of you have some knowledge about this issue. Thanks. - Yes, all repeating decimal numbers are rational. I'm sure someone here will give a good explanation but here's one link that'll", "# RD Sharma Solutions Class 9 Number System Exercise 1.2 ## RD Sharma Solutions Class 9 Chapter 1 Ex 1.2 Q1. Express the following rational numbers as decimals: (i) $\\frac{42}{100}$ (ii) $\\frac{327}{500}$ (iii) $\\frac{15}{4}$ Solution: (i) By long division method 100) $\\overline{42}$ ( 0.42 400 $\\overline{200}$ 200 $\\overline{0}$ Therefore, $\\frac{42}{100}$ = 0.42 (ii) By long division method 500) $\\overline{327.000}$ ( 0.654 3000 $\\overline{2700}$ 2500 $\\overline{2000}$ 2000 $\\overline{0}$ Therefore, $\\frac{327}{500}$ = 0.654 (iii) By long division method 4) $\\overline{15.00}$ ( 3.75 12 $\\overline{30}$ 28 $\\overline{20}$ 20 $\\overline{0}$ Therefore, $\\frac{15}{4}$ = 3.75 Q2. Express the following rational numbers as decimals: (i) $\\frac{2}{3}$ (ii) –$\\frac{4}{9}$ (iii) –$\\frac{2}{15}$ (iv) –$\\frac{22}{13}$ (v) $\\frac{437}{999}$ Solution: (i) By long division method 3) $\\overline{2.0000}$ ( 0.66 18 $\\overline{20}$ 18 $\\overline{2}$ Therefore, $\\frac{2}{3}$ = 0.66 (ii) By long division method 9) $\\overline{4.000}$ ( 0.444 3600 $\\overline{4000}$ 3600 $\\overline{4000}$ 3600 $\\overline{400}$ Therefore, – $\\frac{4}{9}$ = – 0.444 (iii) By long division method 15) $\\overline{2.00}$ ( 1.333 15 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{5}$ Therefore, $\\frac{2}{15}$ = -1.333 (iv) By long division method 13) $\\overline{22.000}$ ( 1.69230769 13 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{30}$ 26 $\\overline{40}$ 39 $\\overline{100}$ 91 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{3}$ Therefore, – $\\frac{22}{13}$ = – 1.69230769 (v) By long division method 999) $\\overline{437.0000}$ ( 0.43743 3996 $\\overline{3740}$ 2997 $\\overline{7430}$ 6993 $\\overline{4370}$ 3996 $\\overline{3740}$ 2997 $\\overline{743}$", ". \\overline{857142}$$ Question 3. Express the following in the form p/q, where p and q are integers and q ≠ 0. (i) $$0 . \\overline{6}$$ (ii) $$0 . 4 \\overline{7}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.666… Let x = 0.666… ∴ 10x = 6.66… ∴ 10x = 6 + 0.66… ∴ 10x = 6 + 0.666… ∴ 10x = 6 + x ∴ 9x = 6 x = $$\\frac{2}{3}$$ (ii) $$0 . 4 \\overline{7}$$ = 0.4777… = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ Let x = 0.777… ∴ 10x = 7.77… ∴ 10x = 7.777… ∴ 10x = 7 + 0.777… ∴ 10x = 7 + x ∴ x = $$\\frac{7}{9}$$ $$0 . 4 \\overline{7}$$ = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ = $$\\frac{4}{10}$$ + $$\\frac{7}{90}$$ = $$\\frac{36}{90}$$ + $$\\frac{7}{90}$$ = $$\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ =0.001001… Let x = 0.001001… ∴ 1000x = 1.001…. ∴ 1000x = 1.001001… ∴ 1000x = 1 + 0.001001 … ∴ l000x= 1 + X ∴ 999x = 1 x = $$\\frac{1}{999}$$ Question 4. Express 0.99999…in the form $$\\frac{p}{q}$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Let x = 0.9999… 10x = 9.999… ∴ 10x = 9.9999… ∴ 10x = 9 + 0.9999… ∴ 10x = 9 + x ∴ 9x = 9 x" ]

[ "# Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \\overline{6}$<br/> <br/> (ii) $0.4 \\overline{7}$ <br/> <br/>(iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. (i) $0 . \\overline{6}$ (ii) $0.4 \\overline{7}$ (iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$ Editor", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$", "+0 # Can I get help with 1 question? 0 107 1 +37 Express $$\\frac{0.\\overline{12}}{0.\\overline{321}}$$as a common fraction. Thanks for the help Jun 19, 2021 #1 +2250 +1 12 with line = 12/99 321 with line = 321/999 (12/99)/(321/999) Can you take it from here? =^._.^= Jun 19, 2021", "# Why is ⅓ = 0.3333....? • ## Why is it true that $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ ? I wish I had a very simple and clever way to show this, but all the simple and clever ways usually use some idea that is already a step more advanced. For example, you could say that $$0.\\overline{999}\\ldots = 1$$ and then divide both sides by $$3$$ (but knowing the fact that $$0.\\overline{999}\\ldots = 1$$ is even harder to prove!). Or you could start from the fact that $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ and multiply both sides by $$3,$$ but wait, isn't that just the reverse of what we did to get to this place? We were originally trying to show $$\\frac{1}{9} = 0.\\overline{111}\\ldots$$ in the first place, assuming that we knew the decimal representation of $$\\frac{1}{3}!$$ (Round and round we go!) I think the best way to show why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is just to show the long division. Then you'll find that $$1$$ divided by $$3$$ equals $$0.\\overline{333}\\ldots.$$ . This isn't hard, but there are some basic concepts to lay down first. : place value and the idea of division as a reverse-multiplication operation. ## Place value I have some affectionate names that I personally like to use when referring to place value digits. You can make up your own names,", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "first by $10^{s}$, where $s$ is the length of the period of the pure recurring decimal $x$. Hence, $$10^{s}x = q_{1}q_{2}\\ldots q_{s-1}q_{s}.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + 0.\\overline{q_{1}q_{2}\\ldots q_{s}} = q_{1}q_{2}\\ldots q_{s} + x.$$ Therefore, $$x = \\frac{q_{1}q_{2}\\ldots q_{s}}{10^{s} - 1} = \\frac{q_{1}q_{2}\\ldots q_{s}}{999....9},$$ where the denominator in the last expression is consisted only of $9$'s and it contains the same number of $9$'s as that of the length $s$ of the period of the pure recurring decimal. If $d = g.c.d((q_{1}q_{2}\\ldots q_{s}), (999....9))$, then $$x = \\frac{\\frac{q_{1}q_{2}\\ldots q_{s}}{d}}{\\frac{999....9}{d}},$$ gives the equivalent irreducible proper fraction form of the recurring decimal $x$. Let us illustrate this through an example. Consider $x = 0.\\overline{285714}$, then $x = \\frac{285714}{10^{6} - 1} = \\frac{285714}{999999} = \\frac{2}{7}.$ Case II (Mixed Recurring Decimal) : Let, $0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}}$ be the mixed recurring decimal and we need to find out its equivalent irreducible proper fraction form. $x = 0.r_{1}r_{2}\\ldots r_{t}\\overline{q_{1}q_{2}\\ldots q_{s}},$ then we will multiply $x$ first by $10^{t},$ $t$ is the length of the starting non-repeating part of the decimal representation. Hence, $$10^{t}x = (r_{1}r_{2}\\ldots r_{t}) + 0.\\overline{q_{1}q_{2}\\ldots q_{s}}.$$ The above expression will then be multiplied by $10^{s},$ where $s$ is the length of the period of the mixed recurring decimal $x$. We, therefore, obtain $$10^{s + t}x = 10^{s}(r_{1}r_{2}\\ldots r_{t}) + q_{1}q_{2}\\ldots q_{s}", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", ". \\overline{857142}$$ Question 3. Express the following in the form p/q, where p and q are integers and q ≠ 0. (i) $$0 . \\overline{6}$$ (ii) $$0 . 4 \\overline{7}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.666… Let x = 0.666… ∴ 10x = 6.66… ∴ 10x = 6 + 0.66… ∴ 10x = 6 + 0.666… ∴ 10x = 6 + x ∴ 9x = 6 x = $$\\frac{2}{3}$$ (ii) $$0 . 4 \\overline{7}$$ = 0.4777… = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ Let x = 0.777… ∴ 10x = 7.77… ∴ 10x = 7.777… ∴ 10x = 7 + 0.777… ∴ 10x = 7 + x ∴ x = $$\\frac{7}{9}$$ $$0 . 4 \\overline{7}$$ = $$\\frac{4}{10}$$ + $$\\frac{0.777}{10}$$ = $$\\frac{4}{10}$$ + $$\\frac{7}{90}$$ = $$\\frac{36}{90}$$ + $$\\frac{7}{90}$$ = $$\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ =0.001001… Let x = 0.001001… ∴ 1000x = 1.001…. ∴ 1000x = 1.001001… ∴ 1000x = 1 + 0.001001 … ∴ l000x= 1 + X ∴ 999x = 1 x = $$\\frac{1}{999}$$ Question 4. Express 0.99999…in the form $$\\frac{p}{q}$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Let x = 0.9999… 10x = 9.999… ∴ 10x = 9.9999… ∴ 10x = 9 + 0.9999… ∴ 10x = 9 + x ∴ 9x = 9 x", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =", "# What is 0.8959595... as a fraction ? Jan 9, 2017 $0.8 \\overline{95} = \\frac{887}{990}$ #### Explanation: Given: $0.8 \\overline{95}$ First multiply by $10 \\left(100 - 1\\right) = 1000 - 10$ to get an integer. The first multiplier $10$ is to shift the repeating pattern to just after the decimal point. Then $\\left(100 - 1\\right)$ shifts the number $2$ places to the left and subtracts the original value to cancel out the repeating tail: $\\left(1000 - 10\\right) \\cdot 0.8 \\overline{95} = 895. \\overline{95} - 8. \\overline{95} = 887$ Then divide both ends by $\\left(1000 - 10\\right)$ to find: $0.8 \\overline{95} = \\frac{887}{1000 - 10} = \\frac{887}{990}$ $887$ and $990$ have no common factor larger than $1$, so this fraction is in simplest form.", "one tenth, the three hundredths would turn into a one hundredth, the three thousandths would turn into a one thousandth, etc. Dividing the equivalent fraction, $$\\frac{1}{3},$$ by $$3,$$ gives us the fraction $$\\frac{1}{9}.$$ This will give us the decimal that we want, $$0.\\overline{111}\\ldots,$$ which is then equal to $$\\frac{1}{9}!$$ The same cool pattern applies to the repeating decimal representations of similar fractions whose denominator is $$9,$$ like the ones below: $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ $$0.\\overline{222}\\ldots = \\frac{2}{9}$$ $$0.\\overline{444}\\ldots = \\frac{4}{9}$$ $$0.\\overline{555}\\ldots = \\frac{5}{9}$$ $$0.\\overline{777}\\ldots = \\frac{7}{9}$$ $$0.\\overline{888}\\ldots = \\frac{8}{9}$$ • could you please further elucidate for the students like myself i understood this part that we divide .333 by 3 to arrive at 0.111 which turns 3 tenths into 1 tenths and the rest accordingly but how did we derive 1 / 3 ? did we derive from the fact that 3 divides .333 by .111 so it becomes 1/3 i understood this part - for dividing the 1/3 by 3 is actually multiplying with its reciprocal 1/3 which gives us 1/9 and if i am correct in my above computation then here's my other queries - For the other numbers - 0.222 if we divide 2 into 0.222 - this would give us .111 same as before, but when we are going to divide 2 by 1/2 which", "Thus, the number of time 9 to be repeated in the denominator becomes three. $$\\begin{array}{l}0.\\overline{125} = \\frac{125}{999}\\end{array}$$ ### Case 2: Fraction of the type $$\\begin{array}{l}0.ab..\\overline{cd}\\end{array}$$ The formula to convert this type of repeating decimal to the fraction is given by: $$\\begin{array}{l}0.ab..\\overline{cd} =\\frac{(ab….cd…..) – ab……}{Number \\; of \\; time \\; 9’s \\; the \\; repeating \\; term \\; followed \\; by \\; the \\; number \\; of \\; times \\; 0’s \\; for \\; the \\; non-repeated \\; terms }\\end{array}$$ Example 3: Convert $$\\begin{array}{l}0.12\\overline{34}\\end{array}$$ to the fractional form. Solution: In the given decimal number, 12 is a non-repeated decimal value, and 34 is in the repeating form. Thus denominator becomes 9900. $$\\begin{array}{l}0.12\\overline{34} = \\frac{1234 – 12 }{9900} = \\frac{1222}{9900}\\end{array}$$ Example 4: Convert $$\\begin{array}{l}0.00\\overline{69}\\end{array}$$ in the fraction form. Solution: In the given decimal number, the number 00 is a non-repeated decimal value, and 69 is in the repeating form. Thus, the denominator becomes 9900. $$\\begin{array}{l}0.00\\overline{69} = \\frac{0069}{9900} = \\frac{69}{9900}\\end{array}$$ It is all about the conversion of repeating decimal to the fraction form. To learn more interesting topics in Maths, download BYJU’S – The Learning App and learn with ease. Test your Knowledge on Repeating Decimal to Fraction", "# RD Sharma Solutions Class 9 Number System Exercise 1.2 ## RD Sharma Solutions Class 9 Chapter 1 Ex 1.2 Q1. Express the following rational numbers as decimals: (i) $\\frac{42}{100}$ (ii) $\\frac{327}{500}$ (iii) $\\frac{15}{4}$ Solution: (i) By long division method 100) $\\overline{42}$ ( 0.42 400 $\\overline{200}$ 200 $\\overline{0}$ Therefore, $\\frac{42}{100}$ = 0.42 (ii) By long division method 500) $\\overline{327.000}$ ( 0.654 3000 $\\overline{2700}$ 2500 $\\overline{2000}$ 2000 $\\overline{0}$ Therefore, $\\frac{327}{500}$ = 0.654 (iii) By long division method 4) $\\overline{15.00}$ ( 3.75 12 $\\overline{30}$ 28 $\\overline{20}$ 20 $\\overline{0}$ Therefore, $\\frac{15}{4}$ = 3.75 Q2. Express the following rational numbers as decimals: (i) $\\frac{2}{3}$ (ii) –$\\frac{4}{9}$ (iii) –$\\frac{2}{15}$ (iv) –$\\frac{22}{13}$ (v) $\\frac{437}{999}$ Solution: (i) By long division method 3) $\\overline{2.0000}$ ( 0.66 18 $\\overline{20}$ 18 $\\overline{2}$ Therefore, $\\frac{2}{3}$ = 0.66 (ii) By long division method 9) $\\overline{4.000}$ ( 0.444 3600 $\\overline{4000}$ 3600 $\\overline{4000}$ 3600 $\\overline{400}$ Therefore, – $\\frac{4}{9}$ = – 0.444 (iii) By long division method 15) $\\overline{2.00}$ ( 1.333 15 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{5}$ Therefore, $\\frac{2}{15}$ = -1.333 (iv) By long division method 13) $\\overline{22.000}$ ( 1.69230769 13 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{30}$ 26 $\\overline{40}$ 39 $\\overline{100}$ 91 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{3}$ Therefore, – $\\frac{22}{13}$ = – 1.69230769 (v) By long division method 999) $\\overline{437.0000}$ ( 0.43743 3996 $\\overline{3740}$ 2997 $\\overline{7430}$ 6993 $\\overline{4370}$ 3996 $\\overline{3740}$ 2997 $\\overline{743}$", "# Express 0.99999…in the form $\\frac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Solution: Let $x=0.9999 \\ldots$ $10 x=9.9999 \\ldots$ $10 x=9+x$ $9 x=9$ $x=1$", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "≠ 0. (p, q ϵ Z, q ≠ 0). (i) $$0 . \\overline{6}$$ (ii) $$0 . \\overline{47}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.6666………. Let x = 0.6666…………. ∴ x = 0.6666……….. Multiplying both sides by 10, 10x = 6.666….. 10x = 6 + 0.666 10x = 6 + x 10x – x = 6 9x = 6 $$x=\\frac{6}{9}=\\text { form of } \\frac{\\mathrm{p}}{\\mathrm{q}}$$ ∴ $$0 . \\overline{6}=\\frac{2}{3}$$ (ii) $$0 . \\overline{47}$$ = 0.4777….. Let x = $$0 . \\overline{47}$$ then x = 0.4777…….. Here digit 4 is not repeating but 7 is repeated. Let x = 0.4777 Multiplying both sides by 10. 10x = 4.3 + x 10x -x = 4.3 9x = 4.3 Multiplying both sides by 10. 90x = 43 ∴ $$0.4 \\overline{7}=\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ x = 0.001001………. Multiplying both sides by 1000 1000x = 1.001001 1000x = 1 + x 1000x – x = 1 999x = 1 $$x=\\frac{1}{999}$$ ∴ $$0 . \\overline{001}=\\frac{1}{999}$$ Question 4. Express 0.99999…….. in the form $$\\frac{p}{q}$$ , Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense. Express 0.99999 in the form of $$\\frac{p}{q}$$, Let x = 0.99999 … Multiplying by 10. 10x = 9.9999…… 10x = 9 + 0.9999……. 10x = 9 + x 10x", "to fractions. $0\\text{.}\\text{11}\\frac{2}{3}$ The $\\frac{2}{3}$ appears to occur in the thousands position, but it is referring to $\\frac{2}{3}$ of a hundredth. So, we read $0\\text{.}\\text{11}\\frac{2}{3}$ as \"eleven and two-thirds hundredths.\" $\\begin{array}{ccc}0.11\\frac{2}{3}=\\frac{\\text{11}\\frac{2}{3}}{\\text{100}}& =& \\frac{\\frac{\\text{11}\\cdot 3+2}{3}}{\\text{100}}\\\\ & =& \\frac{\\frac{\\text{35}}{3}}{\\frac{\\text{100}}{1}}\\hfill \\\\ & =& \\frac{\\text{35}}{3}÷\\frac{\\text{100}}{1}\\hfill \\\\ & =& \\frac{\\stackrel{7}{\\overline{)35}}}{3}\\cdot \\frac{1}{\\underset{\\text{20}}{\\overline{)100}}}\\hfill \\\\ & =& \\frac{7}{\\text{60}}\\hfill \\end{array}$ $4\\text{.}\\text{006}\\frac{1}{4}$ Note that $4\\text{.}\\text{006}\\frac{1}{4}=4+\\text{.}\\text{006}\\frac{1}{4}$ $\\begin{array}{ccc}\\hfill 4+.006\\frac{1}{4}& =& 4+\\frac{6\\frac{1}{4}}{\\text{1000}}\\hfill \\\\ & =& 4+\\frac{\\frac{\\text{25}}{4}}{\\frac{\\text{1000}}{1}}\\hfill \\\\ & =& 4+\\frac{\\stackrel{1}{\\overline{)25}}}{4}\\cdot \\frac{1}{\\underset{40}{\\overline{)1000}}}\\hfill \\\\ & =& 4+\\frac{1\\cdot 1}{4\\cdot \\text{40}}\\hfill \\\\ & =& 4+\\frac{1}{\\text{160}}\\hfill \\\\ & =& 4\\frac{1}{\\text{160}}\\hfill \\end{array}$ ## Practice set b Convert each complex decimal to a fraction or mixed number. Be sure to reduce. $0\\text{.}8\\frac{3}{4}$ $\\frac{7}{8}$ $0\\text{.}\\text{12}\\frac{2}{5}$ $\\frac{\\text{31}}{\\text{250}}$ $6\\text{.}\\text{005}\\frac{5}{6}$ $6\\frac{7}{1,\\text{200}}$ $\\text{18}\\text{.}1\\frac{3}{\\text{17}}$ $\\text{18}\\frac{2}{\\text{17}}$ ## Exercises For the following 20 problems, convert each decimal fraction to a proper fraction or a mixed number. Be sure to reduce. 0.7 $\\frac{7}{\\text{10}}$ 0.1 0.53 $\\frac{\\text{53}}{\\text{100}}$ 0.71 0.219 $\\frac{\\text{219}}{1,\\text{000}}$ 0.811 4.8 $4\\frac{4}{5}$ 2.6 16.12 $\\text{16}\\frac{3}{\\text{25}}$ 25.88 6.0005 $6\\frac{1}{2,\\text{000}}$ 1.355 16.125 $\\text{16}\\frac{1}{8}$ 0.375 3.04 $3\\frac{1}{\\text{25}}$ 21.1875 8.225 $8\\frac{9}{\\text{40}}$ 1.0055 9.99995 $9\\frac{\\text{19},\\text{999}}{\\text{20},\\text{000}}$ 22.110 For the following 10 problems, convert each complex decimal to a fraction. $0\\text{.}7\\frac{1}{2}$ $\\frac{3}{4}$ $0\\text{.}\\text{012}\\frac{1}{2}$ $2\\text{.}\\text{16}\\frac{1}{4}$ $2\\frac{\\text{13}}{\\text{80}}$ $5\\text{.}\\text{18}\\frac{2}{3}$ $\\text{14}\\text{.}\\text{112}\\frac{1}{3}$ $\\text{14}\\frac{\\text{337}}{3,\\text{000}}$ $\\text{80}\\text{.}\\text{0011}\\frac{3}{7}$ $1\\text{.}\\text{40}\\frac{5}{\\text{16}}$ $1\\frac{\\text{129}}{\\text{320}}$ $0\\text{.}8\\frac{5}{3}$ $1\\text{.}9\\frac{7}{5}$ $2\\frac{1}{\\text{25}}$ $1\\text{.}7\\frac{\\text{37}}{9}$ ## Exercises for review ( [link] ) Find the greatest common factor of 70, 182, and 154. 14 ( [link] ) Find the greatest common multiple of 14, 26,", "to fractions. $0\\text{.}\\text{11}\\frac{2}{3}$ The $\\frac{2}{3}$ appears to occur in the thousands position, but it is referring to $\\frac{2}{3}$ of a hundredth. So, we read $0\\text{.}\\text{11}\\frac{2}{3}$ as \"eleven and two-thirds hundredths.\" $\\begin{array}{ccc}0.11\\frac{2}{3}=\\frac{\\text{11}\\frac{2}{3}}{\\text{100}}& =& \\frac{\\frac{\\text{11}\\cdot 3+2}{3}}{\\text{100}}\\\\ & =& \\frac{\\frac{\\text{35}}{3}}{\\frac{\\text{100}}{1}}\\hfill \\\\ & =& \\frac{\\text{35}}{3}÷\\frac{\\text{100}}{1}\\hfill \\\\ & =& \\frac{\\stackrel{7}{\\overline{)35}}}{3}\\cdot \\frac{1}{\\underset{\\text{20}}{\\overline{)100}}}\\hfill \\\\ & =& \\frac{7}{\\text{60}}\\hfill \\end{array}$ $4\\text{.}\\text{006}\\frac{1}{4}$ Note that $4\\text{.}\\text{006}\\frac{1}{4}=4+\\text{.}\\text{006}\\frac{1}{4}$ $\\begin{array}{ccc}\\hfill 4+.006\\frac{1}{4}& =& 4+\\frac{6\\frac{1}{4}}{\\text{1000}}\\hfill \\\\ & =& 4+\\frac{\\frac{\\text{25}}{4}}{\\frac{\\text{1000}}{1}}\\hfill \\\\ & =& 4+\\frac{\\stackrel{1}{\\overline{)25}}}{4}\\cdot \\frac{1}{\\underset{40}{\\overline{)1000}}}\\hfill \\\\ & =& 4+\\frac{1\\cdot 1}{4\\cdot \\text{40}}\\hfill \\\\ & =& 4+\\frac{1}{\\text{160}}\\hfill \\\\ & =& 4\\frac{1}{\\text{160}}\\hfill \\end{array}$ ## Practice set b Convert each complex decimal to a fraction or mixed number. Be sure to reduce. $0\\text{.}8\\frac{3}{4}$ $\\frac{7}{8}$ $0\\text{.}\\text{12}\\frac{2}{5}$ $\\frac{\\text{31}}{\\text{250}}$ $6\\text{.}\\text{005}\\frac{5}{6}$ $6\\frac{7}{1,\\text{200}}$ $\\text{18}\\text{.}1\\frac{3}{\\text{17}}$ $\\text{18}\\frac{2}{\\text{17}}$ ## Exercises For the following 20 problems, convert each decimal fraction to a proper fraction or a mixed number. Be sure to reduce. 0.7 $\\frac{7}{\\text{10}}$ 0.1 0.53 $\\frac{\\text{53}}{\\text{100}}$ 0.71 0.219 $\\frac{\\text{219}}{1,\\text{000}}$ 0.811 4.8 $4\\frac{4}{5}$ 2.6 16.12 $\\text{16}\\frac{3}{\\text{25}}$ 25.88 6.0005 $6\\frac{1}{2,\\text{000}}$ 1.355 16.125 $\\text{16}\\frac{1}{8}$ 0.375 3.04 $3\\frac{1}{\\text{25}}$ 21.1875 8.225 $8\\frac{9}{\\text{40}}$ 1.0055 9.99995 $9\\frac{\\text{19},\\text{999}}{\\text{20},\\text{000}}$ 22.110 For the following 10 problems, convert each complex decimal to a fraction. $0\\text{.}7\\frac{1}{2}$ $\\frac{3}{4}$ $0\\text{.}\\text{012}\\frac{1}{2}$ $2\\text{.}\\text{16}\\frac{1}{4}$ $2\\frac{\\text{13}}{\\text{80}}$ $5\\text{.}\\text{18}\\frac{2}{3}$ $\\text{14}\\text{.}\\text{112}\\frac{1}{3}$ $\\text{14}\\frac{\\text{337}}{3,\\text{000}}$ $\\text{80}\\text{.}\\text{0011}\\frac{3}{7}$ $1\\text{.}\\text{40}\\frac{5}{\\text{16}}$ $1\\frac{\\text{129}}{\\text{320}}$ $0\\text{.}8\\frac{5}{3}$ $1\\text{.}9\\frac{7}{5}$ $2\\frac{1}{\\text{25}}$ $1\\text{.}7\\frac{\\text{37}}{9}$ ## Exercises for review ( [link] ) Find the greatest common factor of 70, 182, and 154. 14 ( [link] ) Find the greatest common multiple of 14, 26," ]

[ "= $\\dfrac{3}{{17}}$", "cos^2 \\dfrac{17π}{6} = \\dfrac{(3 – 4\\sqrt{3})}{2}$$", "and $-\\dfrac{5}{3}$.", "\\dfrac{13}{12}$ , $\\sin{x} = \\dfrac{12}{13}$ , $\\cos{x} = \\dfrac{5}{13}$", "$\\dfrac{2}{3}$.", "$\\dfrac{11}{6} = 1 + \\dfrac{5}{6}$", "to $\\dfrac{4}{5}$", "= \\dfrac{4}{1}$$", "is $\\dfrac {1}{ 5 }$.", "\\dfrac{4A}{\\sqrt 3 } }$", "of $\\dfrac{1}{3}$ comes into picture.", "{\\dfrac {1 + 11 \\paren {1 - p} e^t + 11 \\paren {1 - p}^2 e^{2t} + \\paren {1 - p}^3 e^{3t} } {\\paren {1 - \\paren {1 - p} e^t}^5 } }$", "{17} {12}, \\dfrac {577} {408}, \\dotsc$", "= \\dfrac{\\pi}{4}$$", "\\dfrac{7 \\pi}{12} \\right) \\right) = \\dfrac{7 \\pi}{12}$$ You can try the same method for other as well.", "=130\\dfrac{10}{11}$or ${{130.91}^{\\circ }}$.", "$$\\dfrac{81}{2}$$.", "10.) $4 \\dfrac {1}{4} + 2 \\dfrac {3}{4}$ $= 6 \\dfrac {3}{4}$", "\\dfrac{570}{3^8} = \\dfrac{190}{3^{7}}$. Our answer is the numerator, $\\boxed{190}$.", "\\dfrac{a}{b + c} + \\dfrac{b}{a + c}$?" ]

[ "\\dfrac a1$: We multiply $a = \\dfrac a1$ by $\\dfrac cc = 1$, to get $\\dfrac a1 \\times \\dfrac cc = \\;\\dfrac{a\\times c}{c}\\;$ and then add that to the fraction $\\dfrac bc$. For example $$3 \\frac 58\\; = \\;3 + \\frac 58\\; = \\;\\frac{3 \\times 8}{8} + \\frac 58\\; = \\;\\frac{(3\\times 8)+ 5}{8}\\; = \\;\\frac{24 + 5}{8} \\;= \\;\\frac{29}{8}$$ - Good call, you did the general case followed by an example +1 – Amzoti May 24 '13 at 12:03 Here is an example. $$2 \\frac{3}{11}=\\frac{2 \\cdot 11}{11}+\\frac3{11}=\\frac{22}{11}+\\frac3{11}=\\frac{25}{11}$$ Or conversely, $$\\frac{11}3=\\frac{11-3 \\cdot 3}{3}+\\frac{3 \\cdot 3}{3}=\\frac{11-9}{3}+\\frac{9}{3}=\\frac{2}{3}+3=3 \\frac{2}3$$. In general, $$x+\\frac{y}{z}=\\frac{x \\cdot z}{z}+\\frac{y}{z}=\\frac{x \\cdot z +y}z$$ and vice versa. In most cases you'll encounter $y<z$. -", "# Find all real numbers $x$ for which $\\frac{8^x+27^x}{12^x+18^x}=\\frac76$ Find all real numbers $x$ for which $$\\frac{8^x+27^x}{12^x+18^x}=\\frac76$$ I have tried to fiddle with it as follows: $$2^{3x} \\cdot 6 +3^{3x} \\cdot 6=12^x \\cdot 7+18^x \\cdot 7$$ $$3 \\cdot 2^{3x+1}+ 2 \\cdot 3^{3x+1}=7 \\cdot 6^x(2^x+3^x)$$ Dividing both sides by $6$ gives us $$2^{3x}+3^{3x}=7 \\cdot 6^{x-1}(2^x+3^x)$$ Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated. - Ok, I'll make sure I do that in future – John Marty May 8 '13 at 1:58 Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$ $12^x=(2^2\\cdot3)^x=(2^x)^2\\cdot3^x=a^2b$ and similarly, $18^x=ab^2$ So, the problem reduces to $$\\frac{a^3+b^3}{ab(a+b)}=\\frac76$$ $\\displaystyle\\implies 6(a^2-ab+b^2)=7ab$ as $a+b\\ne0$ $\\displaystyle\\implies 6\\left(\\frac ab\\right)^2-13\\cdot\\frac ab+6=0$ $\\displaystyle\\implies \\frac ab=\\frac32$ or $\\dfrac23$ So, $\\displaystyle\\left(\\frac23\\right)^x=\\frac32$ or $\\dfrac23$ If $\\displaystyle\\left(\\frac23\\right)^x=\\frac32\\implies\\left(\\frac23\\right)^x=\\left(\\frac23\\right)^{-1}\\iff\\left(\\frac23\\right)^{x+1}=1$ Similarly, if $\\displaystyle\\left(\\frac23\\right)^x=\\frac23, \\left(\\frac23\\right)^{x-1}=1$ Now if $\\displaystyle u^m=1,$ either $\\displaystyle m=0,u\\ne0;$ or $\\displaystyle u=1$ or $\\displaystyle u=-1,m$ is even - Thanks so much that's really neat! – John Marty May 7 '13 at 4:34 @JohnMarty, my pleasure. Replacement of numbers with indices by algebraic symbol sometimes helps clarity – lab bhattacharjee May 7 '13 at 4:41 (+1) nice answer. – Mhenni Benghorbal May 7 '13 at 5:11 @lab-bhattacharjee, I'd say your comment (\"Replacement of numbers with indices by algebraic symbol sometimes helps", "Feb 15, 2017 at 22:59 $$x + \\frac{1}{3} x = 1 \\cdot x + \\frac{1}{3} x = \\left( 1 + \\frac{1}{3} \\right) x = \\frac{4}{3} x$$ Your full equation is slighty different, but the same rules (distributivity) apply: $$15 = \\frac{2}{3} b + b = \\left( \\frac{2}{3} + 1 \\right) b = \\frac{5}{3} b \\iff \\\\ b = \\frac{3}{5} \\cdot 15 = 9$$ • Thanks for the help - I can see its process now! Jun 29, 2016 at 13:48 You know that you can: • Add fractions as long as they have the same denominator • Multiply both the nominator and the denominator with the same value and the value of the fraction remains the same Having that in mind we are going to add $x$ to $\\frac{x}{3}$.We know that $x=\\frac{3x}{3}$ (multiply both nominator and denominator by 3) Now we can add them together $\\frac{3x+x}{3}=\\frac{4x}{3}$ You need to grasp what is known as the distributive law: $$a(b+c)=ab+ac$$ In your case it is backwards, in other words $$xb+yb=(x+y)b$$ Can you see how that might help. Put $x=\\frac{2}{3},y=1$ and you have $$\\frac{2}{3}b+1\\cdot b=(\\frac{2}{3}+1)b$$ I guess that also uses $1\\cdot b=b$, which I am sure you are happy with. Why is that law true? Well first think of $b$ as \"apples\" we are saying that $x$ apples + $y$ apples", "try to explain what $1 \\cdot 0$ or $0 \\cdot 1$ is... the evaluation of this will result the same every time as the answer would be $0$ which fits both statements. So let's apply this to fractions $0 \\cdot 1 = \\frac{0}{1} \\cdot \\frac{1}{1} = \\frac{0 \\cdot 1}{1 \\cdot 1}$ ... – Francis Cugler Mar 20 '17 at 23:57 • (...continued) $\\frac{0 \\cdot 1}{1 \\cdot 1} = \\frac{0}{1} = 0$ so far so good. So what happens when we reverse this: Let's try it out. But first we need to understand that $\\frac{1}{0} \\neq 0$ we need to consider this the reciprocal of $0$ or $\\frac{0}{1}$ and we know that the only value that has an equivalent reciprocal is $1$ or $\\frac{1}{1}$. Then it would suggest that $\\frac{1}{0} \\cdot \\frac{1}{1} = \\frac{1}{0}$ and that $\\frac{1}{0} \\cdot \\frac{0}{1} = \\frac{0}{0}$ and since $\\frac{0}{0} = 1$ from my above statement it would state that anything divided by $0$ would result in $1$. – Francis Cugler Mar 21 '17 at 0:10 Definition of Division For every real number a and every nonzero real number b, the quotient a$\\div$b, or $\\dfrac{a}{b}$, is defined by: $$a\\div b=a \\cdot \\frac{1}{b}.$$ Dividing by zero would mean multiplying by the reciprocal of 0. But 0 has no reciprocal (because 0 times any number is 0,", "$$AE \\cdot BC = AB \\cdot CE + AC \\cdot BE \\implies AE = AB + AC$$ $\\implies AI +EI = AB + AC, \\hspace{10mm} AI = AB+ AC – BC$ is integer. $$AI = \\frac {2AB \\cdot AC \\cdot cos \\angle CAI}{AB+AC + BC} = \\frac {AB \\cdot AC}{AB+AC + BC} =$$ $$= AB + AC – BC \\implies AC^2 + AB^2 + AB \\cdot AC = BC^2.$$ A quick $(\\mod9)$ check gives that $3\\mid AC$ and $3\\mid AB$. $$AI \\le A_0I_0 = EA_0 – EI_0 = \\frac{2 BC}{\\sqrt{3}} – BC = \\frac {2 - \\sqrt{3}}{\\sqrt{3}} BC = 33.88.$$ Denote $a= \\frac {BC}{3}= 73, b = \\frac {AC}{3}, c = \\frac {AB}{3}, l = \\frac {AI}{3} \\le 11.$ We have equations in integers $\\frac{bc}{a+b+c} = b + c – a = l \\le 11.$ The solution $(b > c)$ is $$b = \\frac{a + l +\\sqrt{a^2 – 6al – 3l^2}}{2}, c = \\frac{a + l -\\sqrt{a^2 – 6al – 3l^2}}{2}.$$ Suppose, $a^2 – 6al – 3l^2 = (a – 3l – t)^2 \\implies \\frac {12l^2}{t} + t + 6l= 2a = 146.$ Now we check all possible $t = {2,3,4,6,12, ml}.$ Case $t = 2 \\implies 6l^2 + 6l = 146 – 2 \\implies l^2 + l = 24 \\implies \\O$ Case $t = 3 \\implies 4l^2", "# Question #bf504 Oct 12, 2016 $4 \\frac{7}{11} \\div \\frac{1}{11} = 51$ #### Explanation: $4 \\frac{7}{11} \\div \\frac{1}{11} = \\left(4 + \\frac{7}{11}\\right) \\div \\frac{1}{11}$ $= \\left(\\frac{44}{11} + \\frac{7}{11}\\right) \\div \\frac{1}{11}$ $= \\left(\\frac{51}{11}\\right) \\div \\frac{1}{11}$ $= \\frac{\\frac{51}{11}}{\\frac{1}{11}}$ $= \\frac{\\frac{51}{11}}{\\frac{1}{11}} \\cdot \\frac{11}{11}$ $= \\frac{\\frac{51}{11} \\cdot 11}{\\frac{1}{11} \\cdot 11}$ $= \\frac{51}{1}$ $= 51$ A common trick when dividing by a fraction is to take its reciprocal and then multiply, i.e. $x \\div \\frac{a}{b} = x \\cdot \\frac{b}{a}$ The process used above shows why this works: $x \\div \\frac{a}{b} = \\frac{x}{\\frac{a}{b}} = \\frac{x \\cdot \\frac{b}{a}}{\\frac{a}{b} \\cdot \\frac{b}{a}} = \\frac{x \\cdot \\frac{b}{a}}{1} = x \\cdot \\frac{b}{a}$", "a) It's given that $\\frac{dV}{dt}=-1$. We know that $V = \\frac{2\\pi rh}{3}$ and $\\frac{dV}{dh} = \\frac{2\\pi r}{3}$. We also know that, by similar triangles, $h=3$ implies that $r=1$. $\\frac{dV}{dt}=\\frac{dV}{dh}\\cdot\\frac{dh}{dt} \\implies -1 = \\frac{2\\pi\\cdot 1}{3}\\cdot\\frac{dh}{dt} \\implies \\frac{dh}{dt}=-\\frac{3}{2\\pi}$", "$$4/5$$ 5: $$−1 \\frac{3}{7}$$ 7: $$5/6$$ 9: $$−7/4$$ 11: $$1/6$$ 13: $$32/105$$ 15: $$7 \\frac{5}{6}$$ 17: $$1$$ 19: $$29/30$$ 21: $$2 \\frac{2}{3}$$ 23: $$19/24$$ Exercise $$\\PageIndex{8}$$ Perform the operations. Reduce answers to lowest terms. 1. $$\\frac{3}{14} \\cdot \\frac{7}{3} \\div \\frac{1}{8}$$ 2. $$\\frac{1}{2} \\cdot (-\\frac{4}{5}) \\div \\frac{14}{15}$$ 3. $$\\frac{1}{2} \\div \\frac{3}{4} \\cdot \\frac{1}{5}$$ 4. $$-\\frac{5}{9} \\div \\frac{5}{3} \\cdot \\frac{5}{2}$$ 5. $$\\frac{5}{12} - \\frac{9}{21}+\\frac{3}{9}$$ 6. $$-\\frac{3}{10} - \\frac{5}{12} + \\frac{1}{20}$$ 7. $$\\frac{4}{5} \\div 4 \\cdot \\frac{1}{2}$$ 8. $$\\frac{5}{3} \\div 15 \\cdot \\frac{2}{3}$$ 9. What is the product of $$\\frac{3}{16}$$ and $$\\frac{4}{9}$$? 10. What is the product of $$−\\frac{24}{5}$$ and $$\\frac{25}{8}$$? 11. What is the quotient of $$\\frac{5}{9}$$ and $$\\frac{25}{3}$$? 12. What is the quotient of $$−\\frac{16}{5}$$ and $$32$$? 13. Subtract $$\\frac{1}{6}$$ from the sum of $$\\frac{9}{2}$$ and $$\\frac{2}{3}$$. 14. Subtract $$\\frac{1}{4}$$ from the sum of $$\\frac{3}{4}$$ and $$\\frac{6}{5}$$. 15. What is the total width when $$3$$ boards, each with a width of $$2 \\frac{5}{8}$$ inches, are glued together? 16. The precipitation in inches for a particular 3-day weekend was published as $$\\frac{3}{10}$$ inches on Friday, $$1\\frac{1}{2}$$ inches on Saturday, and $$\\frac{3}{4}$$ inches on Sunday. Calculate the total precipitation over this period. 17. A board that is $$5\\frac{1}{4}$$ feet long is to be cut into $$7$$ pieces of equal length. What is length of each piece? 18. How many $$\\frac{3}{4}$$ inch", "operation. $a \\div b$ is really just a shorthand for $a \\cdot b^{-1}$, so it only makes sense to use this shorthand if $b^{-1}$ exists (where $b^{-1}$ is defined to be some element making $b \\cdot b^{-1} = 1$). So $\\div 0$ just doesn't make sense in any equation, ever. As others have mentioned, you can have formulas where you divide by an indeterminate $x$, then take a limit as $x \\to 0$, but that is different as you are never explicitly dividing by $0$. Consider the following: $\\frac{0}{0} \\cdot 0 = \\frac{3 \\cdot 0}{0^2}\\cdot 0 = \\frac{0}{3 \\cdot 0^2} \\cdot 0$ (since $0 = 3 \\cdot 0 = 0^2 = 3 \\cdot 0^2$). Allowing these corrupted operations you are trying to use, we would have $0 = 3 = 1/3$. # Why is $0 \\div 0$ undefined? The definition of $\\div$ $$a \\times b = c \\Longrightarrow b = c \\div a$$ We would get $$0 \\times b = 0 \\stackrel{?\\ }{\\Longrightarrow} b = 0 \\div 0$$ But for $b' \\ne b$ we would also get $$0 \\times b' = 0 \\stackrel{?\\ }{\\Longrightarrow} b' = 0 \\div 0$$ If this would be true, we get a contradiction as $$b \\ne b' \\wedge b = b'$$ So $0 \\div 0$ is simply not defined. # Why is $0", "( 3 x - 4 ) ( x - 7 ) } \\div \\frac { ( 2 x - 5 ) } { ( 3 x - 4 ) }$ See part (b). 1. $\\frac { 3 x - 1 } { x + 4 } + \\frac { x - 5 } { x + 4 }$ See part (b). 1. $\\frac { x - 3 } { x + 4 } \\cdot \\frac { 3 x - 10 } { x + 11 } \\cdot \\frac { x + 4 } { 3 x - 10 }$ See part (b). $\\frac { x - 3 } { x + 11 }$", "\\color{red}{\\times \\frac{4}{13}}=\\frac{64}{13}$$ Combining it all~ $$x=\\frac{64}{13}$$ YAY! DONE. Cheers! -Shahar - We start with the initial equation: $$(3+\\frac 14)x-2=14$$ We simplify: $$\\frac {13}4x-2 = 14$$ We add 2 to both sides: $$\\frac {13}4x = 16$$ Then we multiply both sides by 4: $$13x = 64$$ And then divide both sides by 13: $$x = \\frac{64}{13}$$ - I think OP uses the notation $3\\tfrac{1}{4}$ to mean $3+\\tfrac{1}{4}$, or $\\tfrac{13}{4}$. – Servaes Mar 24 '14 at 21:47 I did, @Servaes is correct. – Jacedc Mar 24 '14 at 21:50 EDIT: Corrected the answer to reflect this. – Disousa Mar 24 '14 at 21:58 It should be $(3+\\frac{1}{4})\\times x$, not $3+\\frac{1}{4}\\times x$. – Eff Mar 24 '14 at 22:01 You are correct. Editing again xD – Disousa Mar 24 '14 at 22:02 $$3 \\dfrac{1}{4} × x - 2 = 14$$ $$\\frac{13}{4} × x - 2 = 14$$ $$4\\cdot\\frac{13}{4} × x - 4\\cdot2 =4\\cdot 14$$ $$13 x - 8 =56$$ $$13x=56+8$$ $$13x=64$$ $$x=\\frac{64}{13}$$ because $$3 \\dfrac{1}{4}=3+\\frac{1}{4}=\\frac{13}{4}$$ - Correction, $14*4\\neq64$ :) – Disousa Mar 24 '14 at 21:57 You did $4⋅\\dfrac{13}{4}x - 4⋅2 = 4⋅14$ thus multiplying both sides by 4 (which is correct) but you did it twice on the left side? ($4⋅2$) – Jacedc Mar 24 '14 at 22:00 it is my mistake – Adi Dani Mar 24 '14 at 22:00", "3 \\cdot 2$ • Associative Property: When three or more numbers are multiplied, the sum is the same regardless of how they are grouped. Example: $2 \\cdot (3 \\cdot 4) = (2 \\cdot 3) \\cdot 4$ • Multiplicative Identity Property: The product of any number and one is the original number. Example: $2 \\cdot 1 = 2$ • Distributive Property: The product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For expressions $a, b,$ and $c$ , $a(b+c)=ab+ac$ . Example: $4(2 + 3) = 4(2) + 4(3)$ The multiplicative inverse of a number is the number which produces 1 when multiplied by the original number. The multiplicative inverse of $x$ is the reciprocal $\\frac{1}{x}$ . To find the multiplicative inverse of a fraction, simply invert the fraction : $\\frac{a}{b}$ inverts to $\\frac{b}{a}$ . To divide fractions, invert the divisor and multiply: $\\frac{a}{b} \\div \\frac{c}{d} = \\frac{a}{b} \\times \\frac{d}{c}$ . An object moving at a certain speed will cover a fixed distance in a set time . The quantities speed, distance and time are related through the equation $\\text{Speed} = \\frac{\\text{Distance}}{\\text{Time}}$ . ### Guided Practice Andrew is driving down the freeway. He passes mile marker 27 at exactly mid-day. At 12:35 he passes mile", "For every non-zero value $x$ ($x \\ne 0$), there is a multiplicative inverse value written $\\div x$ so that $x \\cdot \\div x=1\\text{.}$ When $x \\ne 0\\text{,}$ we have $\\div x = \\frac{1}{x}\\text{,}$ which is why the inverse of $x$ is also called the reciprocal. • Commutative properties: $x+y = y+x$ and $x \\cdot y = y \\cdot x\\text{.}$ • Associative properties: $(x+y)+z = x+(y+z)$ and $(x \\cdot y) \\cdot z = (x \\cdot y) \\cdot z\\text{.}$ • Distributive property of multiplication over addition: $x \\cdot (y+z) = x \\cdot y + x \\cdot z\\text{.}$ Example1.3.4 Our pick-a-number example with the dependent variable \\begin{equation*} y=\\frac{2(x+5)-4}{2}-x \\end{equation*} could be used as a mind-reader trick. If a volunteer does the math correctly, you can always guess the final number $y\\text{.}$ Use algebra properties to determine what you should predict. Solution We start with the distributive property to rewrite $2(x+5) = 2x+10\\text{.}$ Subtracting 4 is equivalent to adding $-4\\text{.}$ We can then use the associative property to rewrite $(2x+10)+-4 = 2x+(10+-4) = 2x+6\\text{.}$ Dividing by two is equivalent to multiplying by $\\frac{1}{2}\\text{.}$ This allows us to use the distributive property again, \\begin{align*} \\frac{2(x+5)-4}{2} = \\frac{2x+6}{2} &= \\frac{1}{2}(2x+6) = \\frac{1}{2}(2x) + \\frac{1}{2}(6)\\\\ &= (\\frac{1}{2} \\cdot 2) x + \\frac{6}{2} = x + 3 \\end{align*} The second line used the associative property and", "the ordinal$5$is the order type$a<b<c<d<e$, the ordinal$3$is the order type$x<y<z$, and$5 \\times 3$is $$a_x < b_x < c_x < d_x < e_x < a_y < b_y < c_y < d_y < e_y < a_z < b_z &... 11 Definition of Division For every real number a and every nonzero real number b, the quotient a\\divb, or \\dfrac{a}{b}, is defined by:$$a\\div b=a \\cdot \\frac{1}{b}.$$Dividing by zero would mean multiplying by the reciprocal of 0. But 0 has no reciprocal (because 0 times any number is 0, not 1.) Therefore, division by 0 has no ... 11 Dividing 1 disk into \\frac{1}{n}-ths (\\frac{1}{3}, \\ldots), leads to \\frac{1}{1/n} = n pieces (3,\\ldots): As \\frac{1}{n} approaches 0, the number of pieces n grows without bound. The result upon division by 0, 1/0, should be this limit. But there is no limit. 11 One way to see if the student understands the commutative property of addition is to have \"fill-in-the-blank\" questions such as$$2+3=3+\\_\\_2+3=\\_\\_+22+\\_\\_=3+2\\_\\_+3=3+2$$10 My immediate response is 'wait a few years'. I've spent a fair amount of time with 3 year olds, and most of them are busy learning how to be a person in their own right, how to have a conversation, what the difference is between real and make-believe, and (often) how to tell when they need the toilet. I've read", "to finish this post: $C^{(\\frac{5}{6}-\\frac{3}{4})}\\cdot D^{(\\frac{1}{6}-\\frac{3}{4})}$ C's exponent is $\\frac{5}{6}-\\frac{3}{4} =\\frac{5}{6} \\cdot \\frac{2}{2}-\\frac{3}{4} \\cdot \\frac{3}{3}= \\frac{10}{12}- \\frac{9}{12}= \\frac{1}{12}$ D's exponent is $\\frac{1}{6}-\\frac{3}{4} =\\frac{1}{6} \\cdot \\frac{2}{2}-\\frac{3}{4} \\cdot \\frac{3}{3}= \\frac{2}{12}- \\frac{9}{12}= -\\frac{7}{12}$ you now hve $C^{\\frac{1}{12}}D^{-\\frac{7}{12}}= (\\frac{C}{D^7})^\\frac{1}{12}$ or $\\sqrt[12]{\\frac{C}{D^7}}$ 36. aroub That's the answer that I actually wanted! @phi thanks a looooottt ^_^", "the exp mean? $exp(\\dots)$ is like $e^{\\dots}$ 11. ah ok 12. anyways, I LOVE YOU GUYS! done for now =D 13. Originally Posted by galactus Let's do it this way then. Rewrite $e^{\\frac{ln(3/5)t}{5}}=(3/5)^{\\frac{t}{5}}$ $50=100(3/5)^{\\frac{t}{5}}$ $1/2=(3/5)^{\\frac{t}{5}}$ $ln(1/2)=\\frac{t}{5}ln(3/5)$ $t=\\frac{5ln(1/2)}{ln(3/5)}=6.78$ btw, how does he get $e^{\\frac{ln(3/5)t}{5}}=(3/5)^{\\frac{t}{5}}$? I get that ln and e basically \"cancel\" each other out, but why is the 3/5 to the power of t/5? 14. Disregard my last post, i figured it out. Originally Posted by Moo Hello, ... Put it in the exponential logarithm... $\\log_x (y)=\\frac{\\ln(y)}{\\ln(x)}$ ---> $\\frac{\\ln x}{\\ln 3}-\\frac{\\ln 7}{\\ln 4}=\\frac{\\ln 5}{\\ln 7}$ --> $\\ln x-\\frac{\\ln 7 \\cdot \\ln 3}{\\ln 4}=\\frac{\\ln 5 \\cdot \\ln 3}{\\ln 7}$ $\\ln x=\\frac{\\ln 5 \\cdot \\ln 3}{\\ln 7}+\\frac{\\ln 7 \\cdot \\ln 3}{\\ln 4}$ ---> $x=exp\\left(\\frac{\\ln 5 \\cdot \\ln 3}{\\ln 7}+\\frac{\\ln 7 \\cdot \\ln 3}{\\ln 4}\\right) \\approx 11.59699220980753$ does multiplying an equation such as $\\ln(x)=....$ by e make it into $x=e*...$? 15. Originally Posted by reynardin Disregard my last post, i figured it out. does multiplying an equation such as $\\ln(x)=....$ by e make it into $x=e*...$? For the last one $a^{bc}=(a^b)^c$ So $e^{\\ln(\\frac{3}{5})\\frac{t}{5}}=(e^{\\ln(\\frac{3}{5 })})^{\\frac{t}{5}}=\\bigg(\\frac{3}{5}\\bigg)^{\\frac{ t}{5}}$ and no multiplying by it does nothign raising e to each side power does for example $\\ln(45x-220)=\\ln(5+17x)$ If you multiplied both sides by e would do nothing but raising e to eachside would give $e^{\\ln*45x-220)}=e^{\\ln(5+17x)}\\Rightarrow{45x-220=5+17}$ Page", "was not actually the question asked for homework. Something like this should have been covered by his teacher.) Just follow the pattern. For non-zero $x$, we have $x^3=x\\cdot x \\cdot x$ $x^2=x\\cdot x$ $x^1=x$ $x^0=1$ Continue dividing by $x$ each time, as above... $x^{-1}=1\\div x =\\frac{1}{x}$ $x^{-2}=\\frac{1}{x}\\div x =\\frac{1}{x^2}$ $x^{-3}=\\frac{1}{x^2}\\div x =\\frac{1}{x^3}$ and so on. - Nice work. The only thing I'd have included is the similarity to the negative ide of the number line, 3,2,1,0,-1,-2, etc. Why does bopping negative not bother OP there but it does for exponents? – JoeTaxpayer Jun 3 at 16:37 This is the way I explain it to people. I think it's the most straightforward way. – daviewales Jun 4 at 17:18 In a vague sense, each integer increment of the exponent in $a^b$ means \"multiply by a\" where $a$ is the base of the exponent. So $2^2$ means \"multiply by two, and multiply by two again\" to get \"multiply by $2*2 = 4$\". So imagine you have $2^{3-1}=2^2=4$. What does the $-1$ mean in that expression? You originally had \"multiply by 2, then multiply by 2, then multiply by 2\" or \"multiply by two three times\", but there was also the $-1$, which somehow undid one of those multiply by 2's. The arithmetic function which is the inverse of multiplication is", "4. $2xy \\cdot \\frac{2y^2} {x^3}$ 5. $\\frac{4y^2 - 1} {y^2 - 9} \\cdot \\frac{y - 3} {2y - 1}$ 6. $\\frac{6ab} {a^2} \\cdot \\frac{a^3b} {3b^2}$ 7. $\\frac{x^2} {x - 1} \\div \\frac{x} {x^2 + x -2}$ 8. $\\frac{33a^2} {-5} \\cdot \\frac{20} {11a^3}$ 9. $\\frac{a^2 + 2ab + b^2} {ab^2 - a^2b} \\div (a + b)$ 10. $\\frac{2x^2 + 2x - 24} {x^2 + 3x} \\cdot \\frac{x^2 + x - 6} {x + 4}$ 11. $\\frac{3 - x} {3x - 5} \\div \\frac{x^2 - 9} {2x^2 - 8x - 10}$ 12. $\\frac{x^2 - 25} {x + 3} \\div (x - 5)$ 13. $\\frac{2x + 1} {2x - 1} \\div \\frac{4x^2 - 1} {1 - 2x}$ 14. $\\frac{x} {x - 5} \\cdot \\frac{x^2 - 8x + 15} {x^2 - 3x}$ 15. $\\frac{3x^2 + 5x - 12} {x^2 - 9} \\div \\frac{3x - 4} {3x+ 4}$ 16. $\\frac{5x^2 + 16x + 3} {36x^2 - 25} \\cdot (6x^2 + 5x)$ 17. $\\frac{x^2 + 7x + 10} {x^2 - 9} \\cdot \\frac{x^2 - 3x} {3x^2 + 4x - 4}$ 18. $\\frac{x^2 + x - 12} {x^2 + 4x + 4} \\div \\frac{x - 3} {x + 2}$ 19. $\\frac{x^4 - 16} {x^2 - 9} \\div \\frac{x^2 + 4} {x^2 + 6x + 9}$ 20. $\\frac{x^2 + 8x + 16} {7x^2 + 9x +", "that it means $(a-b)-c$ and not $a-(b-c)$. (Nowadays, we \"agree\" to this convention only in the sense that it is drummed into us in school and we get bad grades if we disagree.) People have not, as far as I know, agreed to such a convention for division. – Andreas Blass Aug 25 '16 at 3:09 $24\\div\\frac{9}{-3}=\\frac{24}{\\frac{9}{-3}}=24\\times\\frac{-3}{9}=24\\times(-3)\\div9$ Division, unlike addition or multiplication, is not associative. So the result of: $$a \\div b \\div c$$ will depend on the order you perform the operations in. Thus in most cases $$a \\div (b \\div c) \\ne (a \\div b) \\div c$$ In your case, I think you have implied: $$24 \\div (9 \\div -3) = (24 \\div 9) \\div -3$$ which is not the case. $a \\div \\frac{b}{c} = a * \\frac{c}{b}$ So $24 \\div \\frac{9}{-3} = 24 * \\frac{-3}{9}$ Fractions are thought of as being a single object. Here is something a little bit more complicated $\\frac {x^2 - 3x + 2}{x - 1}$ We evaluate everything above the line together and everything below the line together. No parenthesis are written but they are implied. So,we see the $\\frac 9{-3}$ and reduce the fraction before moving along. $\\div$ happens left to right. Most math beyond middle school gets rid of the $\\div$ notation precisely because it is frequently ambiguous.", "# How do you simplify 2/(3sqrt63) ? Mar 18, 2018 Shown below... #### Explanation: We need to rationalise the denominator, in this case multiply both numerator and denominator by $\\sqrt{63}$ $\\implies \\frac{2}{3 \\sqrt{63}} \\cdot \\frac{\\sqrt{63}}{\\sqrt{63}}$ $\\implies \\frac{2 \\sqrt{63}}{3 \\cdot 63}$ $\\implies \\frac{2 \\cdot \\sqrt{3 \\cdot 3 \\cdot 7}}{189}$ $\\implies \\frac{6 \\sqrt{7}}{189}$ $\\implies \\frac{2 \\sqrt{7}}{63}$ Mar 18, 2018 $\\frac{2 \\sqrt{7}}{63}$ #### Explanation: $\\frac{2}{3 \\sqrt{63}}$ $\\therefore = \\frac{2}{3 \\sqrt{3 \\cdot 3 \\cdot 7}}$ $\\therefore \\sqrt{3} \\cdot \\sqrt{3} = 3$ $\\therefore = \\frac{2}{3 \\cdot 3 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}}$ $\\therefore = \\frac{2}{9 \\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}$ $\\therefore \\sqrt{7} \\times \\sqrt{7} = 7$ $\\therefore = \\frac{2 \\sqrt{7}}{9 \\times 7}$ $\\therefore = \\frac{2 \\sqrt{7}}{63}$" ]

[ "example: $\\frac 3 5 \\cdot \\frac 5 6 = \\frac {3 \\cdot 5} {5 \\cdot 6} = \\frac {5 \\cdot 3} {5 \\cdot 6} = \\frac 5 5 \\cdot \\frac 3 6 = 1 \\cdot \\frac 3 {2 \\cdot 3} = 1 \\cdot \\frac 3 3 \\cdot \\frac 1 2 = 1 \\cdot 1 \\cdot \\frac 1 2 = \\frac 1 2$ 8. Originally Posted by tkdvipers Cheers I am getting there, What confuses me is the way that. $ \\frac{3}{7} * \\frac{5}{6} $ becomes $ \\frac{1}{7} * \\frac{5}{2} $ I can see that 6/3 = 2 and 3/3 = 1, however I thought the fractions were unrelated and you could only work within the same fraction. for example 3/7 and 5/6 I can do the sums now but want to understand why it is legal. Look at the cancelation that took place in red. The 3 and the 6 have a common factor of 3. $ \\frac{3}{7} * \\frac{5}{6} = \\color{red}\\frac{\\rlap{---}{3}^1}{7} * \\color{red}\\frac{5}{\\rlap{---}{6}^2} =\\color{black}\\frac{1}{7}\\cdot\\frac{5}{2}= \\frac{1\\cdot5}{7\\cdot2}=\\frac{5}{14} $ Another approach: $\\frac{3}{7} \\cdot \\frac{5}{6}=\\frac{3}{6} \\cdot \\frac{5}{7}=\\color{red} \\frac{\\rlap{---}{3}^1}{\\rlap{---}{6}^2}\\cdot\\frac{5}{7}= \\color{black}\\frac{1\\cdot5}{2\\cdot7}=\\frac{5}{14} $", "\\dfrac a1$: We multiply $a = \\dfrac a1$ by $\\dfrac cc = 1$, to get $\\dfrac a1 \\times \\dfrac cc = \\;\\dfrac{a\\times c}{c}\\;$ and then add that to the fraction $\\dfrac bc$. For example $$3 \\frac 58\\; = \\;3 + \\frac 58\\; = \\;\\frac{3 \\times 8}{8} + \\frac 58\\; = \\;\\frac{(3\\times 8)+ 5}{8}\\; = \\;\\frac{24 + 5}{8} \\;= \\;\\frac{29}{8}$$ - Good call, you did the general case followed by an example +1 – Amzoti May 24 '13 at 12:03 Here is an example. $$2 \\frac{3}{11}=\\frac{2 \\cdot 11}{11}+\\frac3{11}=\\frac{22}{11}+\\frac3{11}=\\frac{25}{11}$$ Or conversely, $$\\frac{11}3=\\frac{11-3 \\cdot 3}{3}+\\frac{3 \\cdot 3}{3}=\\frac{11-9}{3}+\\frac{9}{3}=\\frac{2}{3}+3=3 \\frac{2}3$$. In general, $$x+\\frac{y}{z}=\\frac{x \\cdot z}{z}+\\frac{y}{z}=\\frac{x \\cdot z +y}z$$ and vice versa. In most cases you'll encounter $y<z$. -", "# Why does the law of distributivity use the distributive property to prove it exists? I was looking at the Law of Distributivity and disagreed with how it was proven. It proved that $$\\frac ab\\bigg(\\frac cd + \\frac ef\\bigg) = \\frac ab\\cdot \\frac cd + \\frac ab\\cdot \\frac ef.$$ It multiplied $(c\\div d)$ by $(f\\div f)$ and it multiplied $(e\\div f)$ by $(d\\div d)$. Of course, the left hand side of the equation remains unchanged, i.e. $$\\frac ab\\bigg(\\frac cd+\\frac ef\\bigg)=\\frac ab\\bigg(\\frac cd\\cdot \\frac ff+\\frac ef\\cdot \\frac dd\\bigg)=\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg).$$ This is because $(f\\div f)$ and $(d\\div d)$ each equal $1$. Now notice that $df=fd$, so since the fraction pair in the brackets have the same denominator, it can be combined, i.e. $$\\frac ab\\bigg(\\frac{cf}{df}+\\frac{ed}{fd}\\bigg)=\\frac ab\\cdot\\frac{cf+ed}{df}=\\frac{a(cf+ed)}{bdf}.$$ The proof now follows by stating that $a(cf+ed)=acf+aed$, but this is the kind of property it is trying to prove in the first place. It is trying to prove this sense of distribution, and in it, it uses it to prove itself. How is this allowed? Edit: This post is very much related, but it does not answer my question. • It proves distributivity for fractions (i.e. rationals) assuming the distributivity for integers. – Mauro ALLEGRANZA Jun 15 '18 at 6:45 • @MauroALLEGRANZA ok, thank you.... but aren't integers rational themselves? – Mr Pie Jun", "\\color{red}{\\times \\frac{4}{13}}=\\frac{64}{13}$$ Combining it all~ $$x=\\frac{64}{13}$$ YAY! DONE. Cheers! -Shahar - We start with the initial equation: $$(3+\\frac 14)x-2=14$$ We simplify: $$\\frac {13}4x-2 = 14$$ We add 2 to both sides: $$\\frac {13}4x = 16$$ Then we multiply both sides by 4: $$13x = 64$$ And then divide both sides by 13: $$x = \\frac{64}{13}$$ - I think OP uses the notation $3\\tfrac{1}{4}$ to mean $3+\\tfrac{1}{4}$, or $\\tfrac{13}{4}$. – Servaes Mar 24 '14 at 21:47 I did, @Servaes is correct. – Jacedc Mar 24 '14 at 21:50 EDIT: Corrected the answer to reflect this. – Disousa Mar 24 '14 at 21:58 It should be $(3+\\frac{1}{4})\\times x$, not $3+\\frac{1}{4}\\times x$. – Eff Mar 24 '14 at 22:01 You are correct. Editing again xD – Disousa Mar 24 '14 at 22:02 $$3 \\dfrac{1}{4} × x - 2 = 14$$ $$\\frac{13}{4} × x - 2 = 14$$ $$4\\cdot\\frac{13}{4} × x - 4\\cdot2 =4\\cdot 14$$ $$13 x - 8 =56$$ $$13x=56+8$$ $$13x=64$$ $$x=\\frac{64}{13}$$ because $$3 \\dfrac{1}{4}=3+\\frac{1}{4}=\\frac{13}{4}$$ - Correction, $14*4\\neq64$ :) – Disousa Mar 24 '14 at 21:57 You did $4⋅\\dfrac{13}{4}x - 4⋅2 = 4⋅14$ thus multiplying both sides by 4 (which is correct) but you did it twice on the left side? ($4⋅2$) – Jacedc Mar 24 '14 at 22:00 it is my mistake – Adi Dani Mar 24 '14 at 22:00", "\\\\ \\frac{1 \\cdot \\textcolor{blue}{2}}{2 \\cdot \\textcolor{blue}{2}} = \\frac{2}{4} \\; & so \\; \\frac{1}{2} = \\frac{2}{4} \\\\ \\frac{1 \\cdot \\textcolor{blue}{10}}{2 \\cdot \\textcolor{blue}{10}} = \\frac{10}{20} \\; & so \\; \\frac{1}{2} = \\frac{10}{20} \\end{split}$$ So, we say that $$\\frac{1}{2}, \\frac{2}{4}, \\frac{3}{6}$$, and $$\\frac{10}{20}$$ are equivalent fractions. Example 4.14: Find three fractions equivalent to $$\\frac{2}{5}$$. Solution To find a fraction equivalent to $$\\frac{2}{5}$$, we multiply the numerator and denominator by the same number (but not zero). Let us multiply them by 2, 3, and 5. $$\\frac{2 \\cdot \\textcolor{blue}{2}}{5 \\cdot \\textcolor{blue}{2}} = \\frac{4}{10} \\qquad \\frac{2 \\cdot \\textcolor{blue}{3}}{5 \\cdot \\textcolor{blue}{3}} = \\frac{6}{15} \\qquad \\frac{2 \\cdot \\textcolor{blue}{5}}{5 \\cdot \\textcolor{blue}{5}} = \\frac{10}{25}$$ So, $$\\frac{4}{10}, \\frac{6}{15}$$, and $$\\frac{10}{25}$$ are equivalent to $$\\frac{2}{5}$$. Exercise 4.27: Find three fractions equivalent to $$\\frac{3}{5}$$. Exercise 4.28: Find three fractions equivalent to $$\\frac{4}{5}$$. Example 4.15: Find a fraction with a denominator of 21 that is equivalent to $$\\frac{2}{7}$$. Solution To find equivalent fractions, we multiply the numerator and denominator by the same number. In this case, we need to multiply the denominator by a number that will result in 21. Since we can multiply 7 by 3 to get 21, we can find the equivalent fraction by multiplying both the numerator and denominator by 3. $$\\frac{2}{7} = \\frac{2 \\cdot \\textcolor{blue}{3}}{7 \\cdot \\textcolor{blue}{3}} = \\frac{6}{21}$$ Exercise 4.29: Find a fraction with a denominator", "# Find all real numbers $x$ for which $\\frac{8^x+27^x}{12^x+18^x}=\\frac76$ Find all real numbers $x$ for which $$\\frac{8^x+27^x}{12^x+18^x}=\\frac76$$ I have tried to fiddle with it as follows: $$2^{3x} \\cdot 6 +3^{3x} \\cdot 6=12^x \\cdot 7+18^x \\cdot 7$$ $$3 \\cdot 2^{3x+1}+ 2 \\cdot 3^{3x+1}=7 \\cdot 6^x(2^x+3^x)$$ Dividing both sides by $6$ gives us $$2^{3x}+3^{3x}=7 \\cdot 6^{x-1}(2^x+3^x)$$ Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated. - Ok, I'll make sure I do that in future – John Marty May 8 '13 at 1:58 Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$ $12^x=(2^2\\cdot3)^x=(2^x)^2\\cdot3^x=a^2b$ and similarly, $18^x=ab^2$ So, the problem reduces to $$\\frac{a^3+b^3}{ab(a+b)}=\\frac76$$ $\\displaystyle\\implies 6(a^2-ab+b^2)=7ab$ as $a+b\\ne0$ $\\displaystyle\\implies 6\\left(\\frac ab\\right)^2-13\\cdot\\frac ab+6=0$ $\\displaystyle\\implies \\frac ab=\\frac32$ or $\\dfrac23$ So, $\\displaystyle\\left(\\frac23\\right)^x=\\frac32$ or $\\dfrac23$ If $\\displaystyle\\left(\\frac23\\right)^x=\\frac32\\implies\\left(\\frac23\\right)^x=\\left(\\frac23\\right)^{-1}\\iff\\left(\\frac23\\right)^{x+1}=1$ Similarly, if $\\displaystyle\\left(\\frac23\\right)^x=\\frac23, \\left(\\frac23\\right)^{x-1}=1$ Now if $\\displaystyle u^m=1,$ either $\\displaystyle m=0,u\\ne0;$ or $\\displaystyle u=1$ or $\\displaystyle u=-1,m$ is even - Thanks so much that's really neat! – John Marty May 7 '13 at 4:34 @JohnMarty, my pleasure. Replacement of numbers with indices by algebraic symbol sometimes helps clarity – lab bhattacharjee May 7 '13 at 4:41 (+1) nice answer. – Mhenni Benghorbal May 7 '13 at 5:11 @lab-bhattacharjee, I'd say your comment (\"Replacement of numbers with indices by algebraic symbol sometimes helps", "1. \\begin{aligned} \\frac{1+2}{3\\cdot 4} \\amp = \\frac{3}{12}=\\frac{3}{3\\cdot 4}=\\frac{1}{4} \\end{aligned} 2. \\begin{aligned} 1+\\frac{2}{3\\cdot 4} \\amp = 1+ \\frac{2}{12}=\\frac{12}{12}+\\frac{2}{12}=\\frac{14}{12}=\\frac{2\\cdot 7}{2\\cdot 6}=\\frac{7}{6} \\end{aligned} 3. \\begin{aligned} 1+\\frac{2}{3}\\cdot 4 \\amp = 1+ \\frac{8}{3}=\\frac{3}{3}+\\frac{8}{3}=\\frac{11}{3} \\end{aligned} 4. It is already in simplest form. Two real numbers whose product is $1$ are called reciprocals. Therefore, $\\frac{a}{b}$ and $\\frac{b}{a}$ are reciprocals because $\\frac{a}{b}\\cdot\\frac{b}{a}=\\frac{ab}{ab}=1\\text{.}$ For example, \\begin{equation*} \\frac{2}{3}\\cdot\\frac{3}{2}=\\frac{6}{6}=1\\text{.} \\end{equation*} Since their product is $1,\\frac{2}{3}$ and $\\frac{3}{2}$ are reciprocals. Some other reciprocals are \\begin{equation*} \\frac{5}{8}\\, \\, \\text{ and } \\, \\, \\frac{8}{5},\\, \\, \\, \\, 7\\, \\, \\text{ and }\\, \\, \\frac{1}{7},\\, \\, \\, \\, -\\frac{4}{5}\\, \\, \\text{ and }\\, -\\frac{5}{4}\\text{.} \\end{equation*} This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor. This gives us the fraction multiplication rule: ###### Example9 Simplify $\\frac{5}{4}\\div\\frac{3}{5}\\cdot\\frac{1}{2}\\text{.}$ Solution Perform the multiplication and division left to right. \\begin{equation*} \\begin{aligned} \\frac{5}{4}\\div\\alert{\\frac{3}{5}}\\cdot\\frac{1}{2}\\amp =\\frac{5}{4}\\cdot\\alert{\\frac{5}{3}}\\cdot\\frac{1}{2}\\\\ \\amp =\\frac{5\\cdot 5\\cdot 1}{4\\cdot 3\\cdot 2}\\\\ \\amp =\\frac{25}{24} \\end{aligned} \\end{equation*} In algebra, it is often preferable to work with improper fractions. In this case, we leave the answer expressed as an improper fraction. You may already be familiar with square roots. Every nonnegative number has two square roots, defined as follows: \\begin{equation*} s ~\\text{ is a square root of }~ n ~\\text{ if }~s^2 = n. \\end{equation*} For example, 4 is a square root", "try to explain what $1 \\cdot 0$ or $0 \\cdot 1$ is... the evaluation of this will result the same every time as the answer would be $0$ which fits both statements. So let's apply this to fractions $0 \\cdot 1 = \\frac{0}{1} \\cdot \\frac{1}{1} = \\frac{0 \\cdot 1}{1 \\cdot 1}$ ... – Francis Cugler Mar 20 '17 at 23:57 • (...continued) $\\frac{0 \\cdot 1}{1 \\cdot 1} = \\frac{0}{1} = 0$ so far so good. So what happens when we reverse this: Let's try it out. But first we need to understand that $\\frac{1}{0} \\neq 0$ we need to consider this the reciprocal of $0$ or $\\frac{0}{1}$ and we know that the only value that has an equivalent reciprocal is $1$ or $\\frac{1}{1}$. Then it would suggest that $\\frac{1}{0} \\cdot \\frac{1}{1} = \\frac{1}{0}$ and that $\\frac{1}{0} \\cdot \\frac{0}{1} = \\frac{0}{0}$ and since $\\frac{0}{0} = 1$ from my above statement it would state that anything divided by $0$ would result in $1$. – Francis Cugler Mar 21 '17 at 0:10 Definition of Division For every real number a and every nonzero real number b, the quotient a$\\div$b, or $\\dfrac{a}{b}$, is defined by: $$a\\div b=a \\cdot \\frac{1}{b}.$$ Dividing by zero would mean multiplying by the reciprocal of 0. But 0 has no reciprocal (because 0 times any number is 0,", "$$4/5$$ 5: $$−1 \\frac{3}{7}$$ 7: $$5/6$$ 9: $$−7/4$$ 11: $$1/6$$ 13: $$32/105$$ 15: $$7 \\frac{5}{6}$$ 17: $$1$$ 19: $$29/30$$ 21: $$2 \\frac{2}{3}$$ 23: $$19/24$$ Exercise $$\\PageIndex{8}$$ Perform the operations. Reduce answers to lowest terms. 1. $$\\frac{3}{14} \\cdot \\frac{7}{3} \\div \\frac{1}{8}$$ 2. $$\\frac{1}{2} \\cdot (-\\frac{4}{5}) \\div \\frac{14}{15}$$ 3. $$\\frac{1}{2} \\div \\frac{3}{4} \\cdot \\frac{1}{5}$$ 4. $$-\\frac{5}{9} \\div \\frac{5}{3} \\cdot \\frac{5}{2}$$ 5. $$\\frac{5}{12} - \\frac{9}{21}+\\frac{3}{9}$$ 6. $$-\\frac{3}{10} - \\frac{5}{12} + \\frac{1}{20}$$ 7. $$\\frac{4}{5} \\div 4 \\cdot \\frac{1}{2}$$ 8. $$\\frac{5}{3} \\div 15 \\cdot \\frac{2}{3}$$ 9. What is the product of $$\\frac{3}{16}$$ and $$\\frac{4}{9}$$? 10. What is the product of $$−\\frac{24}{5}$$ and $$\\frac{25}{8}$$? 11. What is the quotient of $$\\frac{5}{9}$$ and $$\\frac{25}{3}$$? 12. What is the quotient of $$−\\frac{16}{5}$$ and $$32$$? 13. Subtract $$\\frac{1}{6}$$ from the sum of $$\\frac{9}{2}$$ and $$\\frac{2}{3}$$. 14. Subtract $$\\frac{1}{4}$$ from the sum of $$\\frac{3}{4}$$ and $$\\frac{6}{5}$$. 15. What is the total width when $$3$$ boards, each with a width of $$2 \\frac{5}{8}$$ inches, are glued together? 16. The precipitation in inches for a particular 3-day weekend was published as $$\\frac{3}{10}$$ inches on Friday, $$1\\frac{1}{2}$$ inches on Saturday, and $$\\frac{3}{4}$$ inches on Sunday. Calculate the total precipitation over this period. 17. A board that is $$5\\frac{1}{4}$$ feet long is to be cut into $$7$$ pieces of equal length. What is length of each piece? 18. How many $$\\frac{3}{4}$$ inch", "a) It's given that $\\frac{dV}{dt}=-1$. We know that $V = \\frac{2\\pi rh}{3}$ and $\\frac{dV}{dh} = \\frac{2\\pi r}{3}$. We also know that, by similar triangles, $h=3$ implies that $r=1$. $\\frac{dV}{dt}=\\frac{dV}{dh}\\cdot\\frac{dh}{dt} \\implies -1 = \\frac{2\\pi\\cdot 1}{3}\\cdot\\frac{dh}{dt} \\implies \\frac{dh}{dt}=-\\frac{3}{2\\pi}$", "the additive inverse is 8, since $(-8)+8=0$. The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted $\\frac{1}{a}$, that, when The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted $\\frac{1}{a}$, that, when when multiplied by the original number, results in the multiplicative identity, $1$. $a \\cdot \\frac{1}{a} = 1$ For example, if $a= -\\frac{2}{3}$, the reciprocal, denoted $\\frac{1}{a}$, is $-\\frac{3}{2}$ because $a \\cdot \\frac{1}{a} = (-\\frac{2}{3}) \\cdot (-\\frac{3}{2}) = 1$ ### PROPERTIES OF REAL NUMBERS The following properties hold for real numbers ab, and c. #### Example 7 Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. • $3 \\cdot 6 + 3 \\cdot 4$ • $(5+8)+(-8)$ • $6 – (15+9)$ • $\\frac{4}{7} \\cdot ( \\frac{2}{3} \\cdot \\frac{7}{4})$ • $100 \\cdot [0.75 + (-2.38)]$ Solution Part 1 $3 \\cdot 6 + 3 \\cdot 4$ $= 3 \\cdot (6+4)$ Distributive property $=3 \\cdot 10$ Simplify. $= 30$ Simplify. Part", "2 & \\frac{3}{4} \\\\ \\\\[-4mm] 3 & \\frac{4}{6}\\\\ \\\\ [-4mm] 4 & \\frac{5}{8}\\\\ \\\\[-4mm] 5 & \\frac{6}{10}\\\\ \\\\[-4mm] 6 & \\frac{7}{12} \\\\ \\\\[-4mm] 7 & \\frac{8}{14} \\\\ \\vdots & \\vdots \\end{array}$ It appears that: . $P(n) \\:=\\:\\dfrac{n+1}{2n}$ I'll let you supply the proof. • July 12th 2010, 04:12 PM melese Direct computation. I assume \"find and prove\" means that induction is involved. But anyway you could find the value of $\\prod_{i=2}^{n}(1-\\frac{1}{i^2})$ by direct manipulation of the terms of the product $(1-\\frac{1}{2^2})(1-\\frac{1}{3^2})...(1-\\frac{1}{n^2})$. • July 13th 2010, 01:21 AM Opalg Notice that $1-\\dfrac1{i^2} = \\dfrac{i^2-1}{i^2} = \\dfrac{(i-1)(i+1)}{i^2}$. Therefore $\\left(1-\\dfrac{1}{2^2}\\right)\\left(1-\\dfrac{1}{3^2}\\right)\\ldots\\left(1-\\dfrac{1}{n^2}\\right) = \\dfrac{1\\cdot3}{2^2} \\cdot \\dfrac{2\\cdot4}{3^2} \\cdot \\dfrac{3\\cdot5}{4^2} \\cdots\\dfrac{(n-1)(n+1)}{n^2}.$ Now start cancelling: each of the two factors in the numerators cancels with a factor in a neighbouring denominator. The only terms left uncancelled are $\\frac12$ at the start and $\\frac{n+1}n$ at the end. • July 13th 2010, 11:03 AM jess0517 Quote: Originally Posted by Soroban Hello, jess0517! I cranked out the first few partial products: . . $\\begin{array}{cccc} n & P(n) \\\\ \\hline \\\\[-4mm] 2 & \\frac{3}{4} \\\\ \\\\[-4mm] 3 & \\frac{4}{6}\\\\ \\\\ [-4mm] 4 & \\frac{5}{8}\\\\ \\\\[-4mm] 5 & \\frac{6}{10}\\\\ \\\\[-4mm] 6 & \\frac{7}{12} \\\\ \\\\[-4mm] 7 & \\frac{8}{14} \\\\ \\vdots & \\vdots \\end{array}$ It appears that: . $P(n) \\:=\\:\\dfrac{n+1}{2n}$ I'll let you supply the proof. So using induction would this be", "to finish this post: $C^{(\\frac{5}{6}-\\frac{3}{4})}\\cdot D^{(\\frac{1}{6}-\\frac{3}{4})}$ C's exponent is $\\frac{5}{6}-\\frac{3}{4} =\\frac{5}{6} \\cdot \\frac{2}{2}-\\frac{3}{4} \\cdot \\frac{3}{3}= \\frac{10}{12}- \\frac{9}{12}= \\frac{1}{12}$ D's exponent is $\\frac{1}{6}-\\frac{3}{4} =\\frac{1}{6} \\cdot \\frac{2}{2}-\\frac{3}{4} \\cdot \\frac{3}{3}= \\frac{2}{12}- \\frac{9}{12}= -\\frac{7}{12}$ you now hve $C^{\\frac{1}{12}}D^{-\\frac{7}{12}}= (\\frac{C}{D^7})^\\frac{1}{12}$ or $\\sqrt[12]{\\frac{C}{D^7}}$ 36. aroub That's the answer that I actually wanted! @phi thanks a looooottt ^_^", "$$AE \\cdot BC = AB \\cdot CE + AC \\cdot BE \\implies AE = AB + AC$$ $\\implies AI +EI = AB + AC, \\hspace{10mm} AI = AB+ AC – BC$ is integer. $$AI = \\frac {2AB \\cdot AC \\cdot cos \\angle CAI}{AB+AC + BC} = \\frac {AB \\cdot AC}{AB+AC + BC} =$$ $$= AB + AC – BC \\implies AC^2 + AB^2 + AB \\cdot AC = BC^2.$$ A quick $(\\mod9)$ check gives that $3\\mid AC$ and $3\\mid AB$. $$AI \\le A_0I_0 = EA_0 – EI_0 = \\frac{2 BC}{\\sqrt{3}} – BC = \\frac {2 - \\sqrt{3}}{\\sqrt{3}} BC = 33.88.$$ Denote $a= \\frac {BC}{3}= 73, b = \\frac {AC}{3}, c = \\frac {AB}{3}, l = \\frac {AI}{3} \\le 11.$ We have equations in integers $\\frac{bc}{a+b+c} = b + c – a = l \\le 11.$ The solution $(b > c)$ is $$b = \\frac{a + l +\\sqrt{a^2 – 6al – 3l^2}}{2}, c = \\frac{a + l -\\sqrt{a^2 – 6al – 3l^2}}{2}.$$ Suppose, $a^2 – 6al – 3l^2 = (a – 3l – t)^2 \\implies \\frac {12l^2}{t} + t + 6l= 2a = 146.$ Now we check all possible $t = {2,3,4,6,12, ml}.$ Case $t = 2 \\implies 6l^2 + 6l = 146 – 2 \\implies l^2 + l = 24 \\implies \\O$ Case $t = 3 \\implies 4l^2", "Feb 15, 2017 at 22:59 $$x + \\frac{1}{3} x = 1 \\cdot x + \\frac{1}{3} x = \\left( 1 + \\frac{1}{3} \\right) x = \\frac{4}{3} x$$ Your full equation is slighty different, but the same rules (distributivity) apply: $$15 = \\frac{2}{3} b + b = \\left( \\frac{2}{3} + 1 \\right) b = \\frac{5}{3} b \\iff \\\\ b = \\frac{3}{5} \\cdot 15 = 9$$ • Thanks for the help - I can see its process now! Jun 29, 2016 at 13:48 You know that you can: • Add fractions as long as they have the same denominator • Multiply both the nominator and the denominator with the same value and the value of the fraction remains the same Having that in mind we are going to add $x$ to $\\frac{x}{3}$.We know that $x=\\frac{3x}{3}$ (multiply both nominator and denominator by 3) Now we can add them together $\\frac{3x+x}{3}=\\frac{4x}{3}$ You need to grasp what is known as the distributive law: $$a(b+c)=ab+ac$$ In your case it is backwards, in other words $$xb+yb=(x+y)b$$ Can you see how that might help. Put $x=\\frac{2}{3},y=1$ and you have $$\\frac{2}{3}b+1\\cdot b=(\\frac{2}{3}+1)b$$ I guess that also uses $1\\cdot b=b$, which I am sure you are happy with. Why is that law true? Well first think of $b$ as \"apples\" we are saying that $x$ apples + $y$ apples", "# Determine $p$ for one to have $\\displaystyle\\frac{1}{5}+\\frac{1}{6}=\\frac{1+1}{5+6}=\\frac{2}{11}$, in $\\mathbb{Z_p}$. Determine $p$ for one to have $\\displaystyle\\frac{1}{5}+\\frac{1}{6}=\\frac{1+1}{5+6}=\\frac{2}{11}$, in $\\mathbb{Z_p}$. I know that $p$ must, $13, 43, 61, 101,103$. Try multiplying both sides by $11 \\cdot 5 \\cdot 6$, to get $$121 = 11 \\cdot 6 + 11 \\cdot 5 = 2 \\cdot 5 \\cdot 6 = 60$$ in $\\mathbb{Z}/p\\mathbb{Z}$. In addition to $p=121-60=61$ we can use the extended Euclidean algorithm to obtain $5^{-1}=-12$, $6^{-1}=-10$ and $11^{-1}=-11$ in $\\mathbb{Z}/61 \\mathbb{Z}$. Hence $$\\frac{1}{5}+\\frac{1}{6}=-12-10=-22=2\\cdot (-11)=\\frac{2}{11}.$$", "$$0$$ does not have a reciprocal. Example $$\\PageIndex{11}$$: reciprocal Find the reciprocal of each number. Then check that the product of each number and its reciprocal is $$1$$. 1. $$\\frac{4}{9}$$ 2. $$− \\frac{1}{6}$$ 3. $$− \\frac{14}{5}$$ 4. $$7$$ Solution To find the reciprocals, we keep the sign and invert the fractions. Find the reciprocal of $$\\frac{4}{9}$$. The reciprocal of $$\\frac{4}{9}$$ is $$\\frac{9}{4}$$. Check: Multiply the number and its reciprocal. $$\\frac{4}{9} \\cdot \\frac{9}{4}$$ Multiply numerators and denominators. $$\\frac{36}{36}$$ Simplify. $$1 \\; \\checkmark$$ Find the reciprocal of $$- \\frac{1}{6}$$. The reciprocal of $$- \\frac{1}{6}$$ is $$\\frac{6}{1}$$. Simplify. $$-6$$ Check. $$- \\frac{1}{6} \\cdot (-6) = 1 \\; \\checkmark$$ Find the reciprocal of $$- \\frac{14}{5}$$. $$- \\frac{5}{14}$$ Check. $$- \\frac{14}{5} \\cdot \\left(- \\dfrac{5}{14}\\right) = \\frac{70}{70} = 1 \\; \\checkmark$$ Find the reciprocal of 7. Write 7 as a fraction. $$\\frac{7}{1}$$ Write the reciprocal of $$\\frac{7}{1}$$. $$\\frac{1}{7}$$ Check. $$7 \\cdot \\left(\\dfrac{1}{7}\\right) = 1 \\; \\checkmark$$ Exercise $$\\PageIndex{21}$$ Find the reciprocal: 1. $$\\frac{5}{7}$$ 2. $$− \\frac{1}{8}$$ 3. $$− \\frac{11}{4}$$ 4. $$14$$ $$\\frac{7}{5}$$ $$-8$$ $$-\\frac{4}{11}$$ $$\\frac{1}{14}$$ Exercise $$\\PageIndex{22}$$ Find the reciprocal: 1. $$\\frac{3}{7}$$ 2. $$− \\frac{1}{12}$$ 3. $$− \\frac{14}{9}$$ 4. $$21$$ $$\\frac{7}{3}$$ $$-12$$ $$-\\frac{9}{14}$$ $$\\frac{1}{21}$$ In a previous chapter, we worked with opposites and absolute values. Table $$\\PageIndex{1}$$ compares opposites, absolute values, and reciprocals. Table $$\\PageIndex{1}$$ Opposite Absolute Value Reciprocal has opposite sign is never", "# Why is $\\frac{1}{\\frac{1}{X}}=X$? Can someone help me understand in basic terms why $$\\frac{1}{\\frac{1}{X}} = X$$ And my book says that \"to simplify the reciprocal of a fraction, invert the fraction\"...I don't get this because isn't reciprocal by definition the invert of the fraction? - The reciprocal is by definition the inverse of the fraction --- and you calculate the inverse of the fraction by inverting the fraction. – Gerry Myerson Jun 16 '13 at 3:53 The reciprocal of $a/b$ is defined to be the unique $x$ such that $a/b\\cdot x=1$, denoted by $(a/b)^{-1}$ or $1/(a/b)$. As it happens, $x=b/a$ is a formula for it, which can be easily verified, but there is a difference between a definition and an obvious formula. – anon Jun 16 '13 at 3:58 @user82306 I always ask my students, how many half pizzas in a pizza? The answer, $1/(1/2)=2$. – James S. Cook Jun 16 '13 at 4:10 Maybe this will help you see why $\\;\\dfrac 1{\\large \\frac 1X}= X.\\;$ We multiply numerator and denominator by $X$, which we can do because we can multiply any number by $\\dfrac XX = 1$ without changing the actual value of the number: $$\\frac 1{\\Large \\frac 1X}\\cdot \\frac XX = \\frac X1 = X$$ - $1/x$ is, by definition, the number that you multiply $x$", "# Question #bf504 Oct 12, 2016 $4 \\frac{7}{11} \\div \\frac{1}{11} = 51$ #### Explanation: $4 \\frac{7}{11} \\div \\frac{1}{11} = \\left(4 + \\frac{7}{11}\\right) \\div \\frac{1}{11}$ $= \\left(\\frac{44}{11} + \\frac{7}{11}\\right) \\div \\frac{1}{11}$ $= \\left(\\frac{51}{11}\\right) \\div \\frac{1}{11}$ $= \\frac{\\frac{51}{11}}{\\frac{1}{11}}$ $= \\frac{\\frac{51}{11}}{\\frac{1}{11}} \\cdot \\frac{11}{11}$ $= \\frac{\\frac{51}{11} \\cdot 11}{\\frac{1}{11} \\cdot 11}$ $= \\frac{51}{1}$ $= 51$ A common trick when dividing by a fraction is to take its reciprocal and then multiply, i.e. $x \\div \\frac{a}{b} = x \\cdot \\frac{b}{a}$ The process used above shows why this works: $x \\div \\frac{a}{b} = \\frac{x}{\\frac{a}{b}} = \\frac{x \\cdot \\frac{b}{a}}{\\frac{a}{b} \\cdot \\frac{b}{a}} = \\frac{x \\cdot \\frac{b}{a}}{1} = x \\cdot \\frac{b}{a}$", "3 \\cdot 2$ • Associative Property: When three or more numbers are multiplied, the sum is the same regardless of how they are grouped. Example: $2 \\cdot (3 \\cdot 4) = (2 \\cdot 3) \\cdot 4$ • Multiplicative Identity Property: The product of any number and one is the original number. Example: $2 \\cdot 1 = 2$ • Distributive Property: The product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For expressions $a, b,$ and $c$ , $a(b+c)=ab+ac$ . Example: $4(2 + 3) = 4(2) + 4(3)$ The multiplicative inverse of a number is the number which produces 1 when multiplied by the original number. The multiplicative inverse of $x$ is the reciprocal $\\frac{1}{x}$ . To find the multiplicative inverse of a fraction, simply invert the fraction : $\\frac{a}{b}$ inverts to $\\frac{b}{a}$ . To divide fractions, invert the divisor and multiply: $\\frac{a}{b} \\div \\frac{c}{d} = \\frac{a}{b} \\times \\frac{d}{c}$ . An object moving at a certain speed will cover a fixed distance in a set time . The quantities speed, distance and time are related through the equation $\\text{Speed} = \\frac{\\text{Distance}}{\\text{Time}}$ . ### Guided Practice Andrew is driving down the freeway. He passes mile marker 27 at exactly mid-day. At 12:35 he passes mile" ]

[ "the distinction new acquaintance? . . Did the couples just meet for the first time? . . (No, $A_1$ already knows the other five people.) Give us the original wording of the problem . . . please!", "attended the wedding, and averaged three drinks per person.", "which $3$ other couples are also attending. Several handshakes took place. Nobody shook hands with themselves or their partners. Nobody shook hands with anyone else more than once. $(1): \\quad$ How many times did you shake hands? $(2): \\quad$ How many times did your partner shake hands?", "a party. In that party, some of them shook hands with others. But no one shook hands with their spouse. After the party, you asked everyone(including your wife) how many person they shook hands with. You found that everyone shook hands with a different number of people. How many people did your wide shake hand with? Frankly, my dear, I don't give a damn. ahmedittihad Posts: 133 Joined: Mon Mar 28, 2016 6:21 pm PreviousNext Share with your friends: • Similar topics Replies Views Author Return to Junior Level ### Who is online Users browsing this forum: No registered users and 2 guests", "different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 VP Joined: 19 Oct 2018 Posts: 1174 Location: India Re: Chris and his wife invited a total of ten families on their marriage [#permalink] ### Show Tags 24 Nov 2019, 13:36 Each member of host family shakes hands with 40 persons, and each member of invited family shakes hand with 38 members. Total number of handshakes=$$\\frac{2*40+ 40*38}{2}=800$$ CaptainLevi wrote: Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 Re: Chris and his wife invited a total of ten families on their marriage [#permalink] 24 Nov 2019, 13:36 Display posts from previous: Sort by", "22/06/16 Common ways of introducing strangers Common ways of introducing strangers to each other are: • John, do you know Helen? Helen, this is my friend John. • Sally, I don’t think you’ve met Elaine. • I don’t think you two know each other, do you? • Can/May I introduce John Willis? (more formal) When people are introduced, they usually say How do you do? (formal), Hello, or Hi (very informal). Americans often say How are you? Note that How do you do? is not a question, and the normal reply is How do you do? (It does not mean the same, in British English, as How are you?) People who are introduced often shake hands.", "## If I had three arms, do you suppose I’d want four? How many times have I tried to carry more than my two hands can hold? How many times I have said, “If I only had another arm…” So, the question is, ‘would I really be satisfied with three or would I then wish for four?’ The truth is that we can always think of things that we want more of… more time more money more hands more friends more food (I’m thinking chocolate) So, I may want more, but do I really need more? Perhaps 2013 is the year to simply grateful for what I already have….", "have been widely debated and slightly controversial (the arguments have actually led to hurt feelings and mild name calling), but most historians and experts believe it started with African Americans in slums and ghettos. It was originally a quick, dramatic alternative to the traditional hand shake, but with a shocking twist: Instead of simply shaking hands, the two “High Five-ees” would slap the other's hand. It was exhilarating (and strangely arousing) to watch. Its current name derives from the archaic term \"slapping five,\" an action often accompanied by the phrases \"Gimmie' Five\" or \"Gimmie' some skin my Brotha'!\" In the 1970's, the High Five rose in status in popular culture, and could often be seen at large sporting events, such as not only baseball, but golf (most notably that time when Tiger Woods slapped his caddy in the face), volleyball and extreme yodeling. In the 1980's, its popularity increased, with the gesture's inclusion in dictionaries, movies, television, and half hour radio dramas. Despite, or perhaps in spite of its popularity, it was at this time that many people felt the High Five was \"selling out.\" People no longer High Fived to greet someone or to celebrate, but simply to be \"cool.\" What ever one's opinion on the matter, its cultural significance, or at least popularity, is undeniable. Now", "and my thanks to o_O (Handshake) note to self, read the entire question! words can be boring yes, but if you don't, you'll end up looking like an idiot", "# Party acquaintances There are $$n$$ people at a party. Suppose they are friendly people, and each person has 8 acquaintances. Now, Bob, a passer-by notices this special property: (i) Each pair of people who are mutually acquainted has exactly 4 common acquaintances. (ii) Each pair of people who are mutually unacquainted has exactly 2 common acquaintances Find the number of possible values of $$n$$.", "photo editor and photographer?... ### Three friends went to the movies. They shared 4/5 of a candy bar. If they share the bar equally how much of the candy bar will each person get? Three friends went to the movies. They shared 4/5 of a candy bar. If they share the bar equally how much of the candy bar will each person get?... ### Define the Encomienda system Define the Encomienda system...", "have a case of $Q_r$ which is already established, and putting the extreme couple back in adds one handshake to the husband and $H_{k+1} = P_k$ – Mark Bennet May 26 '12 at 4:00", "not with their spouse. After the party, Jim asks each guest how many people they shook hands with and got answers 0,1,2,3,4,5,6,7,8. How many people did Jeri shake hands with?", "I already found a protocol to find out who is richer (older) between two parties, but Is there any protocol to find the youngest person among 3 parties, without revealing actual ages?", "ever had anyone else misunderstand the purpose of love.event raises hand weakly... Cheers for the info guys. I'll look into those suggestions tonight. Thanks! ### Who is online Users browsing this forum: Geecko, Google [Bot], steVeRoll and 16 guests", "Tim Black's Blog ## Noble Knights can shake hands without having to move very much - and without repeats Posted November 26, 2017 This week, FiveThirtyEight's Riddler featured four puzzles from students. One that really pulled me in was a puzzle about the League of Interesting and Noble Knights (Puzzle 1), which came from the advanced math problem-solving class at Central Academy in Des Moines, Iowa, under the guidance of Mike Marcketti and Brian Reece. Here's the puzzle: After years of hard work as a devoted tower guard, you have been knighted and are set to become a member of a brand-new group dubbed the League of Interesting and Noble Knights, or LINK. Upon coming to the first meeting, you and the other five members are seated at a round table. There, your boss, King Scott, gives a grand introduction in which he welcomes you all to LINK. The king also outlines a greeting ceremony that must take place at the beginning of every LINK meeting. • The ceremony concludes when every member of LINK has met every other member with a handshake. • During each round of handshakes, every member of LINK must be shaking one and only one hand. • Arms cannot cross during the handshake ceremony as it is in violation of the Universal Law", "party., 2: t${host} invites${guest} and one other person to her party., other: t${host} invites${guest} and # other people to her party. }), male: plural({...}), other: plural({...}), }))", "it simply explains what a hug is, which is difficult for some people because it comes so natural for us. We were able to choose a type of hug for our concept namely the Bear hug thanks to this book (Definition of the bear hug follows below). This book also stated the importance of stroking once back during the hug and the warmth of a hug, which we read before in other articles as well. The Bear Hug: \" In the traditional bear hug (named for members of the family Ursidae, who do it best), one hugger usually is taller and broader than the other, but this is not necessary to sustain the emotional quality of bear-hugging. The taller hugger may stand straight or slightly curved over the shorter one, arms wrapped firmly around the other’s shoulders. The shorter of the pair stands straight with head against the taller hugger’s shoulder or chest, arms wrapped—also firmly!—around whatever area between waist and chest that they will reach. Bodies are touching in a powerful, strong squeeze that can last five to ten seconds or more. We suggest you use skill and forbearance in making the hug firm rather than breathless. Always be considerate of your partner, no matter what style of hug you are sharing. The feeling during a bear", "sweets received by people sitting in different positions.", "# English question regarding pigeonhole principle classic question. Mr. and Mrs. Smith invited four couples to their home. Some guests were friends of Mr. Smith, and some others were friends of Mrs. Smith. When the guests arrived, people who knew each other beforehand shook hands, those who did not know each other just greeted each other. After all this took place, the observant Mr. Smith said \"How interesting. If you disregard me, there are no two people present who shook hands the same number of times\". How many times did Mrs. Smith shake hands? My question is regarding the use of \"some\" and \"guests\" Some is defined as being at least one —used to indicate that a logical proposition is asserted only of a subclass or certain members of the class denoted by the term which it modifies. But would \"guests\" imply more than one? • No; \"some elephants are blue\" means the same as \"some elephant is blue\". – mjqxxxx Oct 28 '13 at 23:00 • I suppose Mr. and Mr. Smith knew each other beforehand, so they shook hands with each other? Similarly, any two people who came as a couple shook hands with each other? – bof Oct 28 '13 at 23:13 • The story is obscure and confusing, but I think the intended mathematical" ]

[ "# Handshakes in a party Here's the question which I'm stuck with - There are $20$ married couples at a party. Every man shakes hands with everyone except himself and his spouse. Half of the women refuse to shake hands with any other women. The other $10$ women all shake hands with each other (but not with themselves). How many handshakes are there at the party? My Solution: 1. Handshakes done by men: There are $20$ ways to pick a man and $38$ ways to pick the other person, totaling to $20 \\cdot 38 = 760$. But since every handshake is counted twice, the answer is $\\frac{760}{2} = 380$ 2. Handshakes done by the women who refuse to shake hand with any other women: these are already counted in $380$ handshakes (Because the women can only shake hands with men. This was taken care of above). 3. Handshakes done by the other $10$ women and men: These are counted in $380$ handshakes. 4. Handshakes done by women and women in the group of $10$: are $\\binom{10}{2} = 45$. Hence the total handshakes are $380 + 45 = 425$. However, the total handshakes are $615$ according to my textbook. Can anyone please help me to find out the mistake? • After you compute the number of handshakes done by", "different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 VP Joined: 19 Oct 2018 Posts: 1174 Location: India Re: Chris and his wife invited a total of ten families on their marriage [#permalink] ### Show Tags 24 Nov 2019, 13:36 Each member of host family shakes hands with 40 persons, and each member of invited family shakes hand with 38 members. Total number of handshakes=$$\\frac{2*40+ 40*38}{2}=800$$ CaptainLevi wrote: Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 Re: Chris and his wife invited a total of ten families on their marriage [#permalink] 24 Nov 2019, 13:36 Display posts from previous: Sort by", "Party August 09, 2009 • Math • Puzzles This is very simple but pretty interesting, like all good math should be. You go to a party. There could be any number of people at the party. You run into a friend at the party and you decide to make a little wager. You bet that you can find two people at the party who have shaken hands with the same number of people. Your friend realizes agrees to the idea of the bet but demands that you give him odds. You say, \"Okay, I'll give you good odds. If I find two people who have shaken the same number of hands as each other, you give me $20. If I can't, I'll give you nine hundred trillion dollars. Do we have a bet?\" Your friend quickly agrees because, after all, what is$20 next to the possibility of getting nine hundred trillion? You walk around the room, ask everybody how many people they've shaken hands with, eventually find two people that have the same number, and collect your \\$20. Of course, you knew going in that, at any party of any number of people with any rate of hand shaking going on, you can always find two people who have shaken the same amount of hands as each other.", "+0 # Combinatorics ') +2 167 2 +46 There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there? Aug 20, 2018 #1 +21848 +3 There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there? I assume $$\\dfrac{12\\cdot 10}{2}=60 \\text{ hand shakes}$$ Aug 21, 2018 #2 +98128 +3 Every person shakes hands with 10 other people So... 12 * 10 = 120 But handshake A - C is the same as handshake C - A So...we need to divide this by 2 So 120 / 2 = 60 handshakes Aug 21, 2018", "attended the wedding, and averaged three drinks per person.", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour", "people working in the company is estimated to be $6000$ while the actual strength of employees is $7000$. Which of the following shows the percent error between these two values? 5. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the absolute error? 6. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the relative error? 7. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the percent error?", "There are $44$ people coming to a dinner party. There are $15$ square tables that seat 4 people - one at each side - like this: Or the tables can be joined together to make groups like this: And so on... Find a way to seat the $44$ people using all $15$ tables, with no empty places.", "(5, 25) (B) (6, 25) (C) (25, 5) (D) (5, 20) (A) (5, 25), Explanation: Number pairs extends the pattern on the graph is as w = 5d after d = 4 the next one is d = 5 so w = 5 x 5 = 25 therefore next number pair extends the pattern on the graph is (5,25) matches with (A). Question 9. Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p, where t is the total cost in dollars and p is the number of party favors bought. Which point on the graph shows the total cost of 4 party favors? (TEKS 5.4.C) (A) Point A (B) Point B (C) Point C (D) Point D (D) Point D Explanation: Given Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p, where t is the total cost in dollars and p is the number of party favors bought. the point on the graph shows the total cost of 4 party favors is if p = 4 then t = 5 + 4 = 9 so point D, (4,9) matches with (D). Question 10. The rule for a pattern is e = 6w. What is the output, e, when the input, w, is 6? (TEKS 5.4.C)", "For this activity, you'll need to work with a partner, so the first thing to do is find a friend! Together count from $1$ up to $20$, clapping on each number, but clapping more loudly and speaking loudly on the numbers in the two times table, and quietly on the other numbers. Now clap the five times table together up to about $30$, so this time you are clapping more loudly and speaking loudly on the multiples of five and quietly on the others. If one of you claps the twos in this way and one of you claps the fives, at the same time, can you predict what you would hear? Which numbers would be quiet? Which numbers would be fairly loud and which would be very loud? Now try it - what did you hear? Were you right? Choose another pair of tables and repeat what you have just done. How about the twos and tens? Why not try the fives and tens? Each time predict what you will hear before you clap - which numbers will be loud, which fairly loud and which quiet? If you've enjoyed this activity and would like more of a challenge, try Music to my Ears .", "# At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? I have no idea how to approach this problem. • How many ways can a women choose 3 men in a set of 9 ? that would help. – user451844 Jul 13 '17 at 13:22 If you assume that a man dances with a woman, how many dances will be performed? 9 men -> dance 9 * 4 times with a woman, so 36 dances Each woman performs 3 dances, so 12 women. Count dance pairs . . . Since there are $9$ men and each man dances with $4$ women (presumably this means each man dances exactly $4$ times), there are exactly $36$ dance pairs. Let $w$ be the number of women. Since each woman dances with $3$ men (presumably this means each woman dances exactly $3$ times), there are exactly $3w$ dance pairs. Thus, $3w=36$, so $w=12$. • Obviously, we agree, but I don't think once each really matters. 9 women can dance 3 times with the same men,", "There are $44$ people coming to a dinner party. There are $15$ square tables that seat 4 people - one at each side - like this: Find a way to seat the $44$ people using all $15$ tables, with no empty places.", "Explanation: Let's look how much does the party cost for some numbers of children: 1. For 1 child the price would be $40 + 4 = 44$ 2. For 2 children $40 + 8 = 48$ 3. For $n$ children $40 + 4 n$ The formula in point 3. is a general formula to calculate cost of the party for $n$ children, so we can use it in our task. We have to calculate how many children can take part in a party for $76. So we have to solve inequality: $40 + 4 n \\le 76$$4 n \\le 76 - 40$$4 n \\le 36$$n \\le 9\\$", "is as if he gathered all $900$ people in a room and asked those who were there at $7$ to raise a hand. There would be $450$ hands raised, $50\\%$ of the crowd.", "have a case of $Q_r$ which is already established, and putting the extreme couple back in adds one handshake to the husband and $H_{k+1} = P_k$ – Mark Bennet May 26 '12 at 4:00", "if this is not obvious). Now use the cross-products to obtain: 4m - 4 = 3m + 12 Solve for m and you'll see that there are 16 men. Substitute back into the second equation and you'll find the number of women. Using my versions of the equations (to avoid fractions), we might have chosen instead to eliminate m first, replacing m in the first equation with $$m = 4(w-1)$$ to obtain $$4(w-1)-1=3w$$. This simplifies to $$4w-5=3w$$, so that $$w=5$$, and $$m = 4(w-1) = 4(5-1) = 16$$. Therefore there are 16 men and 5 women (including the Li’s), for a total of 21 people at the party. Here I only have to show Mr. and Mrs. Li’s handshakes, with men in blue and women in red: Mr. Li shook hands with 15 men and 5 women (a 3:1 ratio), while Mrs. Li shook hands with 16 men and 4 women (a 4:1 ratio). It worked. You might also want to use this information to determine the TOTAL number of handshakes. You can find it by examining Pascal's Triangle. I had fun with this one! Thanks! The mention of Pascal’s Triangle refers to the fact that it contains binomial coefficients, which count combinations; in particular, we need $${{21}\\choose{2}} = \\frac{21\\cdot 20}{2\\cdot 1} = 210$$ handshakes in all. If", "Question: At a party with $30$ people, $10$ know nobody there and the remaining $20$ people all know each other. Those who know each other greet with a hug, while those who do not greet with a handshake. After everyone has been introduced to everyone, how many handshakes have occurred? Attempt: Since the $10$ people don't know anyone of the $20$ people who know each other, so each of these $10$ people shakes hand with each of these $20$ people, giving a total of $200$ handshakes. Also, since these $10$ people also shake hands among themselves, we get $\\binom{10}{2}=45$ more handshakes. Thus, the total number of handshakes at this party is $245$. Doubt: Since the number of handshakes is odd, does this not contradict the Handshaking Lemma? If not, what is wrong with my argument. • Could you state the handshaking lemma (there are two things often called the handshaking lemma) and how you have interpreted it? – Air Conditioner Jan 27 '18 at 23:03 • You've counted total edges, not total valences. A person in the group of 10 each shake $29$ hands. The people in the twenty each shake $10$ hands. So the total is $10\\cdot 29 + 20\\cdot 10=490=2\\cdot 245.$ – Thomas Andrews Jan 27 '18 at 23:03 • In any event, the handshake lemma", "# Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? My book gave the answer as $$24$$. I do not understand why. I thought of it like this: You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $$4\\times 6$$ handshakes, but in my answer, you are double counting. How do I approach this problem? • In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 \\times (\\text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes. – M. Vinay Mar 17 at 4:49 • And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you", "# Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? My book gave the answer as $$24$$. I do not understand why. I thought of it like this: You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $$4\\times 6$$ handshakes, but in my answer, you are double counting. How do I approach this problem? • In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 \\times (\\text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes. – M. Vinay Mar 17 at 4:49 • And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you" ]

[ "different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 VP Joined: 19 Oct 2018 Posts: 1174 Location: India Re: Chris and his wife invited a total of ten families on their marriage [#permalink] ### Show Tags 24 Nov 2019, 13:36 Each member of host family shakes hands with 40 persons, and each member of invited family shakes hand with 38 members. Total number of handshakes=$$\\frac{2*40+ 40*38}{2}=800$$ CaptainLevi wrote: Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 Re: Chris and his wife invited a total of ten families on their marriage [#permalink] 24 Nov 2019, 13:36 Display posts from previous: Sort by", "Party August 09, 2009 • Math • Puzzles This is very simple but pretty interesting, like all good math should be. You go to a party. There could be any number of people at the party. You run into a friend at the party and you decide to make a little wager. You bet that you can find two people at the party who have shaken hands with the same number of people. Your friend realizes agrees to the idea of the bet but demands that you give him odds. You say, \"Okay, I'll give you good odds. If I find two people who have shaken the same number of hands as each other, you give me $20. If I can't, I'll give you nine hundred trillion dollars. Do we have a bet?\" Your friend quickly agrees because, after all, what is$20 next to the possibility of getting nine hundred trillion? You walk around the room, ask everybody how many people they've shaken hands with, eventually find two people that have the same number, and collect your \\$20. Of course, you knew going in that, at any party of any number of people with any rate of hand shaking going on, you can always find two people who have shaken the same amount of hands as each other.", "# Handshakes in a party Here's the question which I'm stuck with - There are $20$ married couples at a party. Every man shakes hands with everyone except himself and his spouse. Half of the women refuse to shake hands with any other women. The other $10$ women all shake hands with each other (but not with themselves). How many handshakes are there at the party? My Solution: 1. Handshakes done by men: There are $20$ ways to pick a man and $38$ ways to pick the other person, totaling to $20 \\cdot 38 = 760$. But since every handshake is counted twice, the answer is $\\frac{760}{2} = 380$ 2. Handshakes done by the women who refuse to shake hand with any other women: these are already counted in $380$ handshakes (Because the women can only shake hands with men. This was taken care of above). 3. Handshakes done by the other $10$ women and men: These are counted in $380$ handshakes. 4. Handshakes done by women and women in the group of $10$: are $\\binom{10}{2} = 45$. Hence the total handshakes are $380 + 45 = 425$. However, the total handshakes are $615$ according to my textbook. Can anyone please help me to find out the mistake? • After you compute the number of handshakes done by", "which $3$ other couples are also attending. Several handshakes took place. Nobody shook hands with themselves or their partners. Nobody shook hands with anyone else more than once. $(1): \\quad$ How many times did you shake hands? $(2): \\quad$ How many times did your partner shake hands?", "attended the wedding, and averaged three drinks per person.", "+0 # Combinatorics ') +2 167 2 +46 There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there? Aug 20, 2018 #1 +21848 +3 There are 6 married couples at a party. At the start of the party, every person shakes hands once with every other person except his or her spouse. How many handshakes are there? I assume $$\\dfrac{12\\cdot 10}{2}=60 \\text{ hand shakes}$$ Aug 21, 2018 #2 +98128 +3 Every person shakes hands with 10 other people So... 12 * 10 = 120 But handshake A - C is the same as handshake C - A So...we need to divide this by 2 So 120 / 2 = 60 handshakes Aug 21, 2018", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour", "# Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? My book gave the answer as $$24$$. I do not understand why. I thought of it like this: You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $$4\\times 6$$ handshakes, but in my answer, you are double counting. How do I approach this problem? • In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 \\times (\\text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes. – M. Vinay Mar 17 at 4:49 • And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you", "# Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes? My book gave the answer as $$24$$. I do not understand why. I thought of it like this: You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $$4\\times 6$$ handshakes, but in my answer, you are double counting. How do I approach this problem? • In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 \\times (\\text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes. – M. Vinay Mar 17 at 4:49 • And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you", "For this activity, you'll need to work with a partner, so the first thing to do is find a friend! Together count from $1$ up to $20$, clapping on each number, but clapping more loudly and speaking loudly on the numbers in the two times table, and quietly on the other numbers. Now clap the five times table together up to about $30$, so this time you are clapping more loudly and speaking loudly on the multiples of five and quietly on the others. If one of you claps the twos in this way and one of you claps the fives, at the same time, can you predict what you would hear? Which numbers would be quiet? Which numbers would be fairly loud and which would be very loud? Now try it - what did you hear? Were you right? Choose another pair of tables and repeat what you have just done. How about the twos and tens? Why not try the fives and tens? Each time predict what you will hear before you clap - which numbers will be loud, which fairly loud and which quiet? If you've enjoyed this activity and would like more of a challenge, try Music to my Ears .", "for my Advanced Math Monsters workshop at the APACHE homeschool conference: The median American family has a net worth of about $100 thousand. Bill Gates has a net worth of$56 billion. If Average Jane Homeschooler spends \\$100 in the vendor hall, what would be the equivalent expense for Gates? ## Reblog: The Handshake Problem [Feature photo above by Tobias Wolter (CC-BY-SA-3.0) via Wikimedia Commons.] Seven years ago, our homeschool co-op held an end-of-semester assembly. Each class was supposed to demonstrate something they had learned. I threatened to hand out a ten question pop quiz on integer arithmetic, but instead my pre-algebra students voted to perform a skit. I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives: If seven people meet at a party, and each person shakes the hand of everyone else exactly once, how many handshakes are there in all? In general, if n people meet and shake hands all around, how many handshakes will there be? 1-3 narrators ### Props Each friend will need a sheet of paper with a number written on it big and bold enough to be read by the audience. The numbers needed are 0, 1, 2, 3, … up to one less than the number of friends. Each friend keeps his paper in a pocket", "if this is not obvious). Now use the cross-products to obtain: 4m - 4 = 3m + 12 Solve for m and you'll see that there are 16 men. Substitute back into the second equation and you'll find the number of women. Using my versions of the equations (to avoid fractions), we might have chosen instead to eliminate m first, replacing m in the first equation with $$m = 4(w-1)$$ to obtain $$4(w-1)-1=3w$$. This simplifies to $$4w-5=3w$$, so that $$w=5$$, and $$m = 4(w-1) = 4(5-1) = 16$$. Therefore there are 16 men and 5 women (including the Li’s), for a total of 21 people at the party. Here I only have to show Mr. and Mrs. Li’s handshakes, with men in blue and women in red: Mr. Li shook hands with 15 men and 5 women (a 3:1 ratio), while Mrs. Li shook hands with 16 men and 4 women (a 4:1 ratio). It worked. You might also want to use this information to determine the TOTAL number of handshakes. You can find it by examining Pascal's Triangle. I had fun with this one! Thanks! The mention of Pascal’s Triangle refers to the fact that it contains binomial coefficients, which count combinations; in particular, we need $${{21}\\choose{2}} = \\frac{21\\cdot 20}{2\\cdot 1} = 210$$ handshakes in all. If", "How old am I? The age difference is 23 years, so I must be 23 if my father is twice as old as me. If seven people meet each other and each shakes hands only once, how many handshakes will there have been? 21 (most people would think there were 42 handshakes). ### Tough Math Riddles In a pond, there are some flowers with some bees hovering over them. How many flowers and bees are there if both the following statements are true: 1. If each bee lands on a flower, one bee doesn’t get a flower. 2. If two bees share each flower, there is one flower left out. 4 bees and 3 flowers. The ages of a father and son add up to 66. The father’s age is the son’s age reversed. How old could they be (there are 3 possible solutions)? 42 and 24. 51 and 15. 60 and 6. Janie’s friends were chipping in to buy her a wedding shower present. At first, 10 friends chipped in, but 2 of them dropped out. Each of the 8 had to chip in another dollar to bring the amount back up. How much money did they plan to collect?$40 (10 at $4, or 8 at$5). There are two ducks in front of two other ducks. And,", "have a case of $Q_r$ which is already established, and putting the extreme couple back in adds one handshake to the husband and $H_{k+1} = P_k$ – Mark Bennet May 26 '12 at 4:00", "## Handshakes A group of $200$ high school students, $105$ girls and $95$ boys, is randomly divided into two rows of $100$ students each. Each student in one row is directly opposite a student in the other row, and all the opposite pairs of students shake hands. Prove that the number of girl-girl handshakes is $5$ more than the number of boy-boy handshakes. Source: NCTM Mathematics Teacher 2008 SOLUTION Let $bb$ be the number of boy-boy handshakes; $gg$ the number of girl-girl handshakes; $bg$ the number of boy-girl handshakes. $2bb+bg=95$ $2gg+bg=105$ $95-2bb=105-2gg$ $2gg-2bb=10$ $gg-bb=5$ $gg=bb+5$ The number of girl-girl handshakes is $5$ more than the number of boy-boy handshakes.", "# Giving Presents Level pending $$4$$ couples are trading presents as follows: the couples give each other a present, then each person gives another person of the opposite gender (not including his/her significant other) a present such that every person gets $$2$$ presents, including the present that couples traded each other. How many ways can this be done? ×", "(5, 25) (B) (6, 25) (C) (25, 5) (D) (5, 20) (A) (5, 25), Explanation: Number pairs extends the pattern on the graph is as w = 5d after d = 4 the next one is d = 5 so w = 5 x 5 = 25 therefore next number pair extends the pattern on the graph is (5,25) matches with (A). Question 9. Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p, where t is the total cost in dollars and p is the number of party favors bought. Which point on the graph shows the total cost of 4 party favors? (TEKS 5.4.C) (A) Point A (B) Point B (C) Point C (D) Point D (D) Point D Explanation: Given Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p, where t is the total cost in dollars and p is the number of party favors bought. the point on the graph shows the total cost of 4 party favors is if p = 4 then t = 5 + 4 = 9 so point D, (4,9) matches with (D). Question 10. The rule for a pattern is e = 6w. What is the output, e, when the input, w, is 6? (TEKS 5.4.C)", "$$10^{10}$$ at the latest count! How do people decide who to pair up with? Well, in that case it might have been due to a chance encounter at a mixer at which both got slightly tipsy (and believe it or not he happened to be the nicest guy there).", "B, we have $4$ kisses. For A and C, $4$ more, and finally for B and C, $4$ more, for a total of $12$. If you have a certain amount of patience, you can deal with $4$ people, A, B, C, and D. First A exchanges kisses with B, C, and D. That's $12$ kisses. Now B, C, and D exchange kisses with each other. That's the case of $3$ people, which we have already solved! So they account for $12$ more, for a total of $24$. Now deal with $5$ people, A, B, C, D, and E. First A does the kiss thing with B, C, D, and E. That's $16$ kisses. Then B, C, D, and E exchange kisses. That's already been computed, it is the $4$ people case, we get $24$. so the total is $16+24$. which is $40$. Finally, deal with $6$ people, A to F. First A exchanges kisses with the $5$ others, total $20$. Then the remaining $5$ exchange kisses, total $40$. The sum is $60$. To get a general formula, there are many approaches. We could follow up on our analysis above. Or we could start again. For every pair of friends, there will be $4$ kisses. Do you know a simple formula for the number of pairs there are if" ]

[ "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]", "to do with TeX. A crucial part of Sketch is the projection of 3D objects to 2D. The classic depth sort algorithm (also known as painter's algorithm) is used to determine which polygons are shown . The algorithm also involves splitting overlapping polygons and it is useful to understand what this means in practice. I'll illustrate this with the following example of two intersecting polygons. def scene{ polygon[fill=red!20,draw=red] (0,0,1)(1,0,0)(0,1,0) put {rotate(25)} {polygon[fill=blue!20,draw=blue] (0,0,1)(1,0,0)(0,1,0)} } put { scale(2) then view((5,4,8)) } {scene} global { language tikz } The result is two intersecting polygons. The generated code looks like this: \\filldraw[fill=red!20,draw=none](-.976,-.61)--(1.561,-.381)--(.608,.887)--cycle; \\draw[draw=red](-.976,-.61)--(1.561,-.381)--(.608,.887); \\filldraw[fill=blue!20,draw=blue](-.976,-.61)--(1.415,.371)--(-.66,1.697)--cycle; \\filldraw[fill=red!20,draw=none](-.976,-.61)--(.608,.887)--(0,1.695)--cycle; \\draw[draw=red](.608,.887)--(0,1.695)--(-.976,-.61); To draw the figure, five different paths are drawn: Note that the red triangle is split into two different polygons and two line segments, while the blue triangle is drawn using a single filldraw operation. This is the depth sort and polygon splitting in action. PGF/TikZ options Stroke and fill parameters are set using the well known option syntax: line[arrows=<->,draw=red,line width=ultra thick](0,0,0)(1,1,1) polygon[fill=lightgray,style=semitransparent](0,0,1)(1,0,0)(0,1,0) From the example in the previous section, it is evidently that Sketch needs to know the difference between fill and stroke parameters. The PGF/TikZ parameters recognized by Sketch are listed in the TikZ/PGF options section of the manual. Custom styles TikZ makes it easy to mix stroke and fill parameters in a", "can essentially just place the smaller squares next to each other and cut out two equal straight triangles which corresponds to the triangle in the Pythagorean theorem. Here is the very inoptimized Tikzcode for generating the triangle and squares. \\a and \\b determines the sizes of the squares. \\begin{tikzpicture}[very thick,scale = 1] \\pgfmathsetmacro{\\a}{3}; %Side of bigger square \\pgfmathsetmacro{\\b}{2};%Side of smaller square \\draw [fill = yellow] (0,0) -- (\\b, 0) -- (0, \\a) -- (0,0); \\draw [fill = red](-\\a, 0) -- (\\b - \\a, 0) -- (-\\a, \\a) -- (-\\a,0); \\draw [fill = blue](\\b - \\a, 0) --(0,{\\b * (1 - \\b / \\a)}) -- (0,0) -- (\\b - \\a,0); \\draw [fill = green](\\b - \\a, 0) --(0,{\\b * (1 - \\b / \\a)}) -- (0, \\a) -- (-\\a,\\a) -- (\\b - \\a,0); \\draw [fill = cyan](0,0) -- (0,{\\b * (1 - \\b / \\a) - \\b}) -- (\\b,0) -- (0,0); \\draw [fill = magenta](0,{\\b * (1 - \\b / \\a) - \\b}) -- (0,-\\b) -- (\\b,-\\b) -- (\\b,0) -- (0,{\\b * (1 - \\b / \\a) - \\b}); \\draw [fill = red](\\a, \\b) -- (\\a + \\b, \\b) -- (\\a, \\a + \\b) -- (\\a,\\b + \\b); \\draw [fill = blue] (0,\\a) -- (\\a - \\b,{\\a + \\b * (1 - \\b / \\a)}) -- (\\a - \\b,", "that region, and a resulting probability distribution for the point $C$. Finally, let me mention that one might hope to improve the accuracy of a construction, simply by repeating it and averaging the result, or by some other convergence algorithm. For example, as a first step, we might simply perform the Apollonius bisector construction twice, producing midpoint candidates $C_0$ and $C_1$, and we could proceed simply find the midpoint of $C_0C_1$ as a further presumably more accurate midpoint. Or we could iterate in some other manner and hope to converge to the actual midpoint. For example, we could produce seven midpoint candidates, and take the resulting median point. # Draw an infinite chessboard in perspective, using straightedge only I’d like to explain to you how to draw chessboards by hand in perfect perspective, using only a straightedge. In this post, I’ll explain how to construct chessboards of any size, starting with the size of the basic unit square. This post follows up on the post I made yesterday about how to draw a chessboard in perspective view, using only a straightedge. That method was a subdivision method, where one starts with the boundary of the desired board, and then subdivides to make a chessboard. Now, we start with the basic square and build up. This method is actually", "the same way, until you construct an octagon. How many sides does the resulting polygon have? $[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]$ $\\textbf{(A)}21\\qquad\\textbf{(B)}23\\qquad\\textbf{(C)}25\\qquad\\textbf{(D)}27\\qquad\\textbf{(E)}29$ Problem 10 On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\frac{1}{16}\\qquad\\textbf{(B)}\\frac{7}{16}\\qquad\\textbf{(C)}\\frac{1}2\\qquad\\textbf{(D)}\\frac{9}{16}\\qquad\\textbf{(E)}\\frac{49}{64}$ Problem 11 The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$. How many more sixth graders than seventh graders bought a pencil? $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ 2\\qquad\\textbf{(C)}\\ 3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 5$ Problem 12 The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime? $[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label(\"1\",(0,.5)); label(\"3\",((cos(pi/6))/2,(-sin(pi/6))/2)); label(\"5\",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]$ $[asy] unitsize(30);", "# Problem #2185 2185 A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral? $[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.25,0.2)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); [/asy]$ $\\mathrm{(A) \\ } 15\\qquad \\mathrm{(B) \\ } 17\\qquad \\mathrm{(C) \\ } \\frac{35}{2}\\qquad \\mathrm{(D) \\ } 18\\qquad \\mathrm{(E) \\ } \\frac{55}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "die roll 5. This new \"halfway point\" becomes the starting point for the next iteration 6. Keep repeating the procedure until a pattern emerges The point of the exercise is to try and predict what pattern will emerge. The lecture continues on to talk about Fibonacci numbers, difference equations (or recurrence relations), bound and unbound trajectories, and the Mandelbrot set before it eventually gets back to the result of the Chaos Game. It takes quite a few iterations for a pattern to emerge, so this is a perfect problem for setting up a simulation. You'll need Python and the Python Imaging Library (PIL) to run the following code. from PIL import Imageimport random# calculate the midpoint between two pointsdef midpoint(p1, p2): return ((p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2)# define the size of the image and a few color constants# the hieght of an equilateral triangle is sqrt(3)/2 * basesize = 410, 356 # width, heightwhite = (255, 255, 255)black = (0, 0, 0)# set the corner positions of the triangleblueCorner = (5, 351)redCorner = (205, 5)greenCorner = (405, 351)corners = (redCorner, blueCorner, greenCorner)triangle = Image.new(\"RGBA\", size, white)pivot = blueCornerfor i in range(50000): randCorner = corners[ random.randint(0, 2) ] pivot = midpoint(pivot, randCorner) triangle.putpixel(pivot, black)triangle.save(\"triangle.gif\") The code above creates a blank Image canvas and puts", "a piece of paper and pencil using the following steps (adapted from ): 1. On a piece of paper, draw three points spaced out across the page; they do not necessarily need to be spaced out evenly to form an equilateral triangle though they might be. These will determine the “gameboard”, where the picture will ultimately lie. 2. Label one of the points with “1, 2”, the other point with “3, 4”, and the last point with “5, 6”. 3. Draw a point anywhere on the page: this will be the starting point. 4. Roll a six-sided die. Draw an imaginary line from the current dot towards to the point with the corresponding number, and put a dot half way on the line. This becomes the new “current dot”. 5. Repeat step 4 (until you get bored). Obviously, one would not want to plot a large number of points by hand4; instead, this can be done by computer; the code is shown below. For this algorithm, we assigned $$(0, 0)$$, $$(0, 1)$$ and $$(1, 1)$$ as the vertices of the triangle; they each are labelled with the value $$0$$, $$1$$, and $$2$$, respectively. Instead of using a die to decide which vertex to choose, similarly to the above code, we use the built-in function random.randint, which picks from", "$[asy] import olympiad; size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); real r = 31, t = -10; pair A = origin, B = dir(r-60), C = dir(r); pair X = -0.8 * dir(t), Y = 2 * dir(t); pair D = foot(B,X,Y), E = foot(C,X,Y); draw(A--B--C--A^^X--Y^^B--D^^C--E); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,dir(B--D)); label(\"E\",E,dir(C--E)); [/asy]$ ### Problem 28 The largest prime factor of $999999999999$ is greater than $2006$. Determine the remainder obtained when this prime factor is divided by $2006$. ### Problem 29 The altitudes in triangle $ABC$ have lengths 10, 12, and 15. The area of $ABC$ can be expressed as $\\tfrac{m\\sqrt n}p$, where $m$ and $p$ are relatively prime positive integers and $n$ is a positive integer not divisible by the square of any prime. Find $m + n + p$. $[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair A = (9,8.5), B = origin, C = (15,0); draw(A--B--C--cycle); pair D = foot(A,B,C), E = foot(B,C,A), F = foot(C,A,B); draw(A--D^^B--E^^C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); [/asy]$ ### Problem 30 Triangle $ABC$ is equilateral. Points $D$ and $E$ are the midpoints of segments $BC$ and $AC$ respectively. $F$ is the point on segment $AB$ such that $2BF=AF$. Let $P$ denote the intersection of $AD$ and $EF$, The value of $EP/PF$ can be expressed as $m/n$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] import olympiad;", "# Scan-line algorithm to fill in a triangle I wrote the following script to fill a Triangle. import pygame import random pygame.init() WIN = pygame.display D = WIN.set_mode((1200, 600)) p1 = (200, 50) p2 = (1000, 600) class Vec2: def __init__(self, x, y): self.x = x self.y = y def getLine(start, end): # RETURNS ALL THE PIXELS THAT NEEDS TO BE FILLED TO FORM A LINE x1, y1 = int(start.x), int(start.y) x2, y2 = int(end.x), int(end.y) dx = x2 - x1 dy = y2 - y1 is_steep = abs(dy) > abs(dx) if is_steep: x1, y1 = y1, x1 x2, y2 = y2, x2 swapped = False if x1 > x2: x1, x2 = x2, x1 y1, y2 = y2, y1 swapped = True dx = x2 - x1 dy = y2 - y1 error = int(dx / 2.0) ystep = 1 if y1 < y2 else -1 y = y1 points = [] for x in range(x1, x2 + 1): coord = Vec2(y, x) if is_steep else Vec2(x, y) points.append(coord) error -= abs(dy) if error < 0: y += ystep error += dx if swapped: points.reverse() return points RED = (255, 0, 0) BLUE = (0, 105, 255) GREEN = (0, 255, 0) def drawLine(p1, p2): # DRAWS A LINE BY FILLING IN THE PIXELS RETURNED BY getLine", "Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). ## Solution 3 $[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot(\"C\", C, dir(0)); dot(\"A\", A, dir(90));dot(\"B\", B, dir(180)); dot(\"D\", D, NE); dot(\"E\", E, dir(90));dot(\"F\", F, dir(270)); dot(\"M\", M, NE); dot(\"N\", N, dir(270));dot(\"L\", L, NW); dot(\"X\", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]$ Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\\sin(60)$. Similarly, $AB=3\\sin(60)$, and $BC=5\\sin(60)$. So $\\triangle ABC$ is a 3-4-5 right triangle", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ### Solution 3 $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\\frac{x}{2}$ and $\\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\\left(\\frac{x}{2}\\right) + 4\\left(\\frac{y}{2}\\right)$. $(4)\\frac{x}{2} + (4)\\frac{y}{2}$ $\\Rightarrow~~4\\left(\\frac{x}{2}+ \\frac{y}{2}\\right)$ $\\Rightarrow~~4\\left(\\frac{x+y}{2}\\right)$ Remember that $x+y=3$, so substituting this in we find that the area of all of the triangles is $4\\left(\\frac{3}{2}\\right) = 6$. The area of the $4$ unit squares is $4$, so the area of the square we need is $25- (4+6) = \\boxed{\\textbf{(C)}~15}$", "line EF • Elections In elections candidate 10 political parties. Calculate how many possible ways can the elections finish, if any two parties will not get the same number of votes. • Parallels and one secant There are two different parallel lines a, b, and a line c, that intersect the two parallel lines. Draw a circle that touches all lines at the same time. • Combinations of sweaters I have 4 sweaters two are white, 1 red and 1 green. How many ways can this done? • The camp At the end of the camp a 8 friends exchanged addresses. Any friend gave remaining 7 friends his card. How many addresses they exchanged? • Combi-triangle On each side of the square is marked 10 different points outside the vertices of the square. How many triangles can be constructed from this set of points, where each vertex of the triangle lie on the other side of the square? • Prize How many ways can be rewarded 9 participants with the first, second, and third prizes in a sports competition? • Lie/do not lie The function is given by the rule f(x) = 8x+16. Determine whether point D[-1; 8] lies on this function. Solve graphically or numerically and give reasons for the your answer. • Three-digit numbers How many", "# Problem #2199 2199 The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label(\"6\",(-0.6,1)); label(\"1\",(0.6,1)); label(\"2\",(0.6,-1)); label(\"9\",(-0.6,-1)); label(\"7\",(1.2,0)); label(\"3\",(-1.2,0)); label(\"pointer\",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label(\"1\",midpoint((4,-1)--(4,-2)),W); label(\"2\",midpoint((4,0)--(4,-1)),W); label(\"3\",midpoint((4,1)--(4,0)),W); label(\"4\",midpoint((4,2)--(4,1)),W); label(\"1\",midpoint((4,-2)--(5,-2)),S); label(\"2\",midpoint((5,-2)--(6,-2)),S); label(\"3\",midpoint((7,-2)--(6,-2)),S); [/asy]$ $\\textbf{(A) } \\frac{1}{3} \\qquad\\textbf{(B) } \\frac{4}{9} \\qquad\\textbf{(C) } \\frac{1}{2} \\qquad\\textbf{(D) } \\frac{5}{9} \\qquad\\textbf{(E) } \\frac{2}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Problem #2199 2199 The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label(\"6\",(-0.6,1)); label(\"1\",(0.6,1)); label(\"2\",(0.6,-1)); label(\"9\",(-0.6,-1)); label(\"7\",(1.2,0)); label(\"3\",(-1.2,0)); label(\"pointer\",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label(\"1\",midpoint((4,-1)--(4,-2)),W); label(\"2\",midpoint((4,0)--(4,-1)),W); label(\"3\",midpoint((4,1)--(4,0)),W); label(\"4\",midpoint((4,2)--(4,1)),W); label(\"1\",midpoint((4,-2)--(5,-2)),S); label(\"2\",midpoint((5,-2)--(6,-2)),S); label(\"3\",midpoint((7,-2)--(6,-2)),S); [/asy]$ $\\textbf{(A) } \\frac{1}{3} \\qquad\\textbf{(B) } \\frac{4}{9} \\qquad\\textbf{(C) } \\frac{1}{2} \\qquad\\textbf{(D) } \\frac{5}{9} \\qquad\\textbf{(E) } \\frac{2}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the", "slope of $\\overline{DC}$ as $\\dfrac{0-(-4)}{x-6}=-\\dfrac{4}{3}$ $\\dfrac{4}{x-6}=-\\dfrac{4}{3}$ $x-6=-3$ $x=3$ Thus, $\\overline{BC}$ intersects the $x$-axis at $D(3,0)$. Observe that the four vertical lines $x=-3,x=0,x=3$, and $x=6$ intersect the $x$-axis at equal intervals. By the Proportionality theorem, if parallel lines intersect two transversals, then they divide the transversals proportionally. Given that the four vertical lines are parallel and divide the transversal $x$-axis into thirds, they also divide the transversal $\\overline{BC}$ into thirds. $BE=ED=DC=BC/3=15/3=5$ Answer: $5$ units Alternative solution Using points $E(0,y)$ and $B(-3,8)$ we express the slope of $\\overline{EB}$ as $\\dfrac{y-8}{0-(-3)}=-\\dfrac{4}{3}$ $\\dfrac{y-8}{3}=-\\dfrac{4}{3}$ $y-8=-4$ $y=4$ The triangle with vertices at $(0,0),D(3,0)$ and $E(0,4)$ is a $3,4,5$ right triangle. ## Ten-Card Game A game is played with a deck of ten cards numbered from $1$ to $10$. Shuffle the deck thoroughly. Take the top card. If it is numbered $1$, you win. If it is numbered $k$, where $k>1$, then replace the card into the $k$th position from the top and draw again. You are allowed a maximum of $3$ draws before losing the game. What is the probability of winning? Source: NCTM Mathematics Teacher, September 2006 Solution It helps to build a deck of cards made out of cutout papers and simulate a few draws in order to understand the mechanics of the game. For example, we discover that having the card", "math; import fontsize; defaultpen(fontsize(9pt)); real w=1.2bp; pen BCpen=red+w; pen ACpen=deepgreen+w; pen ABpen=blue+w; pen linepen=deepblue+w; pen xlinepen=orange+w; pen arcpen=gray+0.4bp; pen anglepen=black+0.4bp; real a=6, b=7, c=8; real x; real r; pair p,q,t; guide gp,gq; pair A,B,C,D,Ap,Bp,Cp; pair H4,H2,H1; srand(12345); // Step 1. Select two arbitrary points A and B on the sheet A=rotate(unitrand()*60+30)*E; B=(0,0); // and connect them with a line AB using straight-edge: guide AB=A--B; draw(AB,ABpen); real ticksize=arclength(AB)/32; // Step 2. Split the line AB in two by drawing the two arcs from the endpoints // and the line throught the two intersections r=arclength(AB); gp=Arc(A,r,-60,-40); gq=Arc(B,r,0,20); draw(gp,arcpen); draw(gq,arcpen); p=intersectionpoint(gp,gq); gp=Arc(A,r,180,220); gq=Arc(B,r,120,150); draw(gp,arcpen); draw(gq,arcpen); q=intersectionpoint(gp,gq); H4=intersectionpoint(p--q,AB); draw(arcpoint(H4--p,ticksize)--arcpoint(H4--q,ticksize),arcpen); // Step 3. Similarly, split BH4 in two: r=arclength(B--H4); gp=Arc(H4,r,-60,-40); gq=Arc(B,r,0,20); draw(gp,arcpen); draw(gq,arcpen); p=intersectionpoint(gp,gq); gp=Arc(H4,r,180,220); gq=Arc(B,r,120,150); draw(gp,arcpen); draw(gq,arcpen); q=intersectionpoint(gp,gq); H2=intersectionpoint(p--q,AB); draw(arcpoint(H2--p,ticksize)--arcpoint(H2--q,ticksize),arcpen); // Step 4. Similarly, split BH2 in two: r=arclength(B--H2); gp=Arc(H2,r,-60,-40); gq=Arc(B,r,0,20); draw(gp,arcpen); draw(gq,arcpen); p=intersectionpoint(gp,gq); gp=Arc(H2,r,180,220); gq=Arc(B,r,120,150); draw(gp,arcpen); draw(gq,arcpen); q=intersectionpoint(gp,gq); H1=intersectionpoint(p--q,AB); draw(arcpoint(H1--p,ticksize)--arcpoint(H1--q,ticksize),arcpen); // At this point we have all measures ready: AB=8, AH1=7, AH2=6 // Step 5. Construct point C r=arclength(A--H1); gp=Arc(A,r,-65,-55); r=arclength(A--H2); gq=Arc(B,r,10,20); draw(gp,arcpen); draw(gq,arcpen); C=intersectionpoint(gp,gq); draw(B--C,BCpen); draw(A--C,ACpen); draw(Arc(A,H1,C),arcpen); // Step 6. Construct a line AA' || BC r=arclength(A--C); gp=Arc(B,r,110,130); r=arclength(B--C); gq=Arc(A,r,180,200); draw(gp,arcpen); draw(gq,arcpen); Ap=intersectionpoint(gp,gq); draw(B--Ap,arcpen); drawline(A,Ap,arcpen); // Step 7. Consrtuct point C' on AB' r=arclength(A--C); gp=Arc(A,r,180,240); Cp=intersectionpoint(A--(2Ap-A),gp); // Step 8. Consrtuct point B' r=arclength(A--B); gp=Arc(A,r,140,160); r=arclength(B--C); gq=Arc(Cp,r,70,100); draw(gp,arcpen); draw(gq,arcpen); Bp=intersectionpoint(gp,gq); draw(A--Bp,ABpen);" ]

[ "## Algebra 1 $\\frac{1}{4}$ To find a probability $P(x)$, we want to divide the number of events in the group $x$, and divide that number by the total number of events. In this case, there is $1$ blue section on the spinner, and $4$ sections in total, so P(blue) is $\\frac{1}{4}$.", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? $$11$$ $$12$$ $$13$$ $$14$$ $$15$$ ###### Solution(s): Note that the number of tiles the $$n$$ th square requires is $$n^2$$ since each side has $$n$$ tiles. Therefore, the $$6$$ th square will need $$6^2 = 36$$ tiles and $$7$$ th will need $$7^2 = 49.$$ The difference is $$49 - 36 = 13.$$ Thus, C is the correct answer. 12. A board game spinner is divided into three regions labeled $$A$$, $$B$$ and $$C$$. The probability of the arrow stopping on region $$A$$ is $$\\frac{1}{3}$$ and on region $$B$$ is $$\\frac{1}{2}$$. The probability of the arrow stopping on region $$C$$ is $$\\dfrac{1}{12}$$ $$\\dfrac{1}{6}$$ $$\\dfrac{1}{5}$$ $$\\dfrac{1}{3}$$ $$\\dfrac{2}{5}$$ ###### Solution(s): The total probability is $$1.$$ We need to subtract the probability of the spinner landing on $$B$$ and $$A$$ to get $$C,$$ which is $1 - \\dfrac{1}{3} - \\dfrac{1}{2} = \\dfrac{1}{6}.$ Thus, B is the correct answer. 13. For his birthday, Bert gets a box that holds $$125$$ jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans", "fraction Answer: $$\\frac{2}{5}$$ AMC-8, 2016 problem 21 Challenges and Thrills in Pre College Mathematics ## Try with Hints There are 5 Chips, 3 red and 2 green Can you now finish the problem .......... We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement Can you finish the problem........ There are 5 Chips, 3 red and 2 green we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green but we draw randomly. there are 3 red boxes and 2 green boxes Therefore the probability that the 3 reds are drawn=$$\\frac{2}{5}$$ # Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? • $$\\frac{2}{3}$$ • $$\\frac{1}{3}$$ • $$\\frac{1}{4}$$", "Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12 .$ What should his graph look like? AMC 8, 2006, Problem 17 Jeff rotates spinners $P, Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2006, Problem 20 A singles tournament had six players. Each player played every other player only once, with no ties. If Helen", "## Algebra 1 $\\frac{3}{4}$ To find a probability $P(x)$, we want to divide the number of events in the group $x$, and divide that number by the total number of events. In this case, there are $3$ sections that are not green on the spinner, and $4$ sections in total, so P(not green) is $\\frac{3}{4}$.", "## Algebra 1 $\\frac{1}{2}$ To find a probability $P(x)$, we want to divide the number of events in the group $x$, and divide that number by the total number of events. In this case, there are $2$ orange sections on the spinner, and $4$ sections in total, so P(orange) is $\\frac{2}{4}$, which is equivalent to $\\frac{1}{2}$.", "a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on a flat surface. What is the probability that the spinner stops with a red side facing you? Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for", "of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ $(C) \\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2007, Problem 25 On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has a radius of $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12", "= $\\frac{1}{2}$ • The probability of landing on heads is $P$(tails) = $\\frac{1}{2}$ • Dice: (a six-sided die, if singular) • When rolling a die, it will land with one of its six flat faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$ #### Lessons • Introduction Introduction to Probability Outcomes for Coins, Dice, and Spinners: a) What is probability? b) What are outcomes in probability? 1. Outcomes for flipping a coin 2. Outcomes for rolling", "Home > INT2 > Chapter 3 > Lesson 3.1.4 > Problem3-42 3-42. The spinners at right are spun and the results are added. Assume each of the outcomes on Spinner #2 are equally likely. Homework Help ✎ 1. What is $P$(sum is $4$)? Draw an area model or tree diagram. 2. What is $P$(sum is $8$)? Using the area model or tree diagram you made in part (a), circle all the places on your model where the sum is $8$. $P(\\text{sum is 8})=\\frac{1}{12}+\\frac{1}{4}=\\frac{1}{3}$", "Find the probability that Rick does not eat exactly two servings of fruits and vegetables. 1-0.28 = 0.72 Exercises 3–7 When Jenna goes to the farmers’ market, she also usually buys some broccoli. The possible number of heads of broccoli that she buys and the probabilities are given in the table below. Exercise 3. Find the probability that Jenna: a. $$\\frac{1}{4}$$ b. 1-$$\\frac{1}{4}$$ = $$\\frac{3}{4}$$ c. $$\\frac{5}{12}$$ + $$\\frac{1}{4}$$ + $$\\frac{1}{12}$$ = $$\\frac{3}{4}$$ d. $$\\frac{1}{4}$$ + $$\\frac{1}{12}$$ = $$\\frac{1}{3}$$ The diagram below shows a spinner designed like the face of a clock. The sectors of the spinner are colored red (R), blue (B), green (G), and yellow (Y). Exercise 4. Writing your answers as fractions in lowest terms, find the probability that the pointer stops on the following colors. a. Red: b. Blue: c. Green: d. Yellow: a. Red: $$\\frac{1}{12}$$ b. Blue: $$\\frac{2}{12}$$ = $$\\frac{1}{6}$$ c. Green: $$\\frac{5}{12}$$ d. Yellow: $$\\frac{4}{12}$$ = $$\\frac{1}{3}$$ Exercise 5. Complete the table of probabilities below. Exercise 6. Find the probability that the pointer stops in either the blue region or the green region. $$\\frac{1}{6}$$ + $$\\frac{5}{12}$$ = $$\\frac{7}{12}$$ Exercise 7. Find the probability that the pointer does not stop in the green region. 1-$$\\frac{5}{12}$$ = $$\\frac{7}{12}$$ ### Eureka Math Grade 7 Module 5 Lesson 5 Problem Set Answer Key Question 1. The", "### Home > CCG > Chapter 4 > Lesson 4.2.5 > Problem4-110 4-110. When he was in first grade, Harvey played games with spinners. One game he especially liked had two spinners and several markers that you moved around a board. You were only allowed to move if your color came up on both spinners. 1. Harvey always chose purple because that was his favorite color. What was the probability that Harvey could move his marker? Review the Math Notes box in Lesson 4.2.3. $\\frac{1}{12}$ 2. Is the event that Harvey wins a union or an intersection of events? Refer to the Math Notes box in Lesson 4.2.4. 3. Was purple the best color choice? Explain. Does purple have the greatest probability? 4. If both spinners are spun, what is the probability that no one gets to move because the two colors are not the same? $\\frac{1}{4} \\cdot \\frac{2}{3} = \\frac{1}{6}$ $\\frac{1}{6} + \\frac{1}{6} + \\frac{1}{3} = \\frac{2}{3}$", "A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered $1, 2, 3,4, 5, 6, 7$ or $8$ as shown. Two such disks are used in a game where their arrows are independently spun. What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest? 1. $\\dfrac{1}{16}$ 1. $\\dfrac{5}{64}$ 1. $\\dfrac{3}{32}$ 1. $\\dfrac{7}{64}$ Given the, two disks When we select only one number from any of the two disks, then the probability $= \\frac{1}{8}$ When we select one number from the first disk and the one number from the second disk simultaneously, then the probability $= \\frac{1}{8} \\times \\frac{1}{8} = \\frac{1}{64}$ The numbers on the resulting sectors upon spinning the two disks is equal to $8$ after the arrows come to rest $= \\underbrace{(1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1)}_{\\text{Number of favorable outcomes = 7}}$ $\\therefore$ The required probability $= \\dfrac{7}{64}$ Correct Answer $:\\text{D}$ ${\\color{Magenta}{\\textbf{PS:}}}$ In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event", "# Thread: red followed by a yellow 1. ## red followed by a yellow there are 3 red spaces, 1 blue space, 3 yellow spaces, and one green space for a total of 8 spaces on a spinner. What is the probability of getting a red spin followed by a yellow spin in two spins? I am new to probability, so I wanted to check out a couple of my homework questions. Here is what I think: There are n(s) = 8*8 = 64 possible spins in two total spins n(r,y) = 3C1 * 3C1 = 9 possible reds followed by a yellow so the probability of spinning a red then yellow in two spins is 9/64??? 2. Hello, ihavvaquestion! The probability that the first spin is Red is: $\\frac{3}{8}$ The probability that the second spin is Yellow is: $\\frac{3}{8}$ Therefore: . $P(\\text{Red, then Yellow}) \\;=\\;\\frac{3}{8}\\cdot\\frac{3}{8} \\;=\\;\\frac{9}{64}$", "will have to continually turn back to the centre room every time it gets to a side room (right next to the centre room). The probability of turning back is less than 1 and so the probability of continually turning back is 0. Next you need to show that the symmetry makes getting to each end room equally likely. To do so, check that we can rotate any path to an end room to get an equally likely path that goes to any other end room we want. Only then can you conclude that the first end room reached by the mouse has food with probability $\\frac{1}{2}$. Whenever the mouse leaves the centre room, it must with probability 1 get to the centre room again, because it will continually turn away from the centre room with probability 0. Thus after getting to a food room, the mouse will almost surely get back to the centre, and from there the first end room that the mouse gets to will have food with probability $\\frac{1}{4}$ and a cat with probability $\\frac{2}{4}$ and be empty (the first food room) with probability $\\frac{1}{4}$. But since the mouse will almost surely return to the centre whenever it gets to the empty end room, the probability that it will never get to a non-empty", "+0 0 108 1 At a county fair there is a spinner game with 10 sectors: 2 red sectors, 2 green sectors, 3 blue sectors, and 3 yellow sectors. If the spinner lands on a red sector, the player wins 3 tokens. If the spinner lands on a green sector, the player wins 5 tokens. If the spinner lands on any other sector, the player loses 2 tokens. Is this game fair for the player and how much will the player win or lose on an average over time? A)County fair games are never fair. B)The expected value is zero, and the game is fair. So the player will break even for a single spin over time. C)The expected value is 0.4 and the game is unfair. The player will win about 0.4 tokens for a single spin over time. D)The expected value is -0.4 and the game is not fair. The player will lose about 0.4 tokens for a single spin over time. Mar 10, 2019 #1 +7 +2 There is a $$\\frac{2}{10}$$ chance (probability of landing on a red sector) that the player wins 3 tokens, and $$\\frac{2}{10} \\cdot 3 = \\frac{6}{10} = \\frac35$$. There is a $$\\frac{2}{10}$$ chance (probability of landing on a green sector) that the player wins 5 tokens, and $$\\frac{2}{10} \\cdot 5", "What is the probability that $x<y$? • $\\frac{1}{8}$ • $\\frac{1}{6}$ • $\\frac{2}{3}$ ### Key Concepts Number system Cube Answer: $\\frac{1}{8}$ AMC-10A (2003) Problem 12 Pre College Mathematics ## Try with Hints The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \\times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with the condition $x<y$? can you finish the problem…….. Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\\triangle AOD=\\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$? can you finish the problem…….. Therefore the required probability ($x<y$) is $\\frac{\\frac{1}{2}}{4}$=$\\frac{1}{8}$ Categories ## Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability", "it is put back into the bag. A second ball is then selected at random from the bag. a) A third ball is selected. Find the probability all three balls selected are yellow. Originally Posted by joshuaa for # 2, the Question was asked Like that! If you say so, we must assume that the first ball is returned to the bag BUT the second ball is not returned. So what is the probability that all three are yellow? i.e. $\\mathscr{P}(Y_1,Y_2,Y_3)=\\left(\\dfrac{5}{10}\\cdot \\dfrac{5}{10}\\cdot\\dfrac{4}{9}\\right)$ 5. ## Re: Two Different Spinners I don't know why the book answer is 1/4 for # 2 and for # 1, the answer is 4/15, but when I do calculations, I get different answer! 6. ## Re: Two Different Spinners For #1 (a), here are the outcomes that yield a prime number: Spinner 1: Spinner 2 3 : 4 4 : 3 (2 outcomes yield this result) 5 : 6 (2 outcomes yield this result) So, there are 5 outcomes that yield prime sums. There are a total of 15 possible outcomes. So, the probability is 5/15, not 4/15. (b) Outcomes that yield products greater than 16: 3 : 6 (2 outcomes) 4 : 6 (2 outcomes) 5 : 4 5 : 6 (2 outcomes) That is 7/15 outcomes that yield products greater than 16.", "## nuccioreggie one year ago A spinner is divided into many sections of equal size. Some sections are red, some are blue, and the remaining are green. The probability of the arrow landing on a section colored red is 12 over 20. The probability of the arrow landing on a section colored blue is 6 over 20. What is the probability of the arrow landing on a green-colored section? 2 over 20 6 over 20 8 over 20 18 over 20 1. nuccioreggie @dmart180 2. anonymous What do you think? 3. anonymous $\\frac{ 12 }{ 20 }+\\frac{ 6 }{ 20 }=$ Solve that and it gives the answer" ]

[ "die roll 5. This new \"halfway point\" becomes the starting point for the next iteration 6. Keep repeating the procedure until a pattern emerges The point of the exercise is to try and predict what pattern will emerge. The lecture continues on to talk about Fibonacci numbers, difference equations (or recurrence relations), bound and unbound trajectories, and the Mandelbrot set before it eventually gets back to the result of the Chaos Game. It takes quite a few iterations for a pattern to emerge, so this is a perfect problem for setting up a simulation. You'll need Python and the Python Imaging Library (PIL) to run the following code. from PIL import Imageimport random# calculate the midpoint between two pointsdef midpoint(p1, p2): return ((p1[0] + p2[0]) / 2, (p1[1] + p2[1]) / 2)# define the size of the image and a few color constants# the hieght of an equilateral triangle is sqrt(3)/2 * basesize = 410, 356 # width, heightwhite = (255, 255, 255)black = (0, 0, 0)# set the corner positions of the triangleblueCorner = (5, 351)redCorner = (205, 5)greenCorner = (405, 351)corners = (redCorner, blueCorner, greenCorner)triangle = Image.new(\"RGBA\", size, white)pivot = blueCornerfor i in range(50000): randCorner = corners[ random.randint(0, 2) ] pivot = midpoint(pivot, randCorner) triangle.putpixel(pivot, black)triangle.save(\"triangle.gif\") The code above creates a blank Image canvas and puts", "the same way, until you construct an octagon. How many sides does the resulting polygon have? $[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]$ $\\textbf{(A)}21\\qquad\\textbf{(B)}23\\qquad\\textbf{(C)}25\\qquad\\textbf{(D)}27\\qquad\\textbf{(E)}29$ Problem 10 On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\frac{1}{16}\\qquad\\textbf{(B)}\\frac{7}{16}\\qquad\\textbf{(C)}\\frac{1}2\\qquad\\textbf{(D)}\\frac{9}{16}\\qquad\\textbf{(E)}\\frac{49}{64}$ Problem 11 The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$. How many more sixth graders than seventh graders bought a pencil? $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ 2\\qquad\\textbf{(C)}\\ 3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 5$ Problem 12 The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime? $[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label(\"1\",(0,.5)); label(\"3\",((cos(pi/6))/2,(-sin(pi/6))/2)); label(\"5\",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]$ $[asy] unitsize(30);", "lets them naturally recognize the complimentary pairs of fractions that make up a … 15 Corresponding lengths of similar figures are given. This problem is taken from the UKMT Mathematical Challenges . Wri… Ratio 4-in-a-line Game. A. Another way to prevent getting this page in the future is to use Privacy Pass. 15:7) 7:10 (B) 3:4. Participate in learning and knowledge sharing. Or submit details below for a call back counter of the shaded to unshaded ) of the shaded -! Are 3 unshaded parts of the perimeters and areas cloudflare Ray ID: 6171e18b4e6d6413 • Your IP: •. And 35 unshaded sections regions inside the shaded and unshaded parts of the unshaded region is 3:.. Original ratio of shaded to unshaded area is $1:3$ tollfree ) or submit details below for call... And diagonal BD=5 cm below to see this rule in action the CAPTCHA proves you a. Identity and LCM species x 4 treatments x 5 replicates 20 unshaded sections the future is to use ratio fractions... 4-In-A-Line cards with 80 shaded sections and 25 unshaded sections game for 4 students split into two.. 2.23 and π = 22/7 ] hexagon has an area of the ratio of shaded to unshaded area... Let 's look at the two similar triangles below to see this rule in action", "# Problem #2262 2262 A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is probability that the dart lands within the center square? $[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2)); draw(A--B--C--D--E--F--G--H--cycle); draw(A--D); draw(B--G); draw(C--F); draw(E--H); [/asy]$ $\\textbf{(A)}\\ \\frac{\\sqrt{2} - 1}{2} \\qquad\\textbf{(B)}\\ \\frac{1}{4} \\qquad\\textbf{(C)}\\ \\frac{2 - \\sqrt{2}}{2} \\qquad\\textbf{(D)}\\ \\frac{\\sqrt{2}}{4} \\qquad\\textbf{(E)}\\ 2 - \\sqrt{2}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? $$11$$ $$12$$ $$13$$ $$14$$ $$15$$ ###### Solution(s): Note that the number of tiles the $$n$$ th square requires is $$n^2$$ since each side has $$n$$ tiles. Therefore, the $$6$$ th square will need $$6^2 = 36$$ tiles and $$7$$ th will need $$7^2 = 49.$$ The difference is $$49 - 36 = 13.$$ Thus, C is the correct answer. 12. A board game spinner is divided into three regions labeled $$A$$, $$B$$ and $$C$$. The probability of the arrow stopping on region $$A$$ is $$\\frac{1}{3}$$ and on region $$B$$ is $$\\frac{1}{2}$$. The probability of the arrow stopping on region $$C$$ is $$\\dfrac{1}{12}$$ $$\\dfrac{1}{6}$$ $$\\dfrac{1}{5}$$ $$\\dfrac{1}{3}$$ $$\\dfrac{2}{5}$$ ###### Solution(s): The total probability is $$1.$$ We need to subtract the probability of the spinner landing on $$B$$ and $$A$$ to get $$C,$$ which is $1 - \\dfrac{1}{3} - \\dfrac{1}{2} = \\dfrac{1}{6}.$ Thus, B is the correct answer. 13. For his birthday, Bert gets a box that holds $$125$$ jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans", "color constants# the hieght of an equilateral triangle is sqrt(3)/2 * basesize = 410, 356 # width, heightwhite = (255, 255, 255)black = (0, 0, 0)# set the corner positions of the triangleblueCorner = (5, 351)redCorner = (205, 5)greenCorner = (405, 351)corners = (redCorner, blueCorner, greenCorner)triangle = Image.new(\"RGBA\", size, white)pivot = blueCornerfor i in range(50000): randCorner = corners[ random.randint(0, 2) ] pivot = midpoint(pivot, randCorner) triangle.putpixel(pivot, black)triangle.save(\"triangle.gif\") The code above creates a blank Image canvas and puts random pixels on it, but those pixels form a definite pattern. The pixels are selected by picking a random corner of the triangle and calculating the midpoint between that corner and the previous point drawn. This is called the Sierpinski triangle, a fractal first described by Polish mathematician Wacław Sierpiński. This isn't the only fractal that can be created by the Chaos Game. A more general set of rules for the Chaos Game are to start at one vertex of a regular polygon, then draw the next point a fraction r of the distance between it and a vertex picked at random. Most combinations of polygons and fractions won't produce a fractal of course, but there are several that do. From Wolfram MathWorld's article on the Chaos Game we find that for a regular pentagon the fractions 1/3 and 3/8 both", "a piece of paper and pencil using the following steps (adapted from ): 1. On a piece of paper, draw three points spaced out across the page; they do not necessarily need to be spaced out evenly to form an equilateral triangle though they might be. These will determine the “gameboard”, where the picture will ultimately lie. 2. Label one of the points with “1, 2”, the other point with “3, 4”, and the last point with “5, 6”. 3. Draw a point anywhere on the page: this will be the starting point. 4. Roll a six-sided die. Draw an imaginary line from the current dot towards to the point with the corresponding number, and put a dot half way on the line. This becomes the new “current dot”. 5. Repeat step 4 (until you get bored). Obviously, one would not want to plot a large number of points by hand4; instead, this can be done by computer; the code is shown below. For this algorithm, we assigned $$(0, 0)$$, $$(0, 1)$$ and $$(1, 1)$$ as the vertices of the triangle; they each are labelled with the value $$0$$, $$1$$, and $$2$$, respectively. Instead of using a die to decide which vertex to choose, similarly to the above code, we use the built-in function random.randint, which picks from", "# Problem #2185 2185 A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral? $[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.25,0.2)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); [/asy]$ $\\mathrm{(A) \\ } 15\\qquad \\mathrm{(B) \\ } 17\\qquad \\mathrm{(C) \\ } \\frac{35}{2}\\qquad \\mathrm{(D) \\ } 18\\qquad \\mathrm{(E) \\ } \\frac{55}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "times, then cut in the middle. How many pieces are obtained? ## Problem I11 There are 4 men: A, B, C and D. Each has a son. The four sons are asked to enter a dark room. Then A, B, C and D enter the dark room, and each of them walks out with just one child. If none of them comes out with his own son, in how many ways can this happen? ## Problem I12 In the figure below, $BCDE$ is a parallelogram, points $F$ and $G$ are on the segment $ED$, $BCA$ is a right angled triangle, $AC$ is perpendicular to $BC$.Suppose that $BC=8cm$, $AC=7cm$, and the area of the shaded regions is $12cm^2$ more than that of the triangle $AFG$. What is the length of $CG$? $[asy] fill((0,0)--(1,1)--(3,1)--(2,0)--cycle,grey); fill((0,0)--(2,1.5)--(2,0)--cycle,white); draw((0,0)--(2,1.5)--(2,0)--cycle,dot); draw((0,0)--(1,1)--(3,1)--(2,0)--cycle,dot); MP(\"B\",(0,0),W);MP(\"A\",(2,1.5),N);MP(\"C\",(2,0),S); MP(\"E\",(1,1),NW);MP(\"D\",(3,1),NE); MP(\"F\",(1.5,1),NW);MP(\"G\",(2,1),NE); draw((1.9,0)--(1.9,.1)--(2,.1),linewidth(.75)); [/asy]$ ## Problem I13 Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall? ## Problem I14 On a balance scale, three green balls balance six blue balls, two yellow balls balance five blue balls and six blue", "# Problem #2199 2199 The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label(\"6\",(-0.6,1)); label(\"1\",(0.6,1)); label(\"2\",(0.6,-1)); label(\"9\",(-0.6,-1)); label(\"7\",(1.2,0)); label(\"3\",(-1.2,0)); label(\"pointer\",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label(\"1\",midpoint((4,-1)--(4,-2)),W); label(\"2\",midpoint((4,0)--(4,-1)),W); label(\"3\",midpoint((4,1)--(4,0)),W); label(\"4\",midpoint((4,2)--(4,1)),W); label(\"1\",midpoint((4,-2)--(5,-2)),S); label(\"2\",midpoint((5,-2)--(6,-2)),S); label(\"3\",midpoint((7,-2)--(6,-2)),S); [/asy]$ $\\textbf{(A) } \\frac{1}{3} \\qquad\\textbf{(B) } \\frac{4}{9} \\qquad\\textbf{(C) } \\frac{1}{2} \\qquad\\textbf{(D) } \\frac{5}{9} \\qquad\\textbf{(E) } \\frac{2}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Problem #2199 2199 The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label(\"6\",(-0.6,1)); label(\"1\",(0.6,1)); label(\"2\",(0.6,-1)); label(\"9\",(-0.6,-1)); label(\"7\",(1.2,0)); label(\"3\",(-1.2,0)); label(\"pointer\",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label(\"1\",midpoint((4,-1)--(4,-2)),W); label(\"2\",midpoint((4,0)--(4,-1)),W); label(\"3\",midpoint((4,1)--(4,0)),W); label(\"4\",midpoint((4,2)--(4,1)),W); label(\"1\",midpoint((4,-2)--(5,-2)),S); label(\"2\",midpoint((5,-2)--(6,-2)),S); label(\"3\",midpoint((7,-2)--(6,-2)),S); [/asy]$ $\\textbf{(A) } \\frac{1}{3} \\qquad\\textbf{(B) } \\frac{4}{9} \\qquad\\textbf{(C) } \\frac{1}{2} \\qquad\\textbf{(D) } \\frac{5}{9} \\qquad\\textbf{(E) } \\frac{2}{3}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the", "necessary in later chapters. Still need set.seed(), as we almost # always will when creating random objects. set.seed(1) emp_dist_dice <- tibble(ID = 1:100) %>% mutate(die_1 = map_dbl(ID, ~ sample(c(1:6), size = 1))) %>% mutate(die_2 = map_dbl(ID, ~ sample(c(1:6), size = 1))) %>% mutate(sum = die_1 + die_2) %>% ggplot(aes(x = sum)) + geom_histogram(aes(y = after_stat(count/sum(count))), binwidth = 1, color = \"white\") + labs(title = \"Empirical Probability Distribution\", subtitle = \"Sum from rolling two dice, replicated one hundred times\", x = \"Outcome\\nSum of Two Die\", y = \"Probability\") + scale_x_continuous(breaks = seq(2, 12, 1), labels = 2:12) + scale_y_continuous(labels = scales::percent_format(accuracy = 1)) + theme_classic() emp_dist_dice We might consider labeling the y-axis in plots of empirical distributions as “Proportion” rather than “Probability” since it is an actual proportion, calculated from real (or simulated) data. We will keep it as “Probability” since we want to emphasize the parallels between mathematical, empirical and posterior probability distributions. The posterior distribution for rolling two dice a hundred times depends on your expectations. If you take the dice from your Monopoly set, you have reason to believe that the assumptions underlying the mathematical distribution are true. However, if you walk into a crooked casino and a host asks you to play craps, you might be suspicious. In craps, a come-out roll of 7 and", "【全网首发】2021AIME I 真题及答案 【全网首发】2021AIME I 真题及答案 Problem1 Zou and Chou are practicing their 100-meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\\frac23$ if they won the previous race but only $\\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Problem2 In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.$[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ Problem3 Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ Problem4 Find the number of ways $66$ identical coins can be separated into three nonempty piles so that", "square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \\cdot \\frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\\boxed{{\\textbf{(C)}}~15}$. ### Solution 3 $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\\frac{x}{2}$ and $\\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\\left(\\frac{x}{2}\\right) + 4\\left(\\frac{y}{2}\\right)$. $(4)\\frac{x}{2} + (4)\\frac{y}{2}$ $\\Rightarrow~~4\\left(\\frac{x}{2}+ \\frac{y}{2}\\right)$ $\\Rightarrow~~4\\left(\\frac{x+y}{2}\\right)$ Remember that $x+y=3$, so substituting this in we find that the area of all of the triangles is $4\\left(\\frac{3}{2}\\right) = 6$. The area of the $4$ unit squares is $4$, so the area of the square we need is $25- (4+6) = \\boxed{\\textbf{(C)}~15}$", "pair R = (1.5*A + 3.5*C)/5, S = extension(A0,R,A,B), T = extension(C0,R,B,C); pair P2 = extension(A0,T,B0,R), M = extension(A,B,A0,C), N = extension(A,C0,B,C); /* Draw a couple of the common paths */ D(A--B--C--cycle); D(incircle(A,B,C)); /* Label some of the common points */ D(MP(\"A\", A, NE)); D(MP(\"B\", B)); D(MP(\"C\", C)); D(MP(\"I\", I, W)); Out[2]: 1) (a) Extend $A_{1}X$ to intersect $AC$ at $V$. Show that the lines $AA_{1}$, $BV$, and $DE$ are concurrent. (b) Let $A_{2}$ be the intersection point of $AX$ with the sideline $BC$. Show that $$\\frac{A_2B}{A_2C} = \\frac{s-c}{s-b} \\cdot \\frac{A_1B^2}{A_1C^2},$$ where $s = \\frac{a + b + c}{2}$ is the semiperimeter of $\\triangle ABC$. [Hint: Menelaus's Theorem may come in handy.] (c) Prove that lines $AX, BY, CZ$ are concurrent. In [3]: %%asy pumac_power_round_2011_config.asy size(500); /* Draw lines */ D(A--A1--X1, darkgreen); D(B--B1--Y1, darkgreen); D(C--C1--Z1, darkgreen); D(A--P1--B, darkred); D(C--Z1, darkred); D(X1--V); D(P1--A2, darkred); /* Label points */ D(MP(\"P\",P,dir(60))); D(MP(\"D\",D)); D(MP(\"E\",E,NE)); D(MP(\"F\",F,NW)); D(MP(\"X\",X1,ENE)); D(MP(\"Y\",Y1,1.4*dir(150))); D(MP(\"Z\",Z1,NE)); D(P1); D(MP(\"A_1\",A1)); D(MP(\"B_1\",B1,NE)); D(MP(\"C_1\",C1,NW)); D(MP(\"V\",V,NE)); D(MP(\"A_2\",A2)); Out[3]: 2) Show that the lines $B_1C_1$, $EF$, $YZ$ concur at a point, say $A_{0}$. Similarly, the lines $C_1A_1$, $FD$, $ZX$ and $A_1B_1$, $DE$, $XY$ concur at points $B_{0}$ and $C_{0}$, respectively. [Hint: Pascal's theorem might be useful here.] In [4]: %%asy pumac_power_round_2011_config.asy size(600); /* Draw lines */ D(Y1--A0--E); D(F--D); D(A1--C1); D(B1--C0--E); D(D--E--F--cycle,rgb(0,0.7,0)+linewidth(1)); D(A0--B0--C0--cycle, darkblue); D(A--B0--C, darkblue+linetype(\"3 3\"));", "as four parts. Question 2. Use the game board above to find the area of each color in square inches. Which color has the greatest area? Explanation: Blue color has 4 inches length in to breadth. L*B. 2*2=4 Area of the blue surface is4 into no of blocks is 4*12 is 24. Question 3. Design a spinner that has 8 equal parts. One part is red, two parts are blue, one part is yellow, and the rest is green. a. What fraction of the spinner is green? b. Which colour are you most likely to spin ? Explain ### Understand Fractions Activity Fraction Spin and Cover Directions: 1. Take turns using the spinners to find which fraction model to cover. 2. Use a counter to cover the fraction model. 3. Repeat this process until you cover all of the models. 4. The player with the most fraction models covered wins! ### Understand Fractions Chapter Practice 10.1 Equal Parts of a Whole Tell whether the shape shows equal parts or unequal parts. If the shape shows equal parts, then name them. Question 1. Explanation: the above rectangle is divided into 6 equal parts. and name sixths. Question 2. Explanation: The above triangle is divided into 4 equal parts which are equal in size. they name as fourths. Question 3.", "# 2007 AMC 10B Problems/Problem 19 ## Problem The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label(\"6\",(-0.6,1)); label(\"1\",(0.6,1)); label(\"2\",(0.6,-1)); label(\"9\",(-0.6,-1)); label(\"7\",(1.2,0)); label(\"3\",(-1.2,0)); label(\"pointer\",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label(\"1\",midpoint((4,-1)--(4,-2)),W); label(\"2\",midpoint((4,0)--(4,-1)),W); label(\"3\",midpoint((4,1)--(4,0)),W); label(\"4\",midpoint((4,2)--(4,1)),W); label(\"1\",midpoint((4,-2)--(5,-2)),S); label(\"2\",midpoint((5,-2)--(6,-2)),S); label(\"3\",midpoint((7,-2)--(6,-2)),S); [/asy]$ $\\textbf{(A) } \\frac{1}{3} \\qquad\\textbf{(B) } \\frac{4}{9} \\qquad\\textbf{(C) } \\frac{1}{2} \\qquad\\textbf{(D) } \\frac{5}{9} \\qquad\\textbf{(E) } \\frac{2}{3}$ ## Solutions ### Solution 1 When dividing each number on the wheel by $4,$ the remainders are $1, 1, 2, 2, 3,$ and $3.$ Each column on the checkerboard is equally likely to be chosen. When dividing each number on the wheel by $5,$ the remainders are $1, 1, 2, 2, 3,$ and $4.$ The probability that a shaded square in the $1$st or $3$rd row of the $1$st or $3$rd column is chosen is $$\\frac{2}{3} \\times \\frac{3}{6} = \\frac{1}{3}$$", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$." ]

[ "triangles, no interior angle an equation where the sum of the three of.", "What is the measure of the missing angle $x$ from the triangle shown below? 2. Suppose that the triangle’s interior angles have measures of three consecutive even integers. Which of the following shows the measures of the three interior angles? 3. Suppose that the triangle’s interior angles have the following measures: $x^{\\circ}$, $(x – 20)^{\\circ}$, and $(2x + 40)^{\\circ}$. Which of the following shows the value of $x$? 4. The ratio of the measure of the triangle’s interior angles is $2:4:9$. Which of the following shows the measure of the largest angle? 5. Which of the following shows the value of $x + y$?", "triangle are fraction ) given Find 3 measure of the third angle... -- 0.021196--", "# Unknown Interior Angles Geometry Level 1 The three interior angles in a triangle (in degrees) are $$x$$, $$y$$, and $$x+y$$. What is the measure of the largest angle in the triangle (in degrees)? ×", "Q15 What is the measure of each interior angle of an equilateral triangle? 900 600 450 1800 60s Edit Delete Teachers give this quiz to your class", "As shown in the triangle Angle-Sum Theorem gives the relationship among the interior angles a.", "Back ## Triangles in a Triangle The measure of the exterior angle $c$ is always the sum of the measures of $a$ and $b:$ To see how we know this and learn a couple of other useful facts, keep reading. Or, you can skip ahead to the challenge if you feel ready.", "triangle is equal to sum of two of the remote interior angles of the triangle.)", "of a triangle have measures (2x−5)° , (4x−1)° , and 30°. what is the value of x? The interior angles of a triangle have measures (2x−5)° , (4x−1)° , and 30°. what is the value of x?...", "2. 3. 4. 5. 6. 7. 8. Two interior angles of a triangle measure $$32^{\\circ}$$ and $$64^{\\circ}$$. What is the third interior angle of the triangle? 9. Two interior angles of a triangle measure $$111^{\\circ}$$ and $$12^{\\circ}$$. What is the third interior angle of the triangle? 10. Two interior angles of a triangle measure $$2^{\\circ}$$ and $$157^{\\circ}$$. What is the third interior angle of the triangle? Find the value of $$x$$ and the measure of each angle. 11. 12. 13. 14. 15. ## Vocabulary Term Definition Triangle Sum Theorem The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees. Interactive Element Video: Triangle Sum Theorem Principles - Basic Activities: Triangle Sum Theorem Discussion Questions Study Aids: Triangle Relationships Study Guide Practice: Triangle Angle Sum Theorem Real World: Triangle Sum Theorem", "of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 9-11. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 12-14. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 15-17. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 18-20. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? ### Answers for Explore More Problems To view the Explore More answers, open this PDF file and look for section 5.2. ### Vocabulary Language: English Side Side", "a triangle? Yes No 2. The three angles of a triangle measure $$7^\\circ$$, $$164^\\circ$$, and $$9^\\circ$$. What kind of triangle is it? 3. All three angles of a triangle are equal. What is the measure of each angle? 4. At most, how many obtuse angles can a triangle have?", "triangle then verify that sum of measures of two sides of a triangle is greater than the third side.", "Triangle and its properties - Grade 7 (Question: 1621349452) If two angles in a triangle add up to $103^0$, then find the value of the third angle.", "angle measures.", "? Free Version Easy # Exterior Angles Triangle GEOM-YJSXGV What is the sum of the measures of the exterior angles of a triangle? A $360 ^{\\circ}$ B $180 ^{\\circ}$ C $270 ^{\\circ}$ D $90 ^{\\circ}$", "chord CD [Interior angles test]", "is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 15-17. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle? Use the triangle below to answer questions 18-20. 1. What is the measure of the smallest angle of the triangle? 2. What is the measure of the largest angle of the triangle? 3. What is the measure of the third angle of the triangle?", "sum of interior angles of a square #### Quantity B The sum of any four interior angles of a regular pentagon If the perimeter of a right-angled isosceles triangle is (1+$\\sqrt{2}$), then the area of the triangle is?", "## Calculus (3rd Edition) Since the sum of the interior angles of a triangle is $180$, then in a right triangle any other angle $\\theta$ should satisfy $$0\\lt\\theta\\lt \\pi/2$$ so the right answer is (a)." ]

[ "Browse Questions # 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$ Toolbox: • A sentence is a statement if it is true or false always. The sum of all interior angles of a triangle is $180^\\circ$. This sentence is a statement because it is always true.", "## anonymous one year ago What is the mean of the set of numbers? {–8, –4, –3, –11} A. –6.5 B. –4.5 C. 4.5 D. 6.5 1. whpalmer4 The (arithmetic) mean of a set of numbers is the sum of all those numbers, divided by the number of numbers. 2. texaschic101 add them all up, then divide by how many numbers there are 3. anonymous Go to wolframalpha.com 4. whpalmer4 so the mean of $$2,4,6$$ is $\\frac{2+4+6}{3} = 4$", "sum of the angles on the line drawn by extending one of the triangle's sides.", "Question # Prove that the sum of all angles of a triangle is $180^\\circ$ . In this question we have to use construction and use the properties of parallel lines . Use substitution of angles of the triangle with the angles on the straight line to get to the final answer . $\\angle ABC = \\angle EAB$ ( alternate angles are equal as lines BC and EF are parallel ) $\\angle BCA = \\angle FAC$ ( alternate angles ) Now we know that the sum of all linear angles is $180^\\circ$ Therefore $\\angle EAB + \\angle BAC + \\angle FAC = 180^\\circ$ $\\angle ABC + \\angle CAB + \\angle ACB = 180^\\circ$ Hence proved the sum of all angles of a triangle is $180^\\circ$", "$$360^{\\circ}$$.\" ### 6. What is the sum of 3 angles of a triangle? The sum of 3 angles of a triangle is $$180^{\\circ}$$. ### 7. What is the sum of all angles of a triangle? The sum of all angles of a triangle is $$180^{\\circ}$$. More Important Topics Numbers Algebra Geometry Measurement Money Data Trigonometry Calculus", "sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Practical Problems on Relations and Functions in Set Theory, Relations and Functions Class 11 Questions. 3. 150; 149; 147; 148 Example 2:Find the average of 56, 41, 59, 52, 42 and 44. 4. Then he (instantaneously) increased his speed and, for the next hour, kept it at 30 miles per hour. find the value of x. Mean ADVERTISEMENTS: Continuous series means where frequencies are given along with the value of the variable in the form of class intervals. Arithmetic", "The mean of a set of numbers $x_1, x_2 \\ldots x_n$ is the unique value $\\overline{x}$ which minimizes $\\sum (\\overline{x} - x_i)^2$, the sum of squared distances between it and the numbers. Why? The mean seems very natural — for example, it's related to intuitive ideas about fairness — but the sum of squares less so. For instance, what's so special about squaring? Why not the sum of fourth powers $\\sum (\\overline{x} - x_i)^4$ ? Or the sum of unsigned distances $\\sum |\\overline{x} - x_i|$ ?", "## Friday, November 15, 2002 ### Experience Mathematics # 20 -- The sum of angles in a triangle Euclidean or plane geometry begins with notions of points and lines, and the notion that a point lies on a line. Think of lines as sets, and a point as an element belonging to a set. Points and lines satisfy certain axioms. In Euclidean Geometry (or plane geometry), the axioms are based on Euclid’s original axioms. From these axioms, we can use the rules of logic to derive theorems (or propositions) that can be regarded as truthful statements that apply to the plane. Here a plane is a model, or a mini-universe where those axioms and theorems hold. For example, consider the theorem: The sum of angles in a triangle is $180$ degrees. The various terms in this theorem (angle, triangle etc.) are constructs in the plane that we wish to study. The theorem itself is a property that will hold in our mini-universe. The proof should proceed from the axioms, use the definitions of the various constructs, and follow the rules of logic. Even though the theorem is true, it does not imply that the sum all triangles is $180$ degrees. For example, consider the surface of the earth. Draw a triangle with a right angle at the North", "## Friday, November 15, 2002 ### Experience Mathematics # 20 -- The sum of angles in a triangle Euclidean or plane geometry begins with notions of points and lines, and the notion that a point lies on a line. Think of lines as sets, and a point as an element belonging to a set. Points and lines satisfy certain axioms. In Euclidean Geometry (or plane geometry), the axioms are based on Euclid’s original axioms. From these axioms, we can use the rules of logic to derive theorems (or propositions) that can be regarded as truthful statements that apply to the plane. Here a plane is a model, or a mini-universe where those axioms and theorems hold. For example, consider the theorem: The sum of angles in a triangle is $180$ degrees. The various terms in this theorem (angle, triangle etc.) are constructs in the plane that we wish to study. The theorem itself is a property that will hold in our mini-universe. The proof should proceed from the axioms, use the definitions of the various constructs, and follow the rules of logic. Even though the theorem is true, it does not imply that the sum all triangles is $180$ degrees. For example, consider the surface of the earth. Draw a triangle with a right angle at the North", "## Thinking Mathematically (6th Edition) The mean is the sum of the values given divided by the number of values given. There are six values: 3, 6, 2, 1, 7, 3 The mean is: $\\frac{3 + 6 + 2 + 1 + 7 + 3}{6}$ = $\\frac{22}{6}$ $\\approx$ 3.67", "# Angle sum property of triangle Angle sum property of triangle is one of the most important concepts in triangles. The Angle sum property of triangle says that for any given triangle the sum of all the three angles is equal to ${ 180 }^{ 0 }$. We can prove Angle sum property of triangle in so many different ways. In that, I will discuss some important and easy proofs. Out of these proofs whatever you are feeling easy that proof you can learn. But if you learn all different kinds of proofs it will be good. ## One of the Proof of Angle sum property of triangle: We know that the exterior angle of any given triangle is equal to the sum of opposite interior angles of a triangle. ∠ACD = ∠A + ∠B Now consider the straight line BCD. We know that the total sum of angles on a straight line is equal to ${ 180 }^{ 0 }$. ∠BCA + ∠ACD = ${ 180 }^{ 0 }$. Therefore, ∠C + ∠A + ∠B = ${ 180 }^{ 0 }$. In order, if you write then we will get ∠A + ∠B + ∠C = ${ 180 }^{ 0 }$. Hence it is proved. ### The Practical proof of Angle sum property of triangle: All the", "the original triangle and ____________ . Question 2. WRITING In your own words, explain geometric mean. The geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values. Monitoring progress and Modeling with Mathematics In Exercises 3 and 4, identify the similar triangles. Question 3. Question 4. △LKM ~ △LMN ~ △MKN In Exercises 5 – 10, find the value of x. Question 5. Question 6. x = 9.6 Explanation: $$\\frac { QR }{ SR }$$ = $$\\frac { SR }{ TS }$$ $$\\frac { 20 }{ 16 }$$ = $$\\frac { 12 }{ x }$$ 1.25 = $$\\frac { 12 }{ x }$$ x = 9.6 Question 7. Question 8. x = 14.11 Explanation: $$\\frac { AB }{ AC }$$ = $$\\frac { BD }{ BC }$$ $$\\frac { 16 }{ 34 }$$ = $$\\frac { x }{ 30 }$$ x = 14.11 Question 9. Question 10. x = 2.77 Explanation: $$\\frac { 5.8 }{ 3.5 }$$ = $$\\frac { 4.6 }{ x }$$ x = 2.77 In Exercises 11 – 18, find the geometric mean of the two numbers. Question 11. 8 and 32 Question 12. 9 and 16 x = √(9 x 16) x = 12 Question 13. 14 and 20 Question", "Euclid's proposition $32$ consists of two parts, the first of which is External Angle of Triangle equals Sum of other Internal Angles, and the second part of which is this.", "# 3. What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle? As we can see they both angles forms a straight line, Hence the sum of an exterior angle of a triangle and its adjacent interior angle is always $180^0$.", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Practical Problems on Relations and Functions in Set Theory, Relations and Functions Class 11 Questions. This message, it was detected that a number when given a table a number when given a table number. 73, 84, 91, 87, 77, 94 only if every number the. Observations is the lower and [ … ] Statistics - arithmetic mean of the enormous organized data B. Rate, percent, and measurement problems, but no calculators are allowed on the arithmetic mean of the", "# What is the mean of 24, 36, 48 and 60? Apr 9, 2018 #### Explanation: The mean is the average of numbers, so you have to add all the numbers together and divide all of that by the the total numbers there are. Apr 9, 2018 The mean of a set of numbers is the sum of all the numbers divided by the number of numbers. The mean of $24$, $36$, $48$, and $60$ is: $\\frac{24 + 36 + 48 + 60}{4}$ $= \\frac{168}{4}$ $= 42$", "# Triangle Angle Sum Theorem The Triangle Angle Sum Theorem states that for any triangle lying in the Euclidean plane that the sum of the angles is equal to $180^o$ or $\\pi$ radians. This is one of the facts that characterizes Euclidean geometry: it is not, however, true in other geometries. The proof for this theorem is reliant on the parallel postulate, so does not generalize into other geometries. For example, any triangle constructed on the surface of the earth (known as a spherical triangle) the sum of the angles is always greater than $180^o$.", "is that the sum of all angles on a straight line is ${180^\\circ}$.", "# What Is the Sum of Angles of a Triangle? You know that a triangle is a geometric figure with three sides (edges) and three corners (angles). That’s how it got its name! If you use a protractor and measure all the angles in a triangle, you will find that the sum of the angles is always equal to $180$°. Rule ### SumofAnglesofaTriangle The sum of all three angles in a triangle is always $180$°. There are four types of triangles: equilateral triangles, isosceles triangles, right triangles and all other triangles that can’t be categorized in any of the previous three ways— we call these types of triangles scalene triangles. These types of triangles have three different angles and therefore, also three different side lengths." ]

[ "Browse Questions # 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$ Toolbox: • A sentence is a statement if it is true or false always. The sum of all interior angles of a triangle is $180^\\circ$. This sentence is a statement because it is always true.", "# 3. What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle? As we can see they both angles forms a straight line, Hence the sum of an exterior angle of a triangle and its adjacent interior angle is always $180^0$.", "# Linking Triangle Angle Sum and Inscribed Angle Theorems We have shown that the angle sum of a triangle is $180^\\circ$. We have also shown that the measure of an inscribed angle is half the measure the central angle that intercepts the same arc. In this post, we use the Inscribed Angle Theorem to show that the Triangle Angle Sum Theorem holds. Theorem The angle sum of a triangle is $180^\\circ$. Proof Consider the figures above. In the first figure, the triangle is divided into three central angles. Clearly the three angles add up to a complete rotation about the center so measure of blue angle + measure of red angle + measure of green angle = $360^\\circ$. In the second figure, the colored angles are inscribed angles. The measure of each angle is half the measure of its corresponding central angle (the angle of the same color) in the first figure (see also third figure). That is, the measure of the blue angle in the second figure is half the measure of the blue angle in the first figure. This means that the sum of measures of the inscribed angles equal to 1/2(measure of blue angle) + 1/2(measure of red angle) + 1/2(measure of green angle) of the angles in the first figure or 1/2(measure of blue", "is . #### Example 4 Explain why the Third Angle Theorem works. The Third Angle Theorem is really like an extension of the Triangle Sum Theorem. Once you know two angles in a triangle, you automatically know the third because of the Triangle Sum Theorem. This means that if you have two triangles with two pairs of angles congruent between them, when you use the Triangle Sum Theorem on each triangle to come up with the third angle you will get the same answer both times. Therefore, the third pair of angles must also be congruent. #### Example 5 Determine the measure of all the angles in the triangle: First we can see that . This means that also because they are alternate interior angles. was given. This means by the Triangle Sum Theorem that . This means that also because they are alternate interior angles. Finally, by the Triangle Sum Theorem. ### Review Determine the measures of the unknown angles. You may assume that . ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish Triangle Sum Theorem The Triangle Sum Theorem states that the measure of the three interior angles of any triangle will add up to $180^\\circ$. Third Angle Theorem If two angles in one triangle are congruent to", "# Triangles For any triangle the 3 angles add up to $$180^\\circ$$ $x^\\circ = 180^\\circ - (70^\\circ + 50^\\circ )$ $= 180^\\circ - 120^\\circ = 60^\\circ$ ## Equilateral triangle An equilateral triangle is one with all 3 sides equal in length and all 3 angles equal to $$60^\\circ$$ ## Isosceles triangle An isosceles triangle is one with two sides equal in length and two equal angles. Question In the diagram below the triangle is isosceles. What is the value of $$a^\\circ$$? $a^\\circ = 180^\\circ - (70^\\circ + 70^\\circ )$ $= 180^\\circ - 140^\\circ = 40^\\circ$", "## Trigonometry (11th Edition) Clone The sum of the three angles of a triangle add to $180^{\\circ}$, and angle $ABD = 150^{\\circ}$. Therefore the sum of angle $BAD$ and angle $BDA$ add to $30^{\\circ}$ Since triangle $ABD$ is an isosceles triangle and $AB = BD$, then angle $BAD$ is equal to angle $BDA$. Therefore, angle $BDA = 15^{\\circ}$, and angle $BAD = 15^{\\circ}$ $AB$ is a radius of the circle and $BD$ is a radius of the circle. Therefore, $AB = BD$. In the triangle $ABD$, since the two sides $AB$ and $BD$ are equal, the triangle $ABD$ is an isosceles triangle. The sum of the three angles of a triangle add to $180^{\\circ}$, and angle $ABD = 150^{\\circ}$. Therefore the sum of angle $BAD$ and angle $BDA$ add to $30^{\\circ}$ Since triangle $ABD$ is an isosceles triangle and $AB = BD$, then angle $BAD$ is equal to angle $BDA$. Therefore, angle $BDA = 15^{\\circ}$, and angle $BAD = 15^{\\circ}$", "Answer Comment Share Q) 'The sum of all interior angles of a triangle is $180^\\circ$.' Is this sentence a statement? Give reason. $\\begin{array}{1 1} yes \\\\ no \\end{array}$", "n th geometric means. Chapter 3 Assignments 43 3 Inside Out Triangle Sum, Exterior Angle, and Exterior Angle Inequality Theorems 1. Possible answer: /A and /B 15.", "# Proving the Sum of the Interior Angle Measures of a Triangle Is 180 Proving the Sum of the Interior Angle Measures of a Triangle Is 180 $Proving the Sum of the Interior Angle Measures of a Triangle Is 180$ ## This Post Has 3 Comments 1. froyg1234 says: What Do you need help? 2. Expert says: that is the sum step-by-step explanation: there are no other answers without knowing the variables values 3. Expert says:", "# Angle sum property of triangle Angle sum property of triangle is one of the most important concepts in triangles. The Angle sum property of triangle says that for any given triangle the sum of all the three angles is equal to ${ 180 }^{ 0 }$. We can prove Angle sum property of triangle in so many different ways. In that, I will discuss some important and easy proofs. Out of these proofs whatever you are feeling easy that proof you can learn. But if you learn all different kinds of proofs it will be good. ## One of the Proof of Angle sum property of triangle: We know that the exterior angle of any given triangle is equal to the sum of opposite interior angles of a triangle. ∠ACD = ∠A + ∠B Now consider the straight line BCD. We know that the total sum of angles on a straight line is equal to ${ 180 }^{ 0 }$. ∠BCA + ∠ACD = ${ 180 }^{ 0 }$. Therefore, ∠C + ∠A + ∠B = ${ 180 }^{ 0 }$. In order, if you write then we will get ∠A + ∠B + ∠C = ${ 180 }^{ 0 }$. Hence it is proved. ### The Practical proof of Angle sum property of triangle: All the", "Question # Prove that the sum of all angles of a triangle is $180^\\circ$ . In this question we have to use construction and use the properties of parallel lines . Use substitution of angles of the triangle with the angles on the straight line to get to the final answer . $\\angle ABC = \\angle EAB$ ( alternate angles are equal as lines BC and EF are parallel ) $\\angle BCA = \\angle FAC$ ( alternate angles ) Now we know that the sum of all linear angles is $180^\\circ$ Therefore $\\angle EAB + \\angle BAC + \\angle FAC = 180^\\circ$ $\\angle ABC + \\angle CAB + \\angle ACB = 180^\\circ$ Hence proved the sum of all angles of a triangle is $180^\\circ$", "2. 3. 4. 5. 6. 7. 8. Two interior angles of a triangle measure $$32^{\\circ}$$ and $$64^{\\circ}$$. What is the third interior angle of the triangle? 9. Two interior angles of a triangle measure $$111^{\\circ}$$ and $$12^{\\circ}$$. What is the third interior angle of the triangle? 10. Two interior angles of a triangle measure $$2^{\\circ}$$ and $$157^{\\circ}$$. What is the third interior angle of the triangle? Find the value of $$x$$ and the measure of each angle. 11. 12. 13. 14. 15. ## Vocabulary Term Definition Triangle Sum Theorem The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees. Interactive Element Video: Triangle Sum Theorem Principles - Basic Activities: Triangle Sum Theorem Discussion Questions Study Aids: Triangle Relationships Study Guide Practice: Triangle Angle Sum Theorem Real World: Triangle Sum Theorem", "$$360^{\\circ}$$.\" ### 6. What is the sum of 3 angles of a triangle? The sum of 3 angles of a triangle is $$180^{\\circ}$$. ### 7. What is the sum of all angles of a triangle? The sum of all angles of a triangle is $$180^{\\circ}$$. More Important Topics Numbers Algebra Geometry Measurement Money Data Trigonometry Calculus", "### Theory: Let's see the remarkable property that connects three angles of a triangle. The sum of the measure of three angles of a triangle is 180°. Example: Consider a triangle $$ABC$$ with interior angles measures $$∠1$$, $$∠2$$ and $$∠3$$. By the sum of measure of angles of triangle property, $$∠1 + ∠2 + ∠3 = 180°$$. Important! In a triangle: • If the measure of all angles are different, then all sides are different. • If the measure of two angles are equal, then two sides are equal. • If the measure of three angles are equal, then three sides are equal and each angle measures $$60°$$. • Sum of three angles of a triangle is $$180°$$.", "sum of the angles on the line drawn by extending one of the triangle's sides.", "A geometry problem by Brilliant Member Geometry Level 3 The ratio of the interior angles of a triangle is $$1:2:6$$. If the perimeter of the triangle is $$24$$ feet. Find the area of the triangle in square feet rounded to the nearest integer. ×", ". If the measures of all angles are less than $90^\\circ$ , then it is an acute triangle . If the measure of one angle is greater than $90^\\circ$ , then it is an obtuse triangle . The sum of the measures of the interior angles of any triangle is $180^\\circ$ . If the three angles of a triangle are all the same, then the triangle is an equiangular triangle and each angle measure is $60^\\circ$ . Equilateral triangles are always equiangular and vice versa. In fact, the number of sides that are the same length will always correspond to the number of angles that are the same measure. Example A Classify the triangle below by its sides and angles. Solution: This triangle has two sides that are congruent (the same length), so it is isosceles. It also has one angle that is greater than $90^\\circ$ , so it is obtuse. Example B The measures of two angles of a triangle are $30^\\circ$ and $60^\\circ$ . What type of triangle is it? Solution: You know that the measures of the three angles must add up to $180^\\circ$ . This means that the measure of the third angle is $180^\\circ-60^\\circ-30^\\circ=90^\\circ$ , so it is a right triangle. Because all of the angles are different measures, all of the sides", "## Calculus (3rd Edition) Since the sum of the interior angles of a triangle is $180$, then in a right triangle any other angle $\\theta$ should satisfy $$0\\lt\\theta\\lt \\pi/2$$ so the right answer is (a).", "? Free Version Easy # Exterior Angles Triangle GEOM-YJSXGV What is the sum of the measures of the exterior angles of a triangle? A $360 ^{\\circ}$ B $180 ^{\\circ}$ C $270 ^{\\circ}$ D $90 ^{\\circ}$", "Mathematics # State true or false.The sum of interior angles of a triangle is ${ 180 }^{ \\circ }$. True ##### SOLUTION It is true that the sum of interior angles of a triangle is ${ 180 }^{ \\circ }$. You're just one step away TRUE/FALSE Medium Published on 09th 09, 2020 Questions 120418 Subjects 10 Chapters 88 Enrolled Students 87 #### Realted Questions Q1 TRUE/FALSE Medium Identify and match the following : • A. True • B. False Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q2 Single Correct Medium If two lines intersect such that four vertical angles are equal, then each angle is: • A. $45^{\\circ}$ • B. $100^{\\circ}$ • C. $180^{\\circ}$ • D. $90^{\\circ}$ Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q3 Single Correct Medium An exterior angle of a triangle is equal to the sum of the two interior opposite angles. • A. False • B. Aambiguous • C. Data Insufficient • D. True Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published on 09th 09, 2020 Q4 Subjective Medium Write the measure of supplementary angle of the given angles: $132^{\\circ}$. Asked in: Mathematics - Lines and Angles 1 Verified Answer | Published" ]

[ "# The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2 oranges is 285 gms. What is 1 apple and 1 orange weight individually ? $${}$$", "Pizza Portions My friends and I love pizza. Can you help us share these pizzas equally? Pies Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie? Red Balloons, Blue Balloons Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst? Orange Drink Stage: 2 Challenge Level: This is a $750$ ml bottle of concentrated orange squash. It is enough to make fifteen $250$ ml glasses of diluted orange drink. How much water is needed to make $10$ litres of this drink?", "of 890 grams on a balance scale. He balances the scale by placing two oranges, an apple, and a lemon on the other side. Each orange weighs 280 grams. The lemon weighs 195 grams less than each orange. So the lemon weighs 280 g – 195 g = 85 grams, Now weigh of two oranges is 280 g + 280 g = 560 grams, given 2 oranges + lemon + apple = 890 grams, therefore weigh of apple = 890 g – (560g + 85 g) = 890 g – 645 g = 245 g, therefore, the mass of the apple is 245 g, modeled problem with a tape diagram as shown above and statement is mass of apple is two hundred fortyy five grams. Question 2. Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters shorter than Brian. Betina is 26 centimeters taller than Bonnie. How tall is Betina? Betina is 155 cm or 1 m 55 cm tall, Statement: Betina is one hundred fifty five centimeters or one meter fifty five centimeters tall, Explanation: Given Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters shorter than Brian means Bonnie is 1 m 87 cm – 58 = as 1 m 87 cm = 1 X 100 cm + 87 cm = 187", "So equating these we get $$6a = 5o$$ Thus we see that $6$ apples weights as much as $5$ oranges. - Thank you ever so much, really appreciate it. Now I see where I was going wrong. – Jay Oct 30 '12 at 19:09", "The oranges all weigh $130$ grams. The lemons are also all the same weight, which is less than $\\frac{2}{3}$ of the weight of an orange.", "× 0.51 > 1 Its false 2 × 0.51 > 1 Explanation: 2 × 0.51 = 1.02. Question 7. 4 × 0.49 > 2 Its true 4 × 0.49 > 2. Explanation: 4 × 0.49 = 1.96 Question 8. 5 × 0.97 > 5 Its false 5 × 0.97 > 5. Explanation: 5 × 0.97 = 4.85. Question 9. Jan goes to the grocery store to buy apples and oranges. Jan buys 5 apples. Is it reasonable that the apples weigh less than 5 pounds? Explain how you know. Yes, its reasonable that the apples weigh less than 5 pounds because Weight of 5 apples = 3.71 lb. Explanation: Number of apples Jan buys = 5. Weight of each apple = 0.74 lb. Weight of 5 apples = Number of apples Jan buys × Weight of each apple = 5 × 0.74 lb = 3.70 lb. Jan buys 6 oranges. Is it reasonable that the oranges weigh less than 3 pounds? Explain how you know. Explanation: Number of oranges Jan buys = 6. Weight of each orange = 0.55 lb. Weight of 5 oranges = Number of oranges Jan buys × Weight of each orange = 5 × 0.55 lb = 2.75 lb. Question 10. Reason Suppose you multiply 4 and 0.____7. What are the possible values of", "## 'Oranges and Lemons' printed from http://nrich.maths.org/ The oranges all weigh $130$ grams. The lemons are also all the same weight, which is less than $\\frac{2}{3}$ of the weight of an orange.", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Stage: 2 Short Challenge Level: We had a solution to this problem from someone who didn't give their name or school - but thank you! From the picture of the scales, the apple must weigh less than the four weights altogether. If the apple weighs $180g$, for the scale to be balanced the weights would have to be $45g$ each because there are four of them. But we know from the picture that the apple weighs less than the weights put together, so each weight must be more than $45g$. If the apple weighed $375g$, then for the scale to be balanced, each weight would be $93.75g$. So, for the weights to be heavier than the apple, each", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Age 7 to 11 Short Challenge Level: The picture above shows four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights? If the apple weighs $180g$, how heavy must one weight be? If the apple weighed $375g$, how heavy would one weight be? If the apple was a giant one and weighed a full kilo and the weights were each $250g$, what would the scale look like? How do you know? Can you prove it?", "$= 0.2$ kg Joe eats three. So remaining apples $= 2$ Weight of two apples $= 0.2 \\times 2 = 0.4$ kg 3 ### Joe weighs 35 kg, Lily weighs 90 pounds, and Tom weighs 330,000 dg. Who weighs the most? Joe Lily Tom All are of equal weight CorrectIncorrect Lily’s weight $= 90$ pounds. We know that 1 lbs $= 0.45$ kg So, 90 lbs $= 90 \\times 0.45$ $= 40.5$ kg Tom’s weight $= 330,000$ dg According to the conversion table, 330,000 dg $= 330,000 \\div (10 \\times 10 \\times 10 \\times 10)$ kg $= 33$ kg Joe weighs 35 kg. Therefore, Lily weighs the most. 4 ### 250 ants lift 1 mg of breadcrumbs each and take it to their anthill. The total weight of the collected bread crumbs is: 0.00025 kg 0.0025 kg 0.025 kg 0.25 kg CorrectIncorrect Total weight of the breadcrumbs lifted $= 250 \\times 1$ $= 250$ mg From the conversion table, $250 \\div (10 \\times 10 \\times 10 \\times 10 \\times 10 \\times 10)$ $= 0.00025$ kg 5 ### In Joe’s schoolbag, the books weigh 1 kg. Her tiffin weighs a pound, and she carries a 1l bottle. In addition, she carries a file weighing 200 grams. The total weight of her school bag is approximate: 1.65 kg 2.65 kg", "the way shown. Can you see what the answers always have in common? How Much Did it Cost? Stage: 2 Challenge Level: Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether. What's My Weight? Stage: 2 Short Challenge Level: There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?", "# Free Practice 6.G.A.2 – 9005401 Jimmy is filling a rectangular box with sugar cubes. If the dimensions of the box are $12$\" by $16$\" by $12$\" and the sugar cubes have side length of $4$\", then how many sugar cubes can fit in the box? A. $12$ B. $144$ C. $36$ D. $48$ E. $576$", "# The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2 oranges is 285 gms. What is 1 apple and 1 orange weight individually ? ## Question 1 & 2 Question 1&2: The total weight is 255 gms for ... » All original content on these pages is fingerprinted and certified by Digiprove Insert math as $${}$$", "hold 50 oranges. If Bob needs to ship 932 oranges, how many crates will he need?", "+ 1 $$\\frac{5}{6}$$ = 7 $$\\frac{5}{6}$$ feet Thus the correct answer is option c. Question 2. Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. How many pounds of apples did Harry pick those three days? Options: a. 132 $$\\frac{3}{8}$$ pounds b. 141 $$\\frac{3}{8}$$ pounds c. 142 $$\\frac{1}{8}$$ pounds d. 142 $$\\frac{3}{8}$$ pounds Answer: 142 $$\\frac{3}{8}$$ pounds Explanation: Given, Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. 45 $$\\frac{7}{8}$$ + 42 $$\\frac{3}{8}$$ + 54 $$\\frac{1}{8}$$ 45 + 42 + 54 = 141 141 + $$\\frac{7}{8}$$ + $$\\frac{3}{8}$$ + $$\\frac{1}{8}$$ 141 + 1 $$\\frac{3}{8}$$ = 142 $$\\frac{3}{8}$$ pounds Thus the correct answer is option d. Spiral Review Question 3. There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. How many oranges were left? Options: a. 2 $$\\frac{1}{3}$$ oranges b. 2 $$\\frac{2}{3}$$ oranges c. 3 $$\\frac{1}{3}$$ oranges d. 9 $$\\frac{2}{3}$$ oranges Answer: 9 $$\\frac{2}{3}$$ oranges Explanation: Given, There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. 6 + 3", "## Elementary Technical Mathematics $g$. A typical apple weighs less than a pound, and one pound is 453 grams. Thus, the weight of most apples can be measured with units of grams or $g$.", "than wife i-e he ate x+5 carro... » Question 1 & 2 Question 1&2: The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2... » Insert math as $${}$$", "oranges, 25% bananas, 20% apples 57. 10 straw hats, 50 beanies, 40 cowboy hats 59. Tom ate 6, Joe ate 3, and Albert ate 3. 61. 124 oranges, 10 lemons, 8 pomegranates", "Check the bill and tell, did he pay equal money for both? Solution He got one watermelon for $$5$$ However, he got one pumpkin in $$10$$ As, $$5 \\neq 10$$ Hence, as per the bill, it can be told that he did not pay equal money for both. $$\\therefore$$ He paid unequal money. Example 2 William and Jill got their height's checked. This is how it looks. Do the heights of William and Jill look equal? Solution As per their heights visible in the scales, the height of William is more than the height of Jill. Hence, we can conclude that the heights of William and Jill are not equal. $$\\therefore$$ Their heights are unequal. ## Interactive Questions Here are a few activities for you to practice. Challenging Questions 1. If the price of 4 watermelons is exactly the same as the price of 5 apples, can Jane say that price of 6 watermelons is equal to the price of 10 apples? 2. Mike and his sister are in the same class. He scored 5 marks more than what his sister scored. If the sum of marks scored by Mike and his sister is 25. Is Mike's score equal to 20? ## Let's Summarize! We hope you enjoyed learning about unequal with the practice questions. Now you will", "equal number of candies Number of candies with Jimmy = 4 ✕ 3 = 12 Number of candies with Joe = 3 ✕ 4 = 12 Thus, both have an equal number of candies." ]

[ "into a ratio The fraction $\\frac{11}{14}$ can mean two possibilities • Two quantities are in $11 : 14$ ratio • one quantity is specified as $\\frac{11}{14}$ part of the whole. In the context of ratio, the fraction is taken to be in the form of comparison of two quantities, unless specified in the question. So, the answer is \"$11 : 14$\". The number of apples are $\\frac{2}{3}$ of the number of oranges. What is the ratio of apples to oranges? The given fraction specifies the comparison of apples to oranges. Number of apples is $\\frac{2}{3}$ of the number of oranges. Which means, for every orange, there are $\\frac{2}{3}$ apples. So the ratio of apples to oranges is $\\frac{2}{3} : 1$, which is equivalently $2 : 3$. So, the answer is \"$2 : 3$\". A basket has $\\frac{2}{3}$ apples and rest oranges. What is the ratio of apples to oranges? The given fraction specifies the quantity of apples. Quantity of apples $\\frac{2}{3}$ Quantity of oranges $1 - \\frac{2}{3} = \\frac{1}{3}$ So the ratio of apples to oranges is $\\frac{2}{3} : \\frac{1}{3}$ which is equivalently $2 : 1$. So the answer is \"$2 : 1$\". summary Ratio Vs Fraction : comparison of two quantities given as magnitude of one to magnitude of another is ratio. Fractions are also used to", "× 0.51 > 1 Its false 2 × 0.51 > 1 Explanation: 2 × 0.51 = 1.02. Question 7. 4 × 0.49 > 2 Its true 4 × 0.49 > 2. Explanation: 4 × 0.49 = 1.96 Question 8. 5 × 0.97 > 5 Its false 5 × 0.97 > 5. Explanation: 5 × 0.97 = 4.85. Question 9. Jan goes to the grocery store to buy apples and oranges. Jan buys 5 apples. Is it reasonable that the apples weigh less than 5 pounds? Explain how you know. Yes, its reasonable that the apples weigh less than 5 pounds because Weight of 5 apples = 3.71 lb. Explanation: Number of apples Jan buys = 5. Weight of each apple = 0.74 lb. Weight of 5 apples = Number of apples Jan buys × Weight of each apple = 5 × 0.74 lb = 3.70 lb. Jan buys 6 oranges. Is it reasonable that the oranges weigh less than 3 pounds? Explain how you know. Explanation: Number of oranges Jan buys = 6. Weight of each orange = 0.55 lb. Weight of 5 oranges = Number of oranges Jan buys × Weight of each orange = 5 × 0.55 lb = 2.75 lb. Question 10. Reason Suppose you multiply 4 and 0.____7. What are the possible values of", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Stage: 2 Short Challenge Level: We had a solution to this problem from someone who didn't give their name or school - but thank you! From the picture of the scales, the apple must weigh less than the four weights altogether. If the apple weighs $180g$, for the scale to be balanced the weights would have to be $45g$ each because there are four of them. But we know from the picture that the apple weighs less than the weights put together, so each weight must be more than $45g$. If the apple weighed $375g$, then for the scale to be balanced, each weight would be $93.75g$. So, for the weights to be heavier than the apple, each", "the fruits and the columns represent the number of each fruit. To compare the ratio of apples to oranges, we can create ratios by dividing the number of apples by the number of oranges or vice versa. You can use these examples as a guide, and adjust it to your specific lesson plan. Ratios tables are a great tool to help students understand the relationship between different values, and it can be applied to many different math concepts. ### What people say about \"Ratio Tables\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "4 small pizzas in each box. Mrs. Owens buys 8 boxes. How many pizzas does she have? 1. 4 2. 12 3. 24 4. 32 3 x 3 = 1. 6 2. 0 3. 9 Which expression is the same as 7 + 7 + 7? 1. 21 / 7 2. 3 x 7 3. 7 x 1 4. 21 - 7 Factors are the 2. numbers that you multiply Any number times 1 equals the same number. 1. True 2. False Sydney helps her mother with the garden for 7 days. They picked 3 tomatoes each day. How many tomatoes do they have in total? 1. 12 2. 21 3. 18 4. 73 $[ 5 xx 3 ] =$ Which array would you draw? 1. 5 rows of 3 objects 2. 3 rows of 5 objects 3. 15 objects in one row 4. 2 columns of five Marty bought 3 bags of oranges at the store. There were 9 oranges in each bag. How many oranges did Marty have all together? 1. 9 + 9 = 18 or 2 x 9 = 18 2. 3 x 3 = 9 or 3 + 3 + 3 = 9 3. 3 x 9 = 27 or 9 + 9 + 9 = 27 8 x 2 = 1. 16", "maths > commercial-arithmetics Ratio & Fraction Differences what you'll learn... overview Ratio and Fraction are two methods to specifying relative magnitude of quantities. Eg: 12$12$ and 24$24$ are in 1:2$1 : 2$ ratio OR 12$12$ is 12$\\frac{1}{2}$ of 24$24$. Understanding the differences in these two representation is explained. ratio vs fraction Consider $4$ apples and $6$ oranges. And there are pieces of a cake. The cake was cut into $6$ pieces and only $4$ pieces are remaining. • ratio of apples to oranges is $4 : 6$ which is equivalently $2 : 3$ A ratio represents comparison of two quantities given as the magnitude of one to the magnitude of another. • number of apples is $\\frac{4}{6}$ fraction of number of oranges. which is equivalently $\\frac{2}{3}$. A fraction can be used for comparing two quantities. • number of cakes is $\\frac{4}{6}$ fraction of the whole, which is equivalently $\\frac{2}{3}$. A fraction can be ised to specify a quantity. Note that the fraction can be used both for comparing quantities and to specify a quantity. context Consider $4$ apples and $6$ oranges. • ratio of apples to oranges is $4 : 6$ which is equivalently $2 : 3$ • ratio of apples to fruits is $4 : 10$ which is equivalently $2 : 5$ • number of apples is", "maths > commercial-arithmetics Introduction to Percentages what you'll learn... overview A ratio is represented with two numbers, eg: 3:4$3 : 4$. Comparing two different ratios involve some arithmetics. eg: 4:5$4 : 5$ and 3:4$3 : 4$, which one is larger in ratio? To simplify such comparison, the second term is standardized to a value, say 100$100$, then the given ratios are 80:100$80 : 100$ and 75:100$75 : 100$. It is easier to compare these ratios. Such standard representation is simplified by dropping the known number $100$. and the simplified form is \"percentage\". eg: and . problem with ratios Consider a fruit basket with apples, oranges, and bananas • The ratio of apples to bananas is $1 : 4$ • The ratio of oranges to bananas is $2 : 9$ It is not evident to figure out if apples or oranges are more in the basket from the given two ratios. The ratios are to be converted to similar ratios. Consider a fruit basket with apples, oranges, and bananas • The ratio of apples to bananas is $1 : 4$. This means that, there is $1$ apple for every $4$ bananas. Or $2$ apples for every $8$ bananas. • The ratio of oranges to bananas is $2 : 9$. This means that, there are $2$ oranges for every $9$", "part of the whole. In the context of ratio, the fraction is taken to be in the form of comparison of two quantities, unless specified in the question. Solved Exercise Problem: The number of apples are 2/3 of the number of oranges. What is the ratio of apples to oranges? • 2:3 • 2:3 • 2:1 The answer is \"2:3\". The given fraction specifies the comparison of apples to oranges. Number of apples is 2/3 of the number of oranges. Which means, for every orange, there are 2/3 apples. So the ratio of apples to oranges is 2/3 : 1, which is equivalently 2:3 Solved Exercise Problem: A basket has 2/3 apples and rest oranges. What is the ratio of apples to oranges? • 2:3 • 2:1 • 2:1 The answer is \"2:1\". The given fraction specifies the quantity of apples. Quantity of apples 2/3 Quantity of oranges 1 - 2/3 = 1/3 So the ratio of apples to oranges is 2/3 : 1/3 which is equivalently 2:1 slide-show version coming soon", "+ 1 $$\\frac{5}{6}$$ = 7 $$\\frac{5}{6}$$ feet Thus the correct answer is option c. Question 2. Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. How many pounds of apples did Harry pick those three days? Options: a. 132 $$\\frac{3}{8}$$ pounds b. 141 $$\\frac{3}{8}$$ pounds c. 142 $$\\frac{1}{8}$$ pounds d. 142 $$\\frac{3}{8}$$ pounds Answer: 142 $$\\frac{3}{8}$$ pounds Explanation: Given, Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. 45 $$\\frac{7}{8}$$ + 42 $$\\frac{3}{8}$$ + 54 $$\\frac{1}{8}$$ 45 + 42 + 54 = 141 141 + $$\\frac{7}{8}$$ + $$\\frac{3}{8}$$ + $$\\frac{1}{8}$$ 141 + 1 $$\\frac{3}{8}$$ = 142 $$\\frac{3}{8}$$ pounds Thus the correct answer is option d. Spiral Review Question 3. There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. How many oranges were left? Options: a. 2 $$\\frac{1}{3}$$ oranges b. 2 $$\\frac{2}{3}$$ oranges c. 3 $$\\frac{1}{3}$$ oranges d. 9 $$\\frac{2}{3}$$ oranges Answer: 9 $$\\frac{2}{3}$$ oranges Explanation: Given, There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. 6 + 3", "# At Pruitt's farm, the ratio of oranges to apples is 4 to 8. A total of 96 apples and oranges are grown on his farm. How do you find the number of oranges grown? Mar 28, 2018 No apples $\\setminus \\setminus \\setminus \\setminus 32$ No oranges $\\setminus \\setminus \\setminus 64$ #### Explanation: We have a ratio of $4 : 8$ this can be simplified by dividing by the greatest common factor, in this case 4. So we have: $1 : 2$ Add these together, and divide this into the total number of apples and oranges, which is 96. $1 + 2 = 3$ $\\frac{96}{3} = 32$ Number of apples$= 1 \\times 32 = 32$ Number of oranges$= 2 \\times 32 = 64$ Notice that this as a ratio is: $32 : 64$ This is the same as $1 : 2$", "eleven $44$ $4 \\times 12$ Four times twelve $48$ $4 \\times 13$ Four times thirteen $52$ $4 \\times 14$ Four times fourteen $56$ $4 \\times 15$ Four times fifteen $60$ $4 \\times 16$ Four times sixteen $64$ $4 \\times 17$ Four times seventeen $68$ $4 \\times 18$ Four times eighteen $72$ $4 \\times 19$ Four times nineteen $76$ $4 \\times 20$ Four times twenty $80$ This table also shows an interesting pattern. After the 10th multiple of 4, which is $4\\times 10 = 40$, the last digit of the answers of the next 10 multiples will be equal to the last digit of the first 10 multiples. For example, $4\\times 1= 4$ and $4\\times 11 =44$; the last digit of the 11th multiple is the same as the first multiple of 4. Similarly, $4\\times 3= 12$ and $4\\times 13= 52$, so the last digits of the 3rd and the 13th multiples of 4 are the same. The green digits show this pattern in the figure below. Example 1: Alex has six plastic bags. His father has given him two dozen oranges and asked him to put the oranges in the bags with the condition that all the bags should contain an equal number of oranges. Solution: Alex knows $1$ dozen = $12$. So $2$ dozen oranges will be:", "repeated ? List out the combinations and you will have your answer. – Shailesh Aug 26 '15 at 2:20 • The formula $2A=(A+B)-(B+C)+(C+D)-(D+E)+(E+A)$ might be useful, if you know which sums are which. (It follows because, distributing, we have $A+B-B-C+C+D-D-E+E+A$, and everything cancels out with its neighbor save for $A$.) – Akiva Weinberger Aug 26 '15 at 2:27 • Using my values — 'A' = 53.25, 'B' = 54.75, 'C' = 58.75, 'D' = 60.75, and 'E' = 61.25 — I was unable to provide combinations for 113, 115, 117, and 118. – Sentient Aug 26 '15 at 2:30 • @columbus8myhw I believe 'A' + 'B' = 108 and 'D' + 'E' = 122. These are the only sums I'm certain about. – Sentient Aug 26 '15 at 2:32 • I think it's unsolvable. – Akiva Weinberger Aug 26 '15 at 2:48 First, you’re right in thinking that the total weight is $288.75$; you can also get this by noticing that each pumpkin is weighed four times, and $\\frac{1155}4=288.75$. However, the data are inconsistent. The low total of $108$ must be the sum of the two lightest weights, so the three heaviest pumpkins altogether weigh $288.75-108=180.75$. The two heaviest together weigh $122$, so the middle pumpkin weighs $180.75-122=58.75$. Each pair of pumpkins has an integer total weight, so", "the total weight of the 5 oranges then $${\\text{E}}(X) = 5 \\times 205 = 1025$$ (A1) $${\\text{Var}}(X) = 5 \\times 100 = 500$$ (M1)(A1) $${\\text{P}}(X < 1000) = 0.132$$ A1 [4 marks] b. let Y = B – 3C where B is the weight of a random orange and C the weight of a random lemon (M1) $${\\text{E}}(Y) = 205 – 3 \\times 75 = – 20$$ (A1) $${\\text{Var}}(Y) = 100 + 9 \\times 9 = 181$$ (M1)(A1) $${\\text{P}}(Y > 0) = 0.0686$$ A1 [5 marks] Note: Award A1 for 0.0681 obtained from tables c. ## Examiners report As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and $$\\sum\\limits_{i = 1}^n {{X_i}}$$ . a. As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are", "Sep 2009 Posts: 54434 ### Show Tags 16 Sep 2014, 00:45 2 Official Solution: Because of weight limits, a lorry can carry a maximum of 20 crates of apples. The weight of a crate of oranges is $$\\frac{3}{4}$$ the weight of a crate of apples. If 10 crates of apples were loaded into the lorry, what is the maximum number of crates of oranges that the lorry can still accommodate? A. 8 B. 9 C. 12 D. 13 E. 14 There's still room for 10 crates of apples. The weight of 10 crates of apples is the same as the weight of $$10*(\\frac{4}{3}) = 13\\frac{1}{3}$$ crates of oranges. Because the number of crates has to be an integer, the lorry can accommodate a maximum of 13 crates of oranges. _________________ Manager Joined: 06 Mar 2014 Posts: 231 Location: India GMAT Date: 04-30-2015 ### Show Tags 05 Sep 2015, 10:14 Bunuel wrote: Because of weight limits, a lorry can carry a maximum of 20 crates of apples. The weight of a crate of oranges is $$\\frac{3}{4}$$ the weight of a crate of apples. If 10 crates of apples were loaded into the lorry, what is the maximum number of crates of oranges that the lorry can still accommodate? A. 8 B. 9 C. 12 D. 13 E. 14 VeritasPrepKarishma,", "oranges in a fruit basket containing no other types of fruit, it could be said that \"the whole\" contains five parts, made up of two parts apples and three parts oranges. In this case, $\\tfrac{2}{5}$, or 40% of the whole are apples and $\\tfrac{3}{5}$, or 60% of the whole are oranges. This comparison of a specific quantity to \"the whole\" is sometimes called a proportion. Proportions are sometimes expressed as percentages as demonstrated above. ## Reduction Note that ratios can be reduced (as fractions are) by dividing each quantity by the common factors of all the quantities. Thus the ratio $\\ 40:60$ may be considered equivalent in meaning to the ratio $\\ 2:3$ within contexts concerned only with relative quantities. Mathematically, we can write: \"$\\ 40:60$$\\ =$ \"$\\ 2:3$\" (dividing both quantities by 20). A ratio that has integers for both quantities and that cannot be reduced any further (using integers) is said to be in simplest form or lowest terms. Sometimes it is useful to write a ratio in the form $\\ 1:n$ or $\\ n:1$ to enable comparisons of different ratios. For example, the ratio $\\ 4:5$ can be written as $\\ 1:1.25$ (dividing both sides by 4) Alternatively, $\\ 4:5$ can be written as $\\ 0.8:1$ (dividing both sides by 5) Where the context makes", "So equating these we get $$6a = 5o$$ Thus we see that $6$ apples weights as much as $5$ oranges. - Thank you ever so much, really appreciate it. Now I see where I was going wrong. – Jay Oct 30 '12 at 19:09", "1/2 pounds A bag of apples weighs = 28 ounces we have to find that can tote to hold both bags Convert 28 ounces to pounds 28 ounces = 1.75 pounds ( divide the value of the ounce by 16 ) Given in question that tote can hold up to 4 pounds so, adding 1/2 pounds and 1.75 pounds = 1/2 pounds = 0.5 0.5 pounds + 1.75 pounds = 2.25 pounds Therefore tote holds both the bags of fruit. Precision Which bag of fruit is heavier? Explain. Answer: Bag of apple is heavier ( i.e., a bag of apple weighs 28 ounces ) Explanation: 28 ounces = 1.75 pounds bag of orange is 0.5 pounds 1.74 > 0.5 (So, apple bag weigh is heavier) Think and Grow: Problem Solving: Measurement Example A recipe calls for 2$$\\frac{1}{4}$$ cups of milk. You have $$\\frac{1}{4}$$ pint of whole milk and 1$$\\frac{1}{2}$$ cups of skim milk. Do you have enough milk for the recipe? Answer: Given that , a recipe calls for 2 × 1/4 cups of milk Explanation: 2 × 1/4 => 1/2 = 0.5 cups of milk you have 1/4 pint of milk = 0.25 pints 0.25 pints = 0.5 cups and 1 × 1/2 cups of skim milk => 1/2 = 0.5 cups Recipe calls for 0.5 cups", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Age 7 to 11 Short Challenge Level: The picture above shows four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights? If the apple weighs $180g$, how heavy must one weight be? If the apple weighed $375g$, how heavy would one weight be? If the apple was a giant one and weighed a full kilo and the weights were each $250g$, what would the scale look like? How do you know? Can you prove it?", "fruit. Your manager tells you to maintain a ratio of 2:3 of pears to apples for every size of basket. A small basket gets 2 pears and 3 apples. An extra-large basket must have the same ratio, 2:3, but be five times larger. The ratio of pears:apples is 2:3, so multiply both parts of the ratio times 5 to get the new ratio: 10:15. ## Ratios and proportions practice The class of 10 brown-haired and 6 blonde-haired girls also has boys in it. Of the 12 boys in the class, 4 have blond hair and 8 have brown hair. Write three ratios using this new information. Many ratios can be written from the information. See if you can figure out what these ratios describe: • 4:12 • 8:12 • 4:28 • 8:10 • 28:8 • 4:12 (Blond-haired boys to all boys) • 8:12 (Brown-haired boys to all boys) • 4:28 (Blond-haired boys to all students) • 8:10 (Brown-haired boys to brown-haired girls) • 28:8 (All students to brown-haired boys) ## Lesson summary You have learned that ratios compare values, while proportions compare ratios. Proportions are most often used to ensure ratios are equal when they increase or decrease. You can write ratios as either a fraction or with a colon between them, like this: $\\frac{10}{16}$ or $10:16$. Ratios", "# The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2 oranges is 285 gms. What is 1 apple and 1 orange weight individually ? $${}$$" ]

[ "× 0.51 > 1 Its false 2 × 0.51 > 1 Explanation: 2 × 0.51 = 1.02. Question 7. 4 × 0.49 > 2 Its true 4 × 0.49 > 2. Explanation: 4 × 0.49 = 1.96 Question 8. 5 × 0.97 > 5 Its false 5 × 0.97 > 5. Explanation: 5 × 0.97 = 4.85. Question 9. Jan goes to the grocery store to buy apples and oranges. Jan buys 5 apples. Is it reasonable that the apples weigh less than 5 pounds? Explain how you know. Yes, its reasonable that the apples weigh less than 5 pounds because Weight of 5 apples = 3.71 lb. Explanation: Number of apples Jan buys = 5. Weight of each apple = 0.74 lb. Weight of 5 apples = Number of apples Jan buys × Weight of each apple = 5 × 0.74 lb = 3.70 lb. Jan buys 6 oranges. Is it reasonable that the oranges weigh less than 3 pounds? Explain how you know. Explanation: Number of oranges Jan buys = 6. Weight of each orange = 0.55 lb. Weight of 5 oranges = Number of oranges Jan buys × Weight of each orange = 5 × 0.55 lb = 2.75 lb. Question 10. Reason Suppose you multiply 4 and 0.____7. What are the possible values of", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Stage: 2 Short Challenge Level: We had a solution to this problem from someone who didn't give their name or school - but thank you! From the picture of the scales, the apple must weigh less than the four weights altogether. If the apple weighs $180g$, for the scale to be balanced the weights would have to be $45g$ each because there are four of them. But we know from the picture that the apple weighs less than the weights put together, so each weight must be more than $45g$. If the apple weighed $375g$, then for the scale to be balanced, each weight would be $93.75g$. So, for the weights to be heavier than the apple, each", "the fruits and the columns represent the number of each fruit. To compare the ratio of apples to oranges, we can create ratios by dividing the number of apples by the number of oranges or vice versa. You can use these examples as a guide, and adjust it to your specific lesson plan. Ratios tables are a great tool to help students understand the relationship between different values, and it can be applied to many different math concepts. ### What people say about \"Ratio Tables\"? No one replied yet. X 30% OFF Limited time only! Save Over 30% SAVE $5 It was$16.99 now it is \\$11.99", "+ 1 $$\\frac{5}{6}$$ = 7 $$\\frac{5}{6}$$ feet Thus the correct answer is option c. Question 2. Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. How many pounds of apples did Harry pick those three days? Options: a. 132 $$\\frac{3}{8}$$ pounds b. 141 $$\\frac{3}{8}$$ pounds c. 142 $$\\frac{1}{8}$$ pounds d. 142 $$\\frac{3}{8}$$ pounds Answer: 142 $$\\frac{3}{8}$$ pounds Explanation: Given, Harry works at an apple orchard. He picked 45 $$\\frac{7}{8}$$ pounds of apples on Monday. He picked 42 $$\\frac{3}{8}$$ pounds of apples on Wednesday. He picked 54 $$\\frac{1}{8}$$ pounds of apples on Friday. 45 $$\\frac{7}{8}$$ + 42 $$\\frac{3}{8}$$ + 54 $$\\frac{1}{8}$$ 45 + 42 + 54 = 141 141 + $$\\frac{7}{8}$$ + $$\\frac{3}{8}$$ + $$\\frac{1}{8}$$ 141 + 1 $$\\frac{3}{8}$$ = 142 $$\\frac{3}{8}$$ pounds Thus the correct answer is option d. Spiral Review Question 3. There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. How many oranges were left? Options: a. 2 $$\\frac{1}{3}$$ oranges b. 2 $$\\frac{2}{3}$$ oranges c. 3 $$\\frac{1}{3}$$ oranges d. 9 $$\\frac{2}{3}$$ oranges Answer: 9 $$\\frac{2}{3}$$ oranges Explanation: Given, There were 6 oranges in the refrigerator. Joey and his friends ate 3 $$\\frac{2}{3}$$ oranges. 6 + 3", "into a ratio The fraction $\\frac{11}{14}$ can mean two possibilities • Two quantities are in $11 : 14$ ratio • one quantity is specified as $\\frac{11}{14}$ part of the whole. In the context of ratio, the fraction is taken to be in the form of comparison of two quantities, unless specified in the question. So, the answer is \"$11 : 14$\". The number of apples are $\\frac{2}{3}$ of the number of oranges. What is the ratio of apples to oranges? The given fraction specifies the comparison of apples to oranges. Number of apples is $\\frac{2}{3}$ of the number of oranges. Which means, for every orange, there are $\\frac{2}{3}$ apples. So the ratio of apples to oranges is $\\frac{2}{3} : 1$, which is equivalently $2 : 3$. So, the answer is \"$2 : 3$\". A basket has $\\frac{2}{3}$ apples and rest oranges. What is the ratio of apples to oranges? The given fraction specifies the quantity of apples. Quantity of apples $\\frac{2}{3}$ Quantity of oranges $1 - \\frac{2}{3} = \\frac{1}{3}$ So the ratio of apples to oranges is $\\frac{2}{3} : \\frac{1}{3}$ which is equivalently $2 : 1$. So the answer is \"$2 : 1$\". summary Ratio Vs Fraction : comparison of two quantities given as magnitude of one to magnitude of another is ratio. Fractions are also used to", "of 890 grams on a balance scale. He balances the scale by placing two oranges, an apple, and a lemon on the other side. Each orange weighs 280 grams. The lemon weighs 195 grams less than each orange. So the lemon weighs 280 g – 195 g = 85 grams, Now weigh of two oranges is 280 g + 280 g = 560 grams, given 2 oranges + lemon + apple = 890 grams, therefore weigh of apple = 890 g – (560g + 85 g) = 890 g – 645 g = 245 g, therefore, the mass of the apple is 245 g, modeled problem with a tape diagram as shown above and statement is mass of apple is two hundred fortyy five grams. Question 2. Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters shorter than Brian. Betina is 26 centimeters taller than Bonnie. How tall is Betina? Betina is 155 cm or 1 m 55 cm tall, Statement: Betina is one hundred fifty five centimeters or one meter fifty five centimeters tall, Explanation: Given Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters shorter than Brian means Bonnie is 1 m 87 cm – 58 = as 1 m 87 cm = 1 X 100 cm + 87 cm = 187", "### Back to School Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong? ### The Magic Number and the Hepta-tree Find the exact difference between the largest ball and the smallest ball on the Hepta Tree and then use this to work out the MAGIC NUMBER! ### Shedding Some Light Make an estimate of how many light fittings you can see. Was your estimate a good one? How can you decide? # What's My Weight? ##### Age 7 to 11 Short Challenge Level: The picture above shows four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights? If the apple weighs $180g$, how heavy must one weight be? If the apple weighed $375g$, how heavy would one weight be? If the apple was a giant one and weighed a full kilo and the weights were each $250g$, what would the scale look like? How do you know? Can you prove it?", "So equating these we get $$6a = 5o$$ Thus we see that $6$ apples weights as much as $5$ oranges. - Thank you ever so much, really appreciate it. Now I see where I was going wrong. – Jay Oct 30 '12 at 19:09", "maths > commercial-arithmetics Introduction to Percentages what you'll learn... overview A ratio is represented with two numbers, eg: 3:4$3 : 4$. Comparing two different ratios involve some arithmetics. eg: 4:5$4 : 5$ and 3:4$3 : 4$, which one is larger in ratio? To simplify such comparison, the second term is standardized to a value, say 100$100$, then the given ratios are 80:100$80 : 100$ and 75:100$75 : 100$. It is easier to compare these ratios. Such standard representation is simplified by dropping the known number $100$. and the simplified form is \"percentage\". eg: and . problem with ratios Consider a fruit basket with apples, oranges, and bananas • The ratio of apples to bananas is $1 : 4$ • The ratio of oranges to bananas is $2 : 9$ It is not evident to figure out if apples or oranges are more in the basket from the given two ratios. The ratios are to be converted to similar ratios. Consider a fruit basket with apples, oranges, and bananas • The ratio of apples to bananas is $1 : 4$. This means that, there is $1$ apple for every $4$ bananas. Or $2$ apples for every $8$ bananas. • The ratio of oranges to bananas is $2 : 9$. This means that, there are $2$ oranges for every $9$", "maths > commercial-arithmetics Ratio & Fraction Differences what you'll learn... overview Ratio and Fraction are two methods to specifying relative magnitude of quantities. Eg: 12$12$ and 24$24$ are in 1:2$1 : 2$ ratio OR 12$12$ is 12$\\frac{1}{2}$ of 24$24$. Understanding the differences in these two representation is explained. ratio vs fraction Consider $4$ apples and $6$ oranges. And there are pieces of a cake. The cake was cut into $6$ pieces and only $4$ pieces are remaining. • ratio of apples to oranges is $4 : 6$ which is equivalently $2 : 3$ A ratio represents comparison of two quantities given as the magnitude of one to the magnitude of another. • number of apples is $\\frac{4}{6}$ fraction of number of oranges. which is equivalently $\\frac{2}{3}$. A fraction can be used for comparing two quantities. • number of cakes is $\\frac{4}{6}$ fraction of the whole, which is equivalently $\\frac{2}{3}$. A fraction can be ised to specify a quantity. Note that the fraction can be used both for comparing quantities and to specify a quantity. context Consider $4$ apples and $6$ oranges. • ratio of apples to oranges is $4 : 6$ which is equivalently $2 : 3$ • ratio of apples to fruits is $4 : 10$ which is equivalently $2 : 5$ • number of apples is", "# The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2 oranges is 285 gms. What is 1 apple and 1 orange weight individually ? $${}$$", "Sep 2009 Posts: 54434 ### Show Tags 16 Sep 2014, 00:45 2 Official Solution: Because of weight limits, a lorry can carry a maximum of 20 crates of apples. The weight of a crate of oranges is $$\\frac{3}{4}$$ the weight of a crate of apples. If 10 crates of apples were loaded into the lorry, what is the maximum number of crates of oranges that the lorry can still accommodate? A. 8 B. 9 C. 12 D. 13 E. 14 There's still room for 10 crates of apples. The weight of 10 crates of apples is the same as the weight of $$10*(\\frac{4}{3}) = 13\\frac{1}{3}$$ crates of oranges. Because the number of crates has to be an integer, the lorry can accommodate a maximum of 13 crates of oranges. _________________ Manager Joined: 06 Mar 2014 Posts: 231 Location: India GMAT Date: 04-30-2015 ### Show Tags 05 Sep 2015, 10:14 Bunuel wrote: Because of weight limits, a lorry can carry a maximum of 20 crates of apples. The weight of a crate of oranges is $$\\frac{3}{4}$$ the weight of a crate of apples. If 10 crates of apples were loaded into the lorry, what is the maximum number of crates of oranges that the lorry can still accommodate? A. 8 B. 9 C. 12 D. 13 E. 14 VeritasPrepKarishma,", "part of the whole. In the context of ratio, the fraction is taken to be in the form of comparison of two quantities, unless specified in the question. Solved Exercise Problem: The number of apples are 2/3 of the number of oranges. What is the ratio of apples to oranges? • 2:3 • 2:3 • 2:1 The answer is \"2:3\". The given fraction specifies the comparison of apples to oranges. Number of apples is 2/3 of the number of oranges. Which means, for every orange, there are 2/3 apples. So the ratio of apples to oranges is 2/3 : 1, which is equivalently 2:3 Solved Exercise Problem: A basket has 2/3 apples and rest oranges. What is the ratio of apples to oranges? • 2:3 • 2:1 • 2:1 The answer is \"2:1\". The given fraction specifies the quantity of apples. Quantity of apples 2/3 Quantity of oranges 1 - 2/3 = 1/3 So the ratio of apples to oranges is 2/3 : 1/3 which is equivalently 2:1 slide-show version coming soon", "4 small pizzas in each box. Mrs. Owens buys 8 boxes. How many pizzas does she have? 1. 4 2. 12 3. 24 4. 32 3 x 3 = 1. 6 2. 0 3. 9 Which expression is the same as 7 + 7 + 7? 1. 21 / 7 2. 3 x 7 3. 7 x 1 4. 21 - 7 Factors are the 2. numbers that you multiply Any number times 1 equals the same number. 1. True 2. False Sydney helps her mother with the garden for 7 days. They picked 3 tomatoes each day. How many tomatoes do they have in total? 1. 12 2. 21 3. 18 4. 73 $[ 5 xx 3 ] =$ Which array would you draw? 1. 5 rows of 3 objects 2. 3 rows of 5 objects 3. 15 objects in one row 4. 2 columns of five Marty bought 3 bags of oranges at the store. There were 9 oranges in each bag. How many oranges did Marty have all together? 1. 9 + 9 = 18 or 2 x 9 = 18 2. 3 x 3 = 9 or 3 + 3 + 3 = 9 3. 3 x 9 = 27 or 9 + 9 + 9 = 27 8 x 2 = 1. 16", "describing how these two things are related: • It is positive if higher weight tends to be seen alongside higher diameter; • It is close to 0 if the two don’t seem to be very related; • it is negative if higher weight tends to be seen alongside lower diameter. In the case of apples, we probably expect the covariance between weight and diameter to be positive. Large apples generally weigh more. Aside: astute readers may observe that there are actually two places where weight and diameter intersect in this table, and they will have the same value. This is consistent with the fact that a covariance matrix is always symmetric across its main diagonal. ## Can we use the same $\\Sigma$ for apples and oranges? Credit: Andrew Ng Note that in Andrew’s model, he had separate $\\mu_0$ for apples and $\\mu_1$ for oranges, but he uses the same $\\Sigma$ for both fruit. (Well, Andrew doesn’t refer to any fruit, but we’re using them as a convenient mental device.) This means that in our model, we have the flexibility to allow for the possibility that apples and oranges have different mean prices, weights, diameters, and colours. This makes sense intuitively: oranges are usually a little bigger, apples tend to be redder, etc. But by using the same $\\Sigma$,", "repeated ? List out the combinations and you will have your answer. – Shailesh Aug 26 '15 at 2:20 • The formula $2A=(A+B)-(B+C)+(C+D)-(D+E)+(E+A)$ might be useful, if you know which sums are which. (It follows because, distributing, we have $A+B-B-C+C+D-D-E+E+A$, and everything cancels out with its neighbor save for $A$.) – Akiva Weinberger Aug 26 '15 at 2:27 • Using my values — 'A' = 53.25, 'B' = 54.75, 'C' = 58.75, 'D' = 60.75, and 'E' = 61.25 — I was unable to provide combinations for 113, 115, 117, and 118. – Sentient Aug 26 '15 at 2:30 • @columbus8myhw I believe 'A' + 'B' = 108 and 'D' + 'E' = 122. These are the only sums I'm certain about. – Sentient Aug 26 '15 at 2:32 • I think it's unsolvable. – Akiva Weinberger Aug 26 '15 at 2:48 First, you’re right in thinking that the total weight is $288.75$; you can also get this by noticing that each pumpkin is weighed four times, and $\\frac{1155}4=288.75$. However, the data are inconsistent. The low total of $108$ must be the sum of the two lightest weights, so the three heaviest pumpkins altogether weigh $288.75-108=180.75$. The two heaviest together weigh $122$, so the middle pumpkin weighs $180.75-122=58.75$. Each pair of pumpkins has an integer total weight, so", "# The total weight is 255 gms for 2 apples and 3 oranges and that for 3 apples and 2 oranges is 285 gms. What is 1 apple and 1 orange weight individually ? ## Question 1 & 2 Question 1&2: The total weight is 255 gms for ... » All original content on these pages is fingerprinted and certified by Digiprove Insert math as $${}$$", "fruit. Your manager tells you to maintain a ratio of 2:3 of pears to apples for every size of basket. A small basket gets 2 pears and 3 apples. An extra-large basket must have the same ratio, 2:3, but be five times larger. The ratio of pears:apples is 2:3, so multiply both parts of the ratio times 5 to get the new ratio: 10:15. ## Ratios and proportions practice The class of 10 brown-haired and 6 blonde-haired girls also has boys in it. Of the 12 boys in the class, 4 have blond hair and 8 have brown hair. Write three ratios using this new information. Many ratios can be written from the information. See if you can figure out what these ratios describe: • 4:12 • 8:12 • 4:28 • 8:10 • 28:8 • 4:12 (Blond-haired boys to all boys) • 8:12 (Brown-haired boys to all boys) • 4:28 (Blond-haired boys to all students) • 8:10 (Brown-haired boys to brown-haired girls) • 28:8 (All students to brown-haired boys) ## Lesson summary You have learned that ratios compare values, while proportions compare ratios. Proportions are most often used to ensure ratios are equal when they increase or decrease. You can write ratios as either a fraction or with a colon between them, like this: $\\frac{10}{16}$ or $10:16$. Ratios", "the oranges to the dot plot of the apple weights made in the prior example. The first chart keeps apples and oranges in separate graphs while the second chart combines data with a different marker for apples and orange. This second chart is called a stacked dot plot. From the graphs, we can see that the typical orange weighs about 30 grams more than the typical apple. The spread of weights for apples and oranges is about the same. The shapes of both graphs are symmetric and clustered towards the center. There is a high outlier for apples at 440 grams and a low outlier for oranges at 120 grams. 1.3.5.2: Dot Plots is shared under a CC BY-SA license and was authored, remixed, and/or curated by LibreTexts.", "expert reader. – Julian Kuelshammer Oct 30 '12 at 19:01 How many melons weigh as much as six apples? – Michael Joyce Oct 30 '12 at 19:03 Since $4$ applies weight as much has $2$ melons, $6$ apples will weight as much as $3$ melons. But $5$ oranges weigh as much as $3$ melons. Can you now conclude? - Thank you for time, appreciate it. I can see the pattern behind it. – Jay Oct 30 '12 at 19:11 4 apples = 2 melons ---eq.1 5 oranges = 3 melons---eq.2 Multiply equation 1 with 3/2 for making RHS of both the equations same. Hence, 6 apples= 3 melons---eq.3 Compare eq.2 with eq. 3 Therefore, 6 apples= 5 oranges Thus 6 apples weigh as much as 5 oranges. - Thanks a lot, really appreciate your efforts. Simple simultaneous, but I couldn't get my head around it. – Jay Oct 30 '12 at 19:36 Lets right out what we know in equations. Let $a =$ weight of apple, $m =$weight of melon, $o =$ weight of orange. Then we have the following equations: $$4a = 2m$$ $$5o = 3m$$ So we can eliminate $m$ from these equations by multiplying through on the first equation by $\\frac{3}{2}$ and subtracting from the first. So we get $$6a = 3m$$ $$5o = 3m$$" ]

[ "# ksided 1. ### [Help] The number of possible k-sided polygons in an m by n grid of dots Hello. :spin: This is my first post, so sorry for the potential mistakes below. A grid of dots has the dimensions m by n, where m is greater than or equal to n. Pick k dots from this grid in such a way that they form a polygon with k sides. In other words, no three dots can be collinear...", "## Number of Square Units What is the number of square units enclosed by the polygon $ABCDE$? Source: NCTM Mathematics Teacher 2006 SOLUTION Polygon $HEJC$ encloses polygon $ABCDE$. Number of square units inside polygon $HEJC$ $9\\times 10=90$ Number of square units inside polygon $GBC$ $6\\div 2=3$ Number of square units inside polygon $HIBG$ $4$ Number of square units inside polygon $IAB$ $(4\\times 4)\\div 2=8$ Number of square units inside polygon $EJD$ $(5\\times 10)\\div 2=25$ Number of square units inside polygon $ABCDE$ $90-3-4-8-25=50$ Answer: $50$ square units", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label(\"P\",(4,4),NE); draw((0,4)--(3,4)); draw((0,8)--(3,5)); draw((8,8)--(5,5)); draw((4,8)--(4,5)); draw((4,0)--(4,3)); draw((0,0)--(3,3)); draw((8,0)--(5,3)); draw((5,4)--(8,4)); [/asy]$ Lines of symmetry go through point $P$, and there are $8$ directions the lines could go, and there are $4$ dots at each direction.$\\frac{4\\times8}{80}=\\boxed{\\textbf{(C)} \\frac{2}{5}}$. Video Solution Solution detailing how to solve the problem: https://www.youtube.com/watch?v=4L95z9DwlhI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=7", "# Polygonal numbers A polygonal number is the number of dots in a k-gon of size n. You will be given n and k, and your task is to write a program/function that outputs/prints the corresponding number. # Scoring This is . Shortest solution in bytes wins. # Example The 3rd hexagon number (k=6, n=3) is 28 because there are 28 dots above. # Testcases Can be generated from this Pyth test suite. Usage: two lines per testcase, n above, k below. n k output 10 3 55 10 5 145 100 3 5050 1000 24 10990000 # Further information • Isn't that the 4th hexagonal number in the picture? – Neil May 1 '16 at 13:59 • @Neil We count from zero. – Leaky Nun May 1 '16 at 14:05 • You really are going on a question posting spree, aren't you? – R. Kap May 1 '16 at 17:21 • The example might be off. If you put n=3 and k=6 into your test suite, you get 15. If you put in n=4 and k=6, you get 28. – NonlinearFruit May 3 '16 at 16:37 # Jelly, 7 bytes ’;’;PH+ This uses the formula to compute the nth s-gonal number. Try it online! ### How it works ’;’;PH+ Main link. Arguments: s, n ’ Decrement; yield", "$m$, width $n$ edge squares = $2m + 2n - 4 = 2(m+n-2)$ lines = $2m + 2n + 2(m-1) + 2(n-1) + 2(m-2) + 2(n-2) = 6m + 6n - 12 = 6(m+n-2)$ DOTS AND MORE DOTS side length 3 dots = 25 lines = 24 side length 25 dots = $625 + 676 = 1301$ lines = $25\\times26\\times2 = 1300$ side length 100 dots = $10000 + 10201 = 20201$ lines = $100\\times101\\times2 = 20200$ side length $n$ dots = $n^2 + (n+1)^2 = n^2 + n^2 + n + n + 1 = 2n^2 + 2n + 1$ lines = $2n(n+1) = 2n^2 + 2n$ height $m$, width $n$ dots $= mn + (m+1)(n+1) = mn + m(n+1) + (n+1) = mn + mn + m + n + 1 = 2mn + m + n + 1$ lines $= 2mn + m + n$ RECTANGLE OF DOTS For 2 squares: side length 3 lines = $7\\times3 = 21$ dots = $4\\times7 = 28$ side length 25 lines = $7\\times25 = 175$ dots = $26\\times51 = 1326$ side length 100 lines = 700 dots = $101\\times201 = 20301$ side length $n$ lines $= 7n$ dots $= (n+1)(2n+1)$ Generalising to $p$ squares of side length $n$ lines $= (3p+1)n$ [got $3p+1$ from Tom's matchsticks] dots $=", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $IM=CI\\cdot \\frac{IP}{AI}$. Using a similar method, we can find that $NI=CI\\cdot\\frac{IQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=\\frac{IP}{AI}\\cdot CN+\\frac{IQ}{BI}\\cdot MC=r\\left(\\frac{CN}{AI}+\\frac{MC}{BI}\\right),$$ where $r$ is the inradius of triangle $ABC$. Thus, $r=[ABC]/181$. Also, right triangles $CNA$ and $CBM$ tell us that $CN=117\\sin \\frac{A}{2}$ and $MC=120\\sin\\frac{B}{2}$. But then $[CLA]=\\frac{117\\cdot AL}{2}\\sin\\frac{A}{2}$, and this is equal to $\\frac{117}{242}\\cdot [ABC]$ by the Angle Bisector Theorem. Therefore, solving this for $\\sin\\frac{A}{2}$ and substituting yields $CN=\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL}$. Similarly, $MC=\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK}$. We now replace these in our Ptolemy's expression to get $$MN=\\frac{[ABC]}{181}\\left(\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL\\cdot AI}+\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK\\cdot BI}\\right)$$$$=\\frac{2[ABC]^2}{181}\\left(\\frac{117}{242\\cdot AL\\cdot AI}+\\frac{120}{245\\cdot BK\\cdot BI}\\right).$$ We can also use mass points, assigning masses of $120$, $117$, and $125$ to points $A$, $B$, and $C$, respectively. Then point $L$ has a mass of $242$ and point $K$ has a mass of $245$, so $AI=\\frac{242}{362}\\cdot AL$ and $BI=\\frac{245}{362}\\cdot BK$. This simplifies our expression further to $$MN=4[ABC]^2\\left(\\frac{117}{242^2\\cdot AL^2}+\\frac{120}{245^2\\cdot BK^2}\\right).$$ Then using the angle bisector formula, we find that $AL^2=125\\cdot 117\\left(1-\\frac{120^2}{242^2}\\right)$ and", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $MC=CI\\cdot \\frac{AP}{AI}$. Using a similar method, we can find that $CN=CI\\cdot \\frac{BQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=IM\\cdot \\frac{BQ}{BI}+NI\\cdot\\frac{AP}{AI}=IM\\cdot\\cos \\frac{B}{2}+NI\\cdot\\cos\\frac{A}{2}.$$ Using right triangles $CIM$ and $NCI$, we also know that $IM=CI\\cdot\\sin \\frac{A}{2}$ and $NI=CI\\cdot\\sin\\frac{B}{2}$. Thus $$MN=CI\\cdot\\cos \\frac{B}{2}\\sin\\frac{A}{2}+CI\\cdot\\cos\\frac{A}{2}\\sin\\frac{B}{2}=CI\\cdot\\sin\\left(\\frac{A+B}{2}\\right)=CI\\cdot \\cos \\frac{C}{2}.$$ But this last expression is equal to $CQ$. This a tangent to the incircle, so it has length $s-c=181-125=\\boxed{56}$.", "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "$[asy] currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label(\"O\", O, S); label(\"A\", A, SW); label(\"B\", B, SE); label(\"X\", X, (X-B)/length(X-B)); label(\"Y\", Y, Y); label(\"Z\", Z, (Z-A)/length(Z-A)); label(\"P\", P, (P-T)/length(P-T)); label(\"Q\", Q, SW); label(\"R\", R, SE); label(\"S\", SS, (SS-T)/length(SS-T)); label(\"T\", T, S); [/asy]$ Let $T$ be the foot of the perpendicular from $Y$ to $\\overline{AB}$, let $O$ be the center of the semi-circle. Since we have a semi-circle, if we were to reflect it over $\\overline {AB}$, we would have a full circle, with", "its upper right coordinates are (X, Y). N indicates the number of things on his desk. Then one line follows. This line contains an integer M followed by M pairs of integers. Each of the pair indicates a vertex of the convex polygon the mouse needs at least. The vertices will be given counterclockwise. Then N lines follow, each line contains an integer L followed by L pairs of integers. Each of the pair indicates a vertex of the thing’s convex polygon. The vertices will be given counterclockwise. Technical Specification 1. 1 <= T <= 1000 2. 0 <= N <= 20 3. 3 <= M <= 20 4. 3 <= L <= 20 and all the things are inside the desk. 5. All the coordinates will be greater than or equal to 0 and less than or equal to 10000. 3 10 10 1 4 0 0 2 0 2 2 0 2 4 1 1 9 1 9 9 1 9 10 10 1 3 0 0 2 0 0 2 4 1 1 9 1 9 9 1 9 9 9 4 4 0 0 1 0 1 1 0 1 4 0 0 5 0 5 4 0 4 4 9 0 9 5 5 5 5 0 4 9 9 4 9 4", "be humble! ## Solution 2 [asy] currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label(\"$O$\", O, S); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$X$\", X, (X-B)/length(X-B)); label(\"$Y$\", Y, Y); label(\"$Z$\", Z, (Z-A)/length(Z-A)); label(\"$P$\", P, (P-T)/length(P-T)); label(\"$Q$\", Q, SW); label(\"$R$\", R, SE); label(\"$S$\", SS, (SS-T)/length(SS-T)); label(\"$T$\", T, S); [/asy] Let $T$ be the projection of $Y$ onto $\\seg{AB}$ (Error compiling LaTeX. ! Undefined control sequence.). Notice that $T$ lies on the Simson Line $\\seg{PQ}$ (Error compiling LaTeX. ! Undefined control sequence.) from $Y$ to $\\triangle", "we denote by k Click image to enlarge Total Number of Dots on Sticks The total number of dots on sticks is the product of the number of sticks in a polygon with n sides and the number of dots on each stick. From the table above, a polygon with n sides has (n – 1) sticks and the number of dots on each stick is (k – 1); that is, the number of dots on each side minus the red dot (see third figure). So, the total number of dots on sticks is $(k-1)(n-1)$. Total Number Number of Dots on Triangles The total number of dots on triangles is equal to the number of triangles times the number of dots on each triangle. Again, from the table above, a polygon with n sides has (n-2) triangles. Now, the number of dots in each triangle is the sum of 1 + 2 + 3 + … + (k – 2) as shown above. Recall that that the sum of 1 + 2 + 3 + … + (m – 1) + m = 1/2 m (m + 1). If we let m = k + 2, then the total number of dots in each triangle is 1/2(k-2)(k-1). Therefore, the total number of dots on all the triangles is", "IMu berofbedpam First, as we've said, polygons should not be used with nominal or ordinal data because joining the dots makes the assumption that there is a smooth transition from one datum point to another. For example, imagine that we have a polygon with just two points, as in Figure 2-7. The first point, at a midpoint of 20, shows 100 units on the У-axis, and the second point, which falls at a midpoint of 30, shows 110 units. Even though we may not have gathered any data that correspond to an .X-axis value of 25, we assume they fall on the line, half way between 20 and 30. In this case, they would correspond to 105 units (where the dot is). We can make this assumption only because we're using an interval or ratio level of data; if the distances be- between intervals are variable or unknown, as they are with ordinal data, we couldn't make this as- assumption. A second difference is that bar charts seem to imply that the data are spread equally over the interval. For instance, if we had an interval width of 5 units spanning the numbers 20 through 24, and 10 cases were in that interval, it would appear (and we would assume) that 2 cases fell at 20, 2 at", "coordinates, the first side starts at [0,0], the second at [sqrt(num_squares),sqrt(num_squares)] i.e. [16,16], the third at [2 * sqrt(num_squares), 2 * sqrt(num_squares)] i.e. [32,32], and so on. dot_1A = [[6,7], [6,8], [7,6], [7,7], [7,8], [7,9], [8,6], [8,7], [8,8], [8,9], [9,7], [9,8]]dot_2A = [[1,12], [1,13], [2,11], [2,12], [2,13], [2,14], [3,11], [3,12], [3,13], [3,14], [4,12], [4,13]]dot_2B = [[11,2], [11,3], [12,1], [12,2], [12,3], [12,4], [13,1], [13,2], [13,3], [13,4], [14,2], [14,3]]dot_4A = [[1,2], [1,3], [2,1], [2,2], [2,3], [2,4], [3,1], [3,2], [3,3], [3,4], [4,2], [4,3]]dot_4B = [[11,2], [11,3], [12,1], [12,2], [12,3], [12,4], [13,1], [13,2], [13,3], [13,4], [14,2], [14,3]]dot_4C = [[1,12], [1,13], [2,11], [2,12], [2,13], [2,14], [3,11], [3,12], [3,13], [3,14], [4,12], [4,13]]dot_4D = [[11,12], [11,13], [12,11], [12,12], [12,13], [12,14], [13,11], [13,12], [13,13], [13,14], [14,12], [14,13]]dot_6A = [[1,2], [1,3], [2,1], [2,2], [2,3], [2,4], [3,1], [3,2], [3,3], [3,4], [4,2], [4,3]]dot_6B = [[1,7], [1,8], [2,6], [2,7], [2,8], [2,9], [3,6], [3,7], [3,8], [3,9], [4,7], [4,8]]dot_6C = [[1,12], [1,13], [2,11], [2,12], [2,13], [2,14], [3,11], [3,12], [3,13], [3,14], [4,12], [4,13]]dot_6D = [[11,2], [11,3], [12,1], [12,2], [12,3], [12,4], [13,1], [13,2], [13,3], [13,4], [14,2], [14,3]]dot_6E = [[11,7], [11,8], [12,6], [12,7], [12,8], [12,9], [13,6], [13,7], [13,8], [13,9], [14,7], [14,8]]dot_6F = [[11,12], [11,13], [12,11], [12,12], [12,13], [12,14], [13,11], [13,12], [13,13], [13,14], [14,12], [14,13]]dot_5A = [[1,2], [1,3], [2,1], [2,2], [2,3], [2,4], [3,1], [3,2], [3,3], [3,4], [4,2], [4,3]]dot_5B = [[11,2], [11,3], [12,1], [12,2], [12,3], [12,4], [13,1],", "the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former. The $\\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3),s); draw((0,0)--(3,3)); [/asy]$ The $\\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3)^^(1,0),s); draw((0,0)--(3,3),c); draw((3,3)--(1,0)); [/asy]$ From $(1,0)$, there are two possible ways to move $\\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$. However, from $(0,3)$, there is no way to move $\\sqrt{9}$ away, so we discard it as a possibility. From $(2,3)$, the lengths of $\\sqrt{8},\\ \\sqrt{5},\\ \\sqrt{4},\\ \\sqrt{2}$ fortunately are all", "# Difference between revisions of \"Dodecagon\" A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--G); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$.", "//example polygon with multiple holes a0 = [[0,0],[100,0],[130,50],[30,50]]; // main b0 = [1,0,3,2]; a1 = [[20,20],[40,20],[30,30]]; // hole 1 b1 = [4,5,6]; a2 = [[50,20],[60,20],[40,30]]; // hole 2 b2 = [7,8,9]; a3 = [[65,10],[80,10],[80,40],[65,40]]; // hole 3 b3 = [10,11,12,13]; a4 = [[98,10],[115,40],[85,40],[85,10]]; // hole 4 b4 = [14,15,16,17]; a = concat (a0,a1,a2,a3,a4); b = [b0,b1,b2,b3,b4]; polygon(a,b); //alternate polygon(a,[b0,b1,b2,b3,b4]); #### Extruding a 3D shape from a polygon translate([0,-20,10]) { rotate([90,180,90]) { linear_extrude(50) { polygon( points = [ //x,y /* O . */ [-2.8,0], /* O__X . */ [-7.8,0], /* O \\ X__X . */ [-15.3633,10.30], /* X_______._____O \\ X__X . */ [15.3633,10.30], /* X_______._______X \\ / X__X . O */ [7.8,0], /* X_______._______X \\ / X__X . O__X */ [2.8,0], /* X__________.__________X \\ / \\ O / \\ / / \\ / / X__X . X__X */ [5.48858,5.3], /* X__________.__________X \\ / \\ O__________X / \\ / / \\ / / X__X . X__X */ [-5.48858,5.3], ] ); } } } #### convexity The convexity parameter specifies the maximum number of front sides (back sides) a ray intersecting the object might penetrate. This parameter is needed only for correct display of the object in OpenCSG preview mode and has no effect on the polyhedron rendering. This image shows a 2D shape with a convexity of 2, as the", "A friend ((Thanks to Eli for the suggestion!)), justifiably proudly, shared on Facebook that he’d worked out the number of dots he’d need to draw a hexagon with $n$ dots on each edge. I thought it was a nice puzzle! Before we go anywhere, I’ll clear up what exactly we mean by a hexagon with $n$ dots per side. Here’s a hexagon with $n = 3$: * * * * * * * * * * * * * * * * * * * … which has 19 dots. Can you come up with a general expression for the number of dots? I came up with three methods, including the same one that my friend did. Spoilers below the line. ### Option 1: mechanically When $n=1$, the number of dots is 1. When $n=2$, the number of dots is 7. When $n=3$, the number of dots is 19. We’re looking at an area, so it’s almost certainly a quadratic sequence of the form $an^2 + bn + c$. This gives three equations: • $a + b + c = 1$ [1] • $4a + 2b + c = 7$ [2] • $9a + 3b + c = 19$ [3] Subtracting [1] from [2] gives $3a + b = 6$ Subtracting [2] from [3] gives $5a + b", "3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ \\tikznode{NOV-1-B}{1}, 2, 3, \\dots, \\tikznode{NOV-30-B}{30} & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ & & & & \\\\ & & & & \\end{tabular} \\end{table} \\end{center} \\begin{tikzpicture}[overlay,remember picture] \\node[rounded corners=2mm,draw,fit=(NOV-1-T)(NOV-1-B)] (NOV-L){}; \\node[rounded corners=2mm,draw,fit=(NOV-30-T)(NOV-30-B)] (NOV-R){}; \\draw[thick,-latex] (NOV-L.south) -- ++(0,-0.5) node[below]{$\\displaystyle \\frac{\\sum d_1}{18}$}; \\draw[thick,-latex] (NOV-R.south) -- ++(0,-0.5) node[below]{$\\displaystyle \\frac{\\sum d_1}{18}$}; \\end{tikzpicture} \\end{document} ADDENDUM: Let TeX do" ]

[ "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "area of rectangle $ABGH$ is $4 \\cdot 12 = 48$. Taking the area of rectangle $ABGH$ and subtracting the combined area of $\\triangle AHG$ and $\\triangle CGM$ yields $48 - (24 + \\frac{32}{5}) = \\boxed{\\frac{88}{5}}\\ \\text{(C)}$. ### Solution 2 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label(\"A\",A,NW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",Ep,SE); label(\"F\",F,SE); label(\"G\",G,SE); label(\"H\",H,SW); label(\"I\",I,SW); label(\"J\",J,SW); label(\"K\",K,NW); label(\"L\",L,NW); label(\"M\",M,W); label(\"N\",Z,NE); [/asy]$ Extend $AB$ and $CH$ and call their intersection $N$. The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\\frac{2\\cdot 4}2 = 4$. The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$. Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \\frac 35 \\cdot 12 = \\frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\\frac{1}{\\frac{1}{8}+\\frac{1}{12}} = \\frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\\frac{36}{5}$.) Then the area of $AMN$ is $\\frac 12 \\cdot", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "lines filldraw(PQR,gray(0.8)); draw(P--Pp,linetype(\"8 8\")); draw(Q--Qp,linetype(\"8 8\")); draw(R--Rp,linetype(\"8 8\")); draw(P--Y,linetype(\"8 8\")); draw(Q--Z,linetype(\"8 8\")); //Angle marks draw(rightanglemark(R,Y,P)); draw(rightanglemark(Q,X,P)); draw(rightanglemark(R,Z,Q)); //Length labeling label(\"2\\sqrt{2}\",P--X,fontsize(8pt)); label(\"2\\sqrt{6}\",X--Y,fontsize(8pt)); label(\"2\\sqrt{6}\",Q--Z,fontsize(8pt)); label(\"1\",R--Z,E,fontsize(8pt)); label(\"1\",Z--Y,E,fontsize(8pt)); label(\"3\",(-2.828,1.3)--Q,W,fontsize(8pt)); label(\"5\",Q--R,N,fontsize(8pt)); [/asy]$ The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\\triangle PQX] + [QZYX] + [\\triangle QRZ]$ and $[\\triangle PRY] + [\\triangle PQR]$. Solving, we get: \\begin{aligned} [\\triangle PQR] &= [PQRY] - [\\triangle PRY] \\\\ &= [\\triangle PQX] + [QZYX] + [\\triangle QRZ] - [\\triangle PRY] \\\\ &= \\sqrt{2} + 2\\sqrt{6} + \\sqrt{6}- 2\\sqrt{2}-2\\sqrt{6} \\\\ &= \\boxed{\\textbf{(D)} \\sqrt{6}-\\sqrt{2}} \\end{aligned} (Error compiling LaTeX. ! Package amsmath Error: \\begin{aligned} allowed only in math mode.)", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "$[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,W); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 32 \\qquad \\text{(D)}\\ 64 \\qquad \\text{(E)}\\ 128$ ## Problem 10 A jacket and a shirt originally sold for dollar 80 and dollar 40, respectively. During a sale Chris bought the dollar80 jacket at a $40\\%$ discount and the dollar 40 shirt at a $55\\%$ discount. The total amount saved was what percent of the total of the original prices? $\\text{(A)}\\ 45\\% \\qquad \\text{(B)}\\ 47\\dfrac{1}{2}\\% \\qquad \\text{(C)}\\ 50\\% \\qquad \\text{(D)}\\ 79\\dfrac{1}{6}\\% \\qquad \\text{(E)}\\ 95\\%$. ## Problem 11 Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to $[asy] for(int i = -2; i <= 2; ++i) { draw((i,0)--(i,3),dashed); } draw((-3,1)--(3,1),dashed); draw((-3,2)--(3,2),dashed); draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle); dot((-3,0)); label(\"A\",(-3,0),SW); dot((-3,3)); label(\"B\",(-3,3),NW); dot((0,3)); label(\"C\",(0,3),N); dot((3,3)); label(\"D\",(3,3),NE); dot((3,0)); label(\"E\",(3,0),SE); dot((0,0)); label(\"start\",(0,0),S); label(\"\\longrightarrow\",(0,-0.75),E); label(\"\\longleftarrow\",(0,-0.75),W); label(\"\\textbf{Jane}\",(0,-1.25),W); label(\"\\textbf{Hector}\",(0,-1.25),E); [/asy]$ $\\text{(A)}\\ A \\qquad \\text{(B)}\\ B \\qquad \\text{(C)}\\ C \\qquad \\text{(D)}\\ D \\qquad \\text{(E)}\\ E$ ## Problem", "dot(\"E\", E, NW); dot(\"F\", F, SW); dot(\"G\", G, S); dot(\"H\", H, N); dot(\"I\", I, NE); label(\"X\", X,SE); dot(\"J\", J, SE);[/asy]$ First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $IM=CI\\cdot \\frac{IP}{AI}$. Using a similar method, we can find that $NI=CI\\cdot\\frac{IQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=\\frac{IP}{AI}\\cdot CN+\\frac{IQ}{BI}\\cdot MC=r\\left(\\frac{CN}{AI}+\\frac{MC}{BI}\\right),$$ where $r$ is the inradius of triangle $ABC$. Thus, $r=[ABC]/181$. Also, right triangles $CNA$ and $CBM$ tell us that $CN=117\\sin \\frac{A}{2}$ and $MC=120\\sin\\frac{B}{2}$. But then $[CLA]=\\frac{117\\cdot AL}{2}\\sin\\frac{A}{2}$, and this is equal to $\\frac{117}{242}\\cdot [ABC]$ by the Angle Bisector Theorem. Therefore, solving this for $\\sin\\frac{A}{2}$ and substituting yields $CN=\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL}$. Similarly, $MC=\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK}$. We now replace these in our Ptolemy's expression to get $$MN=\\frac{[ABC]}{181}\\left(\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL\\cdot AI}+\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK\\cdot BI}\\right)$$$$=\\frac{2[ABC]^2}{181}\\left(\\frac{117}{242\\cdot AL\\cdot AI}+\\frac{120}{245\\cdot BK\\cdot BI}\\right).$$ We can also use mass points, assigning masses of $120$, $117$, and $125$ to points $A$, $B$, and $C$, respectively. Then point $L$ has a mass of $242$ and point $K$ has a mass of $245$, so $AI=\\frac{242}{362}\\cdot AL$ and $BI=\\frac{245}{362}\\cdot BK$. This simplifies our expression further to $$MN=4[ABC]^2\\left(\\frac{117}{242^2\\cdot AL^2}+\\frac{120}{245^2\\cdot BK^2}\\right).$$ Then using the angle bisector formula, we find that $AL^2=125\\cdot 117\\left(1-\\frac{120^2}{242^2}\\right)$ and", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $MC=CI\\cdot \\frac{AP}{AI}$. Using a similar method, we can find that $CN=CI\\cdot \\frac{BQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=IM\\cdot \\frac{BQ}{BI}+NI\\cdot\\frac{AP}{AI}=IM\\cdot\\cos \\frac{B}{2}+NI\\cdot\\cos\\frac{A}{2}.$$ Using right triangles $CIM$ and $NCI$, we also know that $IM=CI\\cdot\\sin \\frac{A}{2}$ and $NI=CI\\cdot\\sin\\frac{B}{2}$. Thus $$MN=CI\\cdot\\cos \\frac{B}{2}\\sin\\frac{A}{2}+CI\\cdot\\cos\\frac{A}{2}\\sin\\frac{B}{2}=CI\\cdot\\sin\\left(\\frac{A+B}{2}\\right)=CI\\cdot \\cos \\frac{C}{2}.$$ But this last expression is equal to $CQ$. This a tangent to the incircle, so it has length $s-c=181-125=\\boxed{56}$.", "triangles of area 1 be $x$, then the base length of $\\Delta ADE=4x$. Notice, $\\big(\\frac{DE}{BC}\\big)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$, then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\big(\\frac{4}{\\sqrt{40}}\\big)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\big(1-\\frac{2}{5}\\big)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$ $[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label(\"B\",(0,0),SW); label(\"C\",(60,0),SE); label(\"E\",(50,10),E); label(\"D\",(10,10),W); label(\"A\",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$ Solution by ktong ## Solution 4 The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$. Therefore the area of the trapezoid is $40-16=\\boxed{24}$. ## Solution 5 You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$, so to find the area of such trapezoid $BCED$, we just take $40-16=\\boxed{24}$, like so. ∎ --anna0kear ## Solution 6 The combined area of the small triangles is $7$, and from the fact that each small triangle has an area of $1$, we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$, so will their areas have a proportion that is the square of the", "+0 i need help -2 188 2 In the diagram below, the area of $\\triangle BCD$ is 60. What is the length of side $\\overline{BC}$? [asy] size(5cm); pair a=(9,8); pair b=(0,8); pair c=(0,0); pair d=(15,0); dot(a); dot(b); dot(c); dot(d); draw(b--c--d--a--b--d); label(\"$A$\",a,NE); label(\"$B$\",b,NW); label(\"$C$\",c,SW); label(\"$D$\",d,SE); label(\"9\",(a+b)/2,N); label(\"15\",(c+d)/2,S); draw(b/8--b/8+d/15--d/15); draw(b+(a-b)/9--b+(a-b)/9+(c-b)/8--b+(c-b)/8); [/asy] Apr 8, 2020 #1 +924 0 Can you write this question in laTex so it is readable and you need to attach the diagram aforemention in the question. Thanks! Apr 8, 2020 #2 +924 +1 Your question describes the diagram here but asks for BC instead of the Area. If that is the diagram described, then BC= 8 Hope it helps! Apr 8, 2020", "= Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"\\angle C obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"\\angle B obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: 1. The region in which $\\angle B$ is obtuse is determined by construction. 2. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"A\", A, 1.5*S, linewidth(4.5)); dot(\"B\", B, 1.5*S, linewidth(4.5)); dot(\"D\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"C_2\", C2, 1.5*N, linewidth(4.5)); dot(\"C_1\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"10\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"4\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: • If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. $[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]$ The diagram shows $P$ outside of the grayed locus; notice that the creases [the", "= foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ", "# 2018 AIME II Problems/Problem 14 ## Problem The incircle $\\omega$ of triangle $ABC$ is tangent to $\\overline{BC}$ at $X$. Let $Y \\neq X$ be the other intersection of $\\overline{AX}$ with $\\omega$. Points $P$ and $Q$ lie on $\\overline{AB}$ and $\\overline{AC}$, respectively, so that $\\overline{PQ}$ is tangent to $\\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Diagram $[asy] size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); pair A,B,C,P,Q,Wp,X,Y,Z; B=origin; C=(6.75,0); A=IP(CR(B,7),CR(C,8)); path c=incircle(A,B,C); Wp=IP(c,A--C); Z=IP(c,A--B); X=IP(c,B--C); Y=IP(c,A--X); pair I=incenter(A,B,C); P=extension(A,B,Y,Y+dir(90)*(Y-I)); Q=extension(A,C,P,Y); draw(A--B--C--cycle, black+1); draw(c^^A--X^^P--Q); pen p=4+black; dot(\"A\",A,N,p); dot(\"B\",B,SW,p); dot(\"C\",C,SE,p); dot(\"X\",X,S,p); dot(\"Y\",Y,dir(55),p); dot(\"W\",Wp,E,p); dot(\"Z\",Z,W,p); dot(\"P\",P,W,p); dot(\"Q\",Q,E,p); MA(\"\\beta\",C,X,A,0.3,black); MA(\"\\alpha\",B,A,X,0.7,black); [/asy]$ ## Solution 1 Let the sides $\\overline{AB}$ and $\\overline{AC}$ be tangent to $\\omega$ at $Z$ and $W$, respectively. Let $\\alpha = \\angle BAX$ and $\\beta = \\angle AXC$. Because $\\overline{PQ}$ and $\\overline{BC}$ are both tangent to $\\omega$ and $\\angle YXC$ and $\\angle QYX$ subtend the same arc of $\\omega$, it follows that $\\angle AYP = \\angle QYX = \\angle YXC = \\beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\\triangle APY$ yields $$\\frac{AZ}{AP} = 1 + \\frac{ZP}{AP} = 1 + \\frac{PY}{AP} = 1 + \\frac{\\sin\\alpha}{\\sin\\beta}.$$Similarly, applying the Law of Sines to $\\triangle ABX$ gives $$\\frac{AZ}{AB} = 1" ]

[ "N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $angle EFD=90^circ$, then how many degrees are in the measure of $angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(“$A$”,a, WNW); label(“$B$”,b, ENE); label(“$C$”,c, E); label(“$D$”,d, ESE); label(“$E$”,e, W); label(“$F$”,f, WSW); draw(d–f–a–b–c–d–e); draw(f+(0,0.1)–f+(0.1,0.1)–f+(0.1 ,0)); [/asy] N the diagram below, points $A,$ $E,$ and $F$ lie on the same line. If $ABCDE$ is a regular pentagon, and $\\angle EFD=90^\\circ$, then how many degrees are in the measure of $\\angle FDE$? [asy] size(5.5cm); pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(a rgu*pi/180)); } pair a=cis(1,144); pair b=cis(1,72); pair c=cis(1,0); pair d=cis(1,288); pair e=cis(1,216); pair f=e-(0,2*sin(pi/5)*sin(pi/10)); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); label(\"$A$\",a, WNW); label(\"$B$\",b, ENE); label(\"$C$\",c, E); label(\"$D$\",d, ESE); label(\"$E$\",e, W); label(\"$F$\",f, WSW); draw(d--f--a--b--c--d--e); draw(f+(0,0.1)--f+(0.1,0.1)--f+(0.1 ,0)); [/asy] This Post Has 10 Comments 1. alexabbarker9781 says: :0000000000000000000000 2. annie8348 says: 108 degrees Step-by-step explanation: angle CDE is 108 degrees, which is supplementary to angle EDH, so EDH must be 72 degrees then put it into an equation 90+90+72+x=360 solve x=108 3. sony72 says: 13.9 units Step-by-step explanation: AB = sqrt(2² + 3²) = sqrt(13) BC =", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $IM=CI\\cdot \\frac{IP}{AI}$. Using a similar method, we can find that $NI=CI\\cdot\\frac{IQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=\\frac{IP}{AI}\\cdot CN+\\frac{IQ}{BI}\\cdot MC=r\\left(\\frac{CN}{AI}+\\frac{MC}{BI}\\right),$$ where $r$ is the inradius of triangle $ABC$. Thus, $r=[ABC]/181$. Also, right triangles $CNA$ and $CBM$ tell us that $CN=117\\sin \\frac{A}{2}$ and $MC=120\\sin\\frac{B}{2}$. But then $[CLA]=\\frac{117\\cdot AL}{2}\\sin\\frac{A}{2}$, and this is equal to $\\frac{117}{242}\\cdot [ABC]$ by the Angle Bisector Theorem. Therefore, solving this for $\\sin\\frac{A}{2}$ and substituting yields $CN=\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL}$. Similarly, $MC=\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK}$. We now replace these in our Ptolemy's expression to get $$MN=\\frac{[ABC]}{181}\\left(\\frac{2\\cdot 117\\cdot [ABC]}{242\\cdot AL\\cdot AI}+\\frac{2\\cdot 120\\cdot [ABC]}{245\\cdot BK\\cdot BI}\\right)$$$$=\\frac{2[ABC]^2}{181}\\left(\\frac{117}{242\\cdot AL\\cdot AI}+\\frac{120}{245\\cdot BK\\cdot BI}\\right).$$ We can also use mass points, assigning masses of $120$, $117$, and $125$ to points $A$, $B$, and $C$, respectively. Then point $L$ has a mass of $242$ and point $K$ has a mass of $245$, so $AI=\\frac{242}{362}\\cdot AL$ and $BI=\\frac{245}{362}\\cdot BK$. This simplifies our expression further to $$MN=4[ABC]^2\\left(\\frac{117}{242^2\\cdot AL^2}+\\frac{120}{245^2\\cdot BK^2}\\right).$$ Then using the angle bisector formula, we find that $AL^2=125\\cdot 117\\left(1-\\frac{120^2}{242^2}\\right)$ and", "following picture, which is not to scale. $[asy] size(200); pair A=(0,0),C=(117,0), B=(100,85),I=incenter(A,B,C),K=extension(B,I,A,C),L=extension(A,I,B,C),M=foot(C,B,K),EN=foot(C,A,L); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C)--cycle,black); draw(B--M--C--EN--A); draw(M--K^^EN--L); MP(\"M\",M,WNW);MP(\"N\",EN,NNW);MP(\"L\",L,ENE);MP(\"K\",K,S);MP(\"I\",I,NW); markscalefactor=.75; draw(rightanglemark(C,M,I)^^rightanglemark(I,EN,C)); pair FAC=foot(I,A,C); pair FBC=foot(I,B,C); MP(\"P\",FAC,S);MP(\"Q\",FBC,ESE); draw(I--FAC^^I--FBC,dotted); draw(C--I); draw(rightanglemark(I,FAC,A)); dot(A);dot(B);dot(C);dot(M);dot(EN);dot(K);dot(L);dot(FAC);dot(FBC);dot(I); [/asy]$ We know that $\\frac{\\angle A+\\angle B+\\angle C}{2}=90^\\circ$. In triangle $CBM$, we have $$90^\\circ=\\angle CBM+\\angle BCI+\\angle ICM=\\frac{\\angle B}{2}+\\frac{\\angle C}{2}+\\angle ICM.$$ Therefore, $\\angle ICM=\\frac{\\angle A}{2}$, and $\\triangle CIM\\sim\\triangle AIP$. Thus $MC=CI\\cdot \\frac{AP}{AI}$. Using a similar method, we can find that $CN=CI\\cdot \\frac{BQ}{BI}$. Therefore, our Ptolemy's expression simplifies to $$MN=IM\\cdot \\frac{BQ}{BI}+NI\\cdot\\frac{AP}{AI}=IM\\cdot\\cos \\frac{B}{2}+NI\\cdot\\cos\\frac{A}{2}.$$ Using right triangles $CIM$ and $NCI$, we also know that $IM=CI\\cdot\\sin \\frac{A}{2}$ and $NI=CI\\cdot\\sin\\frac{B}{2}$. Thus $$MN=CI\\cdot\\cos \\frac{B}{2}\\sin\\frac{A}{2}+CI\\cdot\\cos\\frac{A}{2}\\sin\\frac{B}{2}=CI\\cdot\\sin\\left(\\frac{A+B}{2}\\right)=CI\\cdot \\cos \\frac{C}{2}.$$ But this last expression is equal to $CQ$. This a tangent to the incircle, so it has length $s-c=181-125=\\boxed{56}$.", "$[asy] currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label(\"O\", O, S); label(\"A\", A, SW); label(\"B\", B, SE); label(\"X\", X, (X-B)/length(X-B)); label(\"Y\", Y, Y); label(\"Z\", Z, (Z-A)/length(Z-A)); label(\"P\", P, (P-T)/length(P-T)); label(\"Q\", Q, SW); label(\"R\", R, SE); label(\"S\", SS, (SS-T)/length(SS-T)); label(\"T\", T, S); [/asy]$ Let $T$ be the foot of the perpendicular from $Y$ to $\\overline{AB}$, let $O$ be the center of the semi-circle. Since we have a semi-circle, if we were to reflect it over $\\overline {AB}$, we would have a full circle, with", "pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"(0,0)\", A, SW); label(\"(1,0)\", B, SE); label(\"(1,1)\", C, E); label(\"(0,1)\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"P\", P, W); draw(Q -- R, dashed); label(\"\\frac 38\", H--K, E); label(\"\\frac 58\", K--J, W); label(\"\\frac 7{16}\", G--C, E); label(\"\\frac 38\", C--J, N); label(\"\\frac 1{16}\", A--F, dir(160)); [/asy]$ ~IceMatrix ~avn", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "(7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw] That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.", "# Problem #2055 2055 The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$. $[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); [/asy]$ $\\mathrm{(A)}\\ \\frac{4\\sqrt{5}}{3} \\qquad\\mathrm{(B)}\\ \\frac{5\\sqrt{5}}{3} \\qquad\\mathrm{(C)}\\ \\frac{12\\sqrt{5}}{7} \\qquad\\mathrm{(D)}\\ 2\\sqrt{5} \\qquad\\mathrm{(E)}\\ \\frac{5\\sqrt{65}}{9}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "be humble! ## Solution 2 [asy] currentpicture=new picture; size(12cm); pair O, A, B, X, Y, Z, P, Q, R, SS, T; O=(0, 0); A=(-1, 0); B=(1, 0); X=(Cos(144), Sin(144)); Y=(Cos(105), Sin(105)); Z=(Cos(27), Sin(27)); P=foot(Y, A, X); Q=foot(Y, B, X); R=foot(Y, A, Z); SS=foot(Y, B, Z); T=foot(Y, A, B); dot(O); dot(A); dot(B); dot(X); dot(Y); dot(Z); dot(P); dot(Q); dot(R); dot(SS); dot(T); draw(arc(O, 1, 0, 180)); draw(circumcircle(T, A, Y), dotted); draw(circumcircle(T, B, Y), dotted); draw(A -- B); draw(Z -- O -- X -- A -- Z -- B -- X); draw(A -- Y -- B); draw(P -- T -- SS); draw(P -- Y -- Q); draw(R -- Y -- SS); draw(X -- P); draw(Z -- SS); draw(Y -- T); draw(rightanglemark(Y, T, B, 1.25)); draw(rightanglemark(Y, P, A, 1.25)); draw(rightanglemark(Y, Q, X, 1.25)); draw(rightanglemark(Y, R, Z, 1.25)); draw(rightanglemark(Y, SS, B, 1.25)); draw(rightanglemark(A, X, B, 1.25)); draw(rightanglemark(A, Y, B, 1.25)); draw(rightanglemark(A, Z, B, 1.25)); label(\"$O$\", O, S); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$X$\", X, (X-B)/length(X-B)); label(\"$Y$\", Y, Y); label(\"$Z$\", Z, (Z-A)/length(Z-A)); label(\"$P$\", P, (P-T)/length(P-T)); label(\"$Q$\", Q, SW); label(\"$R$\", R, SE); label(\"$S$\", SS, (SS-T)/length(SS-T)); label(\"$T$\", T, S); [/asy] Let $T$ be the projection of $Y$ onto $\\seg{AB}$ (Error compiling LaTeX. ! Undefined control sequence.). Notice that $T$ lies on the Simson Line $\\seg{PQ}$ (Error compiling LaTeX. ! Undefined control sequence.) from $Y$ to $\\triangle", "dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label(\"P\",(4,4),NE); draw((0,4)--(3,4)); draw((0,8)--(3,5)); draw((8,8)--(5,5)); draw((4,8)--(4,5)); draw((4,0)--(4,3)); draw((0,0)--(3,3)); draw((8,0)--(5,3)); draw((5,4)--(8,4)); [/asy]$ Lines of symmetry go through point $P$, and there are $8$ directions the lines could go, and there are $4$ dots at each direction.$\\frac{4\\times8}{80}=\\boxed{\\textbf{(C)} \\frac{2}{5}}$. Video Solution Solution detailing how to solve the problem: https://www.youtube.com/watch?v=4L95z9DwlhI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=7", "black +linewidth(0.6); pen s = fontsize(10); pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); /* construct remaining points */ pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); D(MP(\"A\",A,s)--MP(\"B\",B,N,s)--MP(\"C\",C,s)--cycle); dot(MP(\"D\",D,NE,s));dot(MP(\"E\",E,NW,s));dot(MP(\"F\",F,s));dot(MP(\"D'\",Da,NE,s));dot(MP(\"E'\",Ea,NW,s));dot(MP(\"F'\",Fa,s)); D(D--Ea);D(Da--F);D(Fa--E); MP(\"450\",(B+C)/2,NW);MP(\"425\",(A+B)/2,NE);MP(\"510\",(A+C)/2); [/asy]$ Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($\\triangle ABC \\sim \\triangle DPD' \\sim \\triangle PEE' \\sim \\triangle F'PF$). The remaining three sections are parallelograms. Since $PDAF'$ is a parallelogram, we find $PF' = AD$, and similarly $PE = BD'$. So $d = PF' + PE = AD + BD' = 425 - DD'$. Thus $DD' = 425 - d$. By the same logic, $EE' = 450 - d$. Since $\\triangle DPD' \\sim \\triangle ABC$, we have the proportion: $\\frac{425-d}{425} = \\frac{PD}{510} \\Longrightarrow PD = 510 - \\frac{510}{425}d = 510 - \\frac{6}{5}d$ Doing the same with $\\triangle PEE'$, we find that $PE' =510 - \\frac{17}{15}d$. Now, $d = PD + PE' = 510 - \\frac{6}{5}d + 510 - \\frac{17}{15}d \\Longrightarrow d\\left(\\frac{50}{15}\\right) = 1020 \\Longrightarrow d = \\boxed{306}$. ### Solution 3 Define the points the same as above. Let $[CE'PF] = a$, $[E'EP] = b$, $[BEPD'] = c$, $[D'PD] = d$, $[DAF'P] = e$ and $[F'D'P] = f$ The key theorem we apply here is that", "as shown below: $[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot(\"A\",A,1.5*dir(225),linewidth(4.5)); dot(\"B\",B,1.5*dir(-45),linewidth(4.5)); dot(\"C\",C,1.5*dir(45),linewidth(4.5)); dot(\"D\",D,1.5*dir(135),linewidth(4.5)); dot(\"P\",P,1.5*dir(60),linewidth(4.5)); dot(\"R\",R,1.5*dir(135),linewidth(4.5)); dot(\"O\",O,1.5*dir(45),linewidth(4.5)); dot(\"X\",X,1.5*dir(135),linewidth(4.5)); dot(\"G\",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label(\"9\",midpoint(P--R),dir(A-D),red); label(\"12\",midpoint(R--X),dir(135),red); label(\"15\",midpoint(X--P),dir(0),red); label(\"25\",midpoint(G--D),dir(0),red); [/asy]$ As $\\angle PRX = \\angle AGD = 90^\\circ$ and $\\angle PXR = \\angle ADG$ by the AA Similarity, we conclude that $\\triangle PRX \\sim \\triangle AGD.$ The ratio of similitude is $$\\frac{PX}{AD} = \\frac{RX}{GD}.$$ We get $\\frac{15}{AD} = \\frac{12}{25},$ from which $AD = \\frac{125}{4}.$ Finally, the perimeter of $ABCD$ is $4AD = \\boxed{125}.$ ~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb) ## Solution 2 This solution refers to the Diagram section. Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\\overleftrightarrow{RT}.$ We obtain the", "and $EF$. They are as follows:$$AG = f(x) = -\\frac{3}{5}x + 4$$$$AC = g(x) = -\\frac{4}{5}x + 4$$$$EF = h(x) = 2x - 4$$After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"P'\", (20/7,0),SSW); label(\"Q'\", (40/13,0),SSE); label(\"E'\", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy]$ Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\\frac{20}{7}$ and $\\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \\frac{40}{13} - \\frac{20}{7} = \\frac{20}{91}$. Now we can find $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F} = \\frac{\\frac{20}{91}}{2} = \\boxed{\\textbf{(D)}~\\frac{10}{91}}$ ## Solution 3 (Similar Triangles) $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"H\",H,E); [/asy]$ Extend $AG$ to intersect $CD$ at $H$. Letting $x=\\overline{HC}$, we have that $$\\triangle{HCG}\\sim\\triangle{HDA}\\implies \\dfrac{\\overline{HC}}{\\overline{CG}}=\\dfrac{\\overline{HD}}{\\overline{AD}}\\implies \\dfrac{x}{1}=\\dfrac{x+5}{4}\\implies x=\\dfrac{5}{3}.$$ Then, notice that $\\triangle{AEQ}\\sim\\triangle{HFQ}$ and $\\triangle{AEP}\\sim\\triangle{CFP}$. Thus, we see that", "3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ \\tikznode{NOV-1-B}{1}, 2, 3, \\dots, \\tikznode{NOV-30-B}{30} & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 & 1, 2, 3, \\dots, 30 \\\\ & & & & \\\\ & & & & \\end{tabular} \\end{table} \\end{center} \\begin{tikzpicture}[overlay,remember picture] \\node[rounded corners=2mm,draw,fit=(NOV-1-T)(NOV-1-B)] (NOV-L){}; \\node[rounded corners=2mm,draw,fit=(NOV-30-T)(NOV-30-B)] (NOV-R){}; \\draw[thick,-latex] (NOV-L.south) -- ++(0,-0.5) node[below]{$\\displaystyle \\frac{\\sum d_1}{18}$}; \\draw[thick,-latex] (NOV-R.south) -- ++(0,-0.5) node[below]{$\\displaystyle \\frac{\\sum d_1}{18}$}; \\end{tikzpicture} \\end{document} ADDENDUM: Let TeX do", "side of the cube. $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); [/asy]$ The two triangles on the right and left of the rectangle are also $45-45-90$ triangles because the rectangle is perpendicular to the base, and they share a $45^\\circ$ angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as $s.$ $[asy] unitsize(35mm); defaultpen(linewidth(2pt)+fontsize(12pt)); pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2)); pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1); draw(A--B--C--cycle); draw(W--X--Y--Z--cycle,red); label(\"1\",(A--C),NW); label(\"1\",(B--C),NE); label(\"\\sqrt{2}\",(A--B),S); label(\"s\",(W--Z),E,red); label(\"s\",(X--Y),W,red); label(\"s\\sqrt{2}\",(W--X),N,red); label(\"s\",(A--W),N); label(\"s\",(X--B),N); [/asy]$ Now we can just use segment addition to find the value of $s.$ $$\\sqrt{2}=s+s\\sqrt{2}+s=2s+s\\sqrt{2}=(2+\\sqrt{2})s$$ $$s=\\frac{\\sqrt{2}}{2+\\sqrt{2}}=\\frac{1}{\\sqrt{2}+1}=\\frac{\\sqrt{2}-1}{2-1}=\\sqrt{2}-1$$ The volume of the cube is $s^3 = (\\sqrt{2}-1)^3 = (\\sqrt{2}-1)(3-2\\sqrt{2}) = 3\\sqrt{2}-3-4+2\\sqrt{2} = \\boxed{\\textbf{(A)} 5\\sqrt{2}-7}$ Solution 2 Let $s$, $d$ be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is $$a=\\sqrt{s^2-d^2}=\\sqrt{(\\frac{\\sqrt{3}}{2})^2+(\\frac{1}{2})^2}=\\frac{\\sqrt{2}}{2}$$ Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting $x$ be the edge length of the cube, then the altitude of the smaller pyramid is $\\frac{\\sqrt{2}}{2}x$. Since the altitude of the larger pyramid is the sum of the edge length of the cube and", "= foot(Y, A, X); dot(P); label(\"P\", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label(\"Q\", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label(\"S\", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label(\"R\", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label(\"T\", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l=\"\", real r=40, int n=1, int nm = 0) { string sl = \"\\scriptstyle{\" + l + \"}\"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, \"\\alpha\" ); langle(X, B, A, \"\\beta\", n=2); langle(Y, A, X, \"\\gamma\", nm=1); langle(Y, B, X, \"\\gamma\", nm=1); langle(Z, A, Y, \"\\delta\", nm=2); langle(Z, B, Y, \"\\delta\", nm=2); langle(R, S, Y, \"\\alpha+\\delta\", r=23); langle(Y, Q, P, \"\\beta+\\gamma\", r=23); langle(R, T, P, \"\\chi\", r=15); [/asy]$ Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ", "Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). ## Solution 3 $[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot(\"C\", C, dir(0)); dot(\"A\", A, dir(90));dot(\"B\", B, dir(180)); dot(\"D\", D, NE); dot(\"E\", E, dir(90));dot(\"F\", F, dir(270)); dot(\"M\", M, NE); dot(\"N\", N, dir(270));dot(\"L\", L, NW); dot(\"X\", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]$ Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\\sin(60)$. Similarly, $AB=3\\sin(60)$, and $BC=5\\sin(60)$. So $\\triangle ABC$ is a 3-4-5 right triangle", "the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former. The $\\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3),s); draw((0,0)--(3,3)); [/asy]$ The $\\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3)^^(1,0),s); draw((0,0)--(3,3),c); draw((3,3)--(1,0)); [/asy]$ From $(1,0)$, there are two possible ways to move $\\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$. However, from $(0,3)$, there is no way to move $\\sqrt{9}$ away, so we discard it as a possibility. From $(2,3)$, the lengths of $\\sqrt{8},\\ \\sqrt{5},\\ \\sqrt{4},\\ \\sqrt{2}$ fortunately are all", "(f_{12}, f_{12}), \\dots, (f_{12}, f_{18}), \\dots, (f_{18}, f_{10}), (f_{18}, f_{11}), (f_{18}, f_{12}), \\dots, (f_{18}, f_{18}),}} \\\\ {\\color {Green}{(f_{19}, f_{19}), (f_{19}, f_{20}), (f_{19}, f_{21}), \\dots, (f_{19}, f_{27}), (f_{20}, f_{19}), (f_{20}, f_{20}), (f_{20}, f_{21}), \\dots, (f_{27}, f_{27}), (f_{21}, f_{21}), (f_{21}, f_{20}), (f_{21}, f_{21}), \\dots, (f_{21}, f_{27}), \\dots, (f_{27}, f_{19}), (f_{27}, f_{20}), (f_{27}, f_{21}), \\dots, (f_{27}, f_{27})}} \\right \\}$ Therefore, there are 3 equivalence classes which are as follows- • Class 1 - $[f_1] = [f_2] = [f_3] = \\dots = [f_9] = \\left \\{ f_1, f_2, f_3, \\dots, f_9 \\right \\}$ • Class 2 - $[f_{10}] = [f_{11}] = [f_{12}] = \\dots = [f_{18}] = \\left \\{ f_{10}, f_{11}, f_{12}, \\dots, f_{18} \\right \\}$ • Class 3 - $[f_{19}] = [f_{20}] = [f_{21}] = \\dots = [f_{27}] = \\left \\{ f_{19}, f_{20}, f_{21}, \\dots, f_{27 }\\right \\}$ Therefore, we can see that there are 3 equivalence classes, each having 9 elements.", "C++. If one is new to all of this, asymptote may be easier to learn (it wasn't for me, I prefer metapost for 2d drawing). For example, in asymptote the following code generates the diagram: pair midpoint(pair XXX, pair YYY) { real a,b; a = (XXX.x + YYY.x)/2; b = (XXX.y + YYY.y)/2; return (a,b); } //real u = 14.1732284; // the number of \"size points\" in 0.5 centimeters dotfactor=8; // changes the size of a dot real u = 14; // (u=72)==(u=1 inch); pair z[]; // all the points we are working with pair c1; // select points we are making red pair c2; /////////////////////////////////////////////////////////////////// // this draws one \"factor\" of $!$P$!$ /////////////////////////////////////////////////////////////////// dot((0*u,0*u)); dot((0*u,1*u)); dot((0*u,-1*u)); draw((-1*u,0*u)..(0*u,2*u)..(1*u,0*u)..(0*u,-2*u)..cycle); label(\"$!$C_{1}$!$\",(0*u,-2*u),S); c1 = (0*u,0*u); /////////////////////////////////////////////////////////////////// // this draws the other \"factor\" of $!$P$!$ /////////////////////////////////////////////////////////////////// dot((13*u,0*u)); dot((14*u,0*u)); c2 = (13*u,0*u); draw((12*u,0)..(13.5*u,1*u)..(15*u,0)..(13.5*u,-1*u)..cycle); label(\"$!$C_{2}$!$\",(13.5*u,-1*u),S); /////////////////////////////////////////////////////////////////// // this draws $!$P$!$ /////////////////////////////////////////////////////////////////// label(\"$!$P$!$\",(7*u,0),S); /////////////////////////////////////////////////////////////////// // this draws $!$D$!$ /////////////////////////////////////////////////////////////////// label(\"$!$\\mathbf{1}$!$\",(7*u,5*u),N); // draws $!$\\<f, g\\>$!$ draw((7*u,5*u)--(7*u,0*u),linetype(\"8 8\"),EndArrow); label(\"$!$\\langle f,g\\rangle$!$\",(7*u,2.5*u),E); /////////////////////////////////////////////////////////////////// // draw the morphisms f and g /////////////////////////////////////////////////////////////////// z[0] =(6.5*u,5.25*u); z[1] = c1; //z[1] = (0*u,2*u); draw (z[0]--z[1], EndArrow); label(\"$!$f$!$\", midpoint(z[0],z[1]),NW); z[2] = (7.5*u,5.25*u); z[3] = c2; //z[3] = (13.5*u,1*u); draw (z[2]--z[3], EndArrow); label(\"$!$g$!$\", midpoint(z[2],z[3]),NE); /////////////////////////////////////////////////////////////////// // draw projections /////////////////////////////////////////////////////////////////// z[4] =(6.5*u,-0.5*u); z[5] = c1; //z[5] =(1*u,0); draw (z[4]--z[5], EndArrow); label(\"$!$\\pi_{1}$!$\", midpoint(z[4],z[5]),S); z[6] =(7.5*u,-0.5*u); z[7] =" ]

[ "# Pages have got numbers too! A book have 273 pages. Find the number of digits used for the numbering of all the pages. ×", "# [NTG-context] reverse page order and reverse page numbering in one part of a book Idris Samawi Hamid ادريس سماوي حامد Idris.Hamid at colostate.edu Wed Jul 22 13:58:04 CEST 2015 ```Dear Talal, Salaam. See below: On Wed, 22 Jul 2015 03:49:29 -0600, talazem at fastmail.fm <talazem at fastmail.fm> wrote: > Dear all, > > I am producing a book that is divided into two sections. The book is > published in English, and thus it runs left to right. Part one is an > academic study in English. Part two is a critical edition of an Arabic > text. The book is exactly 100 pages long, with fifty pages going to the > English study, and fifty pages to the Arabic edition. > > In Part One (which includes all the front matter of the book, such as > pagination runs from 1 to 50, and the page numbers should be printed in > Arabic numerals. > > However, in Part Two (the Arabic text): the page ordering should run > right-to-left, the page numbers should run in “reverse” pagination (in > which “page 1\" of the critical edition begins at the “back” of the > book), and the page numbers should be printed in Hindi numerals. > > In other words, the structure of the book’s pages", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers. 10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.) Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "The pages of a book are numbered 1 through n. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in the incorrect sum of 1986. What was the number of the page that was added twice? (第四届AIME1986 第6题)", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: There are N pages, so there will be N unit digits, N - 9 (the number of unit digits) ten digits, and N - 99 hundred digits. N + N - 9 + N - 99 = 930; 3 N = 930 + 108 ; N = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, leave the odd 1 out for now. 40 divided by 2 = 20 and N - 10 + 1 or N - 9 = 20 N = 29 so the last digit you write is the next one, which is \"3\". Solution II: Let there be n numbers: n + (n - 9) = 50, 2n = 59, n = 29.5 which mean the last digit you write is the first digit of the next number, which is \"30\", so 3 is the answer.", "# Page numbers I have a book, each page is numbered from 1,2,3,... in that order, with the odd-numbered pages fall on the right side of the book; while the even-numbered pages fall on the left side of the book. If I multiply 2 consecutive page numbers, I will get a product of 6162. Among these 2 pages, what is the page number of the right side of the book? Try this: All of my problems ×", "[–] 4 points5 points (6 children) I agree with you on the first. I get 580 for the second (but I've been known to derp the underpable) edit: off by one... I also agree with 579. [–] 4 points5 points (1 child) I'm pretty sure it's 579. Because. [–] 0 points1 point (0 children) Yes, I have an off by one error. I solved 9 + 90*2 + (x-100) * 3 = 1629, but if there are 100 pages then there still is at least one page with three digits. [–] 3 points4 points (2 children) I get this: 9 digits for pages 1- 9 180 digits for pages 10- 99 1200 digits for pages 100-499 240 digits for pages 500-579 So I get 579 pages. [–] -2 points-1 points (1 child) the question is do you count the backside of the page even if it's not numbered? [–] 0 points1 point (0 children) I considered it fair to assume the answer would be the same if the pages were all printed one on side or two. [–] 2 points3 points (0 children) I independently got 480, for what it's worth. Edit: Other posters have shown my answer to be wrong in fairly explicit ways. I just did it quickly and stopped trying when it matched another answer.", "solution I) From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers. 189 + 46*3 = 327 digits. #3: \"A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have\"? Similar to one Google interview question. Read the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers +", "• # question_answer The total number of digits used in numbering the pages of a book having 366 pages, is? A) 732 B) 990C) 1098 D) 1305 Required numbers $=9+2\\times 90+267\\times 3$ $=189+801$", "Exactly $195$ digits have been used to number the pages in a book. How many pages does the book have?", "# Reverse page numbering How can one make LaTeX print the pages numbering in reverse order, in the sense that if the document in question has N pages then the first page is numbered N, then N-1, ...? - Out of curiosity: why do you want to do this? – Charles Stewart Nov 27 '10 at 10:58 @Charles: I'd assume it's for an (boring) continuum mechanics textbook or similarly complex discipline, so that the students have an idea how many pages are there left for them to read.. – Martin Tapankov Nov 29 '10 at 12:03 @Martin: This sounds depressingly plausible. – Charles Stewart Nov 29 '10 at 14:05 Too good question, I am that person who likes to have reverse page numbering, it is very motivating to read a book with this numbering. – Kirk Hammett Sep 18 '11 at 18:51 Here's an idea: 1. Run the the *tex compilation as usual, and calculate the number of pages. 2. On the second run, customize the page number to be shown as TotalPageNum-CurrentPagenum+1 (calculated by a formula, that is). 3. There is no step 3. You probably can skip doing this until your document is finished, so that this is a one-off operation. There are so many things that might be unfeasible or outright difficult, but perhaps that", "subtract the addend with the sum which is 46 – 14 and the result will be 32. So we have found out the missing digit which is 2. And we can see in the second addition that one number is missing in the second addend. So here we will subtract the addend with the sum which is 69 – 45 and the result will be 24. So we have found out the missing digit which is 4. And we can see in the third addition that one number is missing in the second addend. So here we will subtract the addend with the sum which is 61 – 87 and the result will be 26. So we have found out the missing digit which is 2. Think and Grow: Modeling Real Life You read 34 pages one day and 23 the next day. How many pages do you read in all? The number of pages that were read is 57 pages. Explanation: Given that 34 pages were read on one day and the other 23 pages were read on the next day. So to find the number of pages that were read, we will add the number of pages which was read on both days. So the number of pages that were read is 34 + 23 which", "## Saturday, June 9, 2012 ### Unit digit, Tenth digit and Digit Sum Word problems on unit digit, tenth digit or digit sum. #1: How many digits are there in the positive integers 1 to 99 inclusive? Solution I: From 1 to 9, there are 9 digits. From 10 to 99, there are 99 - 10 + 1 or 99 - 9 = 90 two digit numbers. 90 x 2 = 180 Solution II: ___ There are 9 one digit numbers (from 1 to 9). ___ ___ There are 9 * 10 = 90 two digit numbers (You can't use \"0\" on the tenth digit but you can use \"0\" on the unit digit.) 90 * 2 + 9 = 189 # 2: A book has 145 pages. How many digits are there if you start counting from page 1? There are 189 digits from page 1 to 99. (See #1, solution I) From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers. 189 + 46*3 = 327 digits. #3: \"A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have\"? Similar to one Google interview question. Read", "-> 6 digits/page -> 5394 start at 852 digits and work your way up through the pages by subtracting the above values to calculate amount of pages with 6 digits (852 - 18 - 3 - 356 - 5)/6 = 80 add all pages = 9 + 1 + 89 + 1 + 80 = 180 13. guido says: pages 1 to 9 : 9 digits #7 : 1 pages 10 to 99 : 180 digits --- total : 189 #7 :10 pages100 to199 : 300 digits --- total : 489 #7 :11 pages200 to299 : 300 digits --- total : 789 #7 :11 pages300 to319 : 60 digits --- total : 849 #7 : 2 pages320 to322 : 3 digits --- total : 852 So book has 322 pages and 35 x digit 7 14. Michael says: (a) We begin by noting the different types of pages: pages containing one digit, two digits, and three digits. Pages 1 through 9 (9 pages, or 1 * 9 total digits) each occupy one of the 852 digits. Pages 10 through 99 (90 pages, or 2 * 90 total digits) each occupy two of the 852 digits. Pages 100 through 999 (900 pages, or 3 * 900 total digits) each occupy three of the 852 digits; however, logic indicates that", "volume = {23}, number = {4}, pages = {501-509} }", "# Pages In a book there are 100 pages . From this book some pages are torn off. The sum of the pages remaining is 4949. How many pages are missing from the book? ×", "page between pages 76-77. 8 unnumbered pages between pages 83-84. Unnumbered page between pages 97-98. 2 unnumbered pages between pages 118-119. 2 unnumbered pages between pages 120-121. 3 unnumbered pages between pages 121-122. 2 unnumbered pages between pages 124-125. 2 unnumbered pages between pages 125-126. Unnumbered page between pages 126-127. 3 unnumbered pages between pages 128-129. 2 unnumbered pages between pages 129-130. 2 unnumbered pages between pages 132-133. Unnumbered page between pages 139-140. Unnumbered page between pages 140-141. Unnumbered page between pages 141-142. Unnumbered page between pages 143-144. Unnumbered page between pages 151-152. https://livrepository.liverpool.ac.uk/id/eprint/3146623", "the odd page which is on the right, this however involves a waste of sheets, since obviously it leaves blank pages to skip the even-numbered page.", "there would be pages missing between 58 and 100. second: The number of digits in the 3 digit pages is 195 - (number of digits in 1 digit pages) - (number of digits in 2 digit pages) =195 - 9 - 90*2 = 6 3. Thanks @SpringFan25. I feel really stupid now. I see how I mixed up no. of pages and no. of digits used! So the digits used are $9 \\times 1 + 90 \\times 2 + 3 \\times 2 = 195$ And hence Total no of pages, $9 + 90 + 3 = 101$." ]

[ "# How many times does the digit 4 occur in the page numbers of a 50 page book? Dec 13, 2016 $15$ #### Explanation: The digit $4$ can occur in the units or tens positions. It occurs in the units position in the numbers: $4 , 14 , 24 , 34 , 44$ It occurs in the tens position in the numbers: $40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49$ So there are a total of $5 + 10 = 15$ occurrences of the digit $4$. If the question asked in how many page numbers the digit $4$ occurs then the answer would be $14$ - avoiding counting $44$ twice. However, since it asks for the number of occurrences of the digit $4$, $44$ should be counted twice.", "+0 # A textbook has pages. How many of the pages have page numbers whose digits add up to exactly ​? 0 37 2 A textbook has $$1000$$ pages. How many of the pages have page numbers whose digits add up to exactly $$4$$? Feb 14, 2023 #1 0 If the textbook has 1000 pages, then the page numbers range from 1 to 1000. To find the number of pages with page numbers whose digits add up to exactly 4, we can use the following formula: Number of pages = $\\sum_{i=1}^{9}$(Number of pages with i as the last digit) For example, the number of pages with 0 as the last digit is the same as the number of pages with 4 as the last digit, as both digits add up to 4. Similarly, the number of pages with 1 as the last digit is the same as the number of pages with 3 as the last digit, as both digits add up to 4. Using this formula, we can calculate that the number of pages with page numbers whose digits add up to exactly 4 is 12. Feb 14, 2023 #2 0 1 == 4 2 == 13 3 == 22 4 == 31 5 == 40 6 == 103 7 == 112 8 == 121 9 ==", "After every two numbers, one is omitted. Because $89 = 2\\times 44 +1$, there must be $44$ page numbers missing and so the number on the last page is $89+44=133$.", "After every two numbers, one is omitted. Because $89 = 2\\times 44 +1$, there must be $44$ page numbers missing and so the number on the last page is $89+44=133$. This problem is taken from the UKMT Mathematical Challenges. View the previous week's solution", "the following is correct: I,$f(x)$and$g(x)\\$ have ... 15 30 50 per page", "there would be pages missing between 58 and 100. second: The number of digits in the 3 digit pages is 195 - (number of digits in 1 digit pages) - (number of digits in 2 digit pages) =195 - 9 - 90*2 = 6 3. Thanks @SpringFan25. I feel really stupid now. I see how I mixed up no. of pages and no. of digits used! So the digits used are $9 \\times 1 + 90 \\times 2 + 3 \\times 2 = 195$ And hence Total no of pages, $9 + 90 + 3 = 101$.", "### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### 14 Divisors What is the smallest number with exactly 14 divisors? ### Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? # Newspaper Sheets ##### Stage: 2 and 3 Short Challenge Level: If page $1$ is on the outside of the first sheet, then page $2$ is on the inside. Pages $3$ and $4$ are on the same sheet, as are pages $5$ and $6$. This therefore shares a sheet with page $7$, so is the inside of the sheet. On the outside on the other side is therefore page $62$. The other three sheets have pages $63$ and $64$, $65$ and $66$ and $67$ and $68$ on them. Therefore there are $68$ pages, and four sheets per page, so $17$ sheets. Alternatively, the sum of the page numbers on the same side of the sheet is always constant, since one side increases by $1$ every time the other decreases by $1$. This means the total is always $8 + 61 = 69$, so page $1$ shares with page $68$. Hence there are", "Karuna-bdc May 26 '13 at 6:22 pages 6 & 27 are mentioned in the question. its means there are $5$ pages on the left of page no. 6, obviously there will be also $5$ pages more on the right of $27$. so, $27 + 5 = 32$. -", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "have an average page number of $19$ which is less than $25.5$, the average page number of all $50$ pages, therefore $k>0$. Since the borrowed sheets start with an odd page number and end with an even page number we have $k \\in \\mathbb N$. We notice that $n < 25$ and $k \\le (49+50)/2-25.5=24<25$. The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$, where the weights are the page number in each group (borrowed vs. remained), therefore $$2nk=2(25-n)(25.5-19)=13(25-n) \\implies 13 | n \\text{ or } 13 | k$$ Since $n, k < 25$ we have either $n=13$ or $k=13$. If $n=13$ then $k=6$. If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\\boxed{\\textbf{(B)} ~13}$. ~asops ## Solution 4 Let $(2k-1)-2n$ be pages be borrowed, the sum of the page numbers on those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$. As $325=25\\cdot13$, we can see $n-k=13$ and that is desired $\\boxed{\\textbf{(B)} ~13}$. ~bluesoul ## Solution 5 Let $c$ be the number of consecutive sheets Hiram’s roommate borrows, and let", "3) 1 = 3 + 3 – 5 2 = 5 – 3 3 = 5 + 3 – 5 4 = 3 + 3 + 3 – 5 5 = 3 + 5 – 3 6 = 5 + 5 + 5 – 3 – 3 – 3 7 = 5 + 5 – 3 8 = 5 + 3 9 = 5 + 5 + 5 – 3 – 3 10 = 5 + 5 + 3 – 3 Question 6. Find all 2-digit numbers each of which is divisible by the sum of its digits. The numbers are 10, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72 and 81. Question 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book? Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number. The remaining number digits i.e., (216 – (9 + 180)) = 27 27 is formed by pages 100 to 108 (9 pages) ∴ The total number of pages = 9 + ’90 + 9 = 108 Question 8. Look at the following pattern. This is called Pascal’s triangle. What is the", "-> 6 digits/page -> 5394 start at 852 digits and work your way up through the pages by subtracting the above values to calculate amount of pages with 6 digits (852 - 18 - 3 - 356 - 5)/6 = 80 add all pages = 9 + 1 + 89 + 1 + 80 = 180 13. guido says: pages 1 to 9 : 9 digits #7 : 1 pages 10 to 99 : 180 digits --- total : 189 #7 :10 pages100 to199 : 300 digits --- total : 489 #7 :11 pages200 to299 : 300 digits --- total : 789 #7 :11 pages300 to319 : 60 digits --- total : 849 #7 : 2 pages320 to322 : 3 digits --- total : 852 So book has 322 pages and 35 x digit 7 14. Michael says: (a) We begin by noting the different types of pages: pages containing one digit, two digits, and three digits. Pages 1 through 9 (9 pages, or 1 * 9 total digits) each occupy one of the 852 digits. Pages 10 through 99 (90 pages, or 2 * 90 total digits) each occupy two of the 852 digits. Pages 100 through 999 (900 pages, or 3 * 900 total digits) each occupy three of the 852 digits; however, logic indicates that", "SEARCH HOME Math Central Quandaries & Queries Question from Chungus: It takes 852 digits to number the pages of a book consecutively. How many pages are there? Hi Chungus, There are 9 one digit pages, pages 1 to 9. Next look at 2 digit numbers. They go from 10 to 99 so that is 90 two digit numbers. Each contains 2 digits so when you get to 99 you have used $9 + 2 \\times 90 = 189$ digits. If we apply a similar argument to three digit numbers we will see that we have gone too far so we need to look at three digit numbers in smaller chunks. From 100 to 199 there are 100 three digit numbers. Since each has 3 digits that's $3 \\times 100 = 300$ digits so when we get to page 199 we have used $9 + 2 \\times 90 + 3 \\times 100 = 489$ digits.", "3 Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$. The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$, the average page number of all $50$ pages, therefore $k>0$. Since the borrowed sheets start with an odd page number and end with an even page number we have $k \\in \\mathbb N$. We notice that $n < 25$ and $k \\le (49+50)/2-25.5=24<25$. The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$, where the weights are the page number in each group (borrowed vs. remained), therefore $$2nk=2(25-n)(25.5-19)=13(25-n) \\implies 13 | n \\text{ or } 13 | k$$ Since $n, k < 25$ we have either $n=13$ or $k=13$. If $n=13$ then $k=6$. If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\\boxed{\\textbf{(B)} ~13}$. ~asops ## Solution 4 Let $(2k-1)-2n$ be pages be borrowed, the sum of digits in those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$. As $325=25\\cdot13$, we can see $n-k=13$ and that is desired $\\boxed{\\textbf{(B)} ~13}$.", "# How many 1's are used in of all the page numbers of a book from $1 - 111$? $\\begin{array}{11} (a)\\; 24 \\\\ (b)\\; 42 \\\\ (c)\\; 39\\\\ (d)\\; 16 \\end{array}$", "years). $8,20,28$ $16,28,36$ $20,35,45$ None of the above options 7 The sequence $s_{0},s_{1},\\dots , s_{9}$ is defined as follows: $s_{0} = s_{1} + 1$ $2s_{i} = s_{i-1} + s_{i+1} + 2 \\text{ for } 1 \\leq i \\leq 8$ $2s_{9} = s_{8} + 2$ What is $s_{0}$? $81$ $95$ $100$ $121$ $190$ 8 A contiguous part, i.e., a set of adjacent sheets, is missing from Tharoor's GRE preparation book. The number on the first missing page is $183$, and it is known that the number on the last missing page has the same three digits, but in a different order. Note that every sheet ... one at the front and one at the back. How many pages are missing from Tharoor's book? $45$ $135$ $136$ $198$ $450$ 9 What is the maximum number of regions that the plane $\\mathbb{R}^{2}$ can be partitioned into using $10$ lines? $25$ $50$ $55$ $56$ $1024$ Hint: Let $A(n)$ be the maximum number of partitions that can be made by $n$ lines. Observe that $A(0) = 1, A(2) = 2, A(2) = 4$ etc. Come up with a recurrence equation for $A(n)$. 10 Please help with this question:- $(\\sqrt{243}+3)^x+(\\sqrt{243}-3)^x=15^x$. 1 vote 11 Oar is to rowboat as foot is to running sneaker skateboard jumping 12 For all integers $y>1, \\: \\langle y \\rangle", "$$\\frac {1}{13x-x^2}=\\frac {1}{40}$$ 40=13x-x2 x2-13x+40=0 x2-5x-8x+40=0 x(x-5)-8(x-5)=0 (x-8)(x-5)=0 x=8, 5 The number is 58 or 85. Note: Join free Sanfoundry classes at Telegram or Youtube 4. The sum of the squares of the left and right pages of a book is 481. What are the page numbers? a) 11, 12 b) 12, 13 c) 17, 18 d) 15, 16 Explanation: Since the pages of books are consecutive numbers, so let the left page number be x. The right page number will be x+1. Sum of the squares of the pages is 481 x2+(x+1)2=481 x2+x2+1+2x=481 2x2+2x-480=0 x2+x-240=0 x2+16x-15x-240=0 x(x+16)-15(x+16)=0 (x-15)(x+16)=0 x=15, -16 Since, page number cannot be negative, so x=15 The two page numbers are 15 and 16. 5. The sum of the squares of two consecutive positive even numbers is 3364. What are the two numbers? a) 40, 42 b) 38, 40 c) 42, 44 d) 44, 46 Explanation: Let one number be x. The other number is x+2 Sum of the squares of the numbers is 3364. x2+(x+2)2=3364 x2+x2+4x+4=3364 2x2+4x-3360=0 x2+2x-1680=0 x2+42x-40x-1680=0 x(x+42)-40(x+42)=0 (x+42)(x-40)=0 x=40, -42 Since we only need positive numbers. Hence, x=40 The two numbers are 40, 42. 6. The sum of the length and the breadth of a rectangle are 97 and the area of the rectangle is 1752. What will be the value", "# Difference between revisions of \"2018 UNCO Math Contest II Problems/Problem 1\" ## Problem A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.) ## Solution We know that we use $1$ digit $9$ times, $2$ digits $90$ times, and $3$ digits $900$ times. So if we have $999$ pages, we have $1 \\cdot 9 + 2 \\cdot 90 + 3 \\cdot 900 = 2889$ digits. Since we want to have $1890$ digits, we do $\\frac{2889 - 1890}{3}=333$ pages less than $999$. So, $999-333=\\boxed {666}$ pages.", "ago..) does XF mean X*F... sorry if im being a pain – Phil Jackson Sep 28 '11 at 20:14 Yes, $XF$ means $X$ times $F$. I suppose I could've written it as $X \\times F$ or $X \\cdot F$, but in mathematics it's common to denote multiplication just by juxtaposition. – Ilmari Karonen Sep 28 '11 at 20:26 show 1 more comment When you don't have categories, the number of pages is $\\frac{L}{F}$, rounded up to the next whole number. So if there are 53 products and you display 8 per page, you will need 7 pages. On page E, you will have objects (E-1)*F+1 to E*F.If you have categories, does each one start on a new page? Then you just have the same calculation for each category. If your 53 objects are in categories of 12,23,5, and 13, you would have 2,3,1, and 2 pages, for a total of 8. You then have to decide whether the showing N to P of M is within the category or over all products. Is this what you were after? - if there are lets say 3 sub cats and 6 prods and each page only allows 6 items per page to be displayed, then 3 cats and 3 prods would be on the first page (showing 1 to 3", "Question 4: If a book has 252 pages, how many digits have been used to number the pages? [1] 650 [2] 648 [3] 660 [4] None of these Option # 2 From page number 1 to age number 9, we will use 1 digit per page $\\Rightarrow$ digits used = 9. From page number 10 to age number 99, we will use 2 digits per page $\\Rightarrow$ digits used = 2 × 90 = 180. From page number 100 to age number 252, we will use 3 digits per page $\\Rightarrow$ digits used = 3 × 153 = 459. Therefore, total number of digits used = 9 + 180 + 459 = 648 Question 5: 1 and 8 are the first two positive integers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4th such number? 1 + 2 + 3 + … + n = $\\frac{n(n+1)}{2}$= M2 (say) $\\Rightarrow$ n(n + 1) = 2 M2 Now n and n + 1 will have no factor in common. Since RHS is twice the square of a positive integer, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square. As twice of a perfect square will be" ]

[ "# How many times does the digit 4 occur in the page numbers of a 50 page book? Dec 13, 2016 $15$ #### Explanation: The digit $4$ can occur in the units or tens positions. It occurs in the units position in the numbers: $4 , 14 , 24 , 34 , 44$ It occurs in the tens position in the numbers: $40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49$ So there are a total of $5 + 10 = 15$ occurrences of the digit $4$. If the question asked in how many page numbers the digit $4$ occurs then the answer would be $14$ - avoiding counting $44$ twice. However, since it asks for the number of occurrences of the digit $4$, $44$ should be counted twice.", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "there would be pages missing between 58 and 100. second: The number of digits in the 3 digit pages is 195 - (number of digits in 1 digit pages) - (number of digits in 2 digit pages) =195 - 9 - 90*2 = 6 3. Thanks @SpringFan25. I feel really stupid now. I see how I mixed up no. of pages and no. of digits used! So the digits used are $9 \\times 1 + 90 \\times 2 + 3 \\times 2 = 195$ And hence Total no of pages, $9 + 90 + 3 = 101$.", "# Difference between revisions of \"2018 UNCO Math Contest II Problems/Problem 1\" ## Problem A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.) ## Solution We know that we use $1$ digit $9$ times, $2$ digits $90$ times, and $3$ digits $900$ times. So if we have $999$ pages, we have $1 \\cdot 9 + 2 \\cdot 90 + 3 \\cdot 900 = 2889$ digits. Since we want to have $1890$ digits, we do $\\frac{2889 - 1890}{3}=333$ pages less than $999$. So, $999-333=\\boxed {666}$ pages.", "After every two numbers, one is omitted. Because $89 = 2\\times 44 +1$, there must be $44$ page numbers missing and so the number on the last page is $89+44=133$.", "Exactly $195$ digits have been used to number the pages in a book. How many pages does the book have?", "# Calculating number of pages from sum of page numbers A novel has 6 chapters. As usual, starting from the first page of the first chapter, the pages of the novel are numbered $1, 2, 3, 4, \\ldots$. Also, each chapter begins on a new page. The last chapter is the longest and the page numbers of its pages add up to 2010. How many pages are there in the first 5 chapters? I got $90$. I factored $4020$, which I got from the summation formula for arithmetic series. Now, $n(a_1+a_n)=4020$, where $n$ is the number of pages of the last chapter and the $a$'s are the starting and ending pages for that chapter. The expression in parenthesis became $(2a_1+n-1)$, because $a_n$ equals $a_1$ plus the number of pages minus one. I then set up some inequalities, tried the suitable factor pairings of $4020$, and got $n$ equal to $20$ and $a_1$ equal to $91$. Is this right? - The approach seems fine --- I haven't checked the details. What contest was this from, please? – Gerry Myerson Nov 10 '13 at 5:17 Also from 2010 Fermat II of Pro2Serve at UTK. – Yadnarav3 Nov 10 '13 at 5:19 Note: it's best to show your work in the question body or as an answer to the question, rather", "Karuna-bdc May 26 '13 at 6:22 pages 6 & 27 are mentioned in the question. its means there are $5$ pages on the left of page no. 6, obviously there will be also $5$ pages more on the right of $27$. so, $27 + 5 = 32$. -", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers. 10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.) Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is", "After every two numbers, one is omitted. Because $89 = 2\\times 44 +1$, there must be $44$ page numbers missing and so the number on the last page is $89+44=133$. This problem is taken from the UKMT Mathematical Challenges. View the previous week's solution", "# Algebra Math Contest Question A novel has 6 chapters. As usual, starting from the first chapter begins on a new page. The last chapter is the longest and the page numbers of its pages add up to 2010: How many pages are there in the first 5 chapter? Let the pages be numbered $$1,\\dots,X_{1}; X_{1}+1,\\dots,X_{2}; X_{2}+1,\\dots,X_{3}; X_{3}+1,\\dots,X_{4}; X_{4}+1,\\dots,X_{5}; X_{5}+1,\\dots,X_{6}$$ Sum of the last chapter pages $= [(X_{6}+X_{5}+1)(X_{6}-X_{5})]/2 = 2010$ (Note sum of arithmetic series with $a_{1} = X_{5}+1$ and $a_{n} = X_{6}$) Now $2\\cdot 2010 = 4020 = 67\\cdot 60 = 134\\cdot 30 = 268\\cdot 15$ Thus $X_{6}+X_{5}+1 = 67$ and $X_{6}-X_{5} =60$, which gives $X_{6} = 63$, and $X_{5} = 3$. This is not possible as there are five chapters. Next $X_{6}+X_{5}+1 = 134$ and $X_{6}-X_{5} = 30$ gives non-integer. So this is not possible. Next $X_{6}+X_{5}+1 = 268$ and $X_{6}-X_{5} = 15$ gives $X_{6} = 141$, and $X_{5} = 126$. This seems plausible but it does not pass the test that chapter 6 is the longest as you could have a longer chapter when $X_{5} = 126$. Any other product of factors would still not hold the condition of the \"Longest Chapter\". What seems to to be answer for this? - I tried MathJax syntax, Seems like it has not worked, Could someone help me", "### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### 14 Divisors What is the smallest number with exactly 14 divisors? ### Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? # Newspaper Sheets ##### Stage: 2 and 3 Short Challenge Level: If page $1$ is on the outside of the first sheet, then page $2$ is on the inside. Pages $3$ and $4$ are on the same sheet, as are pages $5$ and $6$. This therefore shares a sheet with page $7$, so is the inside of the sheet. On the outside on the other side is therefore page $62$. The other three sheets have pages $63$ and $64$, $65$ and $66$ and $67$ and $68$ on them. Therefore there are $68$ pages, and four sheets per page, so $17$ sheets. Alternatively, the sum of the page numbers on the same side of the sheet is always constant, since one side increases by $1$ every time the other decreases by $1$. This means the total is always $8 + 61 = 69$, so page $1$ shares with page $68$. Hence there are", "+0 # A textbook has pages. How many of the pages have page numbers whose digits add up to exactly ​? 0 37 2 A textbook has $$1000$$ pages. How many of the pages have page numbers whose digits add up to exactly $$4$$? Feb 14, 2023 #1 0 If the textbook has 1000 pages, then the page numbers range from 1 to 1000. To find the number of pages with page numbers whose digits add up to exactly 4, we can use the following formula: Number of pages = $\\sum_{i=1}^{9}$(Number of pages with i as the last digit) For example, the number of pages with 0 as the last digit is the same as the number of pages with 4 as the last digit, as both digits add up to 4. Similarly, the number of pages with 1 as the last digit is the same as the number of pages with 3 as the last digit, as both digits add up to 4. Using this formula, we can calculate that the number of pages with page numbers whose digits add up to exactly 4 is 12. Feb 14, 2023 #2 0 1 == 4 2 == 13 3 == 22 4 == 31 5 == 40 6 == 103 7 == 112 8 == 121 9 ==", "Per Manne Jul 21 '14 at 19:09 • Yes, but we are in the realm of pure thought. I was only concerned with the non-uniqueness under the \"rest of the pages\" interpretation. – André Nicolas Jul 23 '14 at 13:42 $$103$$ your book has n=404 pages, so the sum of all pages is n(n+1)/2=81810, since the sum must be 81707, the page missing must be 103 if your book had 405 pages, the sum would be 82215, and the page missing would have to be 508 --impossible if your book had 403 pages, the sum would be 81406, not enough edit: The solution before is assuming 1 number per page. If you have two consecutive numbers per page instead: $51$ and $52$ if your book had 403 pages, the sum would be 81406, not enough your book could have 404 pages, the sum of its pages be 81810, and the missing page have numbers 51 and 52 if your book had 405 or 406 pages, the sum of its pages would be 82215 and 82621 respectively, and you would have 508 and 914 extra in the sum. These are even numbers, and no consecutive numbers add up to an even number, so we rule out these possibilities. your book could have 407 pages, the sum of its pages", "page with page number $n$ in his hand, so we need to solve $$81,707 = p \\Big( 2p + 1 \\Big) - n.$$ As $81,707 \\le p \\Big( 2p + 1 \\Big)$, we obtain $$p \\ge 202,$$ but as $n \\le 2p$, we obtain $$p \\Big(2 p + 1 \\Big) - 81,707 \\le 2 p,$$ whence $$p \\le 202,$$ so the book contains $202$ pages, whence the page number is given by $$202 \\times 405 - 81,707 = 103.$$ The question is: if the father is holding a page $x$ does that mean to exclude the pagenumbers on both sides of the page? Then the page that you father is holding is $51/52$. Hope you get your book back! • A book contains $n$ pages and therefore has pagenumbers from $1$ to $n$. What you are calling \"pages\" are actually leafs. But your end result: pages 51 and 52, is correct. – Christian Blatter Jul 21 '14 at 19:51 • @Christian: You are correct about that, thanks for the correction. I considered a page to be that what has two sided, while is is one. It does not change the outcome. – johannesvalks Jul 21 '14 at 19:56 • @ChristianBlatter I think the correct term is sheet, not leaf. – becko Jul 22 '14 at 17:41 • @becko", "• # question_answer The total number of digits used in numbering the pages of a book having 366 pages, is? A) 732 B) 990C) 1098 D) 1305 Required numbers $=9+2\\times 90+267\\times 3$ $=189+801$", "$$\\frac {1}{13x-x^2}=\\frac {1}{40}$$ 40=13x-x2 x2-13x+40=0 x2-5x-8x+40=0 x(x-5)-8(x-5)=0 (x-8)(x-5)=0 x=8, 5 The number is 58 or 85. Note: Join free Sanfoundry classes at Telegram or Youtube 4. The sum of the squares of the left and right pages of a book is 481. What are the page numbers? a) 11, 12 b) 12, 13 c) 17, 18 d) 15, 16 Explanation: Since the pages of books are consecutive numbers, so let the left page number be x. The right page number will be x+1. Sum of the squares of the pages is 481 x2+(x+1)2=481 x2+x2+1+2x=481 2x2+2x-480=0 x2+x-240=0 x2+16x-15x-240=0 x(x+16)-15(x+16)=0 (x-15)(x+16)=0 x=15, -16 Since, page number cannot be negative, so x=15 The two page numbers are 15 and 16. 5. The sum of the squares of two consecutive positive even numbers is 3364. What are the two numbers? a) 40, 42 b) 38, 40 c) 42, 44 d) 44, 46 Explanation: Let one number be x. The other number is x+2 Sum of the squares of the numbers is 3364. x2+(x+2)2=3364 x2+x2+4x+4=3364 2x2+4x-3360=0 x2+2x-1680=0 x2+42x-40x-1680=0 x(x+42)-40(x+42)=0 (x+42)(x-40)=0 x=40, -42 Since we only need positive numbers. Hence, x=40 The two numbers are 40, 42. 6. The sum of the length and the breadth of a rectangle are 97 and the area of the rectangle is 1752. What will be the value", "3 Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$. The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$, the average page number of all $50$ pages, therefore $k>0$. Since the borrowed sheets start with an odd page number and end with an even page number we have $k \\in \\mathbb N$. We notice that $n < 25$ and $k \\le (49+50)/2-25.5=24<25$. The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$, where the weights are the page number in each group (borrowed vs. remained), therefore $$2nk=2(25-n)(25.5-19)=13(25-n) \\implies 13 | n \\text{ or } 13 | k$$ Since $n, k < 25$ we have either $n=13$ or $k=13$. If $n=13$ then $k=6$. If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\\boxed{\\textbf{(B)} ~13}$. ~asops ## Solution 4 Let $(2k-1)-2n$ be pages be borrowed, the sum of digits in those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$. As $325=25\\cdot13$, we can see $n-k=13$ and that is desired $\\boxed{\\textbf{(B)} ~13}$.", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: There are N pages, so there will be N unit digits, N - 9 (the number of unit digits) ten digits, and N - 99 hundred digits. N + N - 9 + N - 99 = 930; 3 N = 930 + 108 ; N = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, leave the odd 1 out for now. 40 divided by 2 = 20 and N - 10 + 1 or N - 9 = 20 N = 29 so the last digit you write is the next one, which is \"3\". Solution II: Let there be n numbers: n + (n - 9) = 50, 2n = 59, n = 29.5 which mean the last digit you write is the first digit of the next number, which is \"30\", so 3 is the answer.", "0 0 0 0 0 0 1. In a book with 116 pages, what do the page numbers of the middle two pages add up to? 2. What is the largest number that cannot be written in the form $$14n+29m$$, where $$n$$ and $$m$$ are non-negative integers? 3. How many factors does the number $$2^6\\times3^{12}\\times5^2$$ have? 4. How many squares (of any size) are there in a $$15\\times14$$ grid of squares? 5. You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number? 6. What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.) 7. What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.) 8. What is the mean of the answers to questions 6, 7 and 8? 9. How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6? 10. What is the lowest common multiple of 52 and 1066? ### Similar Posts Christmas Card 2016 Christmas (2017) Is Over Christmas (2017) is Coming!" ]

[ "\\ldots . . = 1 - \\frac{1}{2} = \\frac{1}{2}$", "simply $$\\frac{\\partial z}{\\partial x} = f'(x^2 + y^2) \\cdot 2x$$", "will be $$\\frac{488}{3}$$ - extra: derivative of $$2 \\cdot 3^{5x}$$ answer will be $$10 \\cdot \\ln(3) \\cdot 3^{5x}$$ - review topics", "\\cdot x) + C = \\frac{x^2 |x|}{3} + C.$$", "\\cdot \\frac{3}{2} = ...?$ 21. anonymous 3/10 22. anonymous @Michele_Laino", "\\cdot \\frac{x - 3}{1}$ this is the same as $\\frac{x - 3}{x + 2}$", "$x^2(2x^3)^{\\frac{1}{2}} \\;=\\;x^2\\cdot2^{\\frac{1}{2}}\\cdot x^{\\frac{3}{2}} \\;=\\;\\sqrt{2}\\,x^{\\frac{7}{2}}$ Now integrate: . $\\sqrt{2}\\int x^{\\frac{7}{2}}\\,dx$", "i\\cdot\\frac1{2i}=\\pi$.", "+ \\frac{\\partial v}{\\partial h}\\cdot dh$", "- \\ \\frac{1}{3} \\cdot [-1] \\ ) \\ = \\ \\frac{2}{3} \\ \\ ,$$ with the area enclosed by the complete curve being four times this. -", "13}+\\frac{10}{21\\cdot 13}\\approx 0.110$$", "is $$\\frac{2t}{1-t^2}-t=t\\cdot\\frac{2-(1-t^2)}{1-t^2}=t\\cdot\\frac{1+t^2}{1-t^2}$$.", "\\cdot \\frac{m_3}{m_2} \\cdot \\frac{m_1}{m_3} = 1$$", "like: ${\\displaystyle {\\frac {1}{1-2^{-s}}}\\cdot {\\frac {1}{1-3^{-s}}}\\cdot {\\frac {1}{1-5^{-s}}}\\cdot {\\frac {1}{1-7^{-s}}}\\ldots }$", "define $L:=lcm\\{2,3,...,n\\}=2^k3^a5^b\\cdot\\ldots$ , so: $\\sum\\limits_{i=2}^n\\frac{1}{i} = \\frac{(2^k3^a\\cdot\\ldots)\\slash 2}{L}+\\frac{(2^k3^a\\cdot\\ldots)\\slash 3}{L}+\\ldots +\\frac{(2^k3^a\\cdot\\ldots)\\slash n}{L}$ Now pay attention to the fact that all the numerators above are even except precisely one (why?) Because when you get to the $(2^k-1)th$ term, $(2^k3^a\\ldots)/2^k$ reduces to the product of the remaining primes, all odd. Thank you muchly. I see it now.", "\\frac{1}{2 \\cdot {6}^{\\frac{1}{2}}}$", "v - 2\\frac{v\\cdot a - c}{a\\cdot a}a.$", "n - 2 ) } { 3 ! } \\cdot \\frac { 1 } { n ^ { 3 } } , \\frac { ( 1 - \\frac { 1 } { n } ) ( 1 - \\frac { 2 } { n } ) } { 3 ! }$ See part (b).", "$\\frac{1\\cdot\\tfrac{1}{3}}{2} = \\boxed{\\textbf{(E)} \\: \\frac{1}{6}}$ ~Lemma proof by sakshamsethi", "\\cdot \\frac{1}{2^m} \\end{eqnarray}" ]

[ "\\frac{\\frac{\\sqrt{2}}{2}}{1+ \\frac{\\sqrt{2}}{2}}= \\frac{\\sqrt{2}}{2+ \\sqrt{2}}$$. $$AC+ CD= 1+ \\frac{\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{2 + 2\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{4+ 4\\sqrt{2}- 2\\sqrt{2}- 4}{4- 2}= \\frac{2\\sqrt{2}}{2}= \\sqrt{2}$$", "\\frac1{2k+1} = \\frac{n! 2^n}{1 \\cdot 3 \\cdot \\ldots \\cdot (2n+1)} = \\frac{(n!)^2 2^{2n}}{(2n+1)!}$$ as Ira Gessel and T. Amdeberhan were saying. • This is cool, indeed. Nov 26, 2016 at 0:20", "s^{2}}{N\\cdot m}=\\frac{A^{2}\\cdot s^{4}}{kg\\cdot m^2}$$$ $$\\frac{A^{2}\\cdot s^{4}}{kg\\cdot m^2}$$\\$", "$\\frac{1\\cdot\\tfrac{1}{3}}{2} = \\boxed{\\textbf{(E)} \\: \\frac{1}{6}}$ ~Lemma proof by sakshamsethi", "$4^{m-2} (-2 + 9\\cdot 2^m)$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2$ $2^{m-3} (-16 + 5\\cdot 2^{2 m} + 2^{2 + m})/3$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2+1$ $2^{2 m-3} (2 + 5\\cdot 2^m)/3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2$ $2^{ m-5} (-2 + 2^m)^2 /3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2+1$ $2^{2 m-5} (-2 + 2^m)/3$ Weight Multiplicity $0$ $1$ $2^{m}$ $1$ $2^{m-1}$ $9\\cdot 2^{3 m-4} + 5 \\cdot 2^{m-2} - 2^{2 m-2}-2$ $2^{m-1}+1$ $4^{m-2} (-2 + 9\\cdot 2^m)$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2$ $2^{m-3} (-16 + 5\\cdot 2^{2 m} + 2^{2 + m})/3$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2+1$ $2^{2 m-3} (2 + 5\\cdot 2^m)/3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2$ $2^{ m-5} (-2 + 2^m)^2 /3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2+1$ $2^{2 m-5} (-2 + 2^m)/3$ The weight distribution of ${\\mathcal{C}}_{ {\\mathcal{M}}}^{(2)}$ for $m\\geq 6$ and $m\\equiv 2 \\pmod 4$ Weight Multiplicity $0$ $1$ $2^{m-1}$ $-2 + 17 \\cdot 2^{m-3} + 29\\cdot 2^{ 3 m-6} - 2^{2m-5} \\cdot 23$ $2^{m - 1} + 1$ $4^{m-3} (29\\cdot 2^m -20)$ $(2^m \\pm 2^{\\frac{m}{2}})/2$ $2^m (-16 + 3 \\cdot 4^m + 2^m \\cdot 7) /15$ $(2^m \\pm 2^{\\frac{m}{2}})/2+1$ $4^m (1 + 2^m)/5$ $(2^m \\pm 2^{\\frac{m+2}{2}})/2$ $2^{2 m-5} (-10 + 7 \\cdot 2^m) /3$ $(2^m\\pm 2^{\\frac{m+2}{2}})/2+1$ $2^{2 m-5} (-4 + 7 \\cdot 2^m) /3$ $(2^m \\pm 2^{\\frac{m+4}{2}})/2$ $2^{m-7} (8 + 4^m + 2^{1 + m} (-3)) /15$ $(2^m \\pm 2^{\\frac{m+4}{2}})/2+1$ $2^{2 m-7} (-4 + 2^m) /15$ $2^m$ $1$ Weight Multiplicity $0$ $1$", "= \\ sum _ {n = 0} ^ {\\ infty} (- 1) ^ {n} {\\ frac {x ^ {2n + 1} } {(2n + 1)!}} & {\\ Text {for all}} \\ x \\\\ & = x - {\\ frac {x ^ {3}} {6}} + {\\ frac {x ^ {5 }} {120}} - \\ cdots \\\\\\ cos (x) & = \\ sum _ {n = 0} ^ {\\ infty} (- 1) ^ {n} {\\ frac {x ^ {2n}} { (2n)!}} & {\\ Text {for all}} \\ x \\\\ & = 1 - {\\ frac {x ^ {2}} {2}} + {\\ frac {x ^ {4}} {24 }} - \\ cdots \\\\\\ tan (x) & = \\ sum _ {n = 1} ^ {\\ infty} {\\ frac {B_ {2n} (- 4) ^ {n} (1-4 ^ {n })} {(2n)!}} X ^ {2n-1} & {\\ text {for}} | x | <{\\ frac {\\ pi} {2}} \\\\ & = x + {\\ frac {x ^ {3}} {3}} + {\\ frac {2x ^ {5}} {15}} + \\ cdots \\\\\\ sec (x) & = \\ sum _ {n = 0} ^ {\\ infty} ( -1) ^ {n} {\\ frac {E_ {2n}} {(2n)!}} X ^ {2n} & {\\ text {for}} | x | <{\\ frac {\\ pi} {2} } \\\\ & = 1 + {\\ frac {x", "simplify the right hand side and you are done. 3. Is the following correct? ((2.2^k+1)-2+1)/(2^(k+1)=2^(k+2)-1)/2^k+1=2-1/2^(k+1) 4. Originally Posted by annitaz Is the following correct? ((2.2^k+1)-2+1)/(2^(k+1)=2^(k+2)-1)/2^k+1=2-1/2^(k+1) Yes, if you add a few more pairs of parentheses: $\\frac{2 \\cdot 2^{(k+1)} - 2 + 1}{2^{(k+1)}} = \\frac{2^{(k+2)} - 1}{2^{(k+1)}} = 2 - (1/2)^{(k+1)}$ However, it's easier to do this: $2 - (1/2)^k + (1/2)^{(k+1)} = 2 - 2 \\cdot (1/2)^{(k+1)} + (1/2)^{(k+1)} = 2 - (1/2)^{(k+1)}$", "3)^{\\frac{2}{3}}=(3^3)^{\\frac{2}{3}}=(3)^{3\\cdot\\frac{2}{3}}=(3)^{\\frac{6}{3}}=3^2$ Another alternative : (3$^2$)$^{2/3}$.3$^{2/3}$ = 3$^{4/3}$.3$^{2/3}$ = 3$^{6/3}$ = 3$^2$ = $9$ SIMPLE $27^\\frac{2}{3}=(3\\cdot\\ 3\\cdot\\ 3)^{\\frac{2}{3}}=(3^3)^{\\frac{2}{3}}=(3)^{3\\cdot\\frac{2}{3}}=**(3)^{\\frac{6}{3}}**=3^2$ ACTUALLY. $27^2$. BECOMES $3^6$", "\\cdot x) + C = \\frac{x^2 |x|}{3} + C.$$", "&= \\lvert\\sum_{r\\ge0} x^{2^r}(\\log{2} - \\sum_{k=2^r}^{2^{r+1}-1}\\frac{1}{k}) + \\sum_{r\\ge0}\\sum_{k=2^r}^{2^{r+1}-1}\\frac{x^{2^r} - x^k}{k} \\rvert \\\\ &\\le \\sum_{r\\ge0}\\frac{x^{2^r}}{2^r} + \\sum_{r\\ge0}\\sum_{k=2^r}^{2^{r+1}-1}\\frac{x^{2^r} - x^k}{k} \\\\ &< 2 + \\sum_{r\\ge0}x^{2^r}(1-x)\\sum_{k=2^r}^{2^{r+1}-1}\\frac{1+x+\\cdots+x^{k-2^r-1}}{k} \\\\ &\\le 2 + (1-x)\\sum_{r\\ge0}x^{2^r}\\sum_{k=2^r}^{2^{r+1}-1}\\frac{k-2^r}{k} \\\\ &\\le 2 + (1-x)\\sum_{r\\ge0}x^{2^r}(2^r\\cdot 1) \\\\ &\\le 2 + (1-x)(x + 2\\sum_{r\\ge1}\\sum_{k=2^{r-1}+1}^{2^r}x^k) \\\\ &= 2 + x(1-x) + 2(1-x)\\sum_{k\\ge2} x^k \\\\ &= 2 + x + x^2, \\end{align*} so we're done. -", "\\frac{ (2n-1)(2n) }{ n } \\cdot \\frac{n}{2n(2n-1)}$ $= - \\frac{ \\binom{2n}{n} }{ 2^{2n} (2n-1) }$ $= - \\frac{1}{4^n} \\frac{ \\binom{2n}{n} }{ 2n-1}$", "0}(1+x)^{1/x}\\cdot \\lim_{x\\to 0}\\frac{1}{(x+1)^2}$$ $$=-\\frac12(e)\\cdot (1)$$ $$=-\\frac e2$$", "+ 1} + \\frac{C-2}{x + 2} - \\ldots + (-1)^n\\frac{C_n}{x + n} = \\frac{n!}{x(x + 1)\\ldots (x + n)}$$ 69. Show that $$C_0^2 - C_1^2 + C_2^2 - \\ldots + (-1)^nC_n^2 = 0$$ or $$(-1)^{\\frac{n}{2}}. \\frac{n!}{\\left(\\frac{n}{2}\\right)!}^2$$ according as $$n$$ is odd or even. 70. Show that $${}^mC_r{}^nC_0 + {}^mC_{r - 1}{}^nC_1 + {}^mC_{r - 2}{}^nC_2 + \\ldots + {}^mC_0{}^nC_r = {}^{m + n}C_r,$$ where $$m, n, r$$ are positive integers and $$r < m, r < n.$$ 71. $${}^{2n}C_0^2 - {}^{2n}C_1^2 + {}^{2n}C_2^2 - \\ldots + (-1)^{2n}.{}^{2n}C_{2n}^2 = (-1)^n.{}^{2n}C_n.$$ 72. Show that $$C_1^2 + 2.C_2^2 + 3.C_3^2 + \\ldots + n.C_n^2 = \\frac{(2n - 1)!}{[(n - 1)!]^2}$$ 73. Show that $$C_0^2 + \\frac{C_1^2}{2} + \\frac{C_2^2}{3} + \\ldots + \\frac{C_n^2}{n + 1} = \\frac{(2n + 1)!}{[(n + 1)!]^2}$$ 74. If $$n$$ is a positive integer in $$(1 + x)^n,$$ show that $$2.\\frac{\\left(\\frac{n}{2}!\\right)^2}{n!}[c_0^2 -2.C_1^2 + 3.C_2^2 - \\ldots + (-1)^n.(n + 1)C_n^2] = (-1)^{\\frac{n}{2}}(2 + n)$$ 75. Show that $$\\sum_{0\\leq i \\leq n}\\sum_{0\\leq i \\leq j\\leq n}C_i C_j = 2^{2n - 1}- \\frac{(2n)!}{2(n!)^2}$$ 76. Show that $$\\sum_{r=0}^nC_r^3$$ is equal to the coefficient of $$x^ny^n$$ in the expansion of $$[(1 + x)(1 + y)(x + y)]^n.$$ 77. Let $$n$$ be a positive integer and $$(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \\ldots + a_{2n}x^{2n},$$", "\\frac{a^2+2ab + b^2}{2!} + \\frac{a^3+3a^2b+3ab^2 + b^3}{3!}$. $+\\:\\frac{a^4+4a^3b + 6a^2b^2 + 4ab^3 + b^4}{4!} + \\hdots$ $e^a\\!\\cdot\\!e^b\\;=\\;1 + (ab) + \\frac{(a+b)^2}{2!} + \\frac{(a+b)^3}{3!} + \\frac{(a+b)^4}{4!} + \\hdots \\;=\\;e^{a+b}$", "}{\\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}=x-{\\frac {x^{3}}{3!}}+{\\frac {x^{5}}{5!}}-{\\frac {x^{7}}{7!}}+\\cdots }\\end{array}}}$ (...)", "\\frac{2n-3}{a^2(2n-2)}\\ I_{n-1}(a) + \\frac{1}{a^2(2n-2)}\\ \\frac{x}{(a^2+x^2)^{n-1}}\\; .$$ Using (3) recursively you get: $$\\begin{split} I_n(a) &= \\frac{(2n-3)(2n-5)}{a^4 (2n-2)(2n-4)}\\ I_{n-2}(a) + \\frac{(2n-3)}{a^4(2n-2)(2n-4)}\\ \\frac{x}{(a^2+x^2)^{n-2}}\\\\ &\\phantom{=} + \\frac{1}{a^2(2n-2)}\\ \\frac{x}{(a^2+x^2)^{n-1}}\\\\ &= \\frac{(2n-3)(2n-5)(2n-7)}{a^6 (2n-2)(2n-4)(2n-6)}\\ I_{n-3}(a) + \\frac{(2n-3)(2n-5)}{a^6 (2n-2)(2n-4)(2n-6)}\\ \\frac{x}{(a^2+x^2)^{n-3}}\\\\ &\\phantom{=} + \\frac{(2n-3)}{a^4(2n-2)(2n-4)}\\ \\frac{x}{(a^2+x^2)^{n-2}} + \\frac{1}{a^2(2n-2)}\\ \\frac{x}{(a^2+x^2)^{n-1}}\\\\ &=\\cdots\\\\ &= \\frac{(2n-3)!!}{a^{2n-2} (2n-2)!!}\\ I_1(a)+x\\ \\sum_{k=1}^{n-1} \\frac{1}{a^{2(n-k)}}\\ \\frac{(2n-3)!!}{(2k-1)!!}\\ \\frac{(2k-2)!!}{(2n-2)!!}\\ \\frac{1}{(a^2+x^2)^k}\\; . \\end{split}$$ Since: $$I_1(a) = \\int \\frac{1}{a^2+x^2}\\ \\text{d} x = \\frac{1}{a}\\ \\arctan \\left( \\frac{x}{a}\\right)$$ finally you obtain: $$\\tag{4} \\begin{split} I_n(a) &= \\frac{(2n-3)!!}{a^{2n-1} (2n-2)!!}\\ \\arctan \\left( \\frac{x}{a}\\right) \\\\ &\\phantom{=} +x\\ \\sum_{k=1}^{n-1} \\frac{1}{a^{2(n-k)}}\\ \\frac{(2n-3)!!}{(2k-1)!!}\\ \\frac{(2k-2)!!}{(2n-2)!!}\\ \\frac{1}{(a^2+x^2)^k} \\end{split}$$ which is the elementary closed form of $I_n(a)$. If you set $n=2$ in (4) you obtain exactly: $$I_2(a) = \\frac{1}{2a^3}\\ \\arctan \\left( \\frac{x}{a}\\right) + \\frac{x}{2a^2 (a^2+x^2)}$$ which is the correct value of your integral.", "2x + 3, \\frac{dy}{du} = \\frac{1}{\\ln{3}} \\frac{1}{u} = \\frac{1}{(x^2 + 3x + 5)\\ln{3}}$ $\\frac{dy}{dx} = \\frac{2x + 3}{(x^2 + 3x + 5)\\ln{3}}$ 3. thanks hahah but i figured that one out after $ y = (ln \\cdot e^\\frac{1}{x} \\cdot cos^2 \\cdot \\sqrt {x}) $", "– {a^2}} }+{ \\frac{a}{2} \\cdot \\left( {\\sqrt {4{R^2} – {a^2}} } \\right)^\\prime }={ \\frac{1}{2} \\cdot \\sqrt {4{R^2} – {a^2}} }+{ \\frac{a}{2} \\cdot \\frac{{\\left( { – 2a} \\right)}}{{2\\sqrt {4{R^2} – {a^2}} }} }={ \\frac{{\\sqrt {4{R^2} – {a^2}} }}{2} – \\frac{{{a^2}}}{{2\\sqrt {4{R^2} – {a^2}} }} }={ \\frac{{4{R^2} – {a^2} – {a^2}}}{{2\\sqrt {4{R^2} – {a^2}} }} }={ \\frac{{2{R^2} – {a^2}}}{{\\sqrt {4{R^2} – {a^2}} }}.}$ Determine the critical point(s): ${A^\\prime\\left( a \\right) = 0,\\;} \\Rightarrow {\\frac{{2{R^2} – {a^2}}}{{\\sqrt {4{R^2} – {a^2}} }} = 0,\\;} \\Rightarrow {\\left\\{ {\\begin{array}{*{20}{l}} {2{R^2} – {a^2} = 0}\\\\ {\\sqrt {4{R^2} – {a^2}} \\ne 0} \\end{array}} \\right.,\\;} \\Rightarrow {\\left\\{ {\\begin{array}{*{20}{l}} {a = \\sqrt 2 R}\\\\ {a \\ne 2R} \\end{array}} \\right..}$ We have two critical points $$a = \\sqrt{2}R$$ and $$a = 2R,$$ but as at the second point $$a = 2R$$ the area of the rectangle is zero, further we will consider only the first point $$a = \\sqrt{2}R.$$ The derivative is positive to the left of this point, and negative to the right. Therefore $$a = \\sqrt{2}R$$ is a point of maximum. The maximum value of the area of the rectangle is given by ${{A_{\\max }} }={ A\\left( {\\sqrt 2 R} \\right) }={ \\frac{{\\sqrt 2 R}}{2}\\sqrt {4{R^2} – {{\\left( {\\sqrt 2 R} \\right)}^2}} }={ \\frac{{\\sqrt 2 R}}{2}\\sqrt {4{R^2} – 2{R^2}} }={ \\frac{{\\sqrt 2 R}}{2} \\cdot \\sqrt 2 R", "T(2^{2^{k-1}}) \\\\ &= 1 + 2^{2^{k-1}} + 2^{2^{k-1}+2^{k-2}} T(2^{2^{k-2}}) \\\\ &= 1 + 2^{2^{k-1}} + 2^{2^{k-1}+2^{k-2}} + 2^{2^{k-1}+2^{k-2}+2^{k-3}} T(2^{2^{k-3}}) \\\\ &= \\cdots \\\\ &= 1 + 2^{2^{k-1}} + 2^{2^{k-1}+2^{k-2}} + 2^{2^{k-1}+2^{k-2}+2^{k-3}} + \\cdots + 2^{2^{k-1} + \\cdots + 2^0} T(2^{2^0}) \\\\ &= 1 + 2^{2^{k-1}} + 2^{2^{k-1}+2^{k-2}} + 2^{2^{k-1}+2^{k-2}+2^{k-3}} + \\cdots + 2^{2^{k-1} + \\cdots + 2^0} \\\\ &= 2^{2^{k}} \\left[ 2^{-2^k} + 2^{-2^{k-1}} + 2^{-2^{k-2}} + 2^{-2^{k-3}} + \\cdots + 2^{-2^0} \\right] \\\\ &\\approx 2^{2^k} \\sum_{\\ell=0}^\\infty \\frac{1}{2^{2^\\ell}}. \\end{align} The infinite series converges, and we conclude that $$T(2^{2^k}) = \\Theta(2^{2^{k}})$$. • Hey, Thanks for your answer. Could I just substitute back the $\\ n$ and apply $\\ log$ on both sides to get $\\ \\Theta(log(log(n))$ ? – bm1125 Nov 8 '20 at 11:29", "\\cdot \\frac{1}{2^m} \\end{eqnarray}" ]

[ "$4^{m-2} (-2 + 9\\cdot 2^m)$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2$ $2^{m-3} (-16 + 5\\cdot 2^{2 m} + 2^{2 + m})/3$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2+1$ $2^{2 m-3} (2 + 5\\cdot 2^m)/3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2$ $2^{ m-5} (-2 + 2^m)^2 /3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2+1$ $2^{2 m-5} (-2 + 2^m)/3$ Weight Multiplicity $0$ $1$ $2^{m}$ $1$ $2^{m-1}$ $9\\cdot 2^{3 m-4} + 5 \\cdot 2^{m-2} - 2^{2 m-2}-2$ $2^{m-1}+1$ $4^{m-2} (-2 + 9\\cdot 2^m)$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2$ $2^{m-3} (-16 + 5\\cdot 2^{2 m} + 2^{2 + m})/3$ $(2^m \\pm 2^{\\frac{m+1}{2}})/2+1$ $2^{2 m-3} (2 + 5\\cdot 2^m)/3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2$ $2^{ m-5} (-2 + 2^m)^2 /3$ $(2^m \\pm 2^{\\frac{m+3}{2}})/2+1$ $2^{2 m-5} (-2 + 2^m)/3$ The weight distribution of ${\\mathcal{C}}_{ {\\mathcal{M}}}^{(2)}$ for $m\\geq 6$ and $m\\equiv 2 \\pmod 4$ Weight Multiplicity $0$ $1$ $2^{m-1}$ $-2 + 17 \\cdot 2^{m-3} + 29\\cdot 2^{ 3 m-6} - 2^{2m-5} \\cdot 23$ $2^{m - 1} + 1$ $4^{m-3} (29\\cdot 2^m -20)$ $(2^m \\pm 2^{\\frac{m}{2}})/2$ $2^m (-16 + 3 \\cdot 4^m + 2^m \\cdot 7) /15$ $(2^m \\pm 2^{\\frac{m}{2}})/2+1$ $4^m (1 + 2^m)/5$ $(2^m \\pm 2^{\\frac{m+2}{2}})/2$ $2^{2 m-5} (-10 + 7 \\cdot 2^m) /3$ $(2^m\\pm 2^{\\frac{m+2}{2}})/2+1$ $2^{2 m-5} (-4 + 7 \\cdot 2^m) /3$ $(2^m \\pm 2^{\\frac{m+4}{2}})/2$ $2^{m-7} (8 + 4^m + 2^{1 + m} (-3)) /15$ $(2^m \\pm 2^{\\frac{m+4}{2}})/2+1$ $2^{2 m-7} (-4 + 2^m) /15$ $2^m$ $1$ Weight Multiplicity $0$ $1$", "\\frac{\\frac{\\sqrt{2}}{2}}{1+ \\frac{\\sqrt{2}}{2}}= \\frac{\\sqrt{2}}{2+ \\sqrt{2}}$$. $$AC+ CD= 1+ \\frac{\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{2 + 2\\sqrt{2}}{2+ \\sqrt{2}}= \\frac{4+ 4\\sqrt{2}- 2\\sqrt{2}- 4}{4- 2}= \\frac{2\\sqrt{2}}{2}= \\sqrt{2}$$", "inscribed in the circle with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$ Fig. 1 On one hand, we see the area of $\\Delta ABC$ as $\\frac{1}{2}\\cdot AB\\cdot BC = \\frac{1}{2}\\cdot AB\\cdot L_{n+1}$. On the other hand, it is also $\\frac{1}{2}\\cdot AC\\cdot BE = \\frac{1}{2}\\cdot 2r\\cdot \\frac{L_n}{2}=\\frac{1}{2}\\cdot r\\cdot L_n.$ Therefore, $\\frac{1}{2}AB\\cdot L_{n+1}= \\frac{1}{2}r\\cdot L_n.$ Or, $AB^2\\cdot L_{n+1}^2 = r^2\\cdot L_n^2\\quad\\quad\\quad(1)$ where by Pythagorean theorem, $AB^2= (2r)^2 - L_{n+1}^2.\\quad\\quad\\quad(2)$ Substituting (2) into (1) gives $(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.$ That is, $L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.$ Let $p = L_{n+1}^2$, we have $p^2-4r^2 p + r^2 L_n^2=0.\\quad\\quad\\quad(3)$ Solving (3) for $p$ yields $p = 2r^2 \\pm r \\sqrt{4 r^2-L_n^2}.$ Since $L_n^2$ must be greater than $L_{n+1}^2$ (see Exercise 1), it must be true (see Exercise 2) that $L_{n+1}^2=2r^2 - r \\sqrt{4r^2-L_n^2}.\\quad\\quad\\quad(4)$ Notice when $r=\\frac{1}{2}$, we obtain (5) in “Truth vs. Intellect“. With increasing $n$, $L_n\\cdot 2^n \\approx \\pi\\cdot 2r \\implies \\pi \\approx \\frac{L_n 2^n}{2r}.\\quad\\quad\\quad$ We can now compute the approximate value of $\\pi$ from any circle with radius $r$: Fig. 2 $r=2$ Fig. 3 $r=\\frac{1}{8}$ Exercise 1 Explain $L_{n}^2 > L_{n+1}^2$ geometrically. Exercise 2 Show it is $2r^2-r\\sqrt{4r^2-L_n^2}$ that represents $L_{n+1}^2.$ Truth vs. Intellect It was known long ago that $\\pi$, the ratio of the circumference to the diameter of a circle, is a constant. Nearly", "$x^2(2x^3)^{\\frac{1}{2}} \\;=\\;x^2\\cdot2^{\\frac{1}{2}}\\cdot x^{\\frac{3}{2}} \\;=\\;\\sqrt{2}\\,x^{\\frac{7}{2}}$ Now integrate: . $\\sqrt{2}\\int x^{\\frac{7}{2}}\\,dx$", "4. $2xy \\cdot \\frac{2y^2} {x^3}$ 5. $\\frac{4y^2 - 1} {y^2 - 9} \\cdot \\frac{y - 3} {2y - 1}$ 6. $\\frac{6ab} {a^2} \\cdot \\frac{a^3b} {3b^2}$ 7. $\\frac{x^2} {x - 1} \\div \\frac{x} {x^2 + x -2}$ 8. $\\frac{33a^2} {-5} \\cdot \\frac{20} {11a^3}$ 9. $\\frac{a^2 + 2ab + b^2} {ab^2 - a^2b} \\div (a + b)$ 10. $\\frac{2x^2 + 2x - 24} {x^2 + 3x} \\cdot \\frac{x^2 + x - 6} {x + 4}$ 11. $\\frac{3 - x} {3x - 5} \\div \\frac{x^2 - 9} {2x^2 - 8x - 10}$ 12. $\\frac{x^2 - 25} {x + 3} \\div (x - 5)$ 13. $\\frac{2x + 1} {2x - 1} \\div \\frac{4x^2 - 1} {1 - 2x}$ 14. $\\frac{x} {x - 5} \\cdot \\frac{x^2 - 8x + 15} {x^2 - 3x}$ 15. $\\frac{3x^2 + 5x - 12} {x^2 - 9} \\div \\frac{3x - 4} {3x+ 4}$ 16. $\\frac{5x^2 + 16x + 3} {36x^2 - 25} \\cdot (6x^2 + 5x)$ 17. $\\frac{x^2 + 7x + 10} {x^2 - 9} \\cdot \\frac{x^2 - 3x} {3x^2 + 4x - 4}$ 18. $\\frac{x^2 + x - 12} {x^2 + 4x + 4} \\div \\frac{x - 3} {x + 2}$ 19. $\\frac{x^4 - 16} {x^2 - 9} \\div \\frac{x^2 + 4} {x^2 + 6x + 9}$ 20. $\\frac{x^2 + 8x + 16} {7x^2 + 9x +", "3)^{\\frac{2}{3}}=(3^3)^{\\frac{2}{3}}=(3)^{3\\cdot\\frac{2}{3}}=(3)^{\\frac{6}{3}}=3^2$ Another alternative : (3$^2$)$^{2/3}$.3$^{2/3}$ = 3$^{4/3}$.3$^{2/3}$ = 3$^{6/3}$ = 3$^2$ = $9$ SIMPLE $27^\\frac{2}{3}=(3\\cdot\\ 3\\cdot\\ 3)^{\\frac{2}{3}}=(3^3)^{\\frac{2}{3}}=(3)^{3\\cdot\\frac{2}{3}}=**(3)^{\\frac{6}{3}}**=3^2$ ACTUALLY. $27^2$. BECOMES $3^6$", "is $$\\frac{2t}{1-t^2}-t=t\\cdot\\frac{2-(1-t^2)}{1-t^2}=t\\cdot\\frac{1+t^2}{1-t^2}$$.", "+1)}$ $\\Big]$ $=$ $\\frac{n!}{(n-r)!(r-1)!}$ $\\Big[$ $\\frac{n + 1}{r ( n - r +1)}$ $\\Big]$ $=$ $\\frac{(n+1)!}{r!(n-r)! (n-r+1)!}$ $=$ $\\frac{(n+1)!}{(n - r + 1)! r!}$ $=$ RHS. $\\\\$ Question 12: Prove that: $\\frac{(2n+1)!}{n!}$ $= 2^n \\Big\\{ 1\\cdot 3 \\cdot 5 \\ldots (2n-1)(2n+1) \\Big\\}$ LHS $=$ $\\frac{(2n+1)!}{n!}$ $=$ $\\frac{(2n+1) (2n)!}{n!}$ $=$ $\\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\ldots n(n+1) \\ldots (2n-2) \\cdot (2n-1) \\cdot (2n) \\cdot (2n+1)]}{n!}$ $=$ $\\frac{[1 \\cdot 3 \\cdot 5 \\cdot 7 \\ldots (2n-1)(2n+1)][2 \\cdot 4 \\cdot 6 \\cdot 8 \\ldots (2n-2)(2n)]}{n!}$ $=$ $\\frac{[1 \\cdot 3 \\cdot 5 \\cdot 7 \\ldots (2n-1)(2n+1)][2^n 1 \\cdot 2 \\cdot 3 \\cdot 4 \\ldots (n-1)(n)]}{n!}$ $= 2^n [ 1 \\cdot 3 \\cdot 5 \\cdot 7 \\ldots (2n-1)(2n+1)] =$ RHS. Hence proved. $\\\\$", "$\\frac{1\\cdot\\tfrac{1}{3}}{2} = \\boxed{\\textbf{(E)} \\: \\frac{1}{6}}$ ~Lemma proof by sakshamsethi", "true. $1 + \\frac{1}{2} + ~ ... ~ + \\frac{1}{2^k} + \\frac{1}{2^{k + 1}}$ $= \\left ( 2 - \\frac{1}{2^k} \\right ) + \\frac{1}{2^{k + 1}}$ $= 2 - \\left ( \\frac{1}{2^k} - \\frac{1}{2^{k + 1}} \\right )$ $= 2 - \\left ( \\frac{1}{2^k} - \\frac{1}{2} \\cdot \\frac{1}{2^k} \\right )$ $= 2 - \\left ( 1 - \\frac{1}{2} \\right ) \\frac{1}{2^k}$ $= 2 - \\frac{1}{2} \\cdot \\frac{1}{2^k}$ $= 2 - \\frac{1}{2^{k + 1}}$ as we needed to show. -Dan", "\\cdot e^{x^2}$ This is a combination of the quotient and product rules: $y' = \\frac{(2x)(2x + 1) - (x^2 - 7)(2)}{(2x + 1)^2} \\cdot e^{x^2} + \\frac{x^2 - 7}{2x + 1} \\cdot e^{x^2} \\cdot (2x)$ $y' = \\frac{2x^2 + 2x + 14}{(2x + 1)^2} \\cdot e^{x^2} + \\frac{2x(x^2 - 7)}{2x + 1} \\cdot e^{x^2}$ I'll let you do the rest of the simplifying. -Dan", "define $L:=lcm\\{2,3,...,n\\}=2^k3^a5^b\\cdot\\ldots$ , so: $\\sum\\limits_{i=2}^n\\frac{1}{i} = \\frac{(2^k3^a\\cdot\\ldots)\\slash 2}{L}+\\frac{(2^k3^a\\cdot\\ldots)\\slash 3}{L}+\\ldots +\\frac{(2^k3^a\\cdot\\ldots)\\slash n}{L}$ Now pay attention to the fact that all the numerators above are even except precisely one (why?) Because when you get to the $(2^k-1)th$ term, $(2^k3^a\\ldots)/2^k$ reduces to the product of the remaining primes, all odd. Thank you muchly. I see it now.", "\\cdot 767.1 \\sqrt{\\frac{1}{14} + \\frac{1}{11}}$ $\\boxed{910 \\pm 639.47} \\text{ or } \\boxed{(270.53, 1549.47)}$", "\\ldots . . = 1 - \\frac{1}{2} = \\frac{1}{2}$", "C_n^2 = \\frac{(2n)!}{n!n!}$$ 11. $$\\frac{C_1}{C_0} + 2\\frac{C_2}{C_1} + 3\\frac{C_3}{C_2} + \\ldots + n.\\frac{C_n}{C_{n - 1}} = \\frac{n(n + 1)}{2}$$ 12. $$(1 + {}^nC_1 + {}^nC_2 + \\ldots + {}^nC_n)^2 = 1 + {}^{2n}C_1 + {}^{2n}C_2 + \\ldots + {}^{2n}C_{2n}$$ 13. $$(1 + {}^nC_1 + {}^nC_2 + \\ldots + {}^nC_n)^5 = 1 + {}^{5n}C_1 + {}^{5n}C_2 + \\ldots + {}^{5n}C_{5n}$$ 14. $$C_0 + 5.C_1 + 9.C_2 + \\ldots + (4n + 1).C_n = (2n + 1)2^n$$ 15. $$1 - (1 + x)C_1 + (1 + 2x)C_2 - (1 + 3x)C_3 + \\ldots = 0$$ 16. $$3.C_1 + 7.C_2 + 11.C_3 + \\ldots + (4n - 1)C_n = (2n - 1)2^{n + 1}$$ 17. $$C_0 + \\frac{C_2}{3} + \\frac{C_4}{5} + \\ldots = \\frac{2^n}{n + 1}$$ 18. $${}^nC_0{}^{n + 1}C_1 + {}^nC_1{}^{n + 1}C_2 + \\ldots + {}^nC_n{}^{n + 1}C_{n + 1} = \\frac{(2n + 1)!}{(n + 1)!n!}$$ 19. Prove that the sum of coefficients in the expansion of $$(1 + x - 3x^2)^{2163}$$ is $$-1$$ 20. If $$(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + \\ldots + a_{12}x^{12},$$ show that $$a_2 + a_4 + a_6 + \\ldots + a_{12} = 31$$ 21. Find the sum of rational terms in the expansion of $$(2 + \\sqrt[5]{3})^{10}$$ 22. Find the fractional part of $$\\frac{2^{4n}}{15}.$$ 23. Show that the", "$a_{1}^{2} c^{2} - c^{2} + a^{4} v^{2} - a_{1}^{2} v^{2} = a_{1}^{4} v^{2}$ $a_{1}^{2} c^{2} - a_{1}^{2} v^{2} = c^{2}$ $a_{1}^{2} ( c^{2} - v^{2} ) = c^{2}$ $a_{1}^{2} = \\frac{ c^{2} }{ c^{2} - v^{2} } = \\frac{ 1 }{ 1 - v^{2}/c^{2} }$ so $\\boxed{ a_{1} = \\frac{ 1 }{ \\sqrt{ (1 - v^{2}/c^{2} ) } } }$ Thus we can write $\\boxed{ a_{2} = -v \\cdot \\frac{ 1 }{ \\sqrt{ ( 1 - v^{2}/c^{2} ) } } }$ Using equation (8) we can write $b_{1}^{2} c^{2} = \\frac{ 1 }{ (1 - v^{2}/c^{2} ) } - 1$ $b_{1}^{2} c^{2} = \\frac{ 1 - ( 1 - v^{2}/c^{2} ) }{ (1 - v^{2}/c^{2} ) } = \\frac{ v^{2}/c^{2} }{ (1 - v^{2}/c^{2} ) } = \\frac{ v^{2} }{ c^{2} } \\cdot \\frac { 1 }{ (1 - v^{2}/c^{2} ) }$ so $b_{1}^{2} = \\frac{ v^{2} }{ c^{4} } \\cdot \\frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$ Taking the negative square root we can write $\\boxed{ b_{1} = - \\frac{ v }{ c^{2} } \\cdot \\frac{ 1 }{\\sqrt{ (1 - v^{2}/c^{2} ) }} }$ From equation (9) we can write $b_{2}^{2} c^{2} = c^{2} + v^{2} \\cdot \\frac{ 1 }{ ( 1 - v^{2}/c^{2} ) } = \\frac{ c^{2}( 1 - v^{2}/c^{2} ) + v^{2}", "$a_{1}^{2} c^{2} - c^{2} + a^{4} v^{2} - a_{1}^{2} v^{2} = a_{1}^{4} v^{2}$ $a_{1}^{2} c^{2} - a_{1}^{2} v^{2} = c^{2}$ $a_{1}^{2} ( c^{2} - v^{2} ) = c^{2}$ $a_{1}^{2} = \\frac{ c^{2} }{ c^{2} - v^{2} } = \\frac{ 1 }{ 1 - v^{2}/c^{2} }$ so $\\boxed{ a_{1} = \\frac{ 1 }{ \\sqrt{ (1 - v^{2}/c^{2} ) } } }$ Thus we can write $\\boxed{ a_{2} = -v \\cdot \\frac{ 1 }{ \\sqrt{ ( 1 - v^{2}/c^{2} ) } } }$ Using equation (8) we can write $b_{1}^{2} c^{2} = \\frac{ 1 }{ (1 - v^{2}/c^{2} ) } - 1$ $b_{1}^{2} c^{2} = \\frac{ 1 - ( 1 - v^{2}/c^{2} ) }{ (1 - v^{2}/c^{2} ) } = \\frac{ v^{2}/c^{2} }{ (1 - v^{2}/c^{2} ) } = \\frac{ v^{2} }{ c^{2} } \\cdot \\frac { 1 }{ (1 - v^{2}/c^{2} ) }$ so $b_{1}^{2} = \\frac{ v^{2} }{ c^{4} } \\cdot \\frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$ Taking the negative square root we can write $\\boxed{ b_{1} = - \\frac{ v }{ c^{2} } \\cdot \\frac{ 1 }{\\sqrt{ (1 - v^{2}/c^{2} ) }} }$ From equation (9) we can write $b_{2}^{2} c^{2} = c^{2} + v^{2} \\cdot \\frac{ 1 }{ ( 1 - v^{2}/c^{2} ) } = \\frac{ c^{2}( 1 - v^{2}/c^{2} ) + v^{2}", "- 3 \\sqrt{2}$. $\\therefore x = \\frac{1 \\pm \\sqrt{1 - 4 \\cdot \\sqrt{2} \\cdot - 3 \\sqrt{2}}}{2 \\sqrt{2}}$ $= \\frac{1 \\pm \\sqrt{1 + 24}}{2 \\sqrt{2}}$ $= \\frac{1 \\pm \\sqrt{25}}{2 \\sqrt{2}}$ $= \\frac{1 \\pm 5}{2 \\sqrt{2}}$ $\\therefore {x}_{1} = \\frac{1 + 5}{2 \\sqrt{2}} = \\frac{6}{2 \\sqrt{2}} = \\frac{3}{\\sqrt{2}} = \\frac{3 \\sqrt{2}}{2}$ $\\therefore {x}_{2} = \\frac{1 - 5}{2 \\sqrt{2}} = - \\frac{4}{2 \\sqrt{2}} = - \\frac{2}{\\sqrt{2}} = - \\sqrt{2}$", "Math Help - More problems 1. More problems Simplify: $(2 - a^{2}) \\cdot \\frac {\\sqrt {a + \\sqrt {2} }} {\\sqrt {(\\sqrt {2} + a)^{3}}}, a > \\sqrt {2}$ This one, too: $x^{n-1} \\cdot (x^{n})^{2-n}$ 2. Originally Posted by p.numminen Simplify: $(2 - a^{2}) \\cdot \\frac {\\sqrt {a + \\sqrt {2} }} {\\sqrt {(\\sqrt {2} + a)^{3}}}, a > \\sqrt {2}$ $ (2 - a^{2}) \\frac {\\sqrt {a + \\sqrt {2} }} {\\sqrt {(\\sqrt {2} + a)^{3}}}= (2 - a^{2}) \\frac {(a + \\sqrt {2})^{1/2}} {(\\sqrt {2} + a)^{3/2}}= (2 - a^{2}) \\frac {1} {(\\sqrt {2} + a)}=\\sqrt{2}-a $ 3. How exactly do you get $\\sqrt{2}-a$ out of $(2 - a^{2}) \\frac {1} {(\\sqrt {2} + a)}$ ? 4. Hello, p.numminen! Simplify: . $(2 - a^{2}) \\cdot \\frac {\\sqrt {a + \\sqrt {2} }} {\\sqrt {(\\sqrt {2} + a)^{3}}},\\;\\;a > \\sqrt {2}$ We have: . $(2-a^2)\\cdot \\frac{(\\sqrt{2}+a)^{\\frac{1}{2}}} {(\\sqrt{2}+a)^{\\frac{3}{2}}}\\;\\;=\\;\\;\\frac{2-a^2}{\\sqrt{2}+a}$ The numerator factors: . $\\frac{(\\sqrt{2} - a)(\\sqrt{2} + a)}{\\sqrt{2} + a}$. . . . . and we can cancel.", "Its sum is: . $2\\cdot\\frac{2^n-1}{2-1}\\:=\\:2(2^n-1)$ Hence, [3] becomes: . $-S_n \\;=\\;2(2^n-1) - n\\cdot2^{n+1} \\quad\\Rightarrow\\quad -S_n \\;=\\;2^{n+1} - 2 - n\\cdot2^{n+1}$ . . $-S_n \\;=\\;-2 - (n-1)2^{n+2} \\quad\\Rightarrow\\quad \\boxed{S_n \\;=\\;(n-1)2^{n+1} + 2}$" ]

[ "+0 0 126 3 What is the least integer greater than $\\sqrt{300}$? Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Awesomeguy Jul 6, 2021 #2 +1 That was quick, thank you so much! Guest Jul 6, 2021 #3 +857 0 $\\lceil{\\sqrt{300}}\\rceil = \\lceil{\\sqrt{3 \\cdot 100}}\\rceil = \\lceil{10 \\sqrt{3}}\\rceil \\approx \\lceil{10 \\cdot 1.73}\\rceil = \\lceil{17.3}\\rceil = \\boxed{18}$", "+0 # What is the greatest integer less than or equal to \\sqrt[3]{504}? 0 103 2 +182 What is the greatest integer less than or equal to \\sqrt[3]{504}? Aug 26, 2022 #1 +29 -3 What is the less than or equal to $$\\sqrt[3]{504}$$ $$\\sqrt[3]{504} = 7.95811441579$$ The greatest integer less than 7.95811441579 is 7, so the answer is 7. 7 Aug 26, 2022", "must be greater than 0.", "the smallest representable number is around $$10^{-300}$$.", "## What is the greatest integer that is less than $\\sqrt6+\\sqrt7$? please explain it and answer it Question What is the greatest integer that is less than $\\sqrt6+\\sqrt7$?", "is NOT GREATER THAN $x^2$.", "2$$ are all numbers greater than $$2$$:", "2017, 9:41 p.m. edited Each integer will be greater than 0 and less than 180.", "with the greatest possible divisor which is also a divisor of $x$. – user3243499 May 26 '17 at 19:02", "be greater than or equal to $\\mathbf{\\sqrt{5\\;gr}}$.", "so 300 is closer to 500 than 0. Also, 300 - 200 = 100, so 300 is closer to 200 than 500. 2. 250 is half-way between 0 and 500. Any number greater than 250 will be closer to 500 than it is to 0. 350 is half-way between 200 and 500. Any number less than 350 will be closer to 200 than it is to 500. We are looking for numbers that satisfy both of the above conditions. Thus, the answer is all numbers that are greater than 250 and less than 350. Note that $300 \\gt 250$ and $300 \\lt 350$, so this is consistent with our example in part (a).", "less than 400. Only 396 satisfies this condition.", "nearest integer. In particular, this minimum distance is a maximum if $5c$ is half an integer. Thus the largest possible minimum distance is $\\dfrac{1}{2\\sqrt{29}}$.", "# Primes over 300 Computer Science Level 2 Let P be the 30th prime number greater than 300. What is P? ×", "$$p-1$$, which is less than $$1$$ away from $$\\sqrt{m}$$.", "Primes over 300 Let P be the 30th prime number greater than 300. What is P? × Problem Loading... Note Loading... Set Loading...", "+0 # Help me ASAP please +1 33 2 +31 What is the greatest integer that is less than $\\sqrt6+\\sqrt7$? please explain it and answer it Mar 7, 2020 ### 2+0 Answers #1 +21725 +3 sqrt(6)+sqrt(7) = 5.0952410538477687 So '5' would be the biggest integer less than Mar 7, 2020 #2 +288 +2 So... biggest integer that's less than sqrt6 + sqrt7. Sqrt6 + sqrt7 = 5.095... So the greatest integer less than 5.095 is 5. Mar 7, 2020", "\\sqrt{4 - (x - 1)^2}$ and this expression has a maximum of 3 at $x = 1.$ That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6. I’m grateful to Alexander Bogomolny for spotting the error in my original argument.", "# Finding a greatest common divisor What is the greatest common divisor of $$2982$$ and $$2016$$? ×", "is 38 7200 now hear this is 0084 10800 + 18 x is greater than 30 X + 7 2007 2001 800006 10 - 7 is 3 and 30 - 18 at we have 05:00 - 05:5912123 600 it is 300 can say X is less than 300 here now after that we can say the value of x is lying between the 120 S is greater than 120 and less than 300 answer this question is 120" ]

[ "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "(16 - 1)^2 + 1 = \\boxed{406}$.", "+0 0 126 3 What is the least integer greater than $\\sqrt{300}$? Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Awesomeguy Jul 6, 2021 #2 +1 That was quick, thank you so much! Guest Jul 6, 2021 #3 +857 0 $\\lceil{\\sqrt{300}}\\rceil = \\lceil{\\sqrt{3 \\cdot 100}}\\rceil = \\lceil{10 \\sqrt{3}}\\rceil \\approx \\lceil{10 \\cdot 1.73}\\rceil = \\lceil{17.3}\\rceil = \\boxed{18}$", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "+ 31 = \\boxed{058}$.", "add all three cases: $1+17+243=\\boxed{261}$", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "integer less than or equal to the real number x. How many natural numbers n satisfy the equation $$[\\sqrt{n}] = 17?$$ A. 17 B. 34 C. 35 D. 36 E. 38 From above we have that $$17\\leq{\\sqrt{n}}<18$$ --> $$289\\leq{n}<324$$. Thus n can take 35 integer values from 289 to 323, inclusive. Hi Bunuel, I was able to identify 289<n<=324, but I came with answer 36 because (324-289) + 1 = 36. I think my mistake was I included 289 also in my calculation by adding 1. Whereas 289<n, so n should start from 290. Hence the OA is 35. Am I correct? $$289\\leq{n}<324$$. So, n can take all integer value from 289 to 323 inclusive: 323-289+1 = 35. _________________ Re: [x] is the greatest integer less than or equal to the real n [#permalink] 04 May 2019, 09:10 Display posts from previous: Sort by", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "$14$ of these extra pairs. The answer is $28+21+14 = \\boxed{063}$", "300$ . So, the answer is $2\\cdot150+1$ or $\\boxed{\\textbf{(C)}\\ 301}$.", "2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=\\boxed{007}$. ~adyj", "324.1-b2 $$\\Q(\\sqrt{17})$$ 324.1 324.1-b $$\\bigl[1$$ , $$-1$$ , $$1$$ , $$3 a + 4$$ , $$2 a + 3\\bigr]$$ 324.1-d2 $$\\Q(\\sqrt{17})$$ 324.1 324.1-d $$\\bigl[1$$ , $$-1$$ , $$0$$ , $$27 a + 39$$ , $$-81 a - 127\\bigr]$$", "# Rounding to nearest integer If you round to the nearest integer, why do we look at the number it was before instead of the number which it is at that moment? For example: 17.495 5 or higher goes up which makes it 17.50 which becomes 17.5 Why do we look at the four instead of at the moment the five which makes it 17 instead of 18? (sorry for my long build up/question :( ) It's because $$17.495$$ is closer to $$17$$ than $$18$$ so the nearest integer is $$17$$. The difference between $$17$$ and $$17.495$$ is $$0.495$$, but the difference between $$18$$ and $$17.495$$ is $$0.505$$ which is greater than $$0.495$$.", "= 106$, and $AB = 106 + 87 = \\boxed{193}$.", "* 7 * 131$ $2 + 3 + 5 + 7 + 131 = \\boxed{148}$", "that $332.5$ is closer to $385$ than any other answer choices, so our answer is $385$ which is $\\boxed{\\textbf{(A)}}$.", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "odd numbers whose product is 323. Two consecutive odd numbers whose product is 323 are 17 and 19, and $-17$ and $-19.$" ]

[ "+0 0 126 3 What is the least integer greater than $\\sqrt{300}$? Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Jul 6, 2021 #1 +238 +2 Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\\sqrt{300}$ Awesomeguy Jul 6, 2021 #2 +1 That was quick, thank you so much! Guest Jul 6, 2021 #3 +857 0 $\\lceil{\\sqrt{300}}\\rceil = \\lceil{\\sqrt{3 \\cdot 100}}\\rceil = \\lceil{10 \\sqrt{3}}\\rceil \\approx \\lceil{10 \\cdot 1.73}\\rceil = \\lceil{17.3}\\rceil = \\boxed{18}$", "+ 30\\sqrt{3}$, and so our answer is $\\boxed{053}$.", "+0 # What is the greatest integer less than or equal to \\sqrt[3]{504}? 0 103 2 +182 What is the greatest integer less than or equal to \\sqrt[3]{504}? Aug 26, 2022 #1 +29 -3 What is the less than or equal to $$\\sqrt[3]{504}$$ $$\\sqrt[3]{504} = 7.95811441579$$ The greatest integer less than 7.95811441579 is 7, so the answer is 7. 7 Aug 26, 2022", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "integer less than or equal to the real number x. How many natural numbers n satisfy the equation $$[\\sqrt{n}] = 17?$$ A. 17 B. 34 C. 35 D. 36 E. 38 From above we have that $$17\\leq{\\sqrt{n}}<18$$ --> $$289\\leq{n}<324$$. Thus n can take 35 integer values from 289 to 323, inclusive. Hi Bunuel, I was able to identify 289<n<=324, but I came with answer 36 because (324-289) + 1 = 36. I think my mistake was I included 289 also in my calculation by adding 1. Whereas 289<n, so n should start from 290. Hence the OA is 35. Am I correct? $$289\\leq{n}<324$$. So, n can take all integer value from 289 to 323 inclusive: 323-289+1 = 35. _________________ Re: [x] is the greatest integer less than or equal to the real n [#permalink] 04 May 2019, 09:10 Display posts from previous: Sort by", "\\sqrt{129}}{20}$. Thus, our final answer is $3 + 129 + 20 = \\boxed{152}$.", "less than $30$ are $8, 27, 125$, and so the answer is $7122 + 8 + 27 + 125 = \\boxed{7282}$.", "710 \\qquad 147$$ We know that 1470 and 1047 are greater than the three three-digit numbers, since 1470 and 1047 are both greater than one thousand, and the three-digit numbers are less than one thousand. 1470 is greater than 1047 because 1470 contains one thousand and four hundreds, while 1047 contains one thousand and less than one hundred. We know that 847 is greater than 710 because 847 contains eight hundreds, and 710 contains less than eight hundreds. 710 is greater than 147 because 710 contains seven hundreds, and 147 contains less than two hundreds.", "+ 31 = \\boxed{058}$.", "# Rounding to nearest integer If you round to the nearest integer, why do we look at the number it was before instead of the number which it is at that moment? For example: 17.495 5 or higher goes up which makes it 17.50 which becomes 17.5 Why do we look at the four instead of at the moment the five which makes it 17 instead of 18? (sorry for my long build up/question :( ) It's because $$17.495$$ is closer to $$17$$ than $$18$$ so the nearest integer is $$17$$. The difference between $$17$$ and $$17.495$$ is $$0.495$$, but the difference between $$18$$ and $$17.495$$ is $$0.505$$ which is greater than $$0.495$$.", "2$$ are all numbers greater than $$2$$:", "so 300 is closer to 500 than 0. Also, 300 - 200 = 100, so 300 is closer to 200 than 500. 2. 250 is half-way between 0 and 500. Any number greater than 250 will be closer to 500 than it is to 0. 350 is half-way between 200 and 500. Any number less than 350 will be closer to 200 than it is to 500. We are looking for numbers that satisfy both of the above conditions. Thus, the answer is all numbers that are greater than 250 and less than 350. Note that $300 \\gt 250$ and $300 \\lt 350$, so this is consistent with our example in part (a).", "## What is the greatest integer that is less than $\\sqrt6+\\sqrt7$? please explain it and answer it Question What is the greatest integer that is less than $\\sqrt6+\\sqrt7$?", "- 17^2 = (18+17)(18-17) = 35.$$. ##### General Discussion Manager Joined: 15 Apr 2013 Posts: 68 Location: India Concentration: Finance, General Management Schools: ISB '15 WE: Account Management (Other) Re: [x] is the greatest integer less than or equal to the real n [#permalink] ### Show Tags 08 Jun 2013, 03:44 Haven't quite understood the Question;Can someone help what is it exactly asking? Math Expert Joined: 02 Sep 2009 Posts: 55668 Re: [x] is the greatest integer less than or equal to the real n [#permalink] ### Show Tags 08 Jun 2013, 04:52 4 2 Intern Joined: 04 May 2013 Posts: 44 Re: [x] is the greatest integer less than or equal to the real n [#permalink] ### Show Tags 24 Aug 2013, 13:09 1 Bunuel wrote: pavan2185 wrote: Haven't quite understood the Question;Can someone help what is it exactly asking? \"[x] is the greatest integer less than or equal to the real number x\" means that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ... Now, since $$[\\sqrt{n}] = 17$$, then $$17\\leq{\\sqrt{n}}<18$$ --> ANy number from this range when rounded down to the nearest integer is 17. Hope it helps. We are only told that \"[x] is the greatest integer less than or equal to the real number ..\"", "-$\\sqrt{x - 1}$= -17$\\sqrt{x - 1}\\$ = 17 x - 1 = 289 x = 290 Vậy x = 290. • 13 lượt xem Cập nhật: 07/09/2021", "(16 - 1)^2 + 1 = \\boxed{406}$.", "# Between what two consecutive integers do sqrt450 lie? 21&22 ${21}^{2} = 441$ ${22}^{2} = 484$ It lies between 21&22 but it is almost $21$ as you can notice it is only $9$ less than 450", "If $a = 18.36$, what is the approximate value of $$\\sqrt{a^2 \\cdot 123456789}$$ Round to the nearest integer. No calculators!", "+0 +1 201 3 +177 How many positive integers less than 2008 have an even number of divisors? May 6, 2021 #1 +2209 0 If they have an odd number of divisors, they're prime, there are 44 prime numbers less than 2008. sqrt(2008) 2008-44 = 1964 =^._.^= May 6, 2021 #2 +2196 +1 Hi Catmg, Your text should read: “If they have an odd number of divisors, they're perfect squares. How many positive integers less than 2008 have an even number of divisors? So ... $$2007-\\lfloor\\sqrt{2007}\\rfloor = 1963$$ --------- Catmg, you have wonderful math skills. You present excellent solutions to a fairly broad range of math questions. GA GingerAle May 6, 2021 #3 +2209 +1 Aww, thank you. :))", "# What are the two integers immediately less than and greater than sqrt5? Sep 28, 2016 $2 < \\sqrt{5} < 3$ The question asks for the two integers immediately less than and greater than $\\sqrt{5}$ $\\sqrt{5} \\cong 2.236$ Hence: $2 < \\sqrt{5} < 3$" ]

[ "and to the nearest hundredth.", "% Round to the nearest thousands.", "# Definer Round the number $22.563$ to the nearest hundredths place. A. $22.56$ B. $22.57$ C. $22.55$ D. $22.54$ E. $22.58$", "Round 5088 to the nearest ten, nearest hundred, and nearest thousand", "$$157$$ $$160$$ Exercise $$\\PageIndex{16}$$ Round to the nearest ten: $$884$$ $$880$$ Example $$\\PageIndex{9}$$: round a whole number Round each number to the nearest hundred: 1. $$23,658$$ 2. $$3,978$$ Solution Locate the hundreds place. The digit of the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. So 23,658 rounded to the nearest hundred is 23,700. Locate the hundreds place. Underline the digit to the right of the hundreds place. The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by added 1 to the 9. Then replace all digits to the right of the hundreds place with zeros. So 3,978 rounded to the nearest hundred is 4,000. Exercise $$\\PageIndex{17}$$ Round to the nearest hundred: $$17,852$$. $$17,900$$ Exercise $$\\PageIndex{18}$$ Round to the nearest hundred: $$4,951$$. $$5,000$$ Example $$\\PageIndex{10}$$: Round a whole number Round each number to the nearest thousand: 1. $$147,032$$ 2. $$29,504$$ Solution Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the", "# How do you round 2,220 to the nearest hundred? $2200.$ It's $2200$, this is the closest 100 to the number. The number lies between $2200 < 2200 < 2300$, thus it is $2200$.", "# How do you round 364 to the nearest hundred? Mar 4, 2018 $400$. #### Explanation: The number $364$ is bounded by $300$ and $400$. Let's see which one it is closest to: $364 - 300 = 64$ $400 - 364 = 36$ $364$ is closest to $400$, so rounded to the nearest hundred, it will be $400$.", "## Prealgebra (7th Edition) $482.782$ To round to the nearest thousandth, look at the ten-thousands place in $482.7817$. It is 7, so round up. Therefore, the answer is $482.782$", "number of places (two places) to the right. Note that we must add two trailing zeros in the dividend. Thus, the problem becomes: $314 \\overline{ )2000}\\nonumber$ We need to round to the nearest hundredth. This requires that we carry the division one additional place to the right of the hundredths place (i.e., to the thousandths place). For the final step, we must round 6.369 to the nearest hundredth. In the schematic that follows, we’ve boxed the hundredths digit (the “rounding digit”) and the “test digit” that follows the “rounding digit.” Because the “test digit” is greater than or equal to 5, we add 1 to the “rounding digit,” then truncate. Therefore, to the nearest hundredth of a foot, the diameter of the circle is approximately $d ≈ 6.37 \\text{ ft.}\\nonumber$ Exercise Dylan has a circular dog pen with circumference 100 feet. Find the radius of the pen, correct to the nearest tenth of a foot. Use π ≈ 3.14. 15.9 feet ## Exercises In Exercises 1-16, solve the equation. 1. $$5.57x − 2.45x = 5.46$$ 2. $$−0.3x − 6.5x = 3.4$$ 3. $$−5.8x + 0.32 + 0.2x = −6.96$$ 4. $$−2.2x − 0.8 − 7.8x = −3.3$$ 5. $$−4.9x + 88.2 = 24.5$$ 6. $$−0.2x − 32.71 = 57.61$$ 7. $$0.35x − 63.58 = 55.14$$ 8. $$−0.2x −", "round off 198.287 to the nearest 10 cents, we have to round it off to 1 place of decimal. =$ 198.30 (iii) Round off 782.58 to the nearest dollar. To round off 782.58 to the nearest dollar, we have to round it off to the nearest whole number. Therefore, $782.58 =$ 783 rounded to the nearest rupee dollar. (iv) Round off 475.095 to the nearest cents. To round off 475.095 to the nearest cents, we have to round it off to 2 places of decimal. 475.095 rounded off to the nearest cents is 475.10. (v) Round off 18.066 cm to the nearest mm. Since, 1 mm = 0.1 cm, so to round it off to the nearest mm, we have to round it off to one decimal places. 18.066 cm = 18.1 cm rounded off to the nearest mm or 181 mm. (vi) Round off 53.4278 m to the nearest cm. Since, 1 cm = 0.01 m, we round off 53.4278 m to the nearest cm. We have to round it off to two places of decimal. 53.427 m = 53.43 m rounded off to the nearest cm or 5343 cm. (vii) 0.00737 kg to nearest g. Since, 1 g = 0.001 kg so to round off 0.00737 kg to the nearest g, we have to round it", "0.98 ________ The hundredths Explanation: As 2 < 5, We round 0.982 to 0.98. The place value of the digit 8 is hundredths. The hundredths Question 5. 3.695 to 4 ________ The ones Explanation: As 6 > 5, We round 3.695 to 4. The place value of the digit 3 is ones. The ones Question 6. 7.486 to 7.5 ________ The tenths Explanation: As 8 > 5, We round 7.486 to 7.5. The place value of the digit 4 is tenths. The tenths Write the place value of the underlined digit. Round each number to the place of the underlined digit. Question 7. 0.592 Place value: ________ Round: ________ Place value: 5 tenths = 0.5 Round: 0.6 Explanation: 0.592 (0 x 1) + (5 x $$\\frac{1}{10}$$) + (9 x $$\\frac{1}{100}$$) + (2 x $$\\frac{1}{1000}$$) Place Value: 5 x $$\\frac{1}{10}$$ = 5 tenths = 0.5 0.592 9 > 5 0.6 Question 8. 6.518 Place value: ________ Round: ________ Place value: 6 ones = 6 Round: 7 Explanation: 6.518 (6 x 1) + (5 x $$\\frac{1}{10}$$) + (1 x $$\\frac{1}{100}$$) + (8 x $$\\frac{1}{1000}$$) Place Value: 6 x 1 = 6 ones = 6 6.518 5 = 5 7 Question 9. 0.809 Place value: ________ Round: ________ Place value: 0 hundredths = 0 Round: 0.8 Explanation: 0.809 (0 x 1)", "+0 # Round to nearest hundredth.... 3.141*15^2*18/3 0 95 2 Round to nearest hundredth.... 3.141*15^2*18/3 Feb 5, 2022 #1 +1209 +2 3.141 * 15^2 * 18/3 = 4240.35 This is the exact answer, so there is actually no need for rounding. Feb 5, 2022 #2 +17 +2 Since $$3.141*15^2 * \\frac{18}{3} = 3.141 * 225 * 6$$ (1) $$= 3.141*5 * 45 * 2 * 3$$(2) $$= 3.141 * 45 * 3 * 10$$(3) $$= 31.41 * 135$$(4) $$= 4240.35$$(5) Note that from step 3 or 4, we found that we don't need to round it to the nearest hundredth because that is what the answer is in and step 5 tels us the answer. Feb 5, 2022", "## Prealgebra (7th Edition) a) $52.9\\%$ b) $52.86\\%$ a) To round to the nearest tenth, inspect the digit in the hundredths position. $52.8\\underset\\uparrow647\\%$$\\Rightarrow since \\geq5, round up 52.8647\\%\\approx52.9\\% b) To round to the nearest hundredth, inspect the digit in the thousandths position. 52.86\\underset\\uparrow47\\%$$\\Rightarrow$ since $\\lt5$, round down $52.8647\\%\\approx52.86\\%$", "+0 pythagorean theorem 0 185 1 what is the squareroot of 1.4184 round to the nearest tenth Guest Apr 20, 2017 Sort: #1 +116 +2 $$\\sqrt{1.4184}$$ = 1.1909659944767525 Rounded to nearest tenth = 1.19 BlueFire28 Apr 20, 2017", "+0 pythagorean theorem 0 231 1 what is the squareroot of 1.4184 round to the nearest tenth Guest Apr 20, 2017 #1 +116 +2 $$\\sqrt{1.4184}$$ = 1.1909659944767525 Rounded to nearest tenth = 1.19 BlueFire28 Apr 20, 2017", "your answer to the nearest hundredth. A 7 4 B $ZA= Round your answer to the nearest hundredth. A 7 4 B$...", "# How do you found 0.468 to the nearest hundredth? Since we are given the decimal $0.468$, and since the $8$ that comes after the $6$ is greater than $5$, we can add $1$ to the hundredths place. 0.468 translates to $- - - \\to$ 0.47", "round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$ D $$56.0$$ Option A is Correct # Rounding Rule for Decimal Number (Thousandths Place) • To round off a decimal to the nearest hundredth place, analyze the digit at the thousandths place. •", "5$ rounded to the nearest unit is $136$. We have that $136 \\cdotp 5$ is half way between $136$ and $137$, and the convention is to round to the nearest even digit. $2 \\cdotp 484$ to Nearest Hundredth $2 \\cdotp 484$ rounded to the nearest hundredth is $2 \\cdotp 48$. This is because $2 \\cdotp 484$ is closer to $2 \\cdotp 48$ than $2 \\cdotp 49$. $0 \\cdotp 0435$ to Nearest Thousandth $0 \\cdotp 0435$ rounded to the nearest thousandth is $0 \\cdotp 044$. We have that $0 \\cdotp 0435$ is half way between $0 \\cdotp 043$ and $0 \\cdotp 044$, and the convention is to round to the nearest even digit. $4 \\cdotp 500 \\, 01$ to Nearest Unit $4 \\cdotp 500 \\, 01$ rounded to the nearest unit is $5$. This is because $4 \\cdotp 500 \\, 01$ is closer to $5$ than $4$.", "## Prealgebra (7th Edition) $600$ To round $568$ to the nearest hundred, look at the tens place. Since it is $\\gt 50$, round up." ]

[ "$$157$$ $$160$$ Exercise $$\\PageIndex{16}$$ Round to the nearest ten: $$884$$ $$880$$ Example $$\\PageIndex{9}$$: round a whole number Round each number to the nearest hundred: 1. $$23,658$$ 2. $$3,978$$ Solution Locate the hundreds place. The digit of the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. So 23,658 rounded to the nearest hundred is 23,700. Locate the hundreds place. Underline the digit to the right of the hundreds place. The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by added 1 to the 9. Then replace all digits to the right of the hundreds place with zeros. So 3,978 rounded to the nearest hundred is 4,000. Exercise $$\\PageIndex{17}$$ Round to the nearest hundred: $$17,852$$. $$17,900$$ Exercise $$\\PageIndex{18}$$ Round to the nearest hundred: $$4,951$$. $$5,000$$ Example $$\\PageIndex{10}$$: Round a whole number Round each number to the nearest thousand: 1. $$147,032$$ 2. $$29,504$$ Solution Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the", "# How do you round 364 to the nearest hundred? Mar 4, 2018 $400$. #### Explanation: The number $364$ is bounded by $300$ and $400$. Let's see which one it is closest to: $364 - 300 = 64$ $400 - 364 = 36$ $364$ is closest to $400$, so rounded to the nearest hundred, it will be $400$.", "rounded up to $$2,000$$. #### Round off the number: $$2,875$$ A $$3,000$$ B $$2,000$$ C $$1,000$$ D $$4,000$$ × Given: $$2,875$$ The nearest thousands to $$2,875$$ are $$2,000$$ and $$3,000$$. Hundreds place digit $$8$$, is greater than $$5$$ and also $$3,000$$ is closer to $$2,875$$. So, we will round the number up to $$3,000$$. Hence, option (A) is correct. ### Round off the number: $$2,875$$ A $$3,000$$ . B $$2,000$$ C $$1,000$$ D $$4,000$$ Option A is Correct # Rounding Rules for Ten-thousands Place • To round off a whole number to the nearest ten-thousands place, analyze the digit at the thousands place. • If the digit at the thousands place is $$5$$ or greater than $$5$$, round the number up to the nearest ten-thousands place. • If the digit at the thousands place is less than $$5$$, round the number down to the nearest ten-thousands place. Let's take an example. Round off the whole number $$23,750$$ to the nearest ten-thousands place. • We know that the nearest ten thousands to $$23,750$$ are $$20,000$$ and $$30,000$$. • Thousands place digit $$3$$ is less than $$5$$ and also $$23,750$$ is closer to $$20,000$$. • So, we will round the number $$23,750$$ down to $$20,000$$. Note: The values from $$20,001$$ to $$24,999$$ are rounded down to $$20,000$$ and the value", "thousands place is 0. Since 0 is less than 5, we do not change the digit in the thousands place. We then replace all digits to the right of the thousands pace with zeros. So 147,032 rounded to the nearest thousand is 147,000. Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 5. Since 5 is greater than or equal to 5, round up by adding 1 to the 9. Then replace all digits to the right of the thousands place with zeros. So 29,504 rounded to the nearest thousand is 30,000. Notice that in part (b), when we add $$1$$ thousand to the $$9$$ thousands, the total is $$10$$ thousands. We regroup this as $$1$$ ten thousand and $$0$$ thousands. We add the $$1$$ ten thousand to the $$2$$ ten thousands and put a $$0$$ in the thousands place. Exercise $$\\PageIndex{19}$$ Round to the nearest thousand: $$63,921$$. $$64,000$$ Exercise $$\\PageIndex{20}$$ Round to the nearest thousand: $$156,437$$. $$156,000$$ ## Key Concepts Figure $$\\PageIndex{13}$$ • Name a whole number in words. • Starting at the digit on the left, name the number in each period, followed by the period name. Do not include the period name for the ones. • Use commas in the", "round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$ D $$56.0$$ Option A is Correct # Rounding Rule for Decimal Number (Thousandths Place) • To round off a decimal to the nearest hundredth place, analyze the digit at the thousandths place. •", "and $$8$$ is greater than $$5$$. • Thus, the digit of tenths place increases by $$1$$ and the digit at the hundredths place become zero. • So, the nearest tenths to $$49.38$$ after round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$", "0.98 ________ The hundredths Explanation: As 2 < 5, We round 0.982 to 0.98. The place value of the digit 8 is hundredths. The hundredths Question 5. 3.695 to 4 ________ The ones Explanation: As 6 > 5, We round 3.695 to 4. The place value of the digit 3 is ones. The ones Question 6. 7.486 to 7.5 ________ The tenths Explanation: As 8 > 5, We round 7.486 to 7.5. The place value of the digit 4 is tenths. The tenths Write the place value of the underlined digit. Round each number to the place of the underlined digit. Question 7. 0.592 Place value: ________ Round: ________ Place value: 5 tenths = 0.5 Round: 0.6 Explanation: 0.592 (0 x 1) + (5 x $$\\frac{1}{10}$$) + (9 x $$\\frac{1}{100}$$) + (2 x $$\\frac{1}{1000}$$) Place Value: 5 x $$\\frac{1}{10}$$ = 5 tenths = 0.5 0.592 9 > 5 0.6 Question 8. 6.518 Place value: ________ Round: ________ Place value: 6 ones = 6 Round: 7 Explanation: 6.518 (6 x 1) + (5 x $$\\frac{1}{10}$$) + (1 x $$\\frac{1}{100}$$) + (8 x $$\\frac{1}{1000}$$) Place Value: 6 x 1 = 6 ones = 6 6.518 5 = 5 7 Question 9. 0.809 Place value: ________ Round: ________ Place value: 0 hundredths = 0 Round: 0.8 Explanation: 0.809 (0 x 1)", "5$ rounded to the nearest unit is $136$. We have that $136 \\cdotp 5$ is half way between $136$ and $137$, and the convention is to round to the nearest even digit. $2 \\cdotp 484$ to Nearest Hundredth $2 \\cdotp 484$ rounded to the nearest hundredth is $2 \\cdotp 48$. This is because $2 \\cdotp 484$ is closer to $2 \\cdotp 48$ than $2 \\cdotp 49$. $0 \\cdotp 0435$ to Nearest Thousandth $0 \\cdotp 0435$ rounded to the nearest thousandth is $0 \\cdotp 044$. We have that $0 \\cdotp 0435$ is half way between $0 \\cdotp 043$ and $0 \\cdotp 044$, and the convention is to round to the nearest even digit. $4 \\cdotp 500 \\, 01$ to Nearest Unit $4 \\cdotp 500 \\, 01$ rounded to the nearest unit is $5$. This is because $4 \\cdotp 500 \\, 01$ is closer to $5$ than $4$.", "digit at the tens place is $$5$$ or greater than $$5$$, round the number up to the nearest hundreds place. • If the digit at the tens place is less than $$5$$, round the number down to the nearest hundreds place. For example: $$365$$ • After analyzing the tens place of the number, we see that the nearest hundreds to $$365$$ are $$300$$ and $$400$$. Tens place digit, $$6$$ is greater than $$5$$ and the number $$365$$ is closer to $$400$$. So, we shall round the number up to $$400$$. Note: The values from $$301$$ to $$349$$ are round off to $$300$$ and the values from $$350$$ to $$399$$ are round off to $$400$$. #### Calculate the round off value of $$575$$. A $$500$$ B $$600$$ C $$400$$ D $$700$$ × The nearest hundreds to $$575$$ are $$500$$ and $$600$$. Tens place digit, $$7$$ is greater than $$5$$ and also $$600$$ is closer to $$575$$. So, we shall round off to $$600$$. Hence, option (B) is correct. ### Calculate the round off value of $$575$$. A $$500$$ . B $$600$$ C $$400$$ D $$700$$ Option B is Correct", "nearest hundred thousand: The hundred thousands digit is $4$, and the digit behind it is $6$, which is greater than $5$; therefore, we have to increase the hundred thousands digit by $1$, giving us the following number: $19,500,000$ e.Rounding to nearest million: The millions digit is $9$, and the digit behind it is $4$, which is less than $5$; therefore, we have to keep the millions digit the same, giving us the following number: $19,000,000$ f.Rounding to nearest ten million: The ten millions digit is $1$, and the digit behind it is $9$, and therefore we have to increase the ten millions digit by $1$, giving us the following number: $20,000,000$", "# How do you round 2,220 to the nearest hundred? $2200.$ It's $2200$, this is the closest 100 to the number. The number lies between $2200 < 2200 < 2300$, thus it is $2200$.", "and set the hundreds digit equal to $0$: $\\left(1 + 1\\right) 000$ $2000$ and we are done. $1993$, rounded to the nearest hundred, is $2000$.", "Figure 1.8. Figure 1.8 We can see that $72 72$ is closer to $70 , 70 ,$ so $72 72$ rounded to the nearest ten is $70 . 70 .$ How do we round $7575$ to the nearest ten. Find $7575$ in Figure 1.9. Figure 1.9 The number $75 75$ is exactly midway between $70 70$ and $80 . 80 .$ So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, $80.80.$ So, $7575$ rounded to the nearest ten is $80.80.$ Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than $5,5,$ round down. If it is greater than or equal to $5,5,$ round up. So, for example, to round $7676$ to the nearest ten, we look at the digit in the ones place. The digit in the ones place is a $6.6.$ Because $66$ is greater than or equal to $5,5,$ we increase the digit in the tens place by one. So the $77$ in the tens place becomes an $8.8.$ Now, replace any digits to the right of the $88$ with zeros. So,", "rounding decimals are correct? 1. The number $3.053$ rounded to the nearest hundredth is $3.05$. 2. The number $3.807$ rounded to the nearest hundredth is $3.80$. 3. The number $3.102$ rounded to the nearest hundredth is $3.11$. 4. The number $3.769$ rounded to the nearest hundredth is $3.76$. 5. The number $3.296$ rounded to the nearest hundredth is $3.30$. Grade 5 :: Place Value by LBeth Which statement correctly compares the two values? 1. The value of the 4 in 7.49 is 10 times the value of the 4 in 4.85. 2. The value of the 4 in 7.49 is $1/10$ the value of the 4 in 4.85. 3. The value of the 4 in 7.49 is 100 times the value of the 4 in 4.85. 4. The value of the 4 in 7.49 is $1/100$ the value of the 4 in 4.85. Grade 5 :: Place Value by desmondsmithsr1 266,661.56 Which six digit is one-tenth of the six digit in the hundreds place? 1. six hundredths 2. sixty 3. six thousands 4. sixty thousands Grade 5 :: Place Value by valdon19 Grade 5 :: Place Value by valdon19 Write the number 678,425 in expanded form. 1. 600,000 + 70,000 + 8,000 + 400 + 20 + 5 2. 60,000 + 70,000 + 8,000 + 400 + 20", "# How do you round the number 894.754,390 to the near million? $0$ $1$ million is $1 , 000 , 000$ So to round to it the number must be larger than or equal to $500 , 000$", "that $2.72$ is closer to $3$ than to $2$. So, $2.72$ rounded to the nearest whole number is $3$. (b) We see that $2.72$ is closer to $2.70$ than $2.80$. So we say that $2.72$ rounded to the nearest tenth is $2.7$. Can we round decimals without number lines? Yes! We use a method based on the one we used to round whole numbers. ### Round a decimal. 1. Locate the given place value and mark it with an arrow. 2. Underline the digit to the right of the given place value. 3. Is this digit greater than or equal to $5?$ • Yes – add $1$ to the digit in the given place value. • No – do not change the digit in the given place value 4. Rewrite the number, removing all digits to the right of the given place value. ### Tip For Success If the digit to the right of your given place value is 5 or above, give your given place value a shove. If the digit to the right of your given place value is 4 or less, let your given place value rest. ### example Round $18.379$ to the nearest hundredth. Solution $18.379$ Locate the hundredths place and mark it with an arrow. Underline the digit to the right of the", "Prealgebra (7th Edition) $24\\underset{\\uparrow}8982$ To round to the nearest thousands, inspect the digit to the right. It is 5 or greater, so the thousands digit rounds up. Zero out all the digits to the right of the thousands digit.", "rounded-off number to the nearest hundred. Hence, $$7456$$ is rounded off to nearest hundreds as $$7500$$. It follows from the above examples that (i) the numbers ending in $$01\\ to\\ 49$$ are rounded off downwards. (ii) the numbers ending in $$50\\ to\\ 99$$ are rounded off upwards. Important Note: If the digit in the tens place is $$0,\\ 1,\\ 2,\\ 3,\\ or\\ 4$$ we replace the digits in the tens and one’s place by zero. If the digit in the tens place is $$5,\\ 6,\\ 7,\\ 8,\\ or\\ 9$$ we replace the digits in the tens and ones place by zeros. We also increase the digit in the hundreds place by $$1$$. ### Estimate to Nearest Thousands In order to estimate to nearest thousands we follow the following procedure: • Obtain the number. • Examine the digit at hundreds place. • If the digit at hundreds place is less than $$5$$, replace each one of the digits at hundreds, tens and ones or units place by $$0$$ and keep all other digits as they are. If the digit at hundreds place is $$5$$ or greater than $$5$$, increase the digit at thousands place by $$1$$ and replace each one of the digits at hundreds, tens and ones place by $$0$$. The number so obtained is the number rounded", "Noelanijd 2022-07-26 1. Round the following numberls to the nearest ten. 527, 104, 955, 423 2. Round the following numerals to the nearest hundred. 689, 527, 1.365, 421 3. Round the following numerals to the nearest thousand. 1.365, 1.748, 5.231, 4.522. 4. Round the following numerals to the nearest tenth. 527.63, 289.34, 671.57, 15.85. encoplemt5 Expert Step 1 1. Round the following numerals to the nearest 10. $527=530\\phantom{\\rule{0ex}{0ex}}104=100\\phantom{\\rule{0ex}{0ex}}955=960\\phantom{\\rule{0ex}{0ex}}423=420$ Step 2 2. Round the numerals to the nearest hundred. $689=700\\phantom{\\rule{0ex}{0ex}}527=500\\phantom{\\rule{0ex}{0ex}}1365=1400\\phantom{\\rule{0ex}{0ex}}421=400$ Step 3 3. Round the numerals to nearest thousand. $1365=1400\\phantom{\\rule{0ex}{0ex}}1748=1700\\phantom{\\rule{0ex}{0ex}}5231=5200\\phantom{\\rule{0ex}{0ex}}4522=4500$ Step 4 4 . Round the numerals to nearest tenth. $527.63=527.6\\phantom{\\rule{0ex}{0ex}}289.34=289.3\\phantom{\\rule{0ex}{0ex}}671.57=671.6\\phantom{\\rule{0ex}{0ex}}15.85=15.9$ Do you have a similar question?", "So 3,978 rounded to the nearest hundred is 4,000. Round to the nearest hundred: $17,852.$ 17,900 Round to the nearest hundred: $4,951.$ 5,000 Round each number to the nearest thousand: 1. $147,032$ 2. $29,504$ Solution ⓐ Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 0. Since 0 is less than 5, we do not change the digit in the thousands place. We then replace all digits to the right of the thousands pace with zeros. So 147,032 rounded to the nearest thousand is 147,000. ⓑ Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 5. Since 5 is greater than or equal to 5, round up by adding 1 to the 9. Then replace all digits to the right of the thousands place with zeros. So 29,504 rounded to the nearest thousand is 30,000. Notice that in part , when we add $1$ thousand to the $9$ thousands, the total is $10$ thousands. We regroup this as $1$ ten thousand and $0$ thousands. We add the $1$ ten thousand to the $3$ ten thousands and put a $0$ in the thousands place. Round to the nearest thousand:" ]

[ "$$157$$ $$160$$ Exercise $$\\PageIndex{16}$$ Round to the nearest ten: $$884$$ $$880$$ Example $$\\PageIndex{9}$$: round a whole number Round each number to the nearest hundred: 1. $$23,658$$ 2. $$3,978$$ Solution Locate the hundreds place. The digit of the right of the hundreds place is 5. Underline the digit to the right of the hundreds place. Since 5 is greater than or equal to 5, round up by adding 1 to the digit in the hundreds place. Then replace all digits to the right of the hundreds place with zeros. So 23,658 rounded to the nearest hundred is 23,700. Locate the hundreds place. Underline the digit to the right of the hundreds place. The digit to the right of the hundreds place is 7. Since 7 is greater than or equal to 5, round up by added 1 to the 9. Then replace all digits to the right of the hundreds place with zeros. So 3,978 rounded to the nearest hundred is 4,000. Exercise $$\\PageIndex{17}$$ Round to the nearest hundred: $$17,852$$. $$17,900$$ Exercise $$\\PageIndex{18}$$ Round to the nearest hundred: $$4,951$$. $$5,000$$ Example $$\\PageIndex{10}$$: Round a whole number Round each number to the nearest thousand: 1. $$147,032$$ 2. $$29,504$$ Solution Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the", "rounded up to $$2,000$$. #### Round off the number: $$2,875$$ A $$3,000$$ B $$2,000$$ C $$1,000$$ D $$4,000$$ × Given: $$2,875$$ The nearest thousands to $$2,875$$ are $$2,000$$ and $$3,000$$. Hundreds place digit $$8$$, is greater than $$5$$ and also $$3,000$$ is closer to $$2,875$$. So, we will round the number up to $$3,000$$. Hence, option (A) is correct. ### Round off the number: $$2,875$$ A $$3,000$$ . B $$2,000$$ C $$1,000$$ D $$4,000$$ Option A is Correct # Rounding Rules for Ten-thousands Place • To round off a whole number to the nearest ten-thousands place, analyze the digit at the thousands place. • If the digit at the thousands place is $$5$$ or greater than $$5$$, round the number up to the nearest ten-thousands place. • If the digit at the thousands place is less than $$5$$, round the number down to the nearest ten-thousands place. Let's take an example. Round off the whole number $$23,750$$ to the nearest ten-thousands place. • We know that the nearest ten thousands to $$23,750$$ are $$20,000$$ and $$30,000$$. • Thousands place digit $$3$$ is less than $$5$$ and also $$23,750$$ is closer to $$20,000$$. • So, we will round the number $$23,750$$ down to $$20,000$$. Note: The values from $$20,001$$ to $$24,999$$ are rounded down to $$20,000$$ and the value", "# How do you round 364 to the nearest hundred? Mar 4, 2018 $400$. #### Explanation: The number $364$ is bounded by $300$ and $400$. Let's see which one it is closest to: $364 - 300 = 64$ $400 - 364 = 36$ $364$ is closest to $400$, so rounded to the nearest hundred, it will be $400$.", "0.98 ________ The hundredths Explanation: As 2 < 5, We round 0.982 to 0.98. The place value of the digit 8 is hundredths. The hundredths Question 5. 3.695 to 4 ________ The ones Explanation: As 6 > 5, We round 3.695 to 4. The place value of the digit 3 is ones. The ones Question 6. 7.486 to 7.5 ________ The tenths Explanation: As 8 > 5, We round 7.486 to 7.5. The place value of the digit 4 is tenths. The tenths Write the place value of the underlined digit. Round each number to the place of the underlined digit. Question 7. 0.592 Place value: ________ Round: ________ Place value: 5 tenths = 0.5 Round: 0.6 Explanation: 0.592 (0 x 1) + (5 x $$\\frac{1}{10}$$) + (9 x $$\\frac{1}{100}$$) + (2 x $$\\frac{1}{1000}$$) Place Value: 5 x $$\\frac{1}{10}$$ = 5 tenths = 0.5 0.592 9 > 5 0.6 Question 8. 6.518 Place value: ________ Round: ________ Place value: 6 ones = 6 Round: 7 Explanation: 6.518 (6 x 1) + (5 x $$\\frac{1}{10}$$) + (1 x $$\\frac{1}{100}$$) + (8 x $$\\frac{1}{1000}$$) Place Value: 6 x 1 = 6 ones = 6 6.518 5 = 5 7 Question 9. 0.809 Place value: ________ Round: ________ Place value: 0 hundredths = 0 Round: 0.8 Explanation: 0.809 (0 x 1)", "thousands place is 0. Since 0 is less than 5, we do not change the digit in the thousands place. We then replace all digits to the right of the thousands pace with zeros. So 147,032 rounded to the nearest thousand is 147,000. Locate the thousands place. Underline the digit to the right of the thousands place. The digit to the right of the thousands place is 5. Since 5 is greater than or equal to 5, round up by adding 1 to the 9. Then replace all digits to the right of the thousands place with zeros. So 29,504 rounded to the nearest thousand is 30,000. Notice that in part (b), when we add $$1$$ thousand to the $$9$$ thousands, the total is $$10$$ thousands. We regroup this as $$1$$ ten thousand and $$0$$ thousands. We add the $$1$$ ten thousand to the $$2$$ ten thousands and put a $$0$$ in the thousands place. Exercise $$\\PageIndex{19}$$ Round to the nearest thousand: $$63,921$$. $$64,000$$ Exercise $$\\PageIndex{20}$$ Round to the nearest thousand: $$156,437$$. $$156,000$$ ## Key Concepts Figure $$\\PageIndex{13}$$ • Name a whole number in words. • Starting at the digit on the left, name the number in each period, followed by the period name. Do not include the period name for the ones. • Use commas in the", "round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$ D $$56.0$$ Option A is Correct # Rounding Rule for Decimal Number (Thousandths Place) • To round off a decimal to the nearest hundredth place, analyze the digit at the thousandths place. •", "Noelanijd 2022-07-26 1. Round the following numberls to the nearest ten. 527, 104, 955, 423 2. Round the following numerals to the nearest hundred. 689, 527, 1.365, 421 3. Round the following numerals to the nearest thousand. 1.365, 1.748, 5.231, 4.522. 4. Round the following numerals to the nearest tenth. 527.63, 289.34, 671.57, 15.85. encoplemt5 Expert Step 1 1. Round the following numerals to the nearest 10. $527=530\\phantom{\\rule{0ex}{0ex}}104=100\\phantom{\\rule{0ex}{0ex}}955=960\\phantom{\\rule{0ex}{0ex}}423=420$ Step 2 2. Round the numerals to the nearest hundred. $689=700\\phantom{\\rule{0ex}{0ex}}527=500\\phantom{\\rule{0ex}{0ex}}1365=1400\\phantom{\\rule{0ex}{0ex}}421=400$ Step 3 3. Round the numerals to nearest thousand. $1365=1400\\phantom{\\rule{0ex}{0ex}}1748=1700\\phantom{\\rule{0ex}{0ex}}5231=5200\\phantom{\\rule{0ex}{0ex}}4522=4500$ Step 4 4 . Round the numerals to nearest tenth. $527.63=527.6\\phantom{\\rule{0ex}{0ex}}289.34=289.3\\phantom{\\rule{0ex}{0ex}}671.57=671.6\\phantom{\\rule{0ex}{0ex}}15.85=15.9$ Do you have a similar question?", "and $$8$$ is greater than $$5$$. • Thus, the digit of tenths place increases by $$1$$ and the digit at the hundredths place become zero. • So, the nearest tenths to $$49.38$$ after round-off is $$49.4$$. Example (2): $$33.42$$ • Let's use place value chart. Tens Ones Tenths Hundredths 3 3 . 4 2 • In place value chart analyze the hundredths place. • Here, the digit at the hundredths place is $$2$$ and $$2$$ is less than $$5$$. • Thus, the digit at tenths place remains same and the digit at the hundredths place become zero. • So, the nearest tenths to $$33.42$$ after round-off is $$33.4$$. #### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ B $$57.0$$ C $$56.3$$ D $$56.0$$ × Place value chart for given decimal number is Tens Ones Tenths Hundredths 5 6 . 3 7 Here, the digit at the hundredths place is $$7$$ which is greater than $$5$$. So, the digit at the tenths place is increased by $$1$$. The digit at the hundredths place becomes zero. So, the nearest tenths to $$56.37$$ after round off is $$56.4$$. Hence, option (A) is correct. ### What is the round off value of $$56.37$$ up to the tenths place? A $$56.4$$ . B $$57.0$$ C $$56.3$$", "Prealgebra (7th Edition) $24\\underset{\\uparrow}8982$ To round to the nearest thousands, inspect the digit to the right. It is 5 or greater, so the thousands digit rounds up. Zero out all the digits to the right of the thousands digit.", "and set the hundreds digit equal to $0$: $\\left(1 + 1\\right) 000$ $2000$ and we are done. $1993$, rounded to the nearest hundred, is $2000$.", "digit at the tens place is $$5$$ or greater than $$5$$, round the number up to the nearest hundreds place. • If the digit at the tens place is less than $$5$$, round the number down to the nearest hundreds place. For example: $$365$$ • After analyzing the tens place of the number, we see that the nearest hundreds to $$365$$ are $$300$$ and $$400$$. Tens place digit, $$6$$ is greater than $$5$$ and the number $$365$$ is closer to $$400$$. So, we shall round the number up to $$400$$. Note: The values from $$301$$ to $$349$$ are round off to $$300$$ and the values from $$350$$ to $$399$$ are round off to $$400$$. #### Calculate the round off value of $$575$$. A $$500$$ B $$600$$ C $$400$$ D $$700$$ × The nearest hundreds to $$575$$ are $$500$$ and $$600$$. Tens place digit, $$7$$ is greater than $$5$$ and also $$600$$ is closer to $$575$$. So, we shall round off to $$600$$. Hence, option (B) is correct. ### Calculate the round off value of $$575$$. A $$500$$ . B $$600$$ C $$400$$ D $$700$$ Option B is Correct", "Figure 1.8. Figure 1.8 We can see that $72 72$ is closer to $70 , 70 ,$ so $72 72$ rounded to the nearest ten is $70 . 70 .$ How do we round $7575$ to the nearest ten. Find $7575$ in Figure 1.9. Figure 1.9 The number $75 75$ is exactly midway between $70 70$ and $80 . 80 .$ So that everyone rounds the same way in cases like this, mathematicians have agreed to round to the higher number, $80.80.$ So, $7575$ rounded to the nearest ten is $80.80.$ Now that we have looked at this process on the number line, we can introduce a more general procedure. To round a number to a specific place, look at the number to the right of that place. If the number is less than $5,5,$ round down. If it is greater than or equal to $5,5,$ round up. So, for example, to round $7676$ to the nearest ten, we look at the digit in the ones place. The digit in the ones place is a $6.6.$ Because $66$ is greater than or equal to $5,5,$ we increase the digit in the tens place by one. So the $77$ in the tens place becomes an $8.8.$ Now, replace any digits to the right of the $88$ with zeros. So,", "# How do you round the number 894.754,390 to the near million? $0$ $1$ million is $1 , 000 , 000$ So to round to it the number must be larger than or equal to $500 , 000$", "### Theory: Vinay's teacher asked him to round off the decimal number $$3.458$$ to the nearest whole number. Vinay did it correctly, so his teacher appreciated him, and Vinay was informed to teach the same to his friends. He explained the following steps to his friends to round off the decimal number to the nearest whole number. To round a decimal: 1. First underline the digit that is to be rounded. 2. Observe the digit which is right of the underlined number, if it is more than or equal to $$5$$, add $$1$$ to the underlined number. 3. If it is less than $$5$$, then the underlined number remains the same. 4. After rounding off the number, ignore all the digits after underlined number. Example: Let us round off the decimal $$4.689$$ to the nearest whole number. First underlined the digit is to rounded $$=$$ $\\underset{¯}{4}.689$ Here, we can observe that the digit next to the underlined digit is $$6$$, which is greater than $$5$$. So, the underlined number will add to $$1$$. Therefore, the nearest digit of the $$4.689$$ to the nearest whole number is $$5$$.", "number of places (two places) to the right. Note that we must add two trailing zeros in the dividend. Thus, the problem becomes: $314 \\overline{ )2000}\\nonumber$ We need to round to the nearest hundredth. This requires that we carry the division one additional place to the right of the hundredths place (i.e., to the thousandths place). For the final step, we must round 6.369 to the nearest hundredth. In the schematic that follows, we’ve boxed the hundredths digit (the “rounding digit”) and the “test digit” that follows the “rounding digit.” Because the “test digit” is greater than or equal to 5, we add 1 to the “rounding digit,” then truncate. Therefore, to the nearest hundredth of a foot, the diameter of the circle is approximately $d ≈ 6.37 \\text{ ft.}\\nonumber$ Exercise Dylan has a circular dog pen with circumference 100 feet. Find the radius of the pen, correct to the nearest tenth of a foot. Use π ≈ 3.14. 15.9 feet ## Exercises In Exercises 1-16, solve the equation. 1. $$5.57x − 2.45x = 5.46$$ 2. $$−0.3x − 6.5x = 3.4$$ 3. $$−5.8x + 0.32 + 0.2x = −6.96$$ 4. $$−2.2x − 0.8 − 7.8x = −3.3$$ 5. $$−4.9x + 88.2 = 24.5$$ 6. $$−0.2x − 32.71 = 57.61$$ 7. $$0.35x − 63.58 = 55.14$$ 8. $$−0.2x −", "that $2.72$ is closer to $3$ than to $2$. So, $2.72$ rounded to the nearest whole number is $3$. (b) We see that $2.72$ is closer to $2.70$ than $2.80$. So we say that $2.72$ rounded to the nearest tenth is $2.7$. Can we round decimals without number lines? Yes! We use a method based on the one we used to round whole numbers. ### Round a decimal. 1. Locate the given place value and mark it with an arrow. 2. Underline the digit to the right of the given place value. 3. Is this digit greater than or equal to $5?$ • Yes – add $1$ to the digit in the given place value. • No – do not change the digit in the given place value 4. Rewrite the number, removing all digits to the right of the given place value. ### Tip For Success If the digit to the right of your given place value is 5 or above, give your given place value a shove. If the digit to the right of your given place value is 4 or less, let your given place value rest. ### example Round $18.379$ to the nearest hundredth. Solution $18.379$ Locate the hundredths place and mark it with an arrow. Underline the digit to the right of the", "# Definer Round the number $22.563$ to the nearest hundredths place. A. $22.56$ B. $22.57$ C. $22.55$ D. $22.54$ E. $22.58$", "rounding decimals are correct? 1. The number $3.053$ rounded to the nearest hundredth is $3.05$. 2. The number $3.807$ rounded to the nearest hundredth is $3.80$. 3. The number $3.102$ rounded to the nearest hundredth is $3.11$. 4. The number $3.769$ rounded to the nearest hundredth is $3.76$. 5. The number $3.296$ rounded to the nearest hundredth is $3.30$. Grade 5 :: Place Value by LBeth Which statement correctly compares the two values? 1. The value of the 4 in 7.49 is 10 times the value of the 4 in 4.85. 2. The value of the 4 in 7.49 is $1/10$ the value of the 4 in 4.85. 3. The value of the 4 in 7.49 is 100 times the value of the 4 in 4.85. 4. The value of the 4 in 7.49 is $1/100$ the value of the 4 in 4.85. Grade 5 :: Place Value by desmondsmithsr1 266,661.56 Which six digit is one-tenth of the six digit in the hundreds place? 1. six hundredths 2. sixty 3. six thousands 4. sixty thousands Grade 5 :: Place Value by valdon19 Grade 5 :: Place Value by valdon19 Write the number 678,425 in expanded form. 1. 600,000 + 70,000 + 8,000 + 400 + 20 + 5 2. 60,000 + 70,000 + 8,000 + 400 + 20", "nearest hundred thousand: The hundred thousands digit is $4$, and the digit behind it is $6$, which is greater than $5$; therefore, we have to increase the hundred thousands digit by $1$, giving us the following number: $19,500,000$ e.Rounding to nearest million: The millions digit is $9$, and the digit behind it is $4$, which is less than $5$; therefore, we have to keep the millions digit the same, giving us the following number: $19,000,000$ f.Rounding to nearest ten million: The ten millions digit is $1$, and the digit behind it is $9$, and therefore we have to increase the ten millions digit by $1$, giving us the following number: $20,000,000$", "rounded-off number to the nearest hundred. Hence, $$7456$$ is rounded off to nearest hundreds as $$7500$$. It follows from the above examples that (i) the numbers ending in $$01\\ to\\ 49$$ are rounded off downwards. (ii) the numbers ending in $$50\\ to\\ 99$$ are rounded off upwards. Important Note: If the digit in the tens place is $$0,\\ 1,\\ 2,\\ 3,\\ or\\ 4$$ we replace the digits in the tens and one’s place by zero. If the digit in the tens place is $$5,\\ 6,\\ 7,\\ 8,\\ or\\ 9$$ we replace the digits in the tens and ones place by zeros. We also increase the digit in the hundreds place by $$1$$. ### Estimate to Nearest Thousands In order to estimate to nearest thousands we follow the following procedure: • Obtain the number. • Examine the digit at hundreds place. • If the digit at hundreds place is less than $$5$$, replace each one of the digits at hundreds, tens and ones or units place by $$0$$ and keep all other digits as they are. If the digit at hundreds place is $$5$$ or greater than $$5$$, increase the digit at thousands place by $$1$$ and replace each one of the digits at hundreds, tens and ones place by $$0$$. The number so obtained is the number rounded" ]

[ "remove one piece from each remaining column. (Brian can avoid dealing with the column that Anna is eliminating because Anna has to go first.) Thus, if Anna follows this strategy, Brian can ensure that the game is reduced to $n$ by $n-1$. If this process repeats, it will eventually by $2$ by $1$ which Brian wins. What if Anna doesn't focus on one column? Regardless, Anna must still eliminate every column but one before the last column gets too small. If Anna switches to a different column, then Brian can ignore both columns that Anna has made progress on, at least until the max column size catches up to one or the other. This is helpful to Brian if anyone, not Anna. Brian can't necessarily rely on hitting $n$ by $n-1$ in this case, but if Anna is working on two columns, Brian can ensure the game reduces to $n-1$ by $n-2$, before both columns are reduced to 0, or similar for working on more columns. If Anna is working on all the columns, then it's even worse for her.", "# [Solutions] Caucasus Mathematical Olympiad 2021 ## Juniors 1. Let $a$, $b$, $c$ be real numbers such that $a^2+b=c^2$, $b^2+c=a^2$, $c^2+a=b^2$. Find all possible values of $abc$. 2. In a triangle $ABC$ let $K$ be a point on the median $BM$ such that $CK=CM$. It appears that $\\angle CBM = 2 \\angle ABM$. Prove that $BC=MK$. 3. We have $n>2$ non-zero integers such that each one of them is divisible by the sum of the other $n-1$ numbers. Prove that the sum of all the given numbers is zero. 4. A square grid $2n \\times 2n$ is constructed of matches (each match is a segment of length 1). By one move Peter can choose a vertex which (at this moment) is the endpoint of 3 or 4 matches and delete two matches whose union is a segment of length 2. Find the least possible number of matches that could remain after a number of Peter's moves. 5. Let $a, b, c$ be positive integers such that the product$$\\gcd(a,b) \\cdot \\gcd(b,c) \\cdot \\gcd(c,a)$$is a perfect square. Prove that the product$$\\operatorname{lcm}(a,b) \\cdot \\operatorname{lcm}(b,c) \\cdot \\operatorname{lcm}(c,a)$$is also a perfect square. 6. A row of $2021$ balls is given. Pasha and Vova play a game, taking turns to perform moves; Pasha begins. On each turn a boy should paint a non-painted ball", "the two dice is divisible by $\\displaystyle 6$ or $\\displaystyle 9.$ (b) both dice showing the same number. (c) the sum of the score is a multiple of $\\displaystyle 3.$ (d) the total score on the two dice is prime. Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\,\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ {{\\text{2}}^{{\\text{nd}}}}\\ \\text{die}\\\\ \\ {{\\text{1}}^{{\\text{st}}}}\\ \\text{die}\\ \\ \\ \\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \\hline & 1 & 2 & 3 & 4 & 5 & 6 \\\\ \\hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\\\ \\hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\\\ \\hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\\\ \\hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\\\ \\hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\\\ \\hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\\\ \\hline\\end{array}\\end{array}$ $\\displaystyle \\therefore\\ \\$Number of possible outcomes $\\displaystyle =36$ (a) Set of favourable outcomes for the product of the scores", "1. In the diagram, the circle has radius $\\sqrt 7$ and and centre $O.$ Points $A$, $B$ and $C$ are on the circle. If $\\angle BOC=120^\\circ$ and $AC = AB + 1,$ determine the length of $AB.$ 2. Brennan chooses a set $A = \\{a, b,c, d, e \\}$ of five real numbers with $a \\leq b \\leq c \\leq d \\leq e.$ Delaney determines the subsets of $A$ containing three numbers and adds up the numbers in these subsets. She obtains the sums $0, 3; 4, 8; 9, 10, 11, 12, 14, 19.$ What are the five numbers in Brennan's set? 3. Determine all solutions to the system of equations $\\begin{cases}x^2 + y^2 + x + y &= 12 \\\\ xy + x + y &= 3\\end{cases}$ 4. Alphonse and Beryl play a game starting with a blank blackboard. Alphonse goes first and the two players alternate turns. On Alphonse's first turn, he writes the integer $10^{2011}$ on the blackboard. On each subsequent turn, each player can do exactly one of the following two things • Replace any number $x$ that is currently on the blackboard with two integers a and b greater than $1$ such that $x = ab,$ or • Erase one or two copies of a number $y$ that appears at least twice on", "on her $$(n^2-2)$$-th move. Suppose not. Recall that by renaming, this implies that Solver plays $$n^2-1$$ in $$r_1$$. Therefore, Spoiler spoils by playing $$n^2$$ in $$c \\setminus r_1$$, where $$c$$ is the unique column with an unfilled cell of $$r_1$$. Thus, after Solver's $$(n^2-2)$$-th move, $$n^2$$ and $$n^2-1$$ still have not been played in $$r_1$$. Spoiler now plays $$n^2-1$$ in $$c \\setminus r_1$$, where $$c$$ is a column containing an unfilled cell of $$r_1$$. Note that after this move, there is still an unfilled cell in $$c \\setminus r_1$$, since Solver has made at most $$n^2-3$$ moves outside of $$r_1$$ and Spoiler had previously never played in $$c$$. Therefore, Solver must play in $$c$$ on her $$(n^2-1)$$-th move; otherwise, Spoiler can play $$n^2$$ in $$c \\setminus r_1$$ on her next move. But now Spoiler can spoil on her $$(n^2-1)$$-th move by playing $$n^2-1$$ in $$c' \\setminus r_1$$, where $$c'$$ is the other column containing an unfilled cell of $$r_1$$. This is joint work with Tillmann Miltzow. • I think you are assuming $n\\geq 3$? – Joel David Hamkins Apr 23 '18 at 14:15 • No, I am assuming $n \\geq 2$. If $n=2$, Spoiler just has to play in the same row but different box than Solver to meet the above conditions. – Tony Huynh Apr 23 '18", "# Coprime Game I Let $$N \\ge 2$$ be a positive integer. Dan and Sam play the following coprime game: starting with the set $$C_N$$ of coprime positive integers of $$N$$ that are less than $$N$$, the players take turns removing some elements from the set. At a player's turn, he chooses a number $$m$$ that remains, and removes $$m$$ and any integer that is not coprime of $$m$$ that remains. The player who moves last loses. For example: $$N =7, C_N = \\{ 1,2,3,4,5,6\\}$$. Dan chooses $$m=6$$, so he removes $$2$$, $$4$$, $$3$$ and $$6$$. The remaining set is $$\\{1,5\\}$$. Sam chooses $$m=5$$. The remaining set is $$\\{1\\}$$. Dan is forced to take $$1$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Dan and Sam play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$1$$. ×", "Think of a number, and keep a running total starting at 0. Each turn, add your number on to the total. Then, if the old total was a multiple of your number, add one to your number. Otherwise, subtract 1. The game ends when your number is 1. Which starting numbers eventually get to 1? In the video, it looks like starting at 4 doesn't, but starting at 2 does. My new sequence, oeis.org/A338807, lists the numbers that eventually reach 1. I'd love to know if there's a pattern! Complete classification of tetrahedra whose angles are all rational multoples of $$\\pi$$: threadreaderapp.com/thread/133, via aperiodical.com/2020/12/aperio The original paper is \"Space vectors forming rational angles\", arxiv.org/abs/2011.14232, by Kiran S. Kedlaya, Alexander Kolpakov, Bjorn Poonen, and Michael Rubinstein I've got a new sequence in the OEIS. It's to do with a really simple number game I made up. Here's a video explaining the rules The skew dice set by The Dice Lab is out now at mathartfun.com/DiceLabDice.htm, featuring the most cursed d20 you're ever likely to see! Full video at youtu.be/ljyJFsaAE84. An animation for Craig Kaplan’s art project, with Saul Schleimer. Here we fly through a cohomology fractal that happens to contain a checkerboard. You can explore it at henryseg.github.io/cohomology_ (Open controls, go to the \"cool examples\" census, and choose \"t12047\".)", "second player will win by removing $12-n$. - You can make a minimax table, starting with 1, should only take a couple minutes. For row $R$ and column $C$, does the player win if there are $R$ pebbles left and he is allowed to draw up to $C$: $$\\begin{array} {c|c|c|c|c|c|} \\text{Pebbles \\ Draws} & 1 & 2 & 3 & 4 & 5 & 6 \\\\ \\hline 1 & W \\\\ \\hline 2 & L & W \\\\ \\hline 3 & W & W & W \\\\ \\hline 4 & L & L & L & W \\\\ \\hline \\end{array}\\\\ \\vdots \\\\ \\begin{array} {c|c|c|} & ... & 12 \\\\ \\hline 12 & ... & ? \\\\ \\hline \\end{array}\\\\$$ - If the 1st player removes n pebbles with $n\\geq6$ or $n=1$ then 2nd player wins. Assume that 1st player removes n pebbles with $n\\in \\{ 2,3,4,5\\}$. If $n=3 or 5$ then 2nd player removes one pebbles and thus wins in the end since after the 1st player plays there is an odd number of pebbles remaining. If $n=2$ and 2nd player removes 1 pebble during first round then 1st player wins. If 2nd player removes 2 pebbles during the first round then he wins in the end. If $n=4$ then the 1st player wins and its easy to see", "puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) The way I read this. Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation", "$k$, $1\\leq k\\leq 10$, for which this is possible. 7. A solitaire game is played on an $m\\times n$ rectangular board, using $mn$ markers which are white on one side and black on the other. Initially, each square of the board contains a marker with its white side up, except for one corner square, which contains a marker with its black side up. In each move, one may take away one marker with its black side up, but must then turn over all markers which are in squares having an edge in common with the square of the removed marker. Determine all pairs $(m,n)$ of positive integers such that all markers can be removed from the board. Number Theory 1. Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$. 2. Determine all pairs $(a,b)$ of real numbers such that $a \\lfloor bn \\rfloor =b \\lfloor an \\rfloor$ for all positive integers $n$. (Note that $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.) 3. Determine the smallest integer $n\\geq 4$ for which one can choose four different numbers $a,b,c$ and $d$ from any $n$ distinct integers such that $a+b-c-d$ is divisible by $20$. 4. A sequence of integers $a_{1},a_{2},a_{3},\\ldots$ is defined as follows: $a_{1} = 1$ and for $n\\geq 1$,", "# Divisor game Let $$N \\ge 2$$ be a positive integer. Alice and Bob play the following divisor game: starting with the set $$D_N$$ of positive divisors of $$N$$, the players take turns removing some elements from the set. At a player's turn, he or she chooses a divisor $$d$$ that remains, and removes $$d$$ and any of the divisors of $$d$$ that remain. The player who moves last loses. For example: $$N = 18, D_N = \\{ 1,2,3,6,9,18 \\}$$. Alice chooses $$d=2$$, so she removes $$1$$ and $$2$$. The remaining set is $$\\{3,6,9,18\\}$$. Bob chooses $$d = 9$$ and so must also remove $$3$$. The remaining set is $$\\{6,18\\}$$. Alice takes $$6$$, and now Bob is forced to take $$18$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Alice and Bob play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$999$$. × Problem Loading... Note Loading... Set Loading...", "that $\\pi(i)\\not\\equiv i, i+1\\pmod n$. Take $B=\\{(0,0),(1,1),\\ldots,(n-1,n-1), (0,1),(1,2),\\ldots,(n-2,n-1),(n-1,0)\\}$ as the chess board. A permutation $\\pi$ which defines a way of seating the husbands, is a placement of $n$ non-attacking rooks such that none of them is in $B$. a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h We need to compute $r_k$, the number of ways of placing $k$ non-attacking rooks on $B$. For our choice of $B$, $r_k$ is the number of ways of choosing $k$ points, no two consecutive, from a collection of $2n$ points arranged in a circle. We first see how to do this in a line. Lemma The number of ways of choosing $k$ non-consecutive objects from a collection of $m$ objects arranged in a line, is ${m-k+1\\choose k}$. Proof. We draw a line of $m-k$ black points, and then insert $k$ red points into the $m-k+1$ spaces between the black points (including the beginning and end). \\begin{align} &\\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\, \\bullet \\, \\sqcup \\\\ &\\qquad\\qquad\\qquad\\quad\\Downarrow\\\\ &\\sqcup \\, \\bullet \\,\\, {\\color{Red}\\bullet} \\, \\bullet \\,\\, {\\color{Red}\\bullet} \\, \\bullet", "expression $$cdcd$$ is reduced since it is an alternating product of $$c$$'s and blocks in $$\\{d, d^2 \\}$$; I need to convince you that it doesn't cancel out. If it did cancel out, then so would $$d^2cdcd^5$$ because $$d^6$$ cancels out; and for the same reason, if $$d^2cdcd^5$$ did cancel out, then so would the original expression $$cdcd$$. And as $$d^3 = e$$, I can rewrite (reduce) the new expression: $$d^2cdcd^5 = d^2cdcd^2e$$. I prefer this last reduced expression because, ignoring the $$e$$ at the end, it start and ends with blocks in $$\\{d, d^2 \\}$$. Here's a five-shot rally of ping-pong between players c and d/dd: 1. referee e toss the ball up from a blank position to a blank region; 2. dd volleys from the blank region to a shaded region; 3. c returns the volley from a shaded region to a blank region; 4. d hits from blank to shaded region; 5. c hits from shaded to blank region; and finally, 6. dd hits a winner from blank to shaded region. Fig. 5. Another ping-pong rally for the ages. The game started at a blank position and ended at a shaded position; therefore, $$d^2cdcd^2e = d^2cdcd^5$$ doesn't cancel out! This in turn means the original expression $$cdcd$$ doesn't cancel out either. The presentation of $$GL(2,\\mathbb", "2013 AMC 12 B #18 Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove 2 or 4 coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove 1 or 3 coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with 2013 coins and when the game starts with 2014 coins? 2021 AIME I #8 (Note: I have the wrong problem number written in the video whiteboard.) Find the number of integers $c$ such that the equation $||20|x|-x^2|-c|=21$ has $12$ distinct solutions. 2021 AIME I #7 Find the number of pairs $(m,n)$ of positive integers with $1 \\le m < n \\le 30$ such that there exists a real number $x$ satisfying $\\sin(mx)+\\sin(nx)=2$. 2021 AIME #5 Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. 2021 AIME I #4 Find the number", "also called the Josephus problem. How can I apply the same logic here? Currently, I figured out that $3^n$ has position $1$ win every time, so I suspect the pattern still holds, but I can't prove it. In the first passing around the circle, student $i$ $(i$ ranging from $1$ to $n)$, will be excluded if and only if $i \\equiv 0$ or $i\\equiv 2\\pmod 3$. In other words, for the $n$-th student to avoid exclusion, we must have $n\\equiv 1 \\pmod 3$. Suppose hence that $n = 3k+1$. After tagging each student exactly once, we're down to a circle with $k+1$ students, so if $k=0$, the game ends right after this first pass. Suppose hence that $k>0$, so the game will continue. In this case, the $n$-th student is now the $(k+1)$-th. Now, the teacher's pattern begins with Goose-Goose-Duck. This means that if $k+1\\leqslant 3$, the $n$-th student will survive as a result of all others being tagged Goose before him. If $k+1>3$, then we'll do a full pass around the circle. It follows that student $i$ $(i$ ranging from $1$ to $k+1)$ will be excluded if and only if $i \\equiv 1$ or $i\\equiv 2\\pmod 3$. Hence, in order for the $(k+1)$-th student to avoid exclusion, we must have $k+1 \\equiv 0 \\iff$ $k \\equiv2 \\pmod", "does Sergey need to determine for sure the number Xenia thought of? 3. A number of $17$ workers stand in a row. Every contiguous group of at least $2$ workers is a $brigade$. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker’s number of assignments is divisible by $4$. Prove that the number of such ways to assign the leaders is divisible by $17$. 4. Consider an integer $n \\ge 2$ and write the numbers $1, 2, \\ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, then writing down the numbers $a+b$ and $\\vert a-b \\vert$ on the board, and then removing repetitions (e.g., if the board contained the numbers $2, 5, 7, 8$, then one could choose the numbers $a = 5$ and $b = 7$, obtaining the board with numbers $2, 8, 12$). For all integers $n \\ge 2$, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves. 5. Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6n$, and $60^\\circ$ rotational symmetry (that is, there is a point $O$ such that a $60^\\circ$ rotation about", "In the game Chomp, two players begin with a rectangular grid. Player 1 chooses any square and removes it from the grid, together with all the squares above and to the right of it. Player 2 chooses one of the remaining squares and removes that, together with all the squares above it and to its right. The two take turns in this way until one of them is forced to remove the last square. That player loses. If the starting grid is square, can either player force a win? futilitycloset.com/2022/06/13/ Change the colour youtu.be/SoxRi269Slw Particle $$P_0$$ is at $$(0, 0)$$ and particle $$P_1$$ is at $$(0, 1)$$ at time $$t=0$$. $$P_0$$ moves right along the $$x$$-axis at speed $$1$$ and $$P_1$$ always moves directly towards $$P_0$$, also at speed $$1$$. In the long run, $$P_1$$ will be moving right along the $$x$$-axis at speed $$1$$, and so the distance between the two particles will no longer decrease. What is that final distance? @mscroggs (It's equivalent to $$K_{26}$$) @mscroggs Here are all the right-angled triangles for question 12 of the advent calendar. I recommend @mscroggs's maths puzzle advent calendar. mscroggs.co.uk Really recommend the video games exhibition at the V & A. youtu.be/VamFVh_AspM . |￣￣| | 🎃 | |__ | (\\_/) || (•ㅅ•) || / づ Issue 08 of Chalkdust", "point $A_0$ is black, $A_{n+1}$ is white, and the rest points are colored randomly either black or white. Prove: among these $n+1$ line segments $A_kA_{k+1}$, where $k=0, 1, \\cdots, n$, the number of those with different colored ending points is odd. Joe wrote $3$ positive integers on the whiteboard, then erased one number and replaced with the sum of the other two numbers minus $1$. He continued doing this and stopped when there're $17$, $1967$ and $1983$ on the whiteboard. Is it possible that the initial three numbers Joe wrote are $2$, $2$ and $2$? Let $a_1$, $a_2$, $a_3$, $\\cdots$, $a_n$ be a random permutation of $1$, $2$, $3$, .., $n$, where $n$ is an odd number. Prove $$(a_1-1)(a_2-2)\\cdots(a_n-n)$$ is an even number. A league with $12$ teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores $2$ points for every game it wins and $1$ point for every game it draws. Which of the following is NOT a true statement about the list of $12$ scores? David, Hikmet, Jack, Marta, Rand, and Todd were in a $12$-person race with $6$ other people. Rand finished $6$ places ahead of Hikmet. Marta finished $1$ place behind Jack. David finished", "The puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation $$s \\equiv 1-x'\\pmod 5$$.", "some pairs with difference $111$ until the turn is finished. Therefore, by the end of the $(2k+1)$-st turn, only $11k$ pairs of difference $111$ have been killed (and other pairs are completely untouched). If you chose to play first you would have 10 turns. -> 10 x 11 = 110 If you chose to play second you would have 9 turns. -> 9 x 11 = 99 If you are second and remove every high number starting with 211 while the first removes the low numbers you would stop at 211 - 99 = 112. The last left numbers would be 112 and one number (between 1 and 112). The first would then only win if that number is 1, because 112 - 1 = 111 (which is >= 111). If it was the opposite and you would play first you could remove 110 numbers and you would win if the difference is >= 111. There are only 211-111 = 100 numbers with a difference >= 111. Remove all of them and you win. So your chance of winning by playing first is 100%." ]

[ "# Divisor game Let $$N \\ge 2$$ be a positive integer. Alice and Bob play the following divisor game: starting with the set $$D_N$$ of positive divisors of $$N$$, the players take turns removing some elements from the set. At a player's turn, he or she chooses a divisor $$d$$ that remains, and removes $$d$$ and any of the divisors of $$d$$ that remain. The player who moves last loses. For example: $$N = 18, D_N = \\{ 1,2,3,6,9,18 \\}$$. Alice chooses $$d=2$$, so she removes $$1$$ and $$2$$. The remaining set is $$\\{3,6,9,18\\}$$. Bob chooses $$d = 9$$ and so must also remove $$3$$. The remaining set is $$\\{6,18\\}$$. Alice takes $$6$$, and now Bob is forced to take $$18$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Alice and Bob play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$999$$. × Problem Loading... Note Loading... Set Loading...", "puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) The way I read this. Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation", "# Coprime Game I Let $$N \\ge 2$$ be a positive integer. Dan and Sam play the following coprime game: starting with the set $$C_N$$ of coprime positive integers of $$N$$ that are less than $$N$$, the players take turns removing some elements from the set. At a player's turn, he chooses a number $$m$$ that remains, and removes $$m$$ and any integer that is not coprime of $$m$$ that remains. The player who moves last loses. For example: $$N =7, C_N = \\{ 1,2,3,4,5,6\\}$$. Dan chooses $$m=6$$, so he removes $$2$$, $$4$$, $$3$$ and $$6$$. The remaining set is $$\\{1,5\\}$$. Sam chooses $$m=5$$. The remaining set is $$\\{1\\}$$. Dan is forced to take $$1$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Dan and Sam play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$1$$. ×", "The puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation $$s \\equiv 1-x'\\pmod 5$$.", "2016 AMC (Sample Question) Let $$k$$ be a positive integer. Bernado and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $$k+1$$ digits. Every time Bernardo writes a number, Silvia erases the last $$k$$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $$k$$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $$f(k)$$ be the smallest positive integer not written on the board. For example, if $$k=1$$, then the numbers that Bernardo writes are $$16,25,36,49,64$$, and the numbers showing on the board after Silvia erases are $$1,2,3,4,5,6$$, and thus $$f(1) = 5$$. What is the sum of digits of $$f(2) + f(4) + f(6) + \\cdots + f(2016)$$? ×", "really understand is, the statement of the problem looks to be similar to my one, but the process cannot continue when there are less than $p$ numbers in my one (as we need $p-1$ numbers in between), yet in this problem it becomes \"until only the last person remains, who is given freedom\". I do not know how they managed to do this. For example, for number $(123456)$, $p=3$ would eliminate $3,6$, then $4,2$, and left us with $1,5$. – Kerry Mar 19 '11 at 21:27 I think in the Josephus problem you just look at the $p$th guy modulo however many are left... – Igor Rivin Mar 19 '11 at 22:02 Sorry I still do not really understand... – Kerry Mar 19 '11 at 22:20 Another good reference for the Josephus problem is Graham, Knuth, and Patashnik's Concrete Mathematics. And to help unconfuse, using your example, we can count off. So we count 1, 2, eliminate 3, 4, 5, eliminate 6, wrap around, 1, 2, eliminate 4, 5, wrap around 1, eliminate 2, 5, wrap around, 1, eliminate 5. Now only 1 is left. – Aubrey da Cunha Mar 19 '11 at 23:28 Hi! Thanks for the info, but if you eliminate $3,4,5,6$ already, then how could you eliminate $4,5$ and $2,5$ again? I feel even more", "It takes time to read Discrete Mathematics Level pending Let $$k$$ be a positive integer. Bernard and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernard starts by writing the smallest perfect square with $$k+1$$ digits. Every time Bernard writes a number, Silvia erases the last $$k$$ digits of it. Bernard then writes the next perfect square and Silvia erases the last $$k$$ digits of it and the process continues until the last 2 numbers that remain on the board differ by at least 2. Let $$f(k)$$ be the smallest positive integer not written on the board. For example, if $$k=1$$, then the numbers that Bernard writes are $$16,25,36,49,64$$ and the numbers showing on the board after Silvia erases are $$1,2,3,4,6$$ and thus $$f(1)=5$$. Compute the sum of $$f(2)+f(4)+f(6)+\\cdots+f(2016)$$. ×", "2, 3, 4, 1, 5,$ $1, 4, 3, 2, 5, 4, 1, 2, 3, 5, 1, 3, 1, 4, 5, 2, 1, 4, 1, 5, 3, 4, 1, 3, 5, 4, 3, 2, 1, 5$ List of greatest divisors $\\leq k=6$. List length is lcm$(1,\\dots,k)$. $1, 2, 3, 4, 5, 6, 1, 4, 3, 5, 1, 6, 1, 2, 5, 4, 1, 6, 1, 5, 3, 2, 1, 6, 5, 2, 3, 4, 1, 6,$ $1, 4, 3, 2, 5, 6, 1, 2, 3, 5, 1, 6, 1, 4, 5, 2, 1, 6, 1, 5, 3, 4, 1, 6, 5, 4, 3, 2, 1, 6$ When you remove the last element of each list it will read the same in both directions. Note-1: the first $k$ numbers of each list are the divisors $\\leq k$. Note-2: there is at least one $1$ between each $k$ in its respective list. (i.e., a co-prime in every segment of size $k$). How can I prove that these finite lists repeat forever? How can I prove that there is at least one $1$ in each segment of size $k$? - add comment ## 1 Answer Let $k=13$. The segment of length $17$ from $2184$ to $2200$, inclusive, has no $1$; every integer in that range has a prime divisor not exceeding $13$.", "# Which number can I erase? All positive integers greater than $$2$$ are written on a board. First we erase number $$3$$ and $$5$$. With 4 positive integers $$a,b,c,d$$ satisfying $$a+b=c+d$$, if $$ab$$ is erased, then $$cd$$ can be erased, otherwise $$cd$$ cannot be erased. For example, $$3=3 \\times 1$$, $$3+1=4=2+2$$ , then $$2 \\times 2 = 4$$ is erased. a. What are the conditions of a number that can be erased ? b. If not only $$3$$ and $$5$$, but every prime number is erased at the beginning, can all other numbers be erased as well? If not, what are the conditions of a number to be erased ? (Sorry for my last question, English is my second language) • What makes you say you can't erase $7$? Surely you could erase $7=7\\times1$, then say $7+1=6+2$, so you can erase $12$. I don't understand. I also think the question is poorly worded and is missing other relevant information. You haven't specified that $a,b\\neq c,d$. Can we choose any number to erase if it has the specified property? Or are we given a starting number, $ab$, to erase and have to find all other numbers, $cd$, that we are then allowed to erase? – Jam Oct 16 '18 at 12:28 • @Jam : thanks for your advice. I", "8\\ 9\\ 10\\ 11\\ 12\\ 13\\ 14\\ 15\\ 16\\ 17\\ 18\\ 19$$ We then remove all numbers following $2$ that are divisible by $2$: $$\\require{cancel} 2\\ 3\\ \\cancel{4}\\ 5\\ \\cancel{6}\\ 7\\ \\cancel{8}\\ 9\\ \\cancel{10}\\ 11 \\cancel{12}\\ 13\\ \\cancel{14}\\ 15\\ \\cancel{16}\\ 17\\ \\cancel{18}\\ 19$$ We then remove all numbers following $3$ that are divisible by $3$: $$2\\ 3\\ \\cancel{4}\\ 5\\ \\cancel{6}\\ 7\\ \\cancel{8}\\ \\cancel{9}\\ \\cancel{10}\\ 11 \\cancel{12}\\ 13\\ \\cancel{14}\\ \\cancel{15}\\ \\cancel{16}\\ 17\\ \\cancel{18}\\ 19$$ The algorithm continues, but none of the succeeding iterations finds any values to remove. Therefore, $2, 3, 5, 7, 11, 13, 17$, and $19$ are the prime numbers less than $20$. To see why this algorithm gives us exactly the prime numbers less than $n$, first note that because we only remove a number when we find a nontrivial factor, we only remove non-primes from the list. What may be a little less obvious is that we remove all non-primes from the list. To see this, suppose $m$ is a non-prime less than $n$, and let $a$ be its smallest nontrivial factor. Then $a$ must be prime because any nontrivial factor of $a$ would be less than $a$ and would also divide $m$. $a$ therefore will not be removed from the list. When $k = a$ in Step 2, $m$ will be removed. There is", "3, 4, 1, 5,$ $1, 4, 3, 2, 5, 4, 1, 2, 3, 5, 1, 3, 1, 4, 5, 2, 1, 4, 1, 5, 3, 4, 1, 3, 5, 4, 3, 2, 1, 5$ List of greatest divisors $\\leq k=6$. List length is lcm$(1,\\dots,k)$. $1, 2, 3, 4, 5, 6, 1, 4, 3, 5, 1, 6, 1, 2, 5, 4, 1, 6, 1, 5, 3, 2, 1, 6, 5, 2, 3, 4, 1, 6,$ $1, 4, 3, 2, 5, 6, 1, 2, 3, 5, 1, 6, 1, 4, 5, 2, 1, 6, 1, 5, 3, 4, 1, 6, 5, 4, 3, 2, 1, 6$ When you remove the last element of each list it will read the same in both directions. Note-1: the first $k$ numbers of each list are the divisors $\\leq k$. Note-2: there is at least one $1$ between each $k$ in its respective list. (i.e., a co-prime in every segment of size $k$). How can I prove that these finite lists repeat forever? How can I prove that there is at least one $1$ in each segment of size $k$? Edit As to the answer by Gerry Myerson: each k-segment has a multiple of k as the last element (i.e., no derangement.) I knew this, but didn't realize that it must be an assumption", "# The only positive divisor of both $a$ and $a + 1$ is $1$ Prove that if $a \\in \\mathbb Z$ then the only positive divisor of both $a$ and $a + 1$ is $1$. When I saw this statement I didn't understand it. The only way that I can see it being true is if a is a negative number but since a ∈ Z a could also be positive and thus it could have more than one positive divisor. If that's the case would I have to disprove this instead of proving? And what would be the best way to do that? • What common divisors do 2 and 3 have? Or 3 and 4 or 9 and 10? – Paul Sundheim Sep 22 '14 at 0:18 • I think it's asking for a single number that divides both $a$ and $a+1$. – Trurl Sep 22 '14 at 0:19 • Please choose descriptive titles. The original title, Prove or Disprove??? carried virtually no information of what the problem is about. – user147263 Sep 22 '14 at 1:11 You seem to have misunderstood the question. As far as I can tell you are thinking about divisors of $a$ and $a+1$ separately, but the question is asking you to show that the only positive number which is a", "has 2^12 divisors [2, 3, 4, 5, 7, 9, 11, 13, 16, 17, 19, 23, 25] = 3212537328000 has 2^13 divisors I will now go through how my program \"thinks\" 1. It starts with $$2$$ primes: $$2$$ and $$3$$ 2. It selects the next candidate, which is the next prime. In this step, $$5$$ 3. It checks if $$2^{2}$$ and $$3^{2}$$ are less than the candidate. We can see that $$4$$ is less than $$5$$, so we put it in the list of divisors 4. We add $$2^{4}$$ to the heap (to be clear, the heap is a different thing from the divisor list). What this will achieve, if you add this number to the heap, is that the number of divisors will remain a power of two. I construct this by making the exponent of two in our list is one less a power of two. $$1 + 2 + 4 = 2^{3} - 1$$ 5. We can see that our candidate prime, $$5$$, is larger than everything in the heap, so we put it in the divisor list 6. We put $$5^{2}$$ to the heap 7. Our next candidate is $$7$$ 8. It checks if there's a number less than $$7$$ in our heap. There isn't, so we add $$7^{2}$$ to the heap and checks the", "the mental process going through her mind. Since she always places the greater number as the subtrahend and the smaller one as the minuend when computing the subtractions it is not reasonable to think that she discovered the concept of a negative number. Instead of that, it seems that she intuitevely understood the following two things: 1. A two-digit number is just the sum of the number formed by the \"tens digit\" followed by $$0$$ with the \"unit digit\". (For example, $$47 = 40 + 7$$.) 2. The way we count actually orders the numbers, that is, she can tell that a number $$n_1$$ comes first than another number $$n_2$$ by counting from $$1$$ to the latter, $$n_2$$. For example, if she wants to know if $$3$$ comes first than $$7$$, she (mentally) counts $$1, 2, \\color{red} 3, 4, 5, 6, \\color{red} 7$$ and so if she (mentally) said $$3$$ before $$7$$, then $$3$$ comes before $$7$$; or mathematically $$3 < 7$$. Also she understands that $$0$$ comes before every \"counting number\". Furthermore she may have mastered the skill of computing subtractions with the non-negative integers less than $$10$$, namely $$\\{0,1,2,3,4,5,6,7,8,9\\}$$, when the subtrahend is greater than the minuend. (This because she has only been taught doing this way and also one-digit subtractions are teached first). So what", "has already been deleted. The second pair has product $$2\\cdot 4=8$$, which can now be deleted. The third pair has product $$3\\cdot 3=9$$, which can also be deleted. Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, and $$9$$. • Using the deleted $$6$$, you know that $$6$$ factors as $$1\\cdot 6=6$$ and $$2\\cdot 3=6$$. Therefore, the corresponding sums are $$1+6=7$$ or $$2+3=5$$. There are three sums to get $$7$$, $$1+6=7$$, $$2+5=7$$, and $$3+4=7$$. There are two sums to get $$5$$, $$1+4=5$$ and $$2+3=5$$. These, combined allow one to delete $$6$$, $$10$$, $$12$$, $$4$$, and 6\\$. Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$9$$, $$10$$, and $$12$$. Continue this until you find a pattern.", "# Divisible by 5? There is a math teacher in a school whose favorite number is a positive integer (let's call it $$x).$$ She told her 5 students--Alice, Beatrice, Candice, Denise, Eunice--the number $$x,$$ and the following conversation took place: Alice: \"$$x+1$$ is not divisible by 5.\" Beatrice: \"$$x+2$$ is not divisible by 5.\" Candice: \"$$x+3$$ is not divisible by 5.\" Denise: \"$$x+4$$ is not divisible by 5.\" Eunice: \"$$x+5$$ is not divisible by 5.\" Without knowing $$x$$, can we guarantee that at least one of these 5 students is lying? ×", "\\pmod 9$. We continue and see: Alice skips $2 \\pmod 3$, Barbara skips $5 \\pmod 9$, Candice skips $14 \\pmod {27}$, Debbie skips $41 \\pmod {81}$, Eliza skips $122 \\pmod {243}$, and Fatima skips $365 \\pmod {729}$. Since the only number congruent to $365 \\pmod {729}$ and less than $1,000$ is $365$, the correct answer is $\\boxed{365\\ \\mathbf{(C)}}$. ## Solution 2 After Alice says all her numbers, the numbers not mentioned yet are $$\\text{Alice: } 2,5,8,11,14,17,\\cdots,998.$$After Barbara says all her numbers, the numbers that haven't been said yet are $$\\text{Barbara: } 5,14,23,32,41,50,\\cdots,995.$$After Candice, the list is $$\\text{Candice: } 14,41,68,\\cdots,986.$$Notice how each list is an arithmetic sequence where the common ratio is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: $$\\text{Debbie: } 41,122,203,\\cdots,959$$$$\\text{Eliza: } 122,365,608,878$$$$\\text{Fatima: } \\boxed{\\textbf{(C) } 365}$$", "# N people sit at a round table, starting from #1, every other one leaves, who's the last one? For example, there are 10 people sitting there. So the 1st round, such people leave: $$\\#1, \\#3, \\#5, \\#7, \\#9$$ and remains $$\\#2, \\#4, \\#6, \\#8, \\#10$$ Then the 2nd round, such people leave: $$\\#2, \\#6, \\#10$$ and remains $$\\#4, \\#8$$ Then the 3rd round, such people leave: $$\\#8$$ and remains $$\\#4$$ so the last one remained is #4. If we note f(10)=4, how to get a general formula for f(N)? - For a more general problem of the same kind, please see the Josephus Problem. – André Nicolas Sep 26 '13 at 7:23 so there's no close form for the general case $k!=2$? – athos Sep 26 '13 at 7:29 None that I know. There is one for $k=2$. – André Nicolas Sep 26 '13 at 7:32 got it, thanks! :) – athos Sep 26 '13 at 8:44", "without the number line. ${-7} + 13 = 6$ Here, we find the difference between $13$ and $7$ which is $6$. The bigger number is positive so our answer is also positive. In adding unlike signs, we find the difference and we keep the sign of the larger number. Let’s go over subtracting signed numbers. Let’s draw a number line. We have $2 - {-4}$, in the number line we end up with $6$. Let’s try a different one. ${-3} - 2$, we start at $-3$ and subtract $2$ going to the left. Here, we end up with $-5$. What if we have ${-2} - 2$, start off at $-2$ and we’re taking $2$ away. In the number line, we get $-4$. Let’s see the logic first. Let’s say you have $\\2$ and we took away ${-4}$. That’s a good thing. Maybe you owed someone $\\4$ and you took it away. You don’t owe someone $\\4$ anymore. The person said, “forget about it”. That’s like having the next $\\4$ since you don’t have to pay that for. So we increased to $6$. Next, let’s say you have ${-\\3}$ because you owe people $\\3$. Then somehow you spend another $\\2$. Maybe you borrowed this amount. So now, you owe $\\5$. You actually have ${-\\5}$. Here, you’re already negative but", "## Sieve of Eratosthenes ### Divisors of a number - example I For example, consider number $$48$$. What are its divisors? $$48:1=48$$ (remainder $$0$$) $$\\rightarrow$$ hence $$1$$ and $$48$$ are divisors of $$48$$ $$48:2=24$$ (remainder $$0$$) $$\\rightarrow$$ hence $$2$$ and $$24$$ are divisors of $$48$$ $$48:3=16$$ (remainder $$0$$) $$\\rightarrow$$ hence $$3$$ and $$16$$ are divisors of $$48$$ $$48:4=12$$ (remainder $$0$$) $$\\rightarrow$$ hence $$4$$ and $$12$$ are divisors of $$48$$ $$48:5=9$$ (remainder $$3$$) $$\\rightarrow$$ hence $$5$$ is not a divisor of $$48$$ $$48:6=8$$ (remainder $$0$$) $$\\rightarrow$$ hence $$6$$ and $$8$$ are divisors of $$48$$ $$48:7=6$$ (remainder $$6$$) $$\\rightarrow$$ hence $$7$$ is not a divisor of $$48$$ Up to know we have found the following divisors: $$1,2,3,4,6,8,12,16,24,48$$. In the last division the quotient is smaller than the divisor. This guarantees that we have already found all divisors. Indeed, if we continue dividing $$48$$ by numbers greater than $$7$$, the quociente will be less than or equal $$6$$ and we already know what are the exact divisions by numbers less than or equal $$6$$: they are necessarily the divisions by $$8$$, $$12$$, $$16$$, $$24$$ and $$48$$, since these were the quotients obtained by dividing $$48$$ by all natural numbers less than or equal to $$6$$ that produced a remainder $$0$$). We may then conclude that the divisors of $$48$$ are just" ]

[ "# Divisor game Let $$N \\ge 2$$ be a positive integer. Alice and Bob play the following divisor game: starting with the set $$D_N$$ of positive divisors of $$N$$, the players take turns removing some elements from the set. At a player's turn, he or she chooses a divisor $$d$$ that remains, and removes $$d$$ and any of the divisors of $$d$$ that remain. The player who moves last loses. For example: $$N = 18, D_N = \\{ 1,2,3,6,9,18 \\}$$. Alice chooses $$d=2$$, so she removes $$1$$ and $$2$$. The remaining set is $$\\{3,6,9,18\\}$$. Bob chooses $$d = 9$$ and so must also remove $$3$$. The remaining set is $$\\{6,18\\}$$. Alice takes $$6$$, and now Bob is forced to take $$18$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Alice and Bob play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$999$$. × Problem Loading... Note Loading... Set Loading...", "# Coprime Game I Let $$N \\ge 2$$ be a positive integer. Dan and Sam play the following coprime game: starting with the set $$C_N$$ of coprime positive integers of $$N$$ that are less than $$N$$, the players take turns removing some elements from the set. At a player's turn, he chooses a number $$m$$ that remains, and removes $$m$$ and any integer that is not coprime of $$m$$ that remains. The player who moves last loses. For example: $$N =7, C_N = \\{ 1,2,3,4,5,6\\}$$. Dan chooses $$m=6$$, so he removes $$2$$, $$4$$, $$3$$ and $$6$$. The remaining set is $$\\{1,5\\}$$. Sam chooses $$m=5$$. The remaining set is $$\\{1\\}$$. Dan is forced to take $$1$$, so he loses. Let $$n \\ge 2$$ be the largest positive integer $$\\le 200$$ such that if Dan and Sam play the divisor game for $$n$$ and both of them play optimally, the second player wins. Find $$n$$. If no such $$n$$ exists, enter $$1$$. ×", "remove one piece from each remaining column. (Brian can avoid dealing with the column that Anna is eliminating because Anna has to go first.) Thus, if Anna follows this strategy, Brian can ensure that the game is reduced to $n$ by $n-1$. If this process repeats, it will eventually by $2$ by $1$ which Brian wins. What if Anna doesn't focus on one column? Regardless, Anna must still eliminate every column but one before the last column gets too small. If Anna switches to a different column, then Brian can ignore both columns that Anna has made progress on, at least until the max column size catches up to one or the other. This is helpful to Brian if anyone, not Anna. Brian can't necessarily rely on hitting $n$ by $n-1$ in this case, but if Anna is working on two columns, Brian can ensure the game reduces to $n-1$ by $n-2$, before both columns are reduced to 0, or similar for working on more columns. If Anna is working on all the columns, then it's even worse for her.", "puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) The way I read this. Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation", "# Problem #2347 2347 Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$, then the numbers that Bernardo writes are $16, 25, 36, 49, 64$, and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$, and thus $f(1) = 5$. What is the sum of the digits of $f(2) + f(4)+ f(6) + ... + f(2016)$? $\\textbf{(A)}\\ 7986\\qquad\\textbf{(B)}\\ 8002\\qquad\\textbf{(C)}\\ 8030\\qquad\\textbf{(D)}\\ 8048\\qquad\\textbf{(E)}\\ 8064$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5),", "on her $$(n^2-2)$$-th move. Suppose not. Recall that by renaming, this implies that Solver plays $$n^2-1$$ in $$r_1$$. Therefore, Spoiler spoils by playing $$n^2$$ in $$c \\setminus r_1$$, where $$c$$ is the unique column with an unfilled cell of $$r_1$$. Thus, after Solver's $$(n^2-2)$$-th move, $$n^2$$ and $$n^2-1$$ still have not been played in $$r_1$$. Spoiler now plays $$n^2-1$$ in $$c \\setminus r_1$$, where $$c$$ is a column containing an unfilled cell of $$r_1$$. Note that after this move, there is still an unfilled cell in $$c \\setminus r_1$$, since Solver has made at most $$n^2-3$$ moves outside of $$r_1$$ and Spoiler had previously never played in $$c$$. Therefore, Solver must play in $$c$$ on her $$(n^2-1)$$-th move; otherwise, Spoiler can play $$n^2$$ in $$c \\setminus r_1$$ on her next move. But now Spoiler can spoil on her $$(n^2-1)$$-th move by playing $$n^2-1$$ in $$c' \\setminus r_1$$, where $$c'$$ is the other column containing an unfilled cell of $$r_1$$. This is joint work with Tillmann Miltzow. • I think you are assuming $n\\geq 3$? – Joel David Hamkins Apr 23 '18 at 14:15 • No, I am assuming $n \\geq 2$. If $n=2$, Spoiler just has to play in the same row but different box than Solver to meet the above conditions. – Tony Huynh Apr 23 '18", "also called the Josephus problem. How can I apply the same logic here? Currently, I figured out that $3^n$ has position $1$ win every time, so I suspect the pattern still holds, but I can't prove it. In the first passing around the circle, student $i$ $(i$ ranging from $1$ to $n)$, will be excluded if and only if $i \\equiv 0$ or $i\\equiv 2\\pmod 3$. In other words, for the $n$-th student to avoid exclusion, we must have $n\\equiv 1 \\pmod 3$. Suppose hence that $n = 3k+1$. After tagging each student exactly once, we're down to a circle with $k+1$ students, so if $k=0$, the game ends right after this first pass. Suppose hence that $k>0$, so the game will continue. In this case, the $n$-th student is now the $(k+1)$-th. Now, the teacher's pattern begins with Goose-Goose-Duck. This means that if $k+1\\leqslant 3$, the $n$-th student will survive as a result of all others being tagged Goose before him. If $k+1>3$, then we'll do a full pass around the circle. It follows that student $i$ $(i$ ranging from $1$ to $k+1)$ will be excluded if and only if $i \\equiv 1$ or $i\\equiv 2\\pmod 3$. Hence, in order for the $(k+1)$-th student to avoid exclusion, we must have $k+1 \\equiv 0 \\iff$ $k \\equiv2 \\pmod", "# [Solutions] Caucasus Mathematical Olympiad 2021 ## Juniors 1. Let $a$, $b$, $c$ be real numbers such that $a^2+b=c^2$, $b^2+c=a^2$, $c^2+a=b^2$. Find all possible values of $abc$. 2. In a triangle $ABC$ let $K$ be a point on the median $BM$ such that $CK=CM$. It appears that $\\angle CBM = 2 \\angle ABM$. Prove that $BC=MK$. 3. We have $n>2$ non-zero integers such that each one of them is divisible by the sum of the other $n-1$ numbers. Prove that the sum of all the given numbers is zero. 4. A square grid $2n \\times 2n$ is constructed of matches (each match is a segment of length 1). By one move Peter can choose a vertex which (at this moment) is the endpoint of 3 or 4 matches and delete two matches whose union is a segment of length 2. Find the least possible number of matches that could remain after a number of Peter's moves. 5. Let $a, b, c$ be positive integers such that the product$$\\gcd(a,b) \\cdot \\gcd(b,c) \\cdot \\gcd(c,a)$$is a perfect square. Prove that the product$$\\operatorname{lcm}(a,b) \\cdot \\operatorname{lcm}(b,c) \\cdot \\operatorname{lcm}(c,a)$$is also a perfect square. 6. A row of $2021$ balls is given. Pasha and Vova play a game, taking turns to perform moves; Pasha begins. On each turn a boy should paint a non-painted ball", "Hide Problem GGame of Divisibility Viet and Nam are playing a game set up with a list of $n$ positive integers $a_1, a_2, \\ldots , a_ n$ and a positive integer $k$. Two players alternatively take turns removing a number from the list until the list is empty. The final score of each player is the sum of all numbers removed by that player. The winner of the game is the only player having the score divisible by $k$. It is a draw if the scores of both players are divisible by $k$ or not divisible by $k$. Your task is to identify the winner of the game assuming that both players play optimally and Viet plays first. Input The input consists of several datasets. The first line of the input contains the number of datasets, which is a positive number and is not greater than $50$. Each dataset is described by two lines: • The first line contains two positive integers $n$ ($1 \\leq n \\leq 1\\, 024$) and $k$ ($2 \\leq k \\leq 32$). • The second line contains $n$ space-separated integers $a_ i$ ($0 \\leq a_ i \\leq 2^{31}$). Output For each test case, output in one line one word, either FIRST if Viet wins, SECOND if Nam wins or DRAW if the game ends", "The puzzle would be very different if David knew what $$j$$ was. And there's no reason, according to what the puzzle has stated, that David can't watch Dorothy put the cards in order and watch her remove precisely the $$j$$th card...) Let $$j \\equiv c_1+c_2+c_3+c_4+c_5 \\pmod 5$$ and $$1 \\le j\\le 5$$. Dorothy removes $$c_j$$. Let $$s = ( c_1+c_2+c_3+c_4+c_5)- c_j\\equiv j-c_j \\pmod 5$$. David knows the value of $$s$$. David has four cards which he will call $$A_1 < A_2 < A_3 < A_4$$ So the way I see David figures: If the missing card is $$c_k=x$$ where he doesn't know what $$k$$ or what $$x$$, then $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ (if $$k-1=0$$ of $$k=5$$ just ignore the non-defined $$A_{term}$$). He tries to solve these for $$k = 1,2,3,4,5$$. Now if we relabel the numbers $$1....124$$ to the numbers $$1..... 120$$ by omitting $$A_1,...,A_4$$ the $$x$$ will get translated to $$x'$$ and $$x'=x$$ if $$x < A_1$$. And $$x'=x-1$$ if $$A_1 < x < A_2$$ and $$x'=x-2$$ if $$A_2< x < A_3$$ and so on. So $$s \\equiv k - x\\pmod 5$$ or $$x \\equiv k-s \\pmod 5$$ and $$A_{k-1} < x < A_k$$ all be become the equation $$s \\equiv 1-x'\\pmod 5$$.", "100 a_8 \\right) - \\cdots$is. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Josephus problem, penultimate survivor Josephus problem:$n$people sit in a circle. Going around the circle, you alternately say \"skip, dismissed, skip, dismissed\", until finally only one person remains. For instance, with 4 people, you skip person 1, dismiss person 2, skip person 3, dismiss person 4, skip person 1, dismiss person 3, and finally person 1 remains. We denote this as$J(4) = 1$. What is a formula for$I(n)$, the number of the penultimate survivor? That is, where should Josephus's friend sit? In our example,$I(4) = 3$since person 3 is the second-to-last person to be dismissed. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Two queens on a chessboard How many ways can you put 2 queens on a chessboard so that they don't attack each other? (Queens attack both on the rows and on the diagonals of a chessboard.) 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Infinite geometric series exercise Find$\\displaystyle \\sum_{n=1}^{\\infty} \\frac{2006^n}{2007^n}$. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Divisibility test for 13 Write an integer$a$in terms of its decimal expansion:$a = a_0 + 10 a_1 + 100 a_2 + \\cdots + a_d 10^d$. Prove that$a$is divisible by 13 if and only if$\\left( a_0 + 10 a_1 + 100 a_2 \\right) - \\left( a_3 + 10 a_4", "It takes time to read Discrete Mathematics Level pending Let $$k$$ be a positive integer. Bernard and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernard starts by writing the smallest perfect square with $$k+1$$ digits. Every time Bernard writes a number, Silvia erases the last $$k$$ digits of it. Bernard then writes the next perfect square and Silvia erases the last $$k$$ digits of it and the process continues until the last 2 numbers that remain on the board differ by at least 2. Let $$f(k)$$ be the smallest positive integer not written on the board. For example, if $$k=1$$, then the numbers that Bernard writes are $$16,25,36,49,64$$ and the numbers showing on the board after Silvia erases are $$1,2,3,4,6$$ and thus $$f(1)=5$$. Compute the sum of $$f(2)+f(4)+f(6)+\\cdots+f(2016)$$. ×", "expression $$cdcd$$ is reduced since it is an alternating product of $$c$$'s and blocks in $$\\{d, d^2 \\}$$; I need to convince you that it doesn't cancel out. If it did cancel out, then so would $$d^2cdcd^5$$ because $$d^6$$ cancels out; and for the same reason, if $$d^2cdcd^5$$ did cancel out, then so would the original expression $$cdcd$$. And as $$d^3 = e$$, I can rewrite (reduce) the new expression: $$d^2cdcd^5 = d^2cdcd^2e$$. I prefer this last reduced expression because, ignoring the $$e$$ at the end, it start and ends with blocks in $$\\{d, d^2 \\}$$. Here's a five-shot rally of ping-pong between players c and d/dd: 1. referee e toss the ball up from a blank position to a blank region; 2. dd volleys from the blank region to a shaded region; 3. c returns the volley from a shaded region to a blank region; 4. d hits from blank to shaded region; 5. c hits from shaded to blank region; and finally, 6. dd hits a winner from blank to shaded region. Fig. 5. Another ping-pong rally for the ages. The game started at a blank position and ended at a shaded position; therefore, $$d^2cdcd^2e = d^2cdcd^5$$ doesn't cancel out! This in turn means the original expression $$cdcd$$ doesn't cancel out either. The presentation of $$GL(2,\\mathbb", "$k$, • remove the first term, • subtract $1$ from the $k$ following terms. 3. If a negative number is obtained, stop. Otherwise, repeat from step 1. In this example, from $(5,3,3,3,2,2,1,1)$ we: 1. Remove $5$ and then subtract $1$ from the next five terms to get $(2,2,2,1,1,1,1)$, 2. Remove $2$ and then subtract $1$ from the next two terms to get $(1,1,1,1,1,1)$, 3. Remove $1$ and then subtract $1$ from the next term to get $(0,1,1,1,1)$; sort to get $(1,1,1,1,0)$. 4. Remove $1$ and then subtract $1$ from the next term to get $(0,1,1,0)$; sort to get $(1,1,0,0)$. 5. Remove $1$ and then subtract $1$ from the next term to get $(0,0,0)$. 6. This is clearly graphical (it corresponds to a graph on $3$ vertices and no edges) so we can stop here. We can stop when we get all zeroes, or earlier if becomes obvious that a sequence is graphical. For example, we could stop at $(1,1,1,1,1,1)$ by realizing that a graph with $6$ vertices connected by $3$ disjoint edges has this degree sequence. In the case of a sequence such as $(1,0,0)$, we don't stop, so the next step is to remove $1$ and then subtract from the next term to get $(-1,0)$. This contains a negative number, so the sequence is not graphical. •", "2013 AMC 12 B #18 Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove 2 or 4 coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove 1 or 3 coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with 2013 coins and when the game starts with 2014 coins? 2021 AIME I #8 (Note: I have the wrong problem number written in the video whiteboard.) Find the number of integers $c$ such that the equation $||20|x|-x^2|-c|=21$ has $12$ distinct solutions. 2021 AIME I #7 Find the number of pairs $(m,n)$ of positive integers with $1 \\le m < n \\le 30$ such that there exists a real number $x$ satisfying $\\sin(mx)+\\sin(nx)=2$. 2021 AIME #5 Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. 2021 AIME I #4 Find the number", "in the limit there would be no three shots in a row by a single player. But this can happen for positive $$p$$. Try to find such a value! Regarding your question about relating $$p$$ to the number of matching digits, the authors of Greedy Galois Games (mentioned in the post) conjecture that this number is approximately proportional to $$1/p^2$$. Here’s a more detailed description (of Conjecture 1.2 in their paper). To concretely relate these quantities, let’s turn the question around and ask how small a $$p$$ we need to get $$n+1$$ matching terms. If we define the polynomial $$g_n(x) = 1-x-x^2+x^3-\\cdots\\pm x^n$$ where the plus/minuses follow the Thue-Morse sequence, then it can be checked that the sign of the number $$g_n(1-p)$$ (i.e., positive or negative) determines the $$n+1$$st player, assuming $$p$$ matches the Thue-Morse sequence up to round $$n$$. So to guarantee that $$p$$ and all smaller values give the correct next player, we need $$1-p$$ to be large enough so that $$g_n(1-p)$$ doesn’t change sign when $$p$$ is decreased toward 0, i.e., we need $$1-p$$ to be larger than the largest root of $$g_n(x)$$ smaller than 1. The authors conjecture that this largest root $$\\alpha_n$$ has the form $$\\alpha_n = 1-\\beta_n$$ where $$\\beta_n \\propto 1/\\sqrt{n}$$, i.e., we need $$p \\propto 1/\\sqrt{n}$$ (or smaller) to guarantee the", "non-attacking rooks on a standard$8 \\times 8$chessboard? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Divisibility of Fermat numbers Let$F_n = 2^{2^n}+ 1$define the sequence of Fermat numbers. Prove that$F_n | 2^{F_n} -2$for all$n$. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Divisibility test for 11 Write an integer$a$in terms of its decimal expansion:$a = a_0 + 10 a_1 + 100 a_2 + \\cdots + a_d 10^d$. Prove that$a$is divisible by 11 if and only if$a_0 - a_1 + a_2 - \\cdots$is. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Josephus problem, sequence of differences Josephus problem:$n$people sit in a circle. Going around the circle, you alternately say \"skip, dismissed, skip, dismissed\", until finally only one person remains. For instance, with 4 people, you skip person 1, dismiss person 2, skip person 3, dismiss person 4, skip person 1, dismiss person 3, and finally person 1 remains. We denote this as$J(4) = 1$. Let$H(n) = J(n+1) - J(n)$. Show that for any even number$2n$,$H(2n) = 2$. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Combinatorics of assigning people to rooms There are 3 rooms in a dormitory, a single, a double, and a quad. How many ways are there to assign 7 people to the rooms? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Tower of Hanoi with special peg Tower of Hanoi:", "second player will win by removing $12-n$. - You can make a minimax table, starting with 1, should only take a couple minutes. For row $R$ and column $C$, does the player win if there are $R$ pebbles left and he is allowed to draw up to $C$: $$\\begin{array} {c|c|c|c|c|c|} \\text{Pebbles \\ Draws} & 1 & 2 & 3 & 4 & 5 & 6 \\\\ \\hline 1 & W \\\\ \\hline 2 & L & W \\\\ \\hline 3 & W & W & W \\\\ \\hline 4 & L & L & L & W \\\\ \\hline \\end{array}\\\\ \\vdots \\\\ \\begin{array} {c|c|c|} & ... & 12 \\\\ \\hline 12 & ... & ? \\\\ \\hline \\end{array}\\\\$$ - If the 1st player removes n pebbles with $n\\geq6$ or $n=1$ then 2nd player wins. Assume that 1st player removes n pebbles with $n\\in \\{ 2,3,4,5\\}$. If $n=3 or 5$ then 2nd player removes one pebbles and thus wins in the end since after the 1st player plays there is an odd number of pebbles remaining. If $n=2$ and 2nd player removes 1 pebble during first round then 1st player wins. If 2nd player removes 2 pebbles during the first round then he wins in the end. If $n=4$ then the 1st player wins and its easy to see", "does Sergey need to determine for sure the number Xenia thought of? 3. A number of $17$ workers stand in a row. Every contiguous group of at least $2$ workers is a $brigade$. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker’s number of assignments is divisible by $4$. Prove that the number of such ways to assign the leaders is divisible by $17$. 4. Consider an integer $n \\ge 2$ and write the numbers $1, 2, \\ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, then writing down the numbers $a+b$ and $\\vert a-b \\vert$ on the board, and then removing repetitions (e.g., if the board contained the numbers $2, 5, 7, 8$, then one could choose the numbers $a = 5$ and $b = 7$, obtaining the board with numbers $2, 8, 12$). For all integers $n \\ge 2$, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves. 5. Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6n$, and $60^\\circ$ rotational symmetry (that is, there is a point $O$ such that a $60^\\circ$ rotation about", "this case working. Adding up all the probabilities, we get \\begin{align*} \\dfrac{1}{4}\\left(3 \\cdot \\dfrac{1}{2} + \\dfrac{1}{4}\\right) &= \\dfrac{1}{4} \\cdot \\dfrac{7}{4}\\\\&= \\dfrac{7}{16}. \\end{align*} Thus, B is the correct answer. 23. Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $$1,$$ then Todd must say the next two numbers ($$2$$ and $$3$$), then Tucker must say the next three numbers ($$4,$$ $$5,$$ $$6$$), then Tadd must say the next four numbers ($$7,$$ $$8,$$ $$9,$$ $$10$$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $$10,000$$ is reached. What is the $$2019$$th number said by Tadd? $$5743$$ $$5885$$ $$5979$$ $$6001$$ $$6011$$ ###### Solution(s): We can find how many numbers each triplet says in one round. \\begin{align*} &\\text{Tadd: } 1, 4, 7, 10, 13 \\cdots \\\\ &\\text{Todd: } 2, 5, 8, 11, 14 \\cdots \\\\ &\\text{Tucker: } 3, 6, 9, 12, 15 \\cdots \\end{align*} Now we can find a general formula for the number of numbers Tadd says after the $$n$$ th round. \\begin{align*} \\sum_{i = 1}^n 3n - 2 &= -2n + 3 \\sum_{i = 1}^n \\\\ &= -2n + \\dfrac{3n(n + 1)}{2} \\\\&= \\dfrac{3n^2 - n}{2}" ]

[ "not important in daily life, so 1 dollar = 100 yen, 1 yen = 1 penny. Easy. When I was there it hovered between 210 and 230, so I just halved everything. Anaxagoras Posts: 29436 Joined: Wed Mar 19, 2008 5:45 am Location: Yokohama/Tokyo, Japan ### Re: Fukushima one year on Rob Lister wrote: Fri Mar 01, 2019 2:10 pm When I was there it hovered between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were outside the base. 5 bucks for a soft drink at some restaurants, for example. I take it it wasn't like that in your day? It's actually better now, that was right after the infamous \"bubble\" when Japanese people were spending money like drunken sailors, so you could get away with prices like that. A fool thinks himself to be wise, but a wise man knows himself to be a fool. William Shakespeare (probably Socrates originally) Rob Lister Posts: 23535 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake", "who wish to buy dollars can do so at a cost of 101.92 JPY per 1 USD.", "between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were outside the base. 5 bucks for a soft drink at some restaurants, for example. I take it it wasn't like that in your day? It's actually better now, that was right after the infamous \"bubble\" when Japanese people were spending money like drunken sailors, so you could get away with prices like that. Rob Lister Posts: 23535 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake Ed Re: Fukushima one year on Anaxagoras wrote: Fri Mar 01, 2019 2:24 pm Rob Lister wrote: Fri Mar 01, 2019 2:10 pm When I was there it hovered between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were", "$dollar bill at the store. Bob is a 3.6 million dollar German Bank Account holder. Bob hands some one 100$ dollar bill at the store What is the probability that Alice transaction was accepted while Bob’s transaction was rejected? A: In the first case, the chances are 100% that the hand-over of the 100 dollars will be accepted as the owner of the account Alice will be the owner of the 5.1 million dollars. In the second case the chances are 100% that the hand-over of the 100 dollars will be rejected as the owner of the account Bob will be the owner of the 3.6 million dollars. As there is no chance to save money, those probabilities will make any gambler to win. Q: One word for a store or department store? I found out that some store or department store is called a “market”. Is this the only word? A: It’s an Americanism for a store or department store. In British English, it’s simply a department store or large shop, particularly one that is well known for selling a range of items. d41b202975 What’s new in The Mirum: otoi (6,232) are a fictional race from the Online computer game, Star Trek Online, the computer game version of the Star Trek TV show series. They are humanoid,", "both a coffee drink and food. Are buying food and buying a coffee drink independent events? How do you know?... ### The Conventions of 1832 and 1833 escalated tensions between Texas and Mexico because what The Conventions of 1832 and 1833 escalated tensions between Texas and Mexico because what... ### Average movie prices in the United States​ are, in​ general, lower than in other countries. It would cost ​$78.58 to buy three tickets in Japan plus two tickets in Switzerland. Three tickets in Switzerland plus two tickets in Japan would cost ​$74.82. How much does an average movie ticket cost in each of these​ countries? Average movie prices in the United States​ are, in​ general, lower than in other countries. It would cost ​$78.58 to buy three tickets in Japan plus two tickets in Switzerland. Three tickets in Switzerland plus two tickets in Japan would cost ​$74.82. How much does an average movie ticket co...", "exchange$600 into Japanese yen. If each dollar is 94.1 yen, how many yen will they get?", "Brazil is the largest coffee-producing country, exporting about one-third of the world’s coffee.6 When a company purchases raw materials from a supplier in another country, the company needs not just money but the money that is used in that country to make the purchase. Thus, the company is concerned about the exchange rate, or the price of the foreign currency. Figure 20.2 Brazilian Reals to One US Dollar7 The currency used in Brazil is called the Brazilian real. Figure 20.2 shows how many Brazilian reals could be purchased for $1.00 from 2010 through the first quarter of 2021. In March 2021, 5.4377 Brazilian reals could be purchased for$1.00. This will often be written in the form of BRL is an abbreviation for Brazilian real, and USD is an abbreviation for the US dollar. This price is known as a currency exchange rate, or the rate at which you can exchange one currency for another currency. If you know the price of $1.00 is 5.4377 Brazilian reals, you can easily find the price of Brazilian reals in US dollars. Simply divide both sides of the equation by 5.4377, or the price of the US dollar: If you have US dollars and want to purchase Brazilian reals, it will cost you$0.1839 for each Brazilian real you want to buy. The", "but didn't see a shipping quote. I just ordered one. They listed 3200 Yen ($29.45 USD) as the price with 0 yen for taxes and 0 yen for shipping. The automated confirmation email also has 3200 Yen as the total. Sounds good, right? Alas, when I told them I was paying with my Visa card, they didn't actually ask for the credit card number. Somehow I suspect my slide rule isn't actually on the way... Perhaps I will get another email from a human asking me to pay for the slide rule I just ordered. 12-31-2019, 04:48 PM Post: #7 Dave Britten Senior Member Posts: 1,378 Joined: Dec 2013 RE: Someone is still making slide rules (12-31-2019 04:46 PM)Guy Macon Wrote: (12-31-2019 03:27 PM)Dave Britten Wrote: That's amazing, I had been looking at Concise circular slide rules from the '60s/'70s on ebay recently. I had no idea they were still making them. Does anybody know what the shipping costs are for US destinations? I went pretty far through the checkout process but didn't see a shipping quote. I just ordered one. They listed 3200 Yen ($29.45 USD) as the price with 0 yen for taxes and 0 yen for shipping. The automated confirmation email also has 3200 Yen as the total. Sounds good, right? Alas, when I told", "services, and financial assets. In other words, since all U.S. goods, services, and financial assets are sold in USD, you need USD to buy them. However, just as with anything, the more expensive the USD is, the less demand there will be for it. For example, assume you wanted a hamburger, and you had a spaceship that could take you between Japan and the U.S. instantaneously. Assume also that a hamburger costs $1 USD in the U.S. and ¥128 in Japan. If the USD to JPY exchange rate is 1 USD: 128 ¥, then a hamburger would cost the same in both countries. However, what if the exchange rate was$1 USD to ¥200? In this case, you would be smart to want to buy Japanese hamburgers, because with $1, you could buy a hamburger in Japan and still have ¥72 left over. Since we need JPY to buy a hamburger in Japan, people will sell USD in exchange for JPY, thereby demanding less USD (and more JPY). You can see in this example that the more expensive the USD is, the less demand there will be for it with respect to buying goods, services, and financial assets. Figure 1. Demand for USD, StudySmarter Originals Demand for currency slopes downward because when Currency 1 becomes cheaper relative to Currency", "can collect during his trip, and the minimum expenses needed for that number of 500-yen coins. For shopping, he can use an arbitrary number of 1-yen, 5-yen, 10-yen, 50-yen, and 100-yen coins he has, and arbitrarily many 1000-yen bills. The shop always returns the exact change, i.e., the difference between the amount he hands over and the price of the souvenir. The shop has sufficient stock of coins and the change is always composed of the smallest possible number of 1-yen, 5-yen, 10-yen, 50-yen, 100-yen, and 500-yen coins and 1000-yen bills. He may use more money than the price of the souvenir, even if he can put the exact money, to obtain desired coins as change; buying a souvenir of 1000 yen, he can hand over one 1000-yen bill and five 100-yen coins and receive a 500-yen coin. Note that using too many coins does no good; handing over ten 100-yen coins and a 1000-yen bill for a souvenir of 1000 yen, he will receive a 1000-yen bill as the change, not two 500-yen coins. ## Input The input consists of at most 50 datasets, each in the following format. n p1 ... pn n is the number of souvenir shops, which is a positive integer not greater than 100. pi is the price of the souvenir of", "from PPP. ### 6.1.1 A Burgers-and-Sushi World Consider a simple example of the problem of changes in relative prices. Suppose there are only two countries, the United States and Japan, and to keep things really simple, assume that people consume only two goods, hamburgers and sushi. Let the United States produce only hamburgers, with a dollar price of$10, and let Japan produce only sushi, with a yen price of ¥5,000. Assume the exchange rate is ¥100/$. The U.S. price level will put a weight of 60% on the dollar price of hamburgers because U.S. consumers prefer hamburgers to sushi and a weight of 40% on the dollar price of sushi (the yen price of sushi divided by the yen-dollar exchange rate). Thus, the U.S. price level will be $\\begin{eqnarray*} P(t,)=0.60 \\times \\ 10 + 0.40 \\times \\frac{\\yen 5,000}{\\yen 100/\\} = \\26 \\end{eqnarray*}$ Now, suppose the Japanese price level places a weight of 35% on the yen price of ham- burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in", "# The Money Puzzle: Maximum amount of change without a dollar Bob goes to a vending machine to buy a can of soda. He opens his wallet, and, to his surprise, has no dollar bills or any bills for that matter. The soda costs exactly $1 USD. Bob also has change in his wallet, but even still, he can't buy the soda. What is the maximum amount of change Bob can have without exact change for a dollar? I know, there's a lot of extra fluff in that question, but that's roughly how I remember seeing it in a puzzle book a while ago. The answer is fairly easy to find out, but I have another question. Is there a general method, formula, or any other way to solve these puzzles without brute-force? This, at least to me, seems to be a recurring problem in books, yet they are often phrased differently and have different currencies, amounts of change required, or other factors changed. ## 3 Answers The answer to the question is$1.19. He has 3 quarters, 4 dimes, and 4 pennies. The unusual answer to this question relies on the fact that the quarter is not an exact multiple of the dime. If every unit of currency was an exact multiple of the next highest one, then", "the station, $230 if not. Not satisfied? There is a place in Shinagawa in which I am interested which has 91.20 (the brochure is sitting next to me) square meters, 32nd floor (amazing view!!), and costs$1,700 per month. But Tokyo is expensive and Shinagawa is one of the most expensive prefectures in Tokyo, so the same thing in Osaka would probably be only $637.50 per month. And all the prices I have listed include utilities, maintenance, etc. And as for food prices, again, what Hodgman said. But the bit about grapes is flat-out wrong. I buy grapes a lot and they never cost me anything significant—maybe 1 or 2 dollars for a quarter of a pound. While there are Japanese grapes, they taste the same or actually better than standard grapes, so I wouldn’t complain. But the main point is that rare foods cost more and local foods cost less. Do you honestly think the price of watermelons applies to standard foods prices? I pay the exact same amount for McDonald’s as in France (though about 2.5 times as much as in Thailand). A bentos package of rice, fish, noodles, and some side things I don’t even know what they are, costs less than a dollar. That’s a full meal for under a dollar. Overall the prices are", "burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in Japan to the price level in the United States is $\\begin{eqnarray*} \\frac{P(t,\\yen)}{P(t,)} = \\frac{\\yen 3,600}{\\ 26} = \\yen 138.5/ \\ \\end{eqnarray*}$ Thus, even though the law of one price is satisfied in each country, the dollar appears to be 38.5% undervalued on the foreign exchange market. The problem is the difference in consumption shares. You should convince yourself that if the consumption shares were the same in both countries and if the law of one price held, then PPP would be satisfied. ### 7.1.2 A Naive Calculation You might be tempted to make the decision by simply comparing the dollar value of the yen salary offer to the dollar salary of your New York offer by converting the yen salary into dollars at the current exchange rate. If the current exchange rate is ¥100/$, the ¥15,000,000 is worth$150,000. If you used this approach, you would accept the job offer to work in Japan. By now, you should realize that", "# While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The... ## Question: While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The exchange rate at the time of his visit is 1 yen = 1.25 cents. What was the cost of the meal in US dollars? ## Proportions and Variation: In diverse applications in the study of arithmetic and algebra, problems are achieved that imply a proportional relationship that exists between two values. A typical example would be when the price of conversion between two different currencies is established. If we know the conversion value of one currency with respect to another, then we can find the price of a product in both currencies. {eq}\\eqalign{ & {\\text{In this specific case }}{\\text{,we have two proportional values }}\\,x\\,\\left( {yen} \\right){\\text{ and }} \\cr & y\\,\\left( {{\\text{dollar cost}}} \\right){\\text{ that have a variation in directly proportional form}}{\\text{. }} \\cr & {\\text{We can begin by taking into consideration the following:}} \\cr & \\,\\,\\,\\,1\\,cent = \\$\\,0.01\\,\\,\\,\\, \\Rightarrow 1.25\\,cents = \\$ \\,0.0125 \\cr & {\\text{So we have:}} \\cr & \\,\\,\\,\\,{x_1} = 1\\,\\,yen \\cr & \\,\\,\\,\\,{y_1} = \\$\\,0.0125 \\cr & \\,\\,\\,\\,{x_2} = 3428\\,\\,yen \\cr & \\,\\,\\,\\,{y_2} = \\$ \\,\\,? \\cr & {\\text{Since}}{\\text{, }}x{\\text{ and }}y{\\text{ vary directly}}{\\text{, then}}{\\text{,", "9 s 512 MB ## Description \"500-yen Saving\" is one of Japanese famous methods to save money. The method is quite simple; whenever you receive a 500-yen coin in your change of shopping, put the coin to your 500-yen saving box. Typically, you will find more than one million yen in your saving box in ten years. Some Japanese people are addicted to the 500-yen saving. They try their best to collect 500-yen coins efficiently by using 1000-yen bills and some coins effectively in their purchasing. For example, you will give 1320 yen (one 1000-yen bill, three 100-yen coins and two 10-yen coins) to pay 817 yen, to receive one 500-yen coin (and three 1-yen coins) in the change. A friend of yours is one of these 500-yen saving addicts. He is planning a sightseeing trip and wants to visit a number of souvenir shops along his way. He will visit souvenir shops one by one according to the trip plan. Every souvenir shop sells only one kind of souvenir goods, and he has the complete list of their prices. He wants to collect as many 500-yen coins as possible through buying at most one souvenir from a shop. On his departure, he will start with sufficiently many 1000-yen bills and no coins at all. The order of", "# Bob's expensive day Bob leaves home with $50. His birthday approaching, he goes to the neighbourhood bakery to order a birthday cake and pays \\$29.95 for the cake, plus \\$3.08 for the candles. Then, he meets three friends in town and they all go grab something to eat: Bob takes a \\$4.50 sandwich and a \\$0.70 soda. After that, they all go to the movie theater. Tickets cost \\$10, but they can benefit from both an \"under 25yo\" \\$2 discount for each one and a \"Free ticket for every three tickets bought\" offer, and split the bill equally between them. They finish the afternoon at a bar, where Bob has two \\$2.75 beers while chatting with his friends. How old is Bob? • Is he in the USA? – A E Nov 12 '14 at 20:16 • This seems to lead to an unprovable assumption that he is <25, but they split the bill, so could potentially be a combination of ages. Are we able to assume his friends are the same age as him? – Jon Story Nov 13 '14 at 0:15 • wait... 75 cent beers? who cares how old he is – Theodosius Von Richthofen Nov 13 '14 at 16:11 • What does Bob do with the remaining 27 cents? – Michael Nov 13", "How much of a 0.08% solution and a 0.03% solution should be used? If the exchange rate between the American dollar and the Japanese yen is such that$4.00 = 442 yen how many yen could be exchanged for $70.00? If the exchange rate between the American dollar and the Japanese yen is such that$4.00 = 442 yen how many yen could be exchanged for \\$ 70.00?", "shops to visit cannot be changed. As far as he can collect the same number of 500-yen coins, he wants to cut his expenses as much as possible. Let's say that he is visiting shops with their souvenir prices of 800 yen, 700 yen, 1600 yen, and 600 yen, in this order. He can collect at most two 500-yen coins spending 2900 yen, the least expenses to collect two 500-yen coins, in this case. After skipping the first shop, the way of spending 700-yen at the second shop is by handing over a 1000-yen bill and receiving three 100-yen coins. In the next shop, handing over one of these 100-yen coins and two 1000-yen bills for buying a 1600-yen souvenir will make him receive one 500-yen coin. In almost the same way, he can obtain another 500-yen coin at the last shop. He can also collect two 500-yen coins buying at the first shop, but his total expenditure will be at least 3000 yen because he needs to buy both the 1600-yen and 600-yen souvenirs in this case. You are asked to make a program to help his collecting 500-yen coins during the trip. Receiving souvenirs' prices listed in the order of visiting the shops, your program is to find the maximum number of 500-yen coins that he", "doll. And all of his employees want to be paid in yuan. His own rent for the factory, or even his own rent for his own house, all has to be paid in yuan. So this is what he needs to sell his doll for, 10 yuan. And at the current exchange rate, that would be $1. Now, let's go across the Pacific. Let's go to the United States. And let's say that we have another entrepreneur who is making soda, or making cola, for export. So let me draw a can of cola. And similar to this guy in China, he needs to sell his product abroad for the equivalent of$1, so that it can cover shipping costs, manufacturing costs, and the high fructose corn syrup, and all of that. And once again, he cares about dollars. Because he has to pay his own mortgage in dollars. His employees need to be paid in dollars. Maybe the shippers he used, they only accept dollars. So this is how both of these characters think about it. Now, at the current exchange rate, let's say that there's a demand for 100 dolls in the United States. This guy is exporting. And so is this guy. We'll make it very simple. They're only focused on exports. So at current exchange--" ]

[ "$dollar bill at the store. Bob is a 3.6 million dollar German Bank Account holder. Bob hands some one 100$ dollar bill at the store What is the probability that Alice transaction was accepted while Bob’s transaction was rejected? A: In the first case, the chances are 100% that the hand-over of the 100 dollars will be accepted as the owner of the account Alice will be the owner of the 5.1 million dollars. In the second case the chances are 100% that the hand-over of the 100 dollars will be rejected as the owner of the account Bob will be the owner of the 3.6 million dollars. As there is no chance to save money, those probabilities will make any gambler to win. Q: One word for a store or department store? I found out that some store or department store is called a “market”. Is this the only word? A: It’s an Americanism for a store or department store. In British English, it’s simply a department store or large shop, particularly one that is well known for selling a range of items. d41b202975 What’s new in The Mirum: otoi (6,232) are a fictional race from the Online computer game, Star Trek Online, the computer game version of the Star Trek TV show series. They are humanoid,", "# While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The... ## Question: While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The exchange rate at the time of his visit is 1 yen = 1.25 cents. What was the cost of the meal in US dollars? ## Proportions and Variation: In diverse applications in the study of arithmetic and algebra, problems are achieved that imply a proportional relationship that exists between two values. A typical example would be when the price of conversion between two different currencies is established. If we know the conversion value of one currency with respect to another, then we can find the price of a product in both currencies. {eq}\\eqalign{ & {\\text{In this specific case }}{\\text{,we have two proportional values }}\\,x\\,\\left( {yen} \\right){\\text{ and }} \\cr & y\\,\\left( {{\\text{dollar cost}}} \\right){\\text{ that have a variation in directly proportional form}}{\\text{. }} \\cr & {\\text{We can begin by taking into consideration the following:}} \\cr & \\,\\,\\,\\,1\\,cent = \\$\\,0.01\\,\\,\\,\\, \\Rightarrow 1.25\\,cents = \\$ \\,0.0125 \\cr & {\\text{So we have:}} \\cr & \\,\\,\\,\\,{x_1} = 1\\,\\,yen \\cr & \\,\\,\\,\\,{y_1} = \\$\\,0.0125 \\cr & \\,\\,\\,\\,{x_2} = 3428\\,\\,yen \\cr & \\,\\,\\,\\,{y_2} = \\$ \\,\\,? \\cr & {\\text{Since}}{\\text{, }}x{\\text{ and }}y{\\text{ vary directly}}{\\text{, then}}{\\text{,", "not important in daily life, so 1 dollar = 100 yen, 1 yen = 1 penny. Easy. When I was there it hovered between 210 and 230, so I just halved everything. Anaxagoras Posts: 29436 Joined: Wed Mar 19, 2008 5:45 am Location: Yokohama/Tokyo, Japan ### Re: Fukushima one year on Rob Lister wrote: Fri Mar 01, 2019 2:10 pm When I was there it hovered between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were outside the base. 5 bucks for a soft drink at some restaurants, for example. I take it it wasn't like that in your day? It's actually better now, that was right after the infamous \"bubble\" when Japanese people were spending money like drunken sailors, so you could get away with prices like that. A fool thinks himself to be wise, but a wise man knows himself to be a fool. William Shakespeare (probably Socrates originally) Rob Lister Posts: 23535 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake", "burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in Japan to the price level in the United States is $\\begin{eqnarray*} \\frac{P(t,\\yen)}{P(t,)} = \\frac{\\yen 3,600}{\\ 26} = \\yen 138.5/ \\ \\end{eqnarray*}$ Thus, even though the law of one price is satisfied in each country, the dollar appears to be 38.5% undervalued on the foreign exchange market. The problem is the difference in consumption shares. You should convince yourself that if the consumption shares were the same in both countries and if the law of one price held, then PPP would be satisfied. ### 7.1.2 A Naive Calculation You might be tempted to make the decision by simply comparing the dollar value of the yen salary offer to the dollar salary of your New York offer by converting the yen salary into dollars at the current exchange rate. If the current exchange rate is ¥100/$, the ¥15,000,000 is worth$150,000. If you used this approach, you would accept the job offer to work in Japan. By now, you should realize that", "for all three portions of the trip. Notice the total paid is $$\\14 + \\35 + \\77 = \\126$$. Solution 3: A third possibility is that they each pay according to their distance travelled. Andy travels 21 km, Bob travels 42 km, and Curly travels 63 km, a total of $$21 + 42 + 63 = 126$$ km. Thus it would cost $$\\126 \\div 126$$ km $$= \\1.00$$ per km per person. So Andy pays$21, Bob pays $42, and Curly pays$63.", "from PPP. ### 6.1.1 A Burgers-and-Sushi World Consider a simple example of the problem of changes in relative prices. Suppose there are only two countries, the United States and Japan, and to keep things really simple, assume that people consume only two goods, hamburgers and sushi. Let the United States produce only hamburgers, with a dollar price of$10, and let Japan produce only sushi, with a yen price of ¥5,000. Assume the exchange rate is ¥100/$. The U.S. price level will put a weight of 60% on the dollar price of hamburgers because U.S. consumers prefer hamburgers to sushi and a weight of 40% on the dollar price of sushi (the yen price of sushi divided by the yen-dollar exchange rate). Thus, the U.S. price level will be $\\begin{eqnarray*} P(t,)=0.60 \\times \\ 10 + 0.40 \\times \\frac{\\yen 5,000}{\\yen 100/\\} = \\26 \\end{eqnarray*}$ Now, suppose the Japanese price level places a weight of 35% on the yen price of ham- burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in", "# Bob's expensive day Bob leaves home with $50. His birthday approaching, he goes to the neighbourhood bakery to order a birthday cake and pays \\$29.95 for the cake, plus \\$3.08 for the candles. Then, he meets three friends in town and they all go grab something to eat: Bob takes a \\$4.50 sandwich and a \\$0.70 soda. After that, they all go to the movie theater. Tickets cost \\$10, but they can benefit from both an \"under 25yo\" \\$2 discount for each one and a \"Free ticket for every three tickets bought\" offer, and split the bill equally between them. They finish the afternoon at a bar, where Bob has two \\$2.75 beers while chatting with his friends. How old is Bob? • Is he in the USA? – A E Nov 12 '14 at 20:16 • This seems to lead to an unprovable assumption that he is <25, but they split the bill, so could potentially be a combination of ages. Are we able to assume his friends are the same age as him? – Jon Story Nov 13 '14 at 0:15 • wait... 75 cent beers? who cares how old he is – Theodosius Von Richthofen Nov 13 '14 at 16:11 • What does Bob do with the remaining 27 cents? – Michael Nov 13", "then restate the number in numerals — there's too much danger of changing one but not the other (see also § 4.1). Before: More than three hundred million (300,000,000) people live in the United States. After: More than 300 million people live in the United States. Before: Guarantor will pay Bank USD one million seven thousand dollars ($1,700,000.00). After: (In a lawsuit Bank lost the difference, i.e.,$693,000; see § 4.1.) For domestic contracts, there's usually no need to say \"in United States dollars.\" (You can put that in the Definitions & Usages section if you want.) In international contracts, use ISO 4217 currency abbreviations such as USD, as in, \"Buyer will pay USD $30 million.\" (The USD abbreviation goes where indicated, not after the numbers.) It's OK to spell out dollar amounts, but it's customary to just use the numbers. Before: Twenty million dollars After:$20,000,000 Better still: $20 million Omit zero cents unless relevant. Before: Alice will pay Bob$5,000.00. After: Alice will pay Bob $5,000. Not:$5 thousand Spell out a percentage if it's at the beginning of a sentence — or just use numbers and rewrite the sentence to avoid starting with the percentage Before: 30% of the proceeds will be donated to charity. After: Thirty percent of the proceeds will be donated to charity. Or: Of the", "doll. And all of his employees want to be paid in yuan. His own rent for the factory, or even his own rent for his own house, all has to be paid in yuan. So this is what he needs to sell his doll for, 10 yuan. And at the current exchange rate, that would be $1. Now, let's go across the Pacific. Let's go to the United States. And let's say that we have another entrepreneur who is making soda, or making cola, for export. So let me draw a can of cola. And similar to this guy in China, he needs to sell his product abroad for the equivalent of$1, so that it can cover shipping costs, manufacturing costs, and the high fructose corn syrup, and all of that. And once again, he cares about dollars. Because he has to pay his own mortgage in dollars. His employees need to be paid in dollars. Maybe the shippers he used, they only accept dollars. So this is how both of these characters think about it. Now, at the current exchange rate, let's say that there's a demand for 100 dolls in the United States. This guy is exporting. And so is this guy. We'll make it very simple. They're only focused on exports. So at current exchange--", "# Consider the table. Mother Nature Tlaloc Bob Value of sales ($) 2850 7750 22500 Dirt ($) 0 2850 0... Consider the table. Mother Nature Tlaloc Bob Value of sales ($) 2850 7750 22500 Dirt ($) 0 2850 0 Bricks ($) 0 0 7750 Wages ($) 1000 3700 10000 Interest payments ($) 1000 600 900 Rent ($) 600 200 3200 Profits ($) 250 400 650 Total expenditure ($) 2850 7750 22500 Mother Nature produces dirt, Tlaloc turns the dirt into brick, and Bob creates brick houses. The economy is characterized by the table. Note all figures are measured in dollars. What is the value added for Mother Nature?", "total amount Mr. Takaha will pay. 3 4980 7980 6980 ### Sample Output 1 15950 The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen. Note that outputs such as 15950.0 will be judged as Wrong Answer. 4 4320 4320 4320 4320 ### Sample Output 2 15120 Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.", "1 In the initial state, Takahashi can achieve the maximum profit of $150$ yen as follows: 1. Move from town $1$ to town $2$. 2. Buy one apple for $50$ yen at town $2$. 3. Move from town $2$ to town $3$. 4. Sell one apple for $200$ yen at town $3$. If, for example, Aoki changes the price of an apple at town $2$ from $50$ yen to $51$ yen, Takahashi will not be able to achieve the profit of $150$ yen. The cost of performing this operation is $1$, thus the answer is $1$. There are other ways to decrease Takahashi's expected profit, such as changing the price of an apple at town $3$ from $200$ yen to $199$ yen. ### Sample Input 2 5 8 50 30 40 10 20 ### Sample Output 2 2 ### Sample Input 3 10 100 7 10 4 5 9 3 6 8 2 1 ### Sample Output 3 2", "Brazil is the largest coffee-producing country, exporting about one-third of the world’s coffee.6 When a company purchases raw materials from a supplier in another country, the company needs not just money but the money that is used in that country to make the purchase. Thus, the company is concerned about the exchange rate, or the price of the foreign currency. Figure 20.2 Brazilian Reals to One US Dollar7 The currency used in Brazil is called the Brazilian real. Figure 20.2 shows how many Brazilian reals could be purchased for $1.00 from 2010 through the first quarter of 2021. In March 2021, 5.4377 Brazilian reals could be purchased for$1.00. This will often be written in the form of BRL is an abbreviation for Brazilian real, and USD is an abbreviation for the US dollar. This price is known as a currency exchange rate, or the rate at which you can exchange one currency for another currency. If you know the price of $1.00 is 5.4377 Brazilian reals, you can easily find the price of Brazilian reals in US dollars. Simply divide both sides of the equation by 5.4377, or the price of the US dollar: If you have US dollars and want to purchase Brazilian reals, it will cost you$0.1839 for each Brazilian real you want to buy. The", "# Fair payment Alice, Bob, and Charlie are having a picnic. Alice has two loaves of bread, Bob has three loaves of bread, but Charlie doesn't have any. Instead, Charlie has $5. They agree to split the bread equally among the three of them, and Charlie can pay Alice and Bob the appropriate amount of money for the bread. How much should Charlie pay to Alice? Clarification: Charlie pays all of his$5 to Alice and Bob; the problem is how to split it fairly. × Problem Loading... Note Loading... Set Loading...", "yen at night. • On the $5$-th day, the piggy bank gets $5$ yen in the morning and has $15$ yen at night. Thus, on the $5$-th night, AtCoDeer will find out that his piggy bank has $12$ yen or more for the first time. Sample Input 2 100128 Sample Output 2 447", "was £1 = JP¥156.88 . (a) Change £850 into Japanese yen. (b) When she returned she had JP¥180 left which she wanted to exchange back into pounds. She checked the currency exchange rates app on her phone and found that the new rate was £1 = JP¥157.22. Change JP¥180 into British pounds. (4 marks) Show answer (a) 850 \\times 156.88 (1) JP¥133348 (1) (b) 180 \\div 157.22 (1) £1.14 (1) 2. (a) On a flight back to London from Hong Kong, Jackie purchased the following items from the in-flight dining options. • A blueberry muffin costing £2.50 • A diet coke costing £3.50 • A tuna sandwich costing £5.50 She paid part of the bill with the HK\\$60 left from her holiday and paid the rest in British pounds. The exchange rate between the British pound and the Hong Kong dollar is £1= HK\\$10.68. How much does Jackie pay in British pounds? (b) When paying the bill Jackie realised that there was a 10\\% service charge on her bill. What was her total bill in Hong Kong dollars (HKD)? Give your answer to the nearest Hong Kong dollar. (6 marks) Show answer (a) 2.50+3.50+5.50=£11.50 (1) 60 \\div 10.68=£5.62 (1) 11.50-5.62 = £5.88 (1) (b) 11.50 \\times 1.10=£12.65 (1) £12.65 \\times 10.68=135.102 (1) HK\\$135 (1) 3. Paula goes on", "wrote down the lower value and a $2 deduction will be taken from the person who wrote down the higher amount. The challenge is: what strategy should both travelers follow to decide the value they should write down? ## Analysis One might expect a traveler's optimum choice to be$100; that is, the traveler values the antiques at the airline manager's maximum allowed price. Remarkably, and, to many, counter-intuitively, the Nash equilibrium solution is in fact just $2; that is, the traveler values the antiques at the airline manager's minimum allowed price. For an understanding of why$2 is the Nash equilibrium consider the following proof: • Alice, having lost her antiques, is asked their value. Alice's first thought is to quote $100, the maximum permissible value. • On reflection, though, she realizes that her fellow traveler, Bob, might also quote$100. And so Alice changes her mind, and decides to quote $99, which, if Bob quotes$100, will pay $101. • But Bob, being in an identical position to Alice, might also think of quoting$99. And so Alice changes her mind, and decides to quote $98, which, if Bob quotes$99, will pay $100. This is greater than the$99 Alice would receive if both she and Bob quoted $99. • This cycle of thought continues, until Alice finally decides to quote just$2—the", "# The Money Puzzle: Maximum amount of change without a dollar Bob goes to a vending machine to buy a can of soda. He opens his wallet, and, to his surprise, has no dollar bills or any bills for that matter. The soda costs exactly $1 USD. Bob also has change in his wallet, but even still, he can't buy the soda. What is the maximum amount of change Bob can have without exact change for a dollar? I know, there's a lot of extra fluff in that question, but that's roughly how I remember seeing it in a puzzle book a while ago. The answer is fairly easy to find out, but I have another question. Is there a general method, formula, or any other way to solve these puzzles without brute-force? This, at least to me, seems to be a recurring problem in books, yet they are often phrased differently and have different currencies, amounts of change required, or other factors changed. ## 3 Answers The answer to the question is$1.19. He has 3 quarters, 4 dimes, and 4 pennies. The unusual answer to this question relies on the fact that the quarter is not an exact multiple of the dime. If every unit of currency was an exact multiple of the next highest one, then", "Multiplying on both sides of the previous equation by the price level in Japan gives $\\begin{eqnarray*} (\\ 100,000 \\; \\mathsf{salary}) \\times \\frac{P_{t,\\yen}}{P_{t,}} =\\yen 15,000,000 \\; \\mathsf{salary} \\end{eqnarray*}$ This equation states that you would be indifferent between the two jobs if your dollar salary multiplied by the PPP exchange rate, $$[P_{t,\\yen}/P_{t,\\}]$$, equals your yen salary offer. Suppose the PPP exchange rate is ¥160/$. To achieve the same purchasing power in Japan as you would have in the United States, you need a salary of $\\begin{eqnarray*} (\\yen 160/ \\) \\times \\ 100,000 = \\yen 16,000,000 \\end{eqnarray*}$ But your offer is only ¥15,000,000. Alternatively, if you divide your yen salary offer by the PPP exchange rate of yen per dollar, you get a dollar equivalent of your yen salary. Then, when you determine your command over goods and services by mentally dividing the dollar equivalent salary by the dollar price level, the resulting units are consumption bundles in Japan. The implied dollar salary is $\\begin{eqnarray*} \\frac{\\yen 15,000,000}{\\yen 160/ \\ }= \\ 93,750 \\end{eqnarray*}$ This calculation states that the purchasing power you would have in Japan from a ¥15,000,000 salary is equivalent to the purchasing power that you would have in the United States from a$93,750 salary. As you can see, if the PPP exchange rate were ¥160/$, you should turn down", "who wish to buy dollars can do so at a cost of 101.92 JPY per 1 USD." ]

[ "not important in daily life, so 1 dollar = 100 yen, 1 yen = 1 penny. Easy. When I was there it hovered between 210 and 230, so I just halved everything. Anaxagoras Posts: 29436 Joined: Wed Mar 19, 2008 5:45 am Location: Yokohama/Tokyo, Japan ### Re: Fukushima one year on Rob Lister wrote: Fri Mar 01, 2019 2:10 pm When I was there it hovered between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were outside the base. 5 bucks for a soft drink at some restaurants, for example. I take it it wasn't like that in your day? It's actually better now, that was right after the infamous \"bubble\" when Japanese people were spending money like drunken sailors, so you could get away with prices like that. A fool thinks himself to be wise, but a wise man knows himself to be a fool. William Shakespeare (probably Socrates originally) Rob Lister Posts: 23535 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake", "# While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The... ## Question: While vacationing, john has dinner at a restaurant in Japan and his tab comes to 3,428 yen. The exchange rate at the time of his visit is 1 yen = 1.25 cents. What was the cost of the meal in US dollars? ## Proportions and Variation: In diverse applications in the study of arithmetic and algebra, problems are achieved that imply a proportional relationship that exists between two values. A typical example would be when the price of conversion between two different currencies is established. If we know the conversion value of one currency with respect to another, then we can find the price of a product in both currencies. {eq}\\eqalign{ & {\\text{In this specific case }}{\\text{,we have two proportional values }}\\,x\\,\\left( {yen} \\right){\\text{ and }} \\cr & y\\,\\left( {{\\text{dollar cost}}} \\right){\\text{ that have a variation in directly proportional form}}{\\text{. }} \\cr & {\\text{We can begin by taking into consideration the following:}} \\cr & \\,\\,\\,\\,1\\,cent = \\$\\,0.01\\,\\,\\,\\, \\Rightarrow 1.25\\,cents = \\$ \\,0.0125 \\cr & {\\text{So we have:}} \\cr & \\,\\,\\,\\,{x_1} = 1\\,\\,yen \\cr & \\,\\,\\,\\,{y_1} = \\$\\,0.0125 \\cr & \\,\\,\\,\\,{x_2} = 3428\\,\\,yen \\cr & \\,\\,\\,\\,{y_2} = \\$ \\,\\,? \\cr & {\\text{Since}}{\\text{, }}x{\\text{ and }}y{\\text{ vary directly}}{\\text{, then}}{\\text{,", "but didn't see a shipping quote. I just ordered one. They listed 3200 Yen ($29.45 USD) as the price with 0 yen for taxes and 0 yen for shipping. The automated confirmation email also has 3200 Yen as the total. Sounds good, right? Alas, when I told them I was paying with my Visa card, they didn't actually ask for the credit card number. Somehow I suspect my slide rule isn't actually on the way... Perhaps I will get another email from a human asking me to pay for the slide rule I just ordered. 12-31-2019, 04:48 PM Post: #7 Dave Britten Senior Member Posts: 1,378 Joined: Dec 2013 RE: Someone is still making slide rules (12-31-2019 04:46 PM)Guy Macon Wrote: (12-31-2019 03:27 PM)Dave Britten Wrote: That's amazing, I had been looking at Concise circular slide rules from the '60s/'70s on ebay recently. I had no idea they were still making them. Does anybody know what the shipping costs are for US destinations? I went pretty far through the checkout process but didn't see a shipping quote. I just ordered one. They listed 3200 Yen ($29.45 USD) as the price with 0 yen for taxes and 0 yen for shipping. The automated confirmation email also has 3200 Yen as the total. Sounds good, right? Alas, when I told", "burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in Japan to the price level in the United States is $\\begin{eqnarray*} \\frac{P(t,\\yen)}{P(t,)} = \\frac{\\yen 3,600}{\\ 26} = \\yen 138.5/ \\ \\end{eqnarray*}$ Thus, even though the law of one price is satisfied in each country, the dollar appears to be 38.5% undervalued on the foreign exchange market. The problem is the difference in consumption shares. You should convince yourself that if the consumption shares were the same in both countries and if the law of one price held, then PPP would be satisfied. ### 7.1.2 A Naive Calculation You might be tempted to make the decision by simply comparing the dollar value of the yen salary offer to the dollar salary of your New York offer by converting the yen salary into dollars at the current exchange rate. If the current exchange rate is ¥100/$, the ¥15,000,000 is worth$150,000. If you used this approach, you would accept the job offer to work in Japan. By now, you should realize that", "who wish to buy dollars can do so at a cost of 101.92 JPY per 1 USD.", "from PPP. ### 6.1.1 A Burgers-and-Sushi World Consider a simple example of the problem of changes in relative prices. Suppose there are only two countries, the United States and Japan, and to keep things really simple, assume that people consume only two goods, hamburgers and sushi. Let the United States produce only hamburgers, with a dollar price of$10, and let Japan produce only sushi, with a yen price of ¥5,000. Assume the exchange rate is ¥100/$. The U.S. price level will put a weight of 60% on the dollar price of hamburgers because U.S. consumers prefer hamburgers to sushi and a weight of 40% on the dollar price of sushi (the yen price of sushi divided by the yen-dollar exchange rate). Thus, the U.S. price level will be $\\begin{eqnarray*} P(t,)=0.60 \\times \\ 10 + 0.40 \\times \\frac{\\yen 5,000}{\\yen 100/\\} = \\26 \\end{eqnarray*}$ Now, suppose the Japanese price level places a weight of 35% on the yen price of ham- burgers (the dollar price of hamburgers multiplied by the yen-dollar exchange rate) because Japanese prefer sushi and a weight of 65% on the yen price of the sushi. Thus, the Japanese price level will be $\\begin{eqnarray*} P(t,\\yen)\\ = 0.35 \\times (\\yen 100/ \\) \\times \\10 + 0.65 \\times \\yen 5,000=\\yen 3,600 \\end{eqnarray*}$ The ratio of the price level in", "Brazil is the largest coffee-producing country, exporting about one-third of the world’s coffee.6 When a company purchases raw materials from a supplier in another country, the company needs not just money but the money that is used in that country to make the purchase. Thus, the company is concerned about the exchange rate, or the price of the foreign currency. Figure 20.2 Brazilian Reals to One US Dollar7 The currency used in Brazil is called the Brazilian real. Figure 20.2 shows how many Brazilian reals could be purchased for $1.00 from 2010 through the first quarter of 2021. In March 2021, 5.4377 Brazilian reals could be purchased for$1.00. This will often be written in the form of BRL is an abbreviation for Brazilian real, and USD is an abbreviation for the US dollar. This price is known as a currency exchange rate, or the rate at which you can exchange one currency for another currency. If you know the price of $1.00 is 5.4377 Brazilian reals, you can easily find the price of Brazilian reals in US dollars. Simply divide both sides of the equation by 5.4377, or the price of the US dollar: If you have US dollars and want to purchase Brazilian reals, it will cost you$0.1839 for each Brazilian real you want to buy. The", "$dollar bill at the store. Bob is a 3.6 million dollar German Bank Account holder. Bob hands some one 100$ dollar bill at the store What is the probability that Alice transaction was accepted while Bob’s transaction was rejected? A: In the first case, the chances are 100% that the hand-over of the 100 dollars will be accepted as the owner of the account Alice will be the owner of the 5.1 million dollars. In the second case the chances are 100% that the hand-over of the 100 dollars will be rejected as the owner of the account Bob will be the owner of the 3.6 million dollars. As there is no chance to save money, those probabilities will make any gambler to win. Q: One word for a store or department store? I found out that some store or department store is called a “market”. Is this the only word? A: It’s an Americanism for a store or department store. In British English, it’s simply a department store or large shop, particularly one that is well known for selling a range of items. d41b202975 What’s new in The Mirum: otoi (6,232) are a fictional race from the Online computer game, Star Trek Online, the computer game version of the Star Trek TV show series. They are humanoid,", "doll. And all of his employees want to be paid in yuan. His own rent for the factory, or even his own rent for his own house, all has to be paid in yuan. So this is what he needs to sell his doll for, 10 yuan. And at the current exchange rate, that would be $1. Now, let's go across the Pacific. Let's go to the United States. And let's say that we have another entrepreneur who is making soda, or making cola, for export. So let me draw a can of cola. And similar to this guy in China, he needs to sell his product abroad for the equivalent of$1, so that it can cover shipping costs, manufacturing costs, and the high fructose corn syrup, and all of that. And once again, he cares about dollars. Because he has to pay his own mortgage in dollars. His employees need to be paid in dollars. Maybe the shippers he used, they only accept dollars. So this is how both of these characters think about it. Now, at the current exchange rate, let's say that there's a demand for 100 dolls in the United States. This guy is exporting. And so is this guy. We'll make it very simple. They're only focused on exports. So at current exchange--", "services, and financial assets. In other words, since all U.S. goods, services, and financial assets are sold in USD, you need USD to buy them. However, just as with anything, the more expensive the USD is, the less demand there will be for it. For example, assume you wanted a hamburger, and you had a spaceship that could take you between Japan and the U.S. instantaneously. Assume also that a hamburger costs $1 USD in the U.S. and ¥128 in Japan. If the USD to JPY exchange rate is 1 USD: 128 ¥, then a hamburger would cost the same in both countries. However, what if the exchange rate was$1 USD to ¥200? In this case, you would be smart to want to buy Japanese hamburgers, because with $1, you could buy a hamburger in Japan and still have ¥72 left over. Since we need JPY to buy a hamburger in Japan, people will sell USD in exchange for JPY, thereby demanding less USD (and more JPY). You can see in this example that the more expensive the USD is, the less demand there will be for it with respect to buying goods, services, and financial assets. Figure 1. Demand for USD, StudySmarter Originals Demand for currency slopes downward because when Currency 1 becomes cheaper relative to Currency", "# Bob's expensive day Bob leaves home with $50. His birthday approaching, he goes to the neighbourhood bakery to order a birthday cake and pays \\$29.95 for the cake, plus \\$3.08 for the candles. Then, he meets three friends in town and they all go grab something to eat: Bob takes a \\$4.50 sandwich and a \\$0.70 soda. After that, they all go to the movie theater. Tickets cost \\$10, but they can benefit from both an \"under 25yo\" \\$2 discount for each one and a \"Free ticket for every three tickets bought\" offer, and split the bill equally between them. They finish the afternoon at a bar, where Bob has two \\$2.75 beers while chatting with his friends. How old is Bob? • Is he in the USA? – A E Nov 12 '14 at 20:16 • This seems to lead to an unprovable assumption that he is <25, but they split the bill, so could potentially be a combination of ages. Are we able to assume his friends are the same age as him? – Jon Story Nov 13 '14 at 0:15 • wait... 75 cent beers? who cares how old he is – Theodosius Von Richthofen Nov 13 '14 at 16:11 • What does Bob do with the remaining 27 cents? – Michael Nov 13", "can collect during his trip, and the minimum expenses needed for that number of 500-yen coins. For shopping, he can use an arbitrary number of 1-yen, 5-yen, 10-yen, 50-yen, and 100-yen coins he has, and arbitrarily many 1000-yen bills. The shop always returns the exact change, i.e., the difference between the amount he hands over and the price of the souvenir. The shop has sufficient stock of coins and the change is always composed of the smallest possible number of 1-yen, 5-yen, 10-yen, 50-yen, 100-yen, and 500-yen coins and 1000-yen bills. He may use more money than the price of the souvenir, even if he can put the exact money, to obtain desired coins as change; buying a souvenir of 1000 yen, he can hand over one 1000-yen bill and five 100-yen coins and receive a 500-yen coin. Note that using too many coins does no good; handing over ten 100-yen coins and a 1000-yen bill for a souvenir of 1000 yen, he will receive a 1000-yen bill as the change, not two 500-yen coins. ## Input The input consists of at most 50 datasets, each in the following format. n p1 ... pn n is the number of souvenir shops, which is a positive integer not greater than 100. pi is the price of the souvenir of", "Japan to the price level in the United States is $\\begin{eqnarray*} \\frac{P(t,\\yen)}{P(t,)} = \\frac{\\yen 3,600}{\\ 26} = \\yen 138.5/ \\ \\end{eqnarray*}$ Thus, even though the law of one price is satisfied in each country, the dollar appears to be 38.5% undervalued on the foreign exchange market. The problem is the difference in consumption shares. You should convince yourself that if the consumption shares were the same in both countries and if the law of one price held, then PPP would be satisfied. It is now straightforward to understand how a change in relative prices can cause a change in the deviation between the exchange rate and measured PPP even though all goods are traded and all prices satisfy the law of one price. Suppose that there is a shift in demand away from U.S. hamburgers and toward Japanese sushi. With no changes in the supplies of the two goods, the relative price of sushi must rise both in the United States and in Japan. The increase in the relative price can be accomplished by an appreciation of the yen relative to the dollar, with no change in the dollar price of hamburgers and no change in the yen price of sushi. Suppose the yen appreciates to ¥90/$. With unchanged dollar prices of hamburgers and yen prices of sushi,", "9 s 512 MB ## Description \"500-yen Saving\" is one of Japanese famous methods to save money. The method is quite simple; whenever you receive a 500-yen coin in your change of shopping, put the coin to your 500-yen saving box. Typically, you will find more than one million yen in your saving box in ten years. Some Japanese people are addicted to the 500-yen saving. They try their best to collect 500-yen coins efficiently by using 1000-yen bills and some coins effectively in their purchasing. For example, you will give 1320 yen (one 1000-yen bill, three 100-yen coins and two 10-yen coins) to pay 817 yen, to receive one 500-yen coin (and three 1-yen coins) in the change. A friend of yours is one of these 500-yen saving addicts. He is planning a sightseeing trip and wants to visit a number of souvenir shops along his way. He will visit souvenir shops one by one according to the trip plan. Every souvenir shop sells only one kind of souvenir goods, and he has the complete list of their prices. He wants to collect as many 500-yen coins as possible through buying at most one souvenir from a shop. On his departure, he will start with sufficiently many 1000-yen bills and no coins at all. The order of", "55,500 Yen in pounds is: $55,500$ Yen $\\div 142$ Yen $= \\pounds390.85$ If he spends $78\\%$ of the Australian dollars, he has $22\\%$ remaining. $0.22 \\times \\text{ A}\\5,025 = \\text{ A}\\1105.50$ A\\$1105.50 in pounds is: $\\text{ A}\\1105.50 \\div \\text{ A}\\1.97 = \\pounds561.16$ The difference between the values of the remaining currency is: $\\pounds561.16 - \\pounds390.85 = \\pounds170.31$ or $\\pounds170$ to the nearest pound.", "between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were outside the base. 5 bucks for a soft drink at some restaurants, for example. I take it it wasn't like that in your day? It's actually better now, that was right after the infamous \"bubble\" when Japanese people were spending money like drunken sailors, so you could get away with prices like that. Rob Lister Posts: 23535 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake Ed Re: Fukushima one year on Anaxagoras wrote: Fri Mar 01, 2019 2:24 pm Rob Lister wrote: Fri Mar 01, 2019 2:10 pm When I was there it hovered between 210 and 230, so I just halved everything. When the yen was at its strongest a few years ago it was up to about 80/dollar. Now it's back over 100, but it's been on both sides of 100 a few times. When I first got to Japan in 92 I was shocked by how expensive some things were", "5 # Consider this statement: It takes more than 110 yen to buy 1 U.S. dollar and more than 1.2 dollars to buy 1 British pound. These values show that the United States ... ## Question ###### Consider this statement: It takes more than 110 yen to buy 1 U.S. dollar and more than 1.2 dollars to buy 1 British pound. These values show that the United States must be a much wealthier country than Japan and that the United Kingdom must be wealthier than the United States. Do you agree with this reasoning? Briefly explain. Consider this statement: It takes more than 110 yen to buy 1 U.S. dollar and more than 1.2 dollars to buy 1 British pound. These values show that the United States must be a much wealthier country than Japan and that the United Kingdom must be wealthier than the United States. Do you agree with this reasoning? Briefly explain. #### Similar Solved Questions ##### Solve the following systems of linear equations by the augumented mntrix to reduced row echelou form:\"2I1 3I]2IAI0z312 AI? 212 Solve the following systems of linear equations by the augumented mntrix to reduced row echelou form: \"2I1 3I] 2IA I0z 312 AI? 212... ##### Pant BA gaseous mixture of 0z and Nz contains 37.8 % nitrogen by mass_What is", "did she pay for the box of cereal? 182. Coupon Diana bought a can of coffee that cost $7.99.7.99.$ She had a coupon for $0.750.75$ off, and the store doubled the coupon. How much did she pay for the can of coffee? 183. Diet Leo took part in a diet program. He weighed $190190$ pounds at the start of the program. During the first week, he lost $4.34.3$ pounds. During the second week, he had lost $2.82.8$ pounds. The third week, he gained $0.70.7$ pounds. The fourth week, he lost $1.91.9$ pounds. What did Leo weigh at the end of the fourth week? 184. Snowpack On April $1,1,$ the snowpack at the ski resort was $44$ meters deep, but the next few days were very warm. By April $5,5,$ the snow depth was $1.61.6$ meters less. On April $8,8,$ it snowed and added $2.12.1$ meters of snow. What was the total depth of the snow? 185. Coffee Noriko bought $44$ coffees for herself and her co-workers. Each coffee was $3.75.3.75.$ How much did she pay for all the coffees? 186. Subway Fare Arianna spends $4.504.50$ per day on subway fare. Last week she rode the subway $66$ days. How much did she spend for the subway fares? 187. Income Mayra earns $9.259.25$ per hour. Last week she worked", "# The Money Puzzle: Maximum amount of change without a dollar Bob goes to a vending machine to buy a can of soda. He opens his wallet, and, to his surprise, has no dollar bills or any bills for that matter. The soda costs exactly $1 USD. Bob also has change in his wallet, but even still, he can't buy the soda. What is the maximum amount of change Bob can have without exact change for a dollar? I know, there's a lot of extra fluff in that question, but that's roughly how I remember seeing it in a puzzle book a while ago. The answer is fairly easy to find out, but I have another question. Is there a general method, formula, or any other way to solve these puzzles without brute-force? This, at least to me, seems to be a recurring problem in books, yet they are often phrased differently and have different currencies, amounts of change required, or other factors changed. ## 3 Answers The answer to the question is$1.19. He has 3 quarters, 4 dimes, and 4 pennies. The unusual answer to this question relies on the fact that the quarter is not an exact multiple of the dime. If every unit of currency was an exact multiple of the next highest one, then", "buy domestic goods or foreign goods, then real exchange rate will be more relevant, because real exchange rate takes the inflation differential among the countries into account and is also used as an indicator of a country’s competitiveness in the foreign trade. Question 4. Suppose it takes 1.25 yen to buy a rupee, and the price level in Japan is 3 and the price level in India is 1.2. Calculate the real exchange rate between India and Japan (the price of Japanese goods in terms of Indian goods). (Hint: First find out the nominal exchange rate as a price of yen in rupees). Price level in foreign country: (Japan) Pf= 3 Price level in home country: (India) P = 1.2 Now, real exchange rate = e$$\\frac{\\mathrm{P}_{\\mathrm{f}}}{\\mathrm{P}}$$ Price of 1.25 yen = 1 rupee Price of 1 yen = $$\\frac{1}{1.25}=\\frac{100}{125}=\\frac{4}{5}$$ rupee Therefore, e = $$\\frac{1}{1.25}=\\frac{100}{125}=\\frac{4}{5}$$ So, real exchange rate = e$$\\frac{\\mathrm{P}_{\\mathrm{f}}}{\\mathrm{P}}$$ = $$\\frac{4}{5} \\times \\frac{30}{1.2}$$ = 2 Therefore, the real exchange rate is 2. Question 5. Explain the automatic mechanism by which BoP equilibrium was achieved under the gold standard. Under the gold standard, adjustment to BOP surpluses or deficits cannot be brought about through changes in the exchange rates. The adjustment must either come about automatically through the working of the economic system or be brought about by" ]

[ "6 Here, 6 total units of fence are covered with paint, from $x=4$ all the way through $x=10$. Problem credits: Brian Dean Contest has ended. No further submissions allowed.", "take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Intern Joined: 15 Oct 2016 Posts: 2 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:06 1 Ray can paint 40 feet every 160 minutes.. Then Ray can paint 40/160 feets per minute. Taylor can paint 50 feet every 125 minutes.. Then Taylor can paint 50/125 fret per minutes Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute.. Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4033 Location: India GPA: 3.5 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:12 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a", "it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 [Reveal] Spoiler: OA Intern Joined: 14 Oct 2016 Posts: 2 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:06 1 KUDOS Ray can paint 40 feet every 160 minutes.. Then Ray can paint 40/160 feets per minute. Taylor can paint 50 feet every 125 minutes.. Then Taylor can paint 50/125 fret per minutes Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute.. Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3253 Location: India GPA: 3.5 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:12 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take", "cost B) Therefore, despite the fact that we can't estimate how much we would have to pay for paint B, we can see statement 2 is sufficient. Therefore the correct answer is D. Re: A home owner must pick between paint A, which costs$6.00 [#permalink] 08 Jan 2012, 06:00 Similar topics Replies Last post Similar Topics: 4 Peter went to the store to buy paint. Small cans cost $30 3 19 Dec 2013, 10:03 3 What is the distance between Harry s home and his office? 4 10 Nov 2011, 15:50 A homeowner must pick between paint A, which costs$6 per 6 19 Apr 2010, 12:59 10 Twelve jurors must be picked from a pool of n potential 7 20 Sep 2009, 01:32", "+0 # Direct 0 330 6 the time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence? Guest May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N} \\\\\\\\ T=k* \\frac{L}{N}\\qquad but\\\\\\\\ 5=k* \\frac{200}{3}\\qquad so\\\\\\\\ k=\\frac{3}{40}\\qquad so\\\\\\\\ T=\\frac{3L}{40N}\\\\\\\\ When N=5 and L=500\\\\\\\\ T=\\frac{3*500}{40*5}\\\\\\\\ T=7.5 \\;\\;hours$$ Melody May 26, 2015 Sort: #1 +26412 +10 The rate at which fence painting is done is 200/(3*5) feet per painter-hour. In order to paint 500 feet we need 500*3*5/200 painter-hours With 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours . Alan May 26, 2015 #2 +81090 +10 If 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour. Then, 5 painters can paint 200/3 ft every hr........so........500/ (200/3) = 1500/200 = 7.5 hrs. CPhill May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N}", "fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:31 1 RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes CEO Joined: 12 Sep 2015 Posts: 2887 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 17", "olive oil in three different sizes. Which size is the best buy, and what is its unit price? Size Price 17 oz$ 8.69 25 oz $11.79 101 oz$46.99 • Six painters Six painters paint an area of 120 m2 in 4 hours. How long does it take for 12 painters to paint 480 m2 with the same performance? • Cleaning 2 Maggie is looking to hire a carpet cleaning company. Cleaning Crew Inc charges $24 for supplies and then$18 per room. If Maggie has \\$96 to spend towards cleaning the carpets, how many rooms can she have cleaned? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • In the In the national park, the ratio of the wooded area to grassland is 4: 1. The total area is 385km2. What area is wooded?", "Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3515 Location: United States (CA) Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 05 Dec 2017, 18:39 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 We are given that Ray paints at a uniform rate of 40 feet every 160 minutes. Thus, the rate of Ray is 40/160= 1/4 ft/min. We are also given that Taylor paints at a uniform rate of 50 feet every 125 minutes. Thus, the rate of Taylor is 50/125= 2/5 ft/min. We need to determine the time it will take to paint a fence, that is 260 feet long, when Ray and Taylor work simultaneously. To determine the time to paint a 260-foot-long fence, we can use the combined work formula: Work done by Ray + Work done by Taylor = 260 feet (the total work completed) Because", "B RELATED VIDEO _________________ Brent Hanneson – Founder of gmatprepnow.com Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2069 Location: United States (CA) Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 05 Dec 2017, 17:39 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 We are given that Ray paints at a uniform rate of 40 feet every 160 minutes. Thus, the rate of Ray is 40/160= 1/4 ft/min. We are also given that Taylor paints at a uniform rate of 50 feet every 125 minutes. Thus, the rate of Taylor is 50/125= 2/5 ft/min. We need to determine the time it will take to paint a fence, that is 260 feet long, when Ray and Taylor work simultaneously. To determine the time to paint a 260-foot-long fence, we can use the combined work formula: Work done by", "Ray + Work done by Taylor = 260 feet (the total work completed) Because Ray and Taylor are working simultaneously, we can let the time they both work together be t minutes. We now can express the individual work done by Ray and Taylor. We must remember that work = rate x time. Work done by Ray = (1/4)t Work done by Taylor = (2/5)t (1/4)t + (2/5)t = 260 We can multiply the entire equation by 20 to cancel out the fraction and we have: 5t + 8t = 5,200 13t = 5,200 t = 400 minutes _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] 05 Dec 2017, 17:39 Display posts from previous: Sort by", "includes interior and exterior house paints in a wide range of colors formulated to approximate authentic colonial colors. Because of the historical association, the line has been well received in Virginia. Most of these paints are sold through paint and hardware stores as the stores’ second or third line of paint. The large national firms such as Benjamin Moore or Sherwin Williams provide extensive services to paint retailers such as computerized color match- ing equipment. Partly because they lack the resources to provide such amenities and partly because they have always considered the commercial paint a sideline, Meyer has never tried to market the commercial line aggressively. Meyer sells 38,000 gallons of commercial paint per year. Charlie is worried about the future of the company. The firm’s strategic goal is to provide a quality product at the lowest possible cost and in a timely fashion. After absorbing the shock of losing the Virginia contract, Charlie wondered whether the firm should consider increasing pro- duction of commercial paints to lessen the company’s dependence on traffic paint contracts. Carl Bunch, who manages the day-to-day operation of the firm, believes the company can double its sales of commercial paint if it undertakes a promotional campaign estimated to cost $60,000. The average price of traffic paint sold last year was$10 per gallon.", "un [#permalink] ### Show Tags 17 Oct 2016, 07:57 2 KUDOS Expert's post Top Contributor alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Let's determine how much each person can paint in ONE MINUTE Ray paints at a uniform rate of 40 feet every 160 minutes We have (40 feet)/(160 minutes) Divide top and bottom by 160 to get: (0.25 feet)/(1 minute) So, Ray paints 0.25 feet every ONE MINUTE Taylor paints at a uniform rate of 50 feet every 125 minutes We have (50 feet)/(125 minutes) Divide top and bottom by 125 to get: (0.4 feet)/(1 minute) So, Taylor paints 0.4 feet every ONE MINUTE So, working TOGETHER, the can paint 0.25 + 0.4 feet every ONE MINUTE In other words, they can paint 0.65 feet every ONE MINUTE How many minutes will it take for them to paint a fence that is 260 feet long? Time = output/rate = 260/0.65 = 400 minutes [Reveal] Spoiler:", "is. I can get to 31 blue boxes; can you go any lower? – TMM Jul 24 '13 at 23:15 • Yes my solution is 31 , but my problem is how to justify that is the solution. – giancarlo Jul 24 '13 at 23:18 • Hmm. The best lower bound I got was 29. – Michael Biro Jul 25 '13 at 1:00 I got a bit carried away, but the minimum seems to be $31$ indeed. Unfortunately my method does not seem to generalize very well, but maybe that's too ambitious. We have a total of $49$ boxes, each of which must have $2$ edges painted blue, adding up to a total of $98$ edges to be painted blue. As the corner boxes have only $2$ edges available for painting, these two edges must be painted blue. This yields $8$ boxes painted blue, each having $3$ edges painted bue. This makes for a total of $24$ edges painted blue, leaving $74$ edges yet to be painted blue. The image on the left shows these $8$ boxes: $\\hspace{60pt}$$\\hspace{100pt} Note that there are 24 boxes yet to be painted that have 'excess' edges, i.e. edges that do not need to be painted blue. They are shown in green in the second picture. These are the 4 corners, having 2", "for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:31 1 KUDOS RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes SVP Joined: 11 Sep 2015 Posts: 1999 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a", "is$20,066,340, of which $5,216,340 will be covered by available grants. What will it cost to buy ceiling molding to go around a rectangular room with length 16 ft and width 10 ft? The value of these bills is$175. Brandon can paint a fence in 12 hours and elaine can paint the same fence in 11 hours. The total amount of money he has is $3.00. How much of each grade of tea must be used? It will take 27 equal monthly payments for Aimee to pay off her credit card balance of$2053.62. Ahmad served 5 more orders than Jane. Use x for your variable. You have 10 fewer quarters than dimes and 5 fewer nickels than quarters. The length of the longer piece was fifteen feet. The area of a state is 184,044 square miles greater than that of a second state. A) At what height do we need to climb to se... Sue can shovel snow from her driveway in 35 minutes. Get a free answer to a quick problem. We first write this as an equation and then solve. His son Roy mows the yard in 75 minutes. A kindergarten teacher has 15 students in her class, and she wants to order cupcakes. How many adult tickets were sold that day? An investor bought 500 shares", "(Analysis by Nick Wu) The most naive approach to this problem is to simulate applying each coat of paint and then looking at each $1 \\times 1$ square to see if it has the right number of coats of paint applied. This approach should get around half credit. Let us imagine solving the one-dimensional variant of this problem, where we have a fence and can apply coats of paint to some subintervals $[l_i, r_i]$ of the fence. Imagine that we actually had $N$ paint cans and put each can at the location $l_i$ that corresponded to where it started, and we also noted at each location $r_i$ where to stop carrying cans of paint with us. If we then walk along the fence and pick up and drop off cans of paint, at each location in the fence we can count how many coats of paint we would use by counting how many cans of paint we are carrying. In terms of how to implement this, we can enumerate, for each location, how many cans of paint we're picking up and how many cans of paint we're dropping off. Note, therefore, that the number of coats of paint that are used at location $x$ is equal to the sum of the changes in number of coats of paint", "fact that we can't estimate how much we would have to pay for paint B, we can see statement 2 is sufficient. Therefore the correct answer is D. Re: A home owner must pick between paint A, which costs $6.00 [#permalink] 08 Jan 2012, 06:00 Similar topics Replies Last post Similar Topics: 13 The total cost to pick apples at a certain orchard consists of a fixed 4 09 Apr 2016, 03:45 What is the cost of one gallon of paint? 2 08 Feb 2016, 07:20 6 What is the distance between Harry s home and his office? 5 10 Nov 2011, 15:50 A homeowner must pick between paint A, which costs$6 per 6 19 Apr 2010, 12:59", "Nov 2019, 21:43 [quote=\"nick1816\"]$$\\frac{1}{m}+\\frac{1}{m+5}=\\frac{1}{m-4}$$ $$m^2-8m-20=0$$ m=10 Could you provide more detail on how you resolve this equation? Thanks Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9850 Location: Pune, India Re: P, V and G had to paint three identical fences. On the [#permalink] ### Show Tags 27 Nov 2019, 22:23 CaptainLevi wrote: P, V and G had to paint three identical fences. On the first day, only P turned up for work and he completed the work only on the first fence, taking m hours. On the second day, all three of them turned up for work and they completed the work only on the second fence, taking (m –4) hours. On the third day, V and G turned up and they completed the work on the third fence, taking (m+ 5) hours. What is the value of m? a) 6 b) 8 c) 9 d) 10 e) 12 Since P takes m hrs to complete one fence, his rate of work = 1/m fence/hr = p P, V and G take (m - 4) hrs to complete one fence so their rate of work = 1/(m - 4) fence/hr = p + v + g V and G take (m + 5) hrs to complete one fence so their rate of work = 1/(m +", "Oct 2016, 08:57 2 Top Contributor alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Let's determine how much each person can paint in ONE MINUTE Ray paints at a uniform rate of 40 feet every 160 minutes We have (40 feet)/(160 minutes) Divide top and bottom by 160 to get: (0.25 feet)/(1 minute) So, Ray paints 0.25 feet every ONE MINUTE Taylor paints at a uniform rate of 50 feet every 125 minutes We have (50 feet)/(125 minutes) Divide top and bottom by 125 to get: (0.4 feet)/(1 minute) So, Taylor paints 0.4 feet every ONE MINUTE So, working TOGETHER, the can paint 0.25 + 0.4 feet every ONE MINUTE In other words, they can paint 0.65 feet every ONE MINUTE How many minutes will it take for them to paint a fence that is 260 feet long? Time = output/rate = 260/0.65 = 400 minutes RELATED VIDEO _________________ Brent Hanneson – GMATPrepNow.com Target Test Prep Representative", "# Surface area of house This is the image: So the question is: (a) Ralph is painting the barn, including the sides and roof. He wants to know how much paint to purchase. What is the total surface area that he is going to be painting? Round to the nearest hundredth. (b) If one paint can covers 57 square feet, how many paint cans should he purchase? (c) If each paint can costs $\\$23.50$, how much will the paint cost? I've done a little the first part, but since the surface area is off, all the other parts are off as well. Please show me how to get the surface area My work so far (A) Total Surface Area Rectangular Prism 2(wl + hl + hw) 2 (45 x 20 + 15 x 20 + 15 x 45) = 3750 Triangular Prism L + 2B h(a + b + c) + Ab 45(15 + 20 + 15) + 4 x 20 2250 + 80 2330 Total surface area - 3750 + 2330 6080 (B) Paint Cans 6080 / 57 106.6 or 107 paint cans I had to divide the total surface area with how much area one paint can would cover. (C) Price If each paint can costs$\\$23.50$, and Ralph buys 107 paint cans, he will spend $\\$2,514.50\\$" ]

[ "each person gets = 0.029. Explanation: Number of bags of grapes Brian has = $$\\frac{1}{5}$$ Number of people she wants to divide them = 7. Number of bags of grapes each person gets = Number of bags of grapes Brian has ÷ Number of people she wants to divide them = $$\\frac{1}{5}$$ ÷ 7 = 0.2 ÷ 7 = 0.029. Question 9. Sherri, Andy, and Grace are painting $$\\frac{1}{6}$$ of a fence. They want to paint an equal share of the fence. How much of the fence should each person paint? Length of of the fence each person paints = 0.056. Explanation: Length of painting of a fence Sherri, Andy, and Grace are painting = $$\\frac{1}{6}$$ Number of people = 3. Length of of the fence each person paints = Length of painting of a fence Sherri, Andy, and Grace are painting ÷ Number of people = $$\\frac{1}{6}$$ ÷ 3 = 0.17 ÷ 3 = 0.056. Question 10. Mona and Eli are washing $$\\frac{1}{2}$$ of the dishes after dinner. They want to wash an equal share of the dishes. How much should each person wash? Length of dishes for wash each person gets = 0.25. Explanation: Length of dishes are washing Mona and Eli after dinner = $$\\frac{1}{2}$$ Number of people = 2. Length of dishes for wash", "6 Here, 6 total units of fence are covered with paint, from $x=4$ all the way through $x=10$. Problem credits: Brian Dean Contest has ended. No further submissions allowed.", "fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:31 1 RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes CEO Joined: 12 Sep 2015 Posts: 2887 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 17", "cost B) Therefore, despite the fact that we can't estimate how much we would have to pay for paint B, we can see statement 2 is sufficient. Therefore the correct answer is D. Re: A home owner must pick between paint A, which costs$6.00 [#permalink] 08 Jan 2012, 06:00 Similar topics Replies Last post Similar Topics: 4 Peter went to the store to buy paint. Small cans cost $30 3 19 Dec 2013, 10:03 3 What is the distance between Harry s home and his office? 4 10 Nov 2011, 15:50 A homeowner must pick between paint A, which costs$6 per 6 19 Apr 2010, 12:59 10 Twelve jurors must be picked from a pool of n potential 7 20 Sep 2009, 01:32", "Ray + Work done by Taylor = 260 feet (the total work completed) Because Ray and Taylor are working simultaneously, we can let the time they both work together be t minutes. We now can express the individual work done by Ray and Taylor. We must remember that work = rate x time. Work done by Ray = (1/4)t Work done by Taylor = (2/5)t (1/4)t + (2/5)t = 260 We can multiply the entire equation by 20 to cancel out the fraction and we have: 5t + 8t = 5,200 13t = 5,200 t = 400 minutes _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] 05 Dec 2017, 17:39 Display posts from previous: Sort by", "for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:31 1 KUDOS RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes SVP Joined: 11 Sep 2015 Posts: 1999 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a", "# It takes Noah 3 hours to paint one side of a fence. It takes Gilberto 5 hours. How long would it take them if they worked together? Mar 12, 2017 I tried this but check it. #### Explanation: Let us say that the rates at which each one paints the surface (of area $1$ for example) are: $\\frac{1}{3}$ and $\\frac{1}{5}$ to paint the same surface so together they will need a time $t$: $\\frac{1}{3} t + \\frac{1}{5} t = 1$ rearranging: $5 t + 3 t = 15$ $8 t = 15$ $t = \\frac{15}{8} = 1.875$ less than 2 hours.", "+0 # Direct 0 330 6 the time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence? Guest May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N} \\\\\\\\ T=k* \\frac{L}{N}\\qquad but\\\\\\\\ 5=k* \\frac{200}{3}\\qquad so\\\\\\\\ k=\\frac{3}{40}\\qquad so\\\\\\\\ T=\\frac{3L}{40N}\\\\\\\\ When N=5 and L=500\\\\\\\\ T=\\frac{3*500}{40*5}\\\\\\\\ T=7.5 \\;\\;hours$$ Melody May 26, 2015 Sort: #1 +26412 +10 The rate at which fence painting is done is 200/(3*5) feet per painter-hour. In order to paint 500 feet we need 500*3*5/200 painter-hours With 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours . Alan May 26, 2015 #2 +81090 +10 If 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour. Then, 5 painters can paint 200/3 ft every hr........so........500/ (200/3) = 1500/200 = 7.5 hrs. CPhill May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N}", "of the fence. Show how the fence might look after she finishes painting. Explanation: Katya’s painting will look like the above picture when she completes painting one fourth of the whole. A. How many equal parts are in the whole fence? There are 4 equal parts in the whole. B. How many equal parts are being counted? 1 equal part is being counted by Katya for painting. Connect to Vocabulary A fraction is a number that names part of a whole or part of a group. The number of equal parts being counted is compared to the number of equal parts in the whole or in the group. Example: $$\\frac{1}{6}$$ C. What fraction names the painted part of the whole fence? Explanation: The fraction $$\\frac{1}{4}$$ is the painted part of the whole fence. Turn and Talk What if there are 3 equal sections in a fence and Katya paints 1 section? What fraction could you write to represent the painted section? $$\\frac{1}{3}$$ Explanation: If there are 3 equal sections in a fence and Katya paints 1 section then the fraction will be number of parts counted divided by number of equal parts in thw whole The fraction that represents the part Katya paints is $$\\frac{1}{3}$$. Question 2. Lev makes 1 blue T-shirt and 3 yellow T-shirts for the", "# Surface area of house This is the image: So the question is: (a) Ralph is painting the barn, including the sides and roof. He wants to know how much paint to purchase. What is the total surface area that he is going to be painting? Round to the nearest hundredth. (b) If one paint can covers 57 square feet, how many paint cans should he purchase? (c) If each paint can costs $\\$23.50$, how much will the paint cost? I've done a little the first part, but since the surface area is off, all the other parts are off as well. Please show me how to get the surface area My work so far (A) Total Surface Area Rectangular Prism 2(wl + hl + hw) 2 (45 x 20 + 15 x 20 + 15 x 45) = 3750 Triangular Prism L + 2B h(a + b + c) + Ab 45(15 + 20 + 15) + 4 x 20 2250 + 80 2330 Total surface area - 3750 + 2330 6080 (B) Paint Cans 6080 / 57 106.6 or 107 paint cans I had to divide the total surface area with how much area one paint can would cover. (C) Price If each paint can costs$\\$23.50$, and Ralph buys 107 paint cans, he will spend $\\$2,514.50\\$", "olive oil in three different sizes. Which size is the best buy, and what is its unit price? Size Price 17 oz$ 8.69 25 oz $11.79 101 oz$46.99 • Six painters Six painters paint an area of 120 m2 in 4 hours. How long does it take for 12 painters to paint 480 m2 with the same performance? • Cleaning 2 Maggie is looking to hire a carpet cleaning company. Cleaning Crew Inc charges $24 for supplies and then$18 per room. If Maggie has \\$96 to spend towards cleaning the carpets, how many rooms can she have cleaned? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • In the In the national park, the ratio of the wooded area to grassland is 4: 1. The total area is 385km2. What area is wooded?", "total cost to paint the room. Since there is a set-up fee of $20 then you would have to add it to the value of what 0.25p will be. Since p represents every square foot of the walls and you would multiply p by 0.25 every square foot is$0.25 then 0.25p+20. Therefore, the total cost to paint the room is 0.25p + 20. Question 10. A farmer gathered eggs from his chickens. Let e represent the total number of eggs he collected. He gives a third of the eggs to his friend, 7 of the eggs to another friend, and the rest to his cousin. Select all the algebraic expressions that represent the number of eggs he gives to his cousin. (A) $$\\frac{1}{3}$$e + 7 (B) e – $$\\frac{1}{3}$$e – 7 (C) e – $$\\frac{e}{3}$$ – 7 (D) $$\\frac{e}{3}$$ – 7 Answer: Option B and C are correct. The total number of eggs he collected = e The number of eggs he gives to his friend = 1/3 The number of eggs he had given to his other friend was 7 and the rest of the eggs were given to his cousin. The expression will be: e – (1/3 + 7) e – [1/3] -7 Question 11. Let g represent the total number of hours Greg will travel", "both numbers by$5$. It costs$\\$3.30$ per dog bone. ##### Solution To find a unit rate, divide by the dollar amount by the number of items. 1.$\\$29.49$for$95$dog treats • Divide both numbers by$95$. •$\\$0.31$ per dog treat 2. $\\$18$for$9$rubber balls • Divide both numbers by$9$. •$\\$2$ per rubber ball 3. $\\$32.50$for$8$dog bones • Divide both numbers by$8$. •$\\$4.06$ per dog bone 4. $\\$19.99$for$4$dog brushes • Divide both numbers by$4$. •$\\$4.99$ per dog brush 5. $\\$50.99$for$6$teddy bears • Divide both numbers by$6$. •$\\$8.49$ per teddy bear 6. $\\$52.75$for$5$bags of dog food • Divide both numbers by$5$. •$\\$10.55$ per bag of dog food • #### Identify the work problems that can be solved by finding a rate. ##### Hints Work problems can be solved by finding a unit rate and non-work problems don't involve unit rates. A unit rate is expressed as a quantity of $1$. An example of a work problem is: Tony and Sofie are partners for a class project. In the past, Tony completed $3$ projects in $6$ hours and Sofie completed $4$ projects in $2$ hours. How long will it take Tony and Sofie to complete one project together? ##### Solution Work Problems: Problems that are solved by finding a unit rate. 1. Two painters, Sam and Marcus, are hired to paint a new apartment. Sam can", "are sharing a cake, which is cut into $$28$$ pieces. The friends eat, on average, $$1\\frac{6}{7}$$ pieces each. What fraction of the cake is left over? ### Determine how many pieces of cake the $$7$$ friends eat in total. We must multiply the number of people by the number of pieces eaten by each person. Therefore $$13$$ pieces of cake were eaten. ### Determine the number of pieces left over. $28 - 13 = 15$ Therefore $$15$$ pieces of cake are left over. ### Determine the fraction of cake left over. $$15$$ out of $$28$$ pieces are left over. Therefore $$\\frac{15}{28}$$ of the cake is left over. ## Worked example 13.26: Solving word problems with fractions Realeboha is painting a wall with an area of $$5\\frac{1}{2}\\ \\text{m}^{2}$$. He has $$4$$ cans of paint, each containing $$420$$ ml of paint. Each can of paint covers $$1\\frac{3}{4}\\ \\text{m}^{2}$$ of the wall. How many millilitres (ml) of paint are left over after Realeboha has painted the wall (ignoring any potential spillages)? ### Summarise the given information. The area to be painted is $$5\\frac{1}{2}\\ \\text{m}^{2}$$ One can of paint covers $$1\\frac{3}{4}\\ \\text{m}^{2}$$ There are $$4$$ cans of paint $$1$$ can = $$450$$ ml of paint We need to find the amount of paint left over. ### Choose the correct calculation. We need", "### Home > CC2 > Chapter 6 > Lesson 6.2.3 > Problem6-92 6-92. Write and solve an inequality for the following situation. Robert is painting a house. He has $35$ cans of paint. He has used $30$ cans of paint on the walls. Now he needs to paint the trim. If each section of trim takes $\\frac { 1 } { 2 }$ can of paint, how many sections of trim can he paint? Show your answer as an inequality with symbols, in words, and with a number line. Make sure that your solution makes sense for this situation. The total number of cans used must be less than or equal to $35$ cans of paint, and only $5$ cans can be used for the trim. $\\frac{1}{2}\\textit{x}+30\\le35$ where $x$ is the number of sections of trim he can paint.", "# 5 Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint $100 m^{2}$ of area is painted. How many cans of paint will she need to paint the room? Total area painted by Daniel : $= 2(15\\times10+10\\times7+7\\times15) - 15\\times10$ ($\\because$ Bottom surface is excluded.) So, Area $= 650 - 150\\ m^2 = 500\\ m^2$ No. of cans of paint required $=\\frac{ 500\\ m^2}{100\\ m^2} = 5$ Thus 5 cans of paint are required. Exams Articles Questions", "# Billy plans to paint baskets. The paint costs $14.50. The baskets cost$7.25 each. How do you write an equation that gives the total cost as a function of the number of baskets made. Determine the cost of four baskets? $T C = 21.75 x$ Cost of four baskets=$.87 #### Explanation: The paint cost is=$ 14.50 The basket cost is=$7.25 Total cost $= \\left(14.50 + 7.25\\right) \\cdot x = 21.75 x$TC$= 21.75 x$TC_(x=4)=21.75 xx 4=$87", "take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Intern Joined: 15 Oct 2016 Posts: 2 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:06 1 Ray can paint 40 feet every 160 minutes.. Then Ray can paint 40/160 feets per minute. Taylor can paint 50 feet every 125 minutes.. Then Taylor can paint 50/125 fret per minutes Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute.. Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4033 Location: India GPA: 3.5 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:12 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a", "is$20,066,340, of which $5,216,340 will be covered by available grants. What will it cost to buy ceiling molding to go around a rectangular room with length 16 ft and width 10 ft? The value of these bills is$175. Brandon can paint a fence in 12 hours and elaine can paint the same fence in 11 hours. The total amount of money he has is $3.00. How much of each grade of tea must be used? It will take 27 equal monthly payments for Aimee to pay off her credit card balance of$2053.62. Ahmad served 5 more orders than Jane. Use x for your variable. You have 10 fewer quarters than dimes and 5 fewer nickels than quarters. The length of the longer piece was fifteen feet. The area of a state is 184,044 square miles greater than that of a second state. A) At what height do we need to climb to se... Sue can shovel snow from her driveway in 35 minutes. Get a free answer to a quick problem. We first write this as an equation and then solve. His son Roy mows the yard in 75 minutes. A kindergarten teacher has 15 students in her class, and she wants to order cupcakes. How many adult tickets were sold that day? An investor bought 500 shares", "Difference between revisions of \"2009 AIME II Problems/Problem 1\" Problem Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. Solution 1 Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$. After painting the wall, the total amount of paint is $482-4x$. Because the total amount in each color is the same after finishing, we have $$\\frac{482-4x}{3} = 130-x$$ $$482-4x=390-3x$$ $$x=92$$ Our answer is $482-4\\cdot 92 = 482 - 368 = \\boxed{114}$. ~ carolynlikesmath Solution 2 (Simple Solution) Let the stripes be $b, r, w,$ and $p$, respectively. Let the red part of the pink be $\\frac{r_p}{p}$ and the white part be $\\frac{w_p}{p}$ for $\\frac{r_p+w_p}{p}=p$. Since the stripes are of equal size, we have $b=r=w=p$. Since the amounts of paint end equal, we have $130-b=164-r-\\frac{r_p}{p}=188-w-\\frac{w_p}{p}$. Thus, we know that $$130-p=164-p-\\frac{r_p}{p}=188-p-\\frac{w_p}{p}$$ $$130=164-\\frac{r_p}{p}=188-\\frac{w_p}{p}$$ $$r_p=34p, w_p=58p$$ $$\\frac{r_p+w_p}{p}=92=p=b.$$ Each paint must" ]

[ "take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Intern Joined: 15 Oct 2016 Posts: 2 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:06 1 Ray can paint 40 feet every 160 minutes.. Then Ray can paint 40/160 feets per minute. Taylor can paint 50 feet every 125 minutes.. Then Taylor can paint 50/125 fret per minutes Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute.. Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4033 Location: India GPA: 3.5 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:12 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a", "fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 22:31 1 RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes CEO Joined: 12 Sep 2015 Posts: 2887 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 17", "it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 [Reveal] Spoiler: OA Intern Joined: 14 Oct 2016 Posts: 2 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:06 1 KUDOS Ray can paint 40 feet every 160 minutes.. Then Ray can paint 40/160 feets per minute. Taylor can paint 50 feet every 125 minutes.. Then Taylor can paint 50/125 fret per minutes Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute.. Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3253 Location: India GPA: 3.5 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:12 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take", "+0 # Direct 0 330 6 the time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence? Guest May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N} \\\\\\\\ T=k* \\frac{L}{N}\\qquad but\\\\\\\\ 5=k* \\frac{200}{3}\\qquad so\\\\\\\\ k=\\frac{3}{40}\\qquad so\\\\\\\\ T=\\frac{3L}{40N}\\\\\\\\ When N=5 and L=500\\\\\\\\ T=\\frac{3*500}{40*5}\\\\\\\\ T=7.5 \\;\\;hours$$ Melody May 26, 2015 Sort: #1 +26412 +10 The rate at which fence painting is done is 200/(3*5) feet per painter-hour. In order to paint 500 feet we need 500*3*5/200 painter-hours With 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours . Alan May 26, 2015 #2 +81090 +10 If 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour. Then, 5 painters can paint 200/3 ft every hr........so........500/ (200/3) = 1500/200 = 7.5 hrs. CPhill May 26, 2015 #3 +91510 +10 ok we might as well all answer this question! This is how I was trained to do it $$\\\\T\\alpha \\frac{L}{N}", "6 Here, 6 total units of fence are covered with paint, from $x=4$ all the way through $x=10$. Problem credits: Brian Dean Contest has ended. No further submissions allowed.", "for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Efficiency of Ray = $$\\frac{40}{160}$$ => $$\\frac{1}{4}$$ Efficiency of Taylor = $$\\frac{50}{125}$$ => $$\\frac{2}{5}$$ Combined efficiency is $$\\frac{1}{4}$$ + $$\\frac{2}{5}$$ = $$\\frac{13}{20}$$ Time required to paint a 260 meters wall is $$260*\\frac{20}{13}$$ = $$400$$ minutes. Hence answer will be (B) 400 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 20 Mar 2015 Posts: 4 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 16 Oct 2016, 21:31 1 KUDOS RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes Thus Total One day work, 1/4+1/2.5 = 65/100 Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes SVP Joined: 11 Sep 2015 Posts: 1999 Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a", "Ray + Work done by Taylor = 260 feet (the total work completed) Because Ray and Taylor are working simultaneously, we can let the time they both work together be t minutes. We now can express the individual work done by Ray and Taylor. We must remember that work = rate x time. Work done by Ray = (1/4)t Work done by Taylor = (2/5)t (1/4)t + (2/5)t = 260 We can multiply the entire equation by 20 to cancel out the fraction and we have: 5t + 8t = 5,200 13t = 5,200 t = 400 minutes _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] 05 Dec 2017, 17:39 Display posts from previous: Sort by", "the FTOC!). ## 12.1 Addition Let’s start off easy: how does a system with two added components behave? In the real world, this could be sending two friends (Frank and George) to build a fence. Let’s say Frank gets the wood, and George gets the paint. What’s the total cost? $\\textit{Total} = \\textit{Frank's cost} + \\textit{George's cost}$ $t(x) = f(x) + g(x)$ The derivative of the entire system, $$\\frac{dt}{dx}$$, is the cost per additional foot. Intuitively, we suspect the total increase is the sum of the increases in the parts: $\\frac{dt}{dx} = \\frac{df}{dx} + \\frac{dg}{dx}$ That relationship makes sense, right? Let’s say Frank’s cost is \\$3/foot for the wood, and George adds \\$0.50/foot for the paint. If we ask for another foot, the total cost will increase by \\$3.50. Here’s the math for that result: • Original: $$f + g$$ • New: $$(f + df) + (g + dg)$$ • Change: $$(f + df) + (g + dg) - (f + g) = df + dg$$ In my head, I imagine $$x$$, the amount you requested, changing silently in a corner. This creates a visible change in $$f$$ (size $$df$$) and $$g$$ (size $$dg$$), and we see the total change as $$df + dg$$. It seems we should just combine the total up front, writing $$total = 3.5x$$", "olive oil in three different sizes. Which size is the best buy, and what is its unit price? Size Price 17 oz$ 8.69 25 oz $11.79 101 oz$46.99 • Six painters Six painters paint an area of 120 m2 in 4 hours. How long does it take for 12 painters to paint 480 m2 with the same performance? • Cleaning 2 Maggie is looking to hire a carpet cleaning company. Cleaning Crew Inc charges $24 for supplies and then$18 per room. If Maggie has \\$96 to spend towards cleaning the carpets, how many rooms can she have cleaned? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • In the In the national park, the ratio of the wooded area to grassland is 4: 1. The total area is 385km2. What area is wooded?", "B RELATED VIDEO _________________ Brent Hanneson – Founder of gmatprepnow.com Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2069 Location: United States (CA) Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 05 Dec 2017, 17:39 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 We are given that Ray paints at a uniform rate of 40 feet every 160 minutes. Thus, the rate of Ray is 40/160= 1/4 ft/min. We are also given that Taylor paints at a uniform rate of 50 feet every 125 minutes. Thus, the rate of Taylor is 50/125= 2/5 ft/min. We need to determine the time it will take to paint a fence, that is 260 feet long, when Ray and Taylor work simultaneously. To determine the time to paint a 260-foot-long fence, we can use the combined work formula: Work done by", "cost B) Therefore, despite the fact that we can't estimate how much we would have to pay for paint B, we can see statement 2 is sufficient. Therefore the correct answer is D. Re: A home owner must pick between paint A, which costs$6.00 [#permalink] 08 Jan 2012, 06:00 Similar topics Replies Last post Similar Topics: 4 Peter went to the store to buy paint. Small cans cost $30 3 19 Dec 2013, 10:03 3 What is the distance between Harry s home and his office? 4 10 Nov 2011, 15:50 A homeowner must pick between paint A, which costs$6 per 6 19 Apr 2010, 12:59 10 Twelve jurors must be picked from a pool of n potential 7 20 Sep 2009, 01:32", "Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3515 Location: United States (CA) Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un [#permalink] ### Show Tags 05 Dec 2017, 18:39 alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 We are given that Ray paints at a uniform rate of 40 feet every 160 minutes. Thus, the rate of Ray is 40/160= 1/4 ft/min. We are also given that Taylor paints at a uniform rate of 50 feet every 125 minutes. Thus, the rate of Taylor is 50/125= 2/5 ft/min. We need to determine the time it will take to paint a fence, that is 260 feet long, when Ray and Taylor work simultaneously. To determine the time to paint a 260-foot-long fence, we can use the combined work formula: Work done by Ray + Work done by Taylor = 260 feet (the total work completed) Because", "Nov 2019, 21:43 [quote=\"nick1816\"]$$\\frac{1}{m}+\\frac{1}{m+5}=\\frac{1}{m-4}$$ $$m^2-8m-20=0$$ m=10 Could you provide more detail on how you resolve this equation? Thanks Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9850 Location: Pune, India Re: P, V and G had to paint three identical fences. On the [#permalink] ### Show Tags 27 Nov 2019, 22:23 CaptainLevi wrote: P, V and G had to paint three identical fences. On the first day, only P turned up for work and he completed the work only on the first fence, taking m hours. On the second day, all three of them turned up for work and they completed the work only on the second fence, taking (m –4) hours. On the third day, V and G turned up and they completed the work on the third fence, taking (m+ 5) hours. What is the value of m? a) 6 b) 8 c) 9 d) 10 e) 12 Since P takes m hrs to complete one fence, his rate of work = 1/m fence/hr = p P, V and G take (m - 4) hrs to complete one fence so their rate of work = 1/(m - 4) fence/hr = p + v + g V and G take (m + 5) hrs to complete one fence so their rate of work = 1/(m +", "un [#permalink] ### Show Tags 17 Oct 2016, 07:57 2 KUDOS Expert's post Top Contributor alanforde800Maximus wrote: Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long? a) 320 b) 400 c) 450 d) 500 e) 580 Let's determine how much each person can paint in ONE MINUTE Ray paints at a uniform rate of 40 feet every 160 minutes We have (40 feet)/(160 minutes) Divide top and bottom by 160 to get: (0.25 feet)/(1 minute) So, Ray paints 0.25 feet every ONE MINUTE Taylor paints at a uniform rate of 50 feet every 125 minutes We have (50 feet)/(125 minutes) Divide top and bottom by 125 to get: (0.4 feet)/(1 minute) So, Taylor paints 0.4 feet every ONE MINUTE So, working TOGETHER, the can paint 0.25 + 0.4 feet every ONE MINUTE In other words, they can paint 0.65 feet every ONE MINUTE How many minutes will it take for them to paint a fence that is 260 feet long? Time = output/rate = 260/0.65 = 400 minutes [Reveal] Spoiler:", "is$20,066,340, of which $5,216,340 will be covered by available grants. What will it cost to buy ceiling molding to go around a rectangular room with length 16 ft and width 10 ft? The value of these bills is$175. Brandon can paint a fence in 12 hours and elaine can paint the same fence in 11 hours. The total amount of money he has is $3.00. How much of each grade of tea must be used? It will take 27 equal monthly payments for Aimee to pay off her credit card balance of$2053.62. Ahmad served 5 more orders than Jane. Use x for your variable. You have 10 fewer quarters than dimes and 5 fewer nickels than quarters. The length of the longer piece was fifteen feet. The area of a state is 184,044 square miles greater than that of a second state. A) At what height do we need to climb to se... Sue can shovel snow from her driveway in 35 minutes. Get a free answer to a quick problem. We first write this as an equation and then solve. His son Roy mows the yard in 75 minutes. A kindergarten teacher has 15 students in her class, and she wants to order cupcakes. How many adult tickets were sold that day? An investor bought 500 shares", "is. I can get to 31 blue boxes; can you go any lower? – TMM Jul 24 '13 at 23:15 • Yes my solution is 31 , but my problem is how to justify that is the solution. – giancarlo Jul 24 '13 at 23:18 • Hmm. The best lower bound I got was 29. – Michael Biro Jul 25 '13 at 1:00 I got a bit carried away, but the minimum seems to be $31$ indeed. Unfortunately my method does not seem to generalize very well, but maybe that's too ambitious. We have a total of $49$ boxes, each of which must have $2$ edges painted blue, adding up to a total of $98$ edges to be painted blue. As the corner boxes have only $2$ edges available for painting, these two edges must be painted blue. This yields $8$ boxes painted blue, each having $3$ edges painted bue. This makes for a total of $24$ edges painted blue, leaving $74$ edges yet to be painted blue. The image on the left shows these $8$ boxes: $\\hspace{60pt}$$\\hspace{100pt} Note that there are 24 boxes yet to be painted that have 'excess' edges, i.e. edges that do not need to be painted blue. They are shown in green in the second picture. These are the 4 corners, having 2", "# It takes Noah 3 hours to paint one side of a fence. It takes Gilberto 5 hours. How long would it take them if they worked together? Mar 12, 2017 I tried this but check it. #### Explanation: Let us say that the rates at which each one paints the surface (of area $1$ for example) are: $\\frac{1}{3}$ and $\\frac{1}{5}$ to paint the same surface so together they will need a time $t$: $\\frac{1}{3} t + \\frac{1}{5} t = 1$ rearranging: $5 t + 3 t = 15$ $8 t = 15$ $t = \\frac{15}{8} = 1.875$ less than 2 hours.", "Cole has 7×5 which is 35 gallons. You can use your solution from Step 1 in order to find how much paint is in each bucket for Step 2. F. Step 2: How can you use an equation to find the amount of paint in each bucket? Now use a ___ equation to find the amount of paint in each bucket. Solve for g. The amount of paint in each bucket is 7 buckets. Explanation: The equation to find the amount of paint in each bucket is 35÷7 which is 5 buckets. G. What operations can you use to check the reasonableness of your answers? ________________ The operations that we have used are division, subtraction. Turn and Talk What is another strategy you could use to solve the problem? 2. Brie is painting a fence. The fence has 7 sections that are eaçh 9 feet long and one 17-foot-long section. Use steps and equations to find the total length of the fence. A. Think about the information you know. The total length of the fence is 80 feet. Explanation: Given that Brie is painting a fence and the fence has 7 sections that are eaçh 9 feet long which is 7×9 = 63 feet long and one 17-foot-long section. So the total length of the fence is 63+17", "# Surface area of house This is the image: So the question is: (a) Ralph is painting the barn, including the sides and roof. He wants to know how much paint to purchase. What is the total surface area that he is going to be painting? Round to the nearest hundredth. (b) If one paint can covers 57 square feet, how many paint cans should he purchase? (c) If each paint can costs $\\$23.50$, how much will the paint cost? I've done a little the first part, but since the surface area is off, all the other parts are off as well. Please show me how to get the surface area My work so far (A) Total Surface Area Rectangular Prism 2(wl + hl + hw) 2 (45 x 20 + 15 x 20 + 15 x 45) = 3750 Triangular Prism L + 2B h(a + b + c) + Ab 45(15 + 20 + 15) + 4 x 20 2250 + 80 2330 Total surface area - 3750 + 2330 6080 (B) Paint Cans 6080 / 57 106.6 or 107 paint cans I had to divide the total surface area with how much area one paint can would cover. (C) Price If each paint can costs$\\$23.50$, and Ralph buys 107 paint cans, he will spend $\\$2,514.50\\$", "if he painted 49 boxes on a particular day? A.$245 B. $54 C.$257 D. $490 Step: 1 Let n be the amount earned by painting 49 boxes. Step: 2 Amount earned = (cost of painting each box) x (number of boxes painted) n = 5 x 49 [Substitute the values.] Step: 3 n = 245 [Multiply.] Step: 4 Charles earned$245 on that day. Correct Answer is : $245 Q9Three people together earn$1110 by doing a piece of work. How much does each person receive, if the amount is distributed equally? A. $365 B.$374 C. $380 D.$370 Step: 1 Let n be the amount received by each person. Step: 2 Three people together earn $1110. Step: 3 n = 11103 [Original equation.] Step: 4 n = 370 [Simplify.] Step: 5 Each person receives$370. Correct Answer is : $370 Q10At a grocery store,360 pears are arranged in two boxes P and Q. How many pears does box Q have, if box Q can hold twice as much as what box P can hold? A. 240 B. 360 C. 120 D. 270 Step: 1 Let n be the number of pears in the box P. Step: 2 Number of pears in the box Q = 2n Step: 3 n + 2n = 360 [Original equation.] Step: 4 3n = 360 [Add.]" ]

[ "# Simplify : $\\large\\frac{\\sqrt {5625}+\\sqrt {441}}{\\sqrt {5625}-\\sqrt {441}}$ $\\begin{array}{1 1} 16 \\\\ 9 \\\\ \\large\\frac{16}{9} \\\\ \\large\\frac{9}{16} \\end{array}$", "# How do you simplify 4.72/0.2? Apr 23, 2017 $23.6 = 23 \\frac{6}{10} = 23 \\frac{3}{5}$ #### Explanation: \"Simplify\" is a very misleading instruction. It can mean a variety of things. Very often it is \"carry out the operation given\" In this case it is a division - shown as a fraction. Division by a decimal should be changed to division by a whole number: $\\frac{4.72}{0.2} \\times \\frac{10}{10} = \\frac{47.2}{2}$ Now simply divide: $47.2 \\div 2 = 23.6$ In fraction form this would be given as: $23 \\frac{6}{10} = 23 \\frac{3}{5}$", "# How do you simplify 48 - 2/3 ? $47.33$ Here, $48 - \\frac{2}{3}$ $= \\frac{48}{1} - \\frac{2}{3}$ $= \\frac{144 - 2}{3}$ $\\left[\\frac{a}{b} - \\frac{c}{d} = \\frac{a d - b c}{b d}\\right]$ $= \\frac{142}{3}$ $= 47.3$", "# How do you simplify 48 - 2/3 ? May 12, 2018 $47.33$ Here, $48 - \\frac{2}{3}$ $= \\frac{48}{1} - \\frac{2}{3}$ $= \\frac{144 - 2}{3}$ $\\left[\\frac{a}{b} - \\frac{c}{d} = \\frac{a d - b c}{b d}\\right]$ $= \\frac{142}{3}$ $= 47.3$", "Simplifying this to a complex number, $$x = {\\frac{-9}{6}} \\pm {\\sqrt{123}i \\over 6}$$, and simplifying the fractions, the final answer is: $$x = {-\\frac{2}{3}} \\pm {\\sqrt{123} \\over 6 }i$$ . Apr 19, 2020", "# How do you simplify (-9-sqrt(108))/3? It is $\\frac{- 9 - \\sqrt{108}}{3} = \\frac{- 9 - 6 \\sqrt{3}}{3} = - 3 - 2 \\sqrt{3}$", "# Simplifying an answer returned by DSolveValue Consider the IVP: $$x'=\\frac92-\\frac{x}{300+2t},\\quad x(0)=0$$ Working the solution by hand, the answer is: $$x(t)=450+3t-\\frac{4500\\sqrt3}{\\sqrt{300+2t}}$$ Entering code and then attempting to simplify, I tried Simplify and Expand. In[214]:= sol = DSolveValue[{x'[t] == 9/2 - x[t]/(300 + 2 t), x[0] == 0}, x[t], t] Out[214]= -(( 3 (750 Sqrt[6] - 150 Sqrt[150 + t] - t Sqrt[150 + t]))/Sqrt[150 + t]) In[215]:= Simplify[%] Out[215]= 3 (150 - 750/Sqrt[25 + t/6] + t) In[216]:= Expand[%] Out[216]= 450 - 2250/Sqrt[25 + t/6] + 3 t Is there a better way to simplify the answer so it looks more like my hand calculated answer, or if not, is there some way of getting clearing fractions from the radical in the denominator? In this case, you can simply call Apart: Apart[sol] $$450+3 t-\\frac{2250 \\sqrt{6}}{\\sqrt{t+150}}$$", "the $\\sqrt{- 1}$. When we simplify, the two x values are: $x = - \\frac{8 \\sqrt{5} i}{5}$ $x = \\frac{8 \\sqrt{5} i}{5}$", "{L}_{1} + 2 {L}_{2}$ $= \\frac{3 \\sqrt{2}}{2} + \\frac{2 \\sqrt{43}}{27} + \\frac{1}{4} {\\sinh}^{-} 1 \\left(2 \\sqrt{2}\\right) + \\frac{1}{2} {\\sinh}^{-} 1 \\left(\\frac{4}{\\sqrt{27}}\\right)$ $\\approx 3.4022$", "# How do you simplify sqrt4/100? $\\frac{\\sqrt{4}}{100} = \\frac{1}{50}$ or $0.02$ $\\sqrt{4} = 2$ $\\frac{\\sqrt{4}}{100} = \\frac{2}{100} = \\frac{1}{50} = 0.02$", "# How do you simplify (6 + 6i ) / (7 + 6i )? $\\frac{42 + 42 i - 36 i + 36}{49 + 36} = \\frac{78 - 6 i}{85}$ $= \\frac{78}{85} - \\frac{6}{85} i$ This can be put in trignometric form $r \\left(\\cos \\theta - i \\sin \\theta\\right)$ if required, by solving $r \\cos \\theta = \\frac{78}{85}$ and $r \\sin \\theta = - \\frac{6}{85}$.", "# Simplifying an answer returned by DSolveValue Consider the IVP: $$x'=\\frac92-\\frac{x}{300+2t},\\quad x(0)=0$$ Working the solution by hand, the answer is: $$x(t)=450+3t-\\frac{4500\\sqrt3}{\\sqrt{300+2t}}$$ Entering code and then attempting to simplify, I tried Simplify and Expand. In[214]:= sol = DSolveValue[{x'[t] == 9/2 - x[t]/(300 + 2 t), x[0] == 0}, x[t], t] Out[214]= -(( 3 (750 Sqrt[6] - 150 Sqrt[150 + t] - t Sqrt[150 + t]))/Sqrt[150 + t]) In[215]:= Simplify[%] Out[215]= 3 (150 - 750/Sqrt[25 + t/6] + t) In[216]:= Expand[%] Out[216]= 450 - 2250/Sqrt[25 + t/6] + 3 t Is there a better way to simplify the answer so it looks more like my hand calculated answer, or if not, is there some way of getting clearing fractions from the radical in the denominator? In this case, you can simply call Apart: Apart[sol] $$450+3 t-\\frac{2250 \\sqrt{6}}{\\sqrt{t+150}}$$ This is almost identical to your hand-written version. Is this good enough? • It is very good with a very easy command. Nice job. Let me ask one more question: How can I multiply both numerator and denominator of the fractional part by sort(3)? – David Jan 31 '15 at 22:25 • @David: Doing that wouldn't be easy, as it would probably try to simplify back to its original condition during evaluation. In general, manipulating how Mathematica chooses to simplify expressions is a", "# How do you simplify 4+(4(5+8))/(15*4)? Oct 21, 2015 The answer is $\\frac{73}{15}$. #### Explanation: $4 + \\frac{4 \\left(5 + 8\\right)}{15 \\cdot 4}$ Simplify $\\left(5 + 8\\right)$ to $13$. $4 + \\frac{4 \\left(13\\right)}{15 \\cdot 4}$ Simplify $4 \\left(13\\right)$ to $52$. $4 + \\frac{52}{15 \\cdot 4}$ Simplify $\\left(15 \\cdot 4\\right)$ to $60$. $4 + \\frac{52}{60}$ Simplify $\\frac{52}{60}$ to $\\frac{13}{15}$. $4 + \\frac{13}{15}$ The LCD is $15$. Multiply $4$ times $\\frac{15}{15}$ in order to make the denominators the same. $4 \\cdot \\frac{15}{15} + \\frac{13}{15} =$ $\\frac{60}{15} + \\frac{13}{15}$ $\\frac{60 + 13}{15} =$ $\\frac{73}{15}$", "# How do you simplify 21/50 div21/50? $\\frac{21}{50} \\div \\frac{21}{50}$ $= \\frac{21}{50} \\cdot \\frac{50}{21}$ $= {\\cancel{21}}^{\\textcolor{red}{1}} / {\\cancel{50}}^{\\textcolor{g r e e n}{1}} \\cdot {\\cancel{50}}^{\\textcolor{g r e e n}{1}} / {\\cancel{21}}^{\\textcolor{red}{1}}$ $= \\frac{1}{1} \\cdot \\frac{1}{1}$ $= 1$", "estimate, but already we can see that $\\sqrt{50} \\div 10 \\approx 0.71.$ Compare this to your original problem: $1 \\div \\sqrt{2}$. $\\sqrt{50} \\div 10$ was a simple calculation to do, as the square root was relatively easy to approximate, and the divisor was a whole number, i.e. this fraction is already in a rationalised form. You might have also noticed that your answer is very similar, or the same, as your answer for $1 \\div \\sqrt{2}.$ Can you see why this is? What happens if you simplify $\\frac{\\sqrt{50}}{10}$?", "# How do you simplify 50 - (2/25)*100? $50 - \\left(\\frac{2}{25}\\right) \\cdot 100$ $= 50 - \\left(\\frac{2 \\cdot \\stackrel{4}{\\cancel{100}}}{\\cancel{25}}\\right)$ $= 50 - 8$ $= 42$", "\\frac{3}{\\sqrt{27x^3}} \\cdot \\frac{\\sqrt{3x}}{\\sqrt{3x}} = \\frac{3\\sqrt{3x}}{\\sqrt{81x^4}} = \\frac{3\\sqrt{3x}}{9x^2} = \\frac{\\sqrt{3x}}{3x^2}$$ Exercise $$\\PageIndex{56}$$ $$\\frac{5}{\\sqrt{10x^5}}$$", "so let's see what we get when we square $69$: ${69}^{2} = {\\left(70 - 1\\right)}^{2} = {70}^{2} - 2 \\cdot 70 + 1 = 4900 - 140 + 1 = 4761$ So putting $n = 4783$ and $a = 69$ we get: $b = n - {a}^{2} = 4783 - 4761 = 22$ So: $\\sqrt{4783} = 69 + \\frac{22}{138 + \\frac{22}{138 + \\frac{22}{138 + \\ldots}}}$ We can truncate after any number of terms to get a rational approximation for $\\sqrt{4783}$: $\\sqrt{4783} \\approx 69$ $\\sqrt{4783} \\approx 69 + \\frac{22}{138} = \\frac{4772}{69} \\approx 69.1594$ $\\sqrt{4783} \\approx 69 + \\frac{22}{138 + \\frac{22}{138}} = 69 + \\frac{22}{\\frac{9533}{69}} = \\frac{659295}{9533} \\approx 69.1592363$ A calculator tells me: $\\sqrt{4873} \\approx 69.1592365487$", "- 22 + 2 \\sqrt{6}}{121 - 6}$ $= \\frac{13 \\sqrt{6} - 28}{115}$ At this stage, I don't see any more ways to simplify this further, so this must be the answer.", "How do you simplify 1/8 + 2/3 + 7/12? May 15, 2018 $\\frac{11}{8}$ Explanation: $\\frac{1}{8} + \\frac{2}{3} + \\frac{7}{12}$ 1. Make the denominator the same by finding the LCM (lowest common multiple) =$\\frac{3}{24} + \\frac{16}{24} + \\frac{14}{24}$ 1. Add the fractions together. This is possible because your denominator is the same for each fraction =$\\frac{33}{24}$ 1. Simplify =$\\frac{11}{8}$" ]

[ "\\sqrt{\\frac{73}{50}} \\left[\\frac{3}{\\sqrt{73}} \\cdot \\frac{7}{\\sqrt{50}} + \\frac{- 8}{\\sqrt{73}} \\cdot \\frac{- 1}{\\sqrt{50}}\\right]$ $= \\frac{1}{50} \\left[- 56 - 3\\right] + \\frac{i}{50} \\left[21 + 8\\right] = \\frac{- 59}{50} + \\frac{29}{50} i$ We can check this with a Cartesian calculation: $\\frac{- 8 + 3 i}{7 - i} \\cdot \\frac{7 + i}{7 + i} = \\frac{- 56 + 21 i - 3 + 8 i}{50} = \\frac{- 59}{50} + \\frac{29}{50} i$ as expected.", "{L}_{1} + 2 {L}_{2}$ $= \\frac{3 \\sqrt{2}}{2} + \\frac{2 \\sqrt{43}}{27} + \\frac{1}{4} {\\sinh}^{-} 1 \\left(2 \\sqrt{2}\\right) + \\frac{1}{2} {\\sinh}^{-} 1 \\left(\\frac{4}{\\sqrt{27}}\\right)$ $\\approx 3.4022$", "\\frac{3}{\\sqrt{27x^3}} \\cdot \\frac{\\sqrt{3x}}{\\sqrt{3x}} = \\frac{3\\sqrt{3x}}{\\sqrt{81x^4}} = \\frac{3\\sqrt{3x}}{9x^2} = \\frac{\\sqrt{3x}}{3x^2}$$ Exercise $$\\PageIndex{56}$$ $$\\frac{5}{\\sqrt{10x^5}}$$", "= \\frac{x^{3} }{3} +C$ 46. $\\int _{}^{}\\frac{dx}{\\sqrt{1-2x} +\\sqrt{3-2x} } = \\frac{1}{6} (1- x)^{\\frac{3}{2} } -\\frac{1}{6} (3-2x)^{\\frac{3}{2} } +C$ 47. $\\int \\tan ^{-1} (\\cot x)\\, dx = \\frac{\\pi }{2} x-\\frac{x^{2} }{2} +C$ 48. $\\int _{0}^{\\pi }\\, ( \\sin ^{2} \\frac{x}{2} -\\cos ^{2} \\frac{x}{2} )\\, dx = 0$ 49. $\\int _{0}^{1}\\frac{dx}{\\sqrt{1+x} -\\sqrt{x} } = \\frac{4\\sqrt{2} }{3}$", "$$52 = 4! * \\sqrt4 + \\sqrt4 + \\sqrt4$$ $$53 = 4! * \\sqrt4 + \\frac{\\sqrt4}{.4}$$ $$54 = 4! * \\sqrt4 + 4 + \\sqrt4$$ ### Show 55 to 60 55 is unexpectedly difficult. Even though it isn't prime, it is 7 away from 48 and so is hard to get to. However, you can use a similar technique as for 36 and divide 24 by $$0.\\overline4 = \\frac49$$, which is the same as multiplying it by $$\\frac94$$, and thereby get to its neighbour 54. 59 looks particularly impressive but is really just 54 + 5. $$55 = 54 + 1 = \\frac{4!}{.\\overline4} + \\frac44$$ $$56 = 4! * \\sqrt4 + 4 + 4$$ $$57 = 4! * \\sqrt4 + \\frac4{.\\overline4}$$ $$58 = 4! * \\sqrt4 + \\frac4{0.4}$$ $$59 = 54 + 5 = \\frac{4!}{.\\overline4} + \\frac{\\sqrt4}{.4}$$ $$60 = 4 * 4 * 4 - 4$$ ### Show 61 to 66 To get to 61, you have to get to 60. Fortunately you can do that from 24 fairly easily using almost the same technique as above. Notice that $$0.4 = \\frac25$$. If you divide 24 by this, it's the same as multiplying by $$\\frac52$$ and you get 60. You can get to lots of other numbers around here from either 60 or $$64 = 4 * 4", "very much! i see the way you got the unkown terms on one side, very nice Thanks again, Max 4. Originally Posted by darksupernova Hey there, Getting very stuck on solving these, ive tried using substitution but i dont know how to rearrange it... can you help? $137.9 = x + \\frac{y}{\\sqrt{0.05}}$ and $275.8 = x + \\frac{y}{\\sqrt{0.007}}$ Since 275.8 is double 137.9, then you can go this simpler way: x + y/sqrt(.007) = 2x + 2y/sqrt(.05) 5. Seeing \"x\" alone in both equations, the first thing I would think of is subtracting one equation from the other: $275.8 = x + \\frac{y}{\\sqrt{0.007}}$ $137.9 = x + \\frac{y}{\\sqrt{0.05}}$ $137.9= \\frac{y}{\\sqrt{0.007}}- \\frac{y}{\\sqrt{0.05}}= y\\left(\\frac{1}{\\sqrt{0.007}}- \\frac{1}{\\sqrt{0.05}}\\right)$ $137.9= y\\left(\\frac{\\sqrt{0.05}- \\sqrt{0.007}}{\\sqrt{(0.05)(.007)}}\\right)$ $y= \\frac{137.9\\sqrt{0.00035}}{\\sqrt{0.05}- \\sqrt{0.007}}$", "these latter five are the numbers we want to add. $\\frac1{19}+ \\frac2{19}+ \\frac3{19}+ \\frac 1{17}+ \\frac2{17}= \\frac6{19} + \\frac 3{17} = \\frac{6\\cdot17 + 3\\cdot19}{17\\cdot19} = \\frac{159}{323}$ and so the answer is $159 + 323 = \\boxed{482}$.", "=(64)^\\frac{1}{12}$ $\\sqrt [6]3 = 3^\\frac{1}{6} = (3^2)^\\frac {1}{12} = (9)^\\frac {1}{12}$ $\\sqrt[4] 5 =5^{\\frac14}= (5^3)^{ \\frac{1}{12}} =(125)^\\frac{1}{12}$ Q.4: The smallest among the numbers 2250, 3150, 5100, 4200 is a) 2250 b) 3150 c) 5100 d) 4200 Ans : c) 5100 2250 =(25)50 =(32)50 3150 = (33)50 = (27)50 5100 = (52)50 =(25)50 4200 = (44)50 = (256)50 Q.5: Which is greater $\\sqrt[3]2$ or $\\sqrt3$ ? a) $\\sqrt[3]2$ b) $\\sqrt3$ c) Equal d) Can not be compared Ans : $\\sqrt3$ Cube of both the numbers are $(\\sqrt[3]2)^3 =2 \\: \\text {and} \\: (\\sqrt3)^3 = 3 \\sqrt3$ Q.6: The smallest among $\\sqrt[6]{12}, \\sqrt[3]4, \\sqrt[4]5, \\sqrt3$ is a) $\\sqrt[6]{12}$ b) $\\sqrt[3]4$ c) $\\sqrt[4]5$ d) $\\sqrt3$ Ans : c) $\\sqrt[4]5$ LCM of 2,3,4 and 6 is 12 $\\sqrt[6]{12}$ =(12)1/6 =(12)2/12 = (122)1/12 = (144)1/12 $\\sqrt[3]4$ = (256)1/12 $\\sqrt[4]5$ = (125)1/12 $\\sqrt3$ = (729)1/12 Q.7: If X =(0.25)1\\2, Y = (0.4)2, Z=(0.216)1/3, then a) Y>X>Z b) X>Y>z c) Z>X>Y d) X>Z>Y Ans : c) Z>X>Y X=(0.25)1/2 =0.5 Y= (0.4)2 = 0.16 Z = (0.216) = 0.6 ### Simplifying when the Root Values are given Q.8: If $\\sqrt {33} =5.745, \\text {than the value of} \\sqrt {\\frac {3}{11}}$ is approximately. a) 1 b) 0.5223 c) 6.32 d) 2.035 Ans : b) 0.5223 $\\sqrt {\\frac {3}{11}} = \\sqrt {\\frac {3\\times 11}{11 \\times 11}} = \\frac", "faithfully giving the relations between the eight sets/atoms. Further, how might I \"polish\" this plot further (for possible journal/arXiv publication)--including labels for the eight sets displayed, choice of colorings...? As further information, let us mention as reported in the last answer to How can one expand an arbitrary boolean combination into the $2^n$ atoms of the associated boolean algebra of size $2^{2^n}$? that the measures/probabilities assigned to the eight atoms/sets are--in the indicated order (G[0] G2,...)-- $$\\left\\{\\frac{2}{121},\\frac{4 \\left(242 \\sqrt{3} \\pi -1311\\right)}{9801},\\frac{524119}{4247100}+\\frac{4 \\pi }{27 \\sqrt{3}}-\\frac{\\sqrt{3} \\log (2)}{\\log (81)}-\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}},\\frac{7909}{8775}-\\frac{4 \\pi }{27 \\sqrt{3}}-\\frac{\\sqrt{3} \\log (2)}{\\log (81)}+\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}},\\frac{1678081}{4247100}-\\frac{4 \\pi }{27 \\sqrt{3}}+\\frac{\\sqrt{3} \\log (2)}{\\log (81)}+\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}},-\\frac{434}{8775}-\\frac{4 \\pi }{27 \\sqrt{3}}+\\frac{\\sqrt{3} \\log (2)}{\\log (81)}+\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}},\\frac{70064}{1061775}-\\frac{4 \\pi }{27 \\sqrt{3}}+\\frac{\\sqrt{3} \\log (2)}{\\log (81)}-\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}},\\frac{87236}{1061775}+\\frac{4 \\pi }{27 \\sqrt{3}}-\\frac{\\sqrt{3} \\log (2)}{\\log (81)}-\\frac{\\cosh ^{-1}(97)}{54 \\sqrt{3}}\\right\\} \\approx \\{0.01652892562,0.002374589709,0.06259481829,0.4157208527,0.4559237002,0.01135281657,0.0 1415526980,0.02134902704\\}$$. So, atoms with small probability may be difficult to effectively label. When I employ the option PlotLegends -> {\"a&&b&&c\", \"!a&&b&&c\", \"!b&&a&&c\", \"!c&&a&&b\", \"!a&&!b&&c\", \"!a&&!c&&b\", \"!c&&!b&&a\", \"!c&&!b&&!a\"} the result is", "you can get to 36 in just two 4s. Notice that the square root of $$\\sqrt{.\\overline4} = \\sqrt{\\frac49} = \\frac23$$. This is not so useful in itself, but if you divide something by $$\\frac23$$, its the same as multiplying by $$\\frac32$$. And you can multiply it by 4!! (The first exclamation point is a factorial in this case.) So 4! times $$\\frac32$$ gets you to 36 and adding 5 more arrives at 41. Most of the other numbers around here can be reached straight from 48. $$41 = 36 + 5 = \\frac{4!}{\\sqrt{.\\overline4}} + \\frac{\\sqrt4}{.4}$$ $$42 = 4! * \\sqrt4 - 4 - \\sqrt4$$ $$43 = 4! * \\sqrt4 - \\frac{\\sqrt4}{.4}$$ $$44 = 4! * \\sqrt4 - \\sqrt4 - \\sqrt4$$ $$46 = 4! * \\sqrt4 - 4 + \\sqrt4$$ $$47 = 4! * \\sqrt4 - \\frac44$$ $$48 = 4! * \\sqrt4 + 4 - 4$$ $$49 = 4! * \\sqrt4 + \\frac44$$ $$50 = 4! * \\sqrt4 + 4 - \\sqrt4$$ ### Show 51 to 54 Numbers ending in 1 seem to be the trickiest. 51 isn't too difficult though. Not if you use the square root of 9 to get to 3 with just two 4s, $$3 = \\sqrt{\\frac4{.\\overline4}}$$ and then add that to 48. $$51 = 48 + 3 = 4! * \\sqrt4 + \\sqrt {\\frac4{.\\overline4}}$$", "# Question #0e8ab Nov 23, 2016 Assuming you're asking what the value of $\\sqrt{56.25}$ is: The answer is: $\\frac{15}{2} \\mathmr{and} 7.5$ #### Explanation: First: Convert $56.25$ to a fraction. $56 + \\frac{1}{4} = \\frac{224}{4} + \\frac{1}{4} = \\frac{225}{4}$ Then take the square root of this value and simplify: $\\sqrt{\\frac{225}{4}} = \\frac{\\sqrt{225}}{\\sqrt{4}} = \\frac{\\sqrt{9} \\cdot \\sqrt{25}}{2} = \\frac{3 \\cdot 5}{2} = \\frac{15}{2} = 7.5$", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$.", "# Simplify : $\\large\\frac{\\sqrt {5625}+\\sqrt {441}}{\\sqrt {5625}-\\sqrt {441}}$ $\\begin{array}{1 1} 16 \\\\ 9 \\\\ \\large\\frac{16}{9} \\\\ \\large\\frac{9}{16} \\end{array}$", "$s=\\frac{d}{\\sqrt{2}}$, so our side length is $\\frac{2\\sqrt{2}+6}{\\sqrt{2}}$. However, we are trying to look for the area, so squaring $\\frac{2\\sqrt{2}+6}{\\sqrt{2}}$ yields $\\frac{44+24\\sqrt{2}}{2}=\\boxed{\\text{(B)}22+12\\sqrt{2}}$", "2}$ $= \\frac{- 8 \\pm \\sqrt{64 - 56}}{4}$ $= \\frac{- 8 \\pm \\sqrt{8}}{4}$ $= \\frac{- 8 \\pm \\sqrt{{2}^{2} \\cdot 2}}{4}$ $= \\frac{- 8 \\pm 2 \\sqrt{2}}{4}$ $= - 2 \\pm \\frac{\\sqrt{2}}{2}$", "\\frac {28 + 7}{2} ; a_1 = \\frac {35}{2} ; \\boxed {a_1 = 17.5} \\\\ b_{1} = \\sqrt {28 \\cdot 7} ; b_{1} = \\sqrt {196} ; \\boxed {b_1 = 14} \\\\ a_2 = \\frac {17.5 + 14}{2} ; a_2 = \\frac {31.5}{2} ; \\boxed {a_2 = 15.75} \\\\ b_{2} = \\sqrt {17.5 \\cdot 14} ; b_{2} = \\sqrt {245} ; \\boxed {b_2 = 15.6524758425} \\\\ a_3 = \\frac {15.75 + 15.6524758425}{2} ; a_3 = \\frac {31.4024758425}{2} ; \\boxed {a_3 = 15.7012379212} \\\\ b_{3} = \\sqrt {15.75 \\cdot 15.6524758425} \\\\ b_{3} = \\sqrt {246.526494519} \\\\ \\boxed {b_3 = 15.7011622028} \\\\ a_4 = \\frac {15.7012379212 + 15.7011622028}{2} \\\\ a_4 = \\frac {31.402400124}{2} \\\\ \\boxed {a_4 = 15.701200062} \\\\ b_{4} = \\sqrt {15.7012379212 \\cdot 15.7011622028} \\\\ b_{4} = \\sqrt {246.527683386} \\\\ \\boxed {b_4 = 15.701200062} As we can see, they eventually converge into the same number. AGM is often used for calculating mathematical constants, notably π, as well as for calculating fast algorithms and exponential or trigonometric functions. ### Root-Mean Square (RMS) The root-mean square, often called the quadratic mean, is the square root of the mean square. The mean square is the arithmetic mean of the squares of a set of numbers. So, the formula for calculating RMS would be: \\bar x_{RMS} = \\sqrt {\\frac {x_1^2+x_2^2+...+x_n^2}{n}} It is mainly", "\\sqrt{13}=2 \\sqrt{13}$ 46. anonymous see pic above...i got it down to there but then stuck 47. myininaya $\\frac{-4}{2 \\sqrt{13}}=\\frac{-2}{\\sqrt{13}}=\\frac{-2 \\sqrt{13}}{13}$", "\\left ( 7 + \\frac{\\sqrt{3}}{2\\sqrt{x}} \\right )$ 6. Originally Posted by CaptainBlack $\\frac{dy}{dx} = \\frac{1}{2\\sqrt{7x + \\sqrt{3x}}} \\cdot \\left ( 7 + \\frac{\\sqrt{3}}{2\\sqrt{x}} \\right )$ Thanks for the spot. I fixed it in the post. -Dan", "r\\cdot\\frac{\\sqrt3}2=\\frac{r\\sqrt3}4$, and thus $$\\frac{\\frac{\\sqrt3}4(r^2+r+1)}{\\frac{\\sqrt3}4(r^2+r+1)+3\\left(\\frac{r\\sqrt3}4\\right)}=\\frac{r^2+r+1}{r^2+4r+1}=\\frac7{10}$$ This simplifies to $r^2-6r+1=0\\Rightarrow \\boxed{\\textbf{(E)}\\ 6}$.", "by $\\sqrt{3}$ yields $\\frac{s^2\\sqrt{3}}{4}=221\\sqrt{3}-330\\implies{p+q+r=221+3+330=\\boxed{554}}$ -Solution by thecmd999" ]

[ "# Simplify : $\\large\\frac{\\sqrt {5625}+\\sqrt {441}}{\\sqrt {5625}-\\sqrt {441}}$ $\\begin{array}{1 1} 16 \\\\ 9 \\\\ \\large\\frac{16}{9} \\\\ \\large\\frac{9}{16} \\end{array}$", "$$52 = 4! * \\sqrt4 + \\sqrt4 + \\sqrt4$$ $$53 = 4! * \\sqrt4 + \\frac{\\sqrt4}{.4}$$ $$54 = 4! * \\sqrt4 + 4 + \\sqrt4$$ ### Show 55 to 60 55 is unexpectedly difficult. Even though it isn't prime, it is 7 away from 48 and so is hard to get to. However, you can use a similar technique as for 36 and divide 24 by $$0.\\overline4 = \\frac49$$, which is the same as multiplying it by $$\\frac94$$, and thereby get to its neighbour 54. 59 looks particularly impressive but is really just 54 + 5. $$55 = 54 + 1 = \\frac{4!}{.\\overline4} + \\frac44$$ $$56 = 4! * \\sqrt4 + 4 + 4$$ $$57 = 4! * \\sqrt4 + \\frac4{.\\overline4}$$ $$58 = 4! * \\sqrt4 + \\frac4{0.4}$$ $$59 = 54 + 5 = \\frac{4!}{.\\overline4} + \\frac{\\sqrt4}{.4}$$ $$60 = 4 * 4 * 4 - 4$$ ### Show 61 to 66 To get to 61, you have to get to 60. Fortunately you can do that from 24 fairly easily using almost the same technique as above. Notice that $$0.4 = \\frac25$$. If you divide 24 by this, it's the same as multiplying by $$\\frac52$$ and you get 60. You can get to lots of other numbers around here from either 60 or $$64 = 4 * 4", "= \\frac{x^{3} }{3} +C$ 46. $\\int _{}^{}\\frac{dx}{\\sqrt{1-2x} +\\sqrt{3-2x} } = \\frac{1}{6} (1- x)^{\\frac{3}{2} } -\\frac{1}{6} (3-2x)^{\\frac{3}{2} } +C$ 47. $\\int \\tan ^{-1} (\\cot x)\\, dx = \\frac{\\pi }{2} x-\\frac{x^{2} }{2} +C$ 48. $\\int _{0}^{\\pi }\\, ( \\sin ^{2} \\frac{x}{2} -\\cos ^{2} \\frac{x}{2} )\\, dx = 0$ 49. $\\int _{0}^{1}\\frac{dx}{\\sqrt{1+x} -\\sqrt{x} } = \\frac{4\\sqrt{2} }{3}$", "\\sqrt{\\frac{73}{50}} \\left[\\frac{3}{\\sqrt{73}} \\cdot \\frac{7}{\\sqrt{50}} + \\frac{- 8}{\\sqrt{73}} \\cdot \\frac{- 1}{\\sqrt{50}}\\right]$ $= \\frac{1}{50} \\left[- 56 - 3\\right] + \\frac{i}{50} \\left[21 + 8\\right] = \\frac{- 59}{50} + \\frac{29}{50} i$ We can check this with a Cartesian calculation: $\\frac{- 8 + 3 i}{7 - i} \\cdot \\frac{7 + i}{7 + i} = \\frac{- 56 + 21 i - 3 + 8 i}{50} = \\frac{- 59}{50} + \\frac{29}{50} i$ as expected.", "you can get to 36 in just two 4s. Notice that the square root of $$\\sqrt{.\\overline4} = \\sqrt{\\frac49} = \\frac23$$. This is not so useful in itself, but if you divide something by $$\\frac23$$, its the same as multiplying by $$\\frac32$$. And you can multiply it by 4!! (The first exclamation point is a factorial in this case.) So 4! times $$\\frac32$$ gets you to 36 and adding 5 more arrives at 41. Most of the other numbers around here can be reached straight from 48. $$41 = 36 + 5 = \\frac{4!}{\\sqrt{.\\overline4}} + \\frac{\\sqrt4}{.4}$$ $$42 = 4! * \\sqrt4 - 4 - \\sqrt4$$ $$43 = 4! * \\sqrt4 - \\frac{\\sqrt4}{.4}$$ $$44 = 4! * \\sqrt4 - \\sqrt4 - \\sqrt4$$ $$46 = 4! * \\sqrt4 - 4 + \\sqrt4$$ $$47 = 4! * \\sqrt4 - \\frac44$$ $$48 = 4! * \\sqrt4 + 4 - 4$$ $$49 = 4! * \\sqrt4 + \\frac44$$ $$50 = 4! * \\sqrt4 + 4 - \\sqrt4$$ ### Show 51 to 54 Numbers ending in 1 seem to be the trickiest. 51 isn't too difficult though. Not if you use the square root of 9 to get to 3 with just two 4s, $$3 = \\sqrt{\\frac4{.\\overline4}}$$ and then add that to 48. $$51 = 48 + 3 = 4! * \\sqrt4 + \\sqrt {\\frac4{.\\overline4}}$$", "# How do you simplify 48 - 2/3 ? $47.33$ Here, $48 - \\frac{2}{3}$ $= \\frac{48}{1} - \\frac{2}{3}$ $= \\frac{144 - 2}{3}$ $\\left[\\frac{a}{b} - \\frac{c}{d} = \\frac{a d - b c}{b d}\\right]$ $= \\frac{142}{3}$ $= 47.3$", "# Question #0e8ab Nov 23, 2016 Assuming you're asking what the value of $\\sqrt{56.25}$ is: The answer is: $\\frac{15}{2} \\mathmr{and} 7.5$ #### Explanation: First: Convert $56.25$ to a fraction. $56 + \\frac{1}{4} = \\frac{224}{4} + \\frac{1}{4} = \\frac{225}{4}$ Then take the square root of this value and simplify: $\\sqrt{\\frac{225}{4}} = \\frac{\\sqrt{225}}{\\sqrt{4}} = \\frac{\\sqrt{9} \\cdot \\sqrt{25}}{2} = \\frac{3 \\cdot 5}{2} = \\frac{15}{2} = 7.5$", "$ABC$ is $$S=S_{ABC}=\\sqrt{p(p-a)(p-b)(p-c)}.$$ Since $$\\begin{eqnarray*} &&p(p-a)(p-b)(p-c) \\\\ &=&\\left( 7\\sqrt{2}+\\frac{21}{5}\\sqrt{10}+\\frac{28}{5}\\sqrt{5}\\right) \\left( -7\\sqrt{2}+\\frac{21}{5}\\sqrt{10}+\\frac{28}{5}\\sqrt{5}\\right) \\\\ &&\\times \\left( 7\\sqrt{2}-\\frac{21}{5}\\sqrt{10}+\\frac{28}{5}\\sqrt{5}\\right) \\left( 7\\sqrt{2}+\\frac{21}{5}\\sqrt{10}-\\frac{28}{5}\\sqrt{5}\\right) \\\\ &=&\\frac{1382976}{25}, \\end{eqnarray*}$$ we get $$S=\\sqrt{\\frac{1382976}{25}}=\\frac{1176}{5}.$$ The area of $DEF$ is $$K=S_{DEF}=\\frac{EF\\times DF}{2}=\\frac{6\\times 8}{2}=24$$ and the ratio $$\\frac{S}{K}=\\frac{1176/5}{24}=\\frac{49}{5}=9.8,$$ as given. Added: The scale is uniform throughout the following diagram drawn with the calculated equations - Here is a diagram. I may or may not post a solution later. Edit I will not post a solution since it appears to be quite messy. Please direct votes towards an actual solution. - In your diagram are the scales of the triangle $[DEF]$ and the big circle's equal? – Américo Tavares Jan 27 '12 at 19:47 No. The scale of the triangle DEF is scaled by a factor u, whereas the radius of the big circle is 14 (absolutely). – Victor Liu Jan 30 '12 at 3:11 Thanks. It seems to me that the solution would be very difficult. – Américo Tavares Jan 30 '12 at 11:10 The scales of your diagram are uniform for both triangles and the big circle. At least the factor $u=2$ is compatible with my verification. – Américo Tavares Feb 2 '12 at 23:03", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$.", "464If $\\l=\\frac{-1+\\sqrt{5}}{2},\\mu=\\frac{3+\\sqrt{13}}{2}$, then the 465vectors $v_1=e_1+(\\l+1)e_2,v_2=e_1-\\l e_2,v_3=\\mu 466e_3+e_4,v_4=(3-\\mu)e_3+e_4$ corresponding to the 4 newforms 467$f_1,f_2,f_3,f_4$ of weights 2 and level $74$. Using (1) one gets 468the first 83 values of the coefficients of these newforms. For 469example for $f_1$ the list of the first values of $a_l$ is 470$\\begin{array}{ccccccc} 471 l & 2 & 3 & 5 & 7 & 11 & 13 \\\\ 472 a_l & 1 & \\frac{-1+\\sqrt{5}}{2} & \\frac{1-3\\sqrt{5}}{2} & -1+\\sqrt{5} & \\frac{-5-\\sqrt{5}}{2} & \\frac{1+3\\sqrt{5}}{2}\\\\ 473\\end{array}$ 474 475and for $f_3$ one gets 476$\\begin{array}{ccccccc} 477 l & 2 & 3 & 5 & 7 & 11 & 13 \\\\ 478 a_l & -1 & \\frac{3+\\sqrt{13}}{2} & -1-\\sqrt{13} & \\frac{1-\\sqrt{13}}{2} & \\frac{-1-\\sqrt{13}}{2} & \\frac{-1+\\sqrt{13}}{2}\\\\ 479\\end{array}$ 480 481\\ee 482 483\\section{Application to the study of Weil curves} 484 485Let $f=\\sum a_nq^n$ be a newform of weight 2 and level $N$ with 486integer coefficients. They correspond to a strong Weil curve $\\E$ 487of conductor $N$. Unfortunately the coefficients $a_n$ don't give 488too much information on $\\E$ and do not allow us to obtain a 489simple equation for $\\E$. (In \\cite{10} there is a method due to 490Serre that sometimes allows us to get such an equation, but that 491method is not systematic.) Here we give a method that at least 492when $N=p$ is a prime, one can solve the problem. 493", "\\frac{3}{\\sqrt{27x^3}} \\cdot \\frac{\\sqrt{3x}}{\\sqrt{3x}} = \\frac{3\\sqrt{3x}}{\\sqrt{81x^4}} = \\frac{3\\sqrt{3x}}{9x^2} = \\frac{\\sqrt{3x}}{3x^2}$$ Exercise $$\\PageIndex{56}$$ $$\\frac{5}{\\sqrt{10x^5}}$$", "{L}_{1} + 2 {L}_{2}$ $= \\frac{3 \\sqrt{2}}{2} + \\frac{2 \\sqrt{43}}{27} + \\frac{1}{4} {\\sinh}^{-} 1 \\left(2 \\sqrt{2}\\right) + \\frac{1}{2} {\\sinh}^{-} 1 \\left(\\frac{4}{\\sqrt{27}}\\right)$ $\\approx 3.4022$", "=(64)^\\frac{1}{12}$ $\\sqrt [6]3 = 3^\\frac{1}{6} = (3^2)^\\frac {1}{12} = (9)^\\frac {1}{12}$ $\\sqrt[4] 5 =5^{\\frac14}= (5^3)^{ \\frac{1}{12}} =(125)^\\frac{1}{12}$ Q.4: The smallest among the numbers 2250, 3150, 5100, 4200 is a) 2250 b) 3150 c) 5100 d) 4200 Ans : c) 5100 2250 =(25)50 =(32)50 3150 = (33)50 = (27)50 5100 = (52)50 =(25)50 4200 = (44)50 = (256)50 Q.5: Which is greater $\\sqrt[3]2$ or $\\sqrt3$ ? a) $\\sqrt[3]2$ b) $\\sqrt3$ c) Equal d) Can not be compared Ans : $\\sqrt3$ Cube of both the numbers are $(\\sqrt[3]2)^3 =2 \\: \\text {and} \\: (\\sqrt3)^3 = 3 \\sqrt3$ Q.6: The smallest among $\\sqrt[6]{12}, \\sqrt[3]4, \\sqrt[4]5, \\sqrt3$ is a) $\\sqrt[6]{12}$ b) $\\sqrt[3]4$ c) $\\sqrt[4]5$ d) $\\sqrt3$ Ans : c) $\\sqrt[4]5$ LCM of 2,3,4 and 6 is 12 $\\sqrt[6]{12}$ =(12)1/6 =(12)2/12 = (122)1/12 = (144)1/12 $\\sqrt[3]4$ = (256)1/12 $\\sqrt[4]5$ = (125)1/12 $\\sqrt3$ = (729)1/12 Q.7: If X =(0.25)1\\2, Y = (0.4)2, Z=(0.216)1/3, then a) Y>X>Z b) X>Y>z c) Z>X>Y d) X>Z>Y Ans : c) Z>X>Y X=(0.25)1/2 =0.5 Y= (0.4)2 = 0.16 Z = (0.216) = 0.6 ### Simplifying when the Root Values are given Q.8: If $\\sqrt {33} =5.745, \\text {than the value of} \\sqrt {\\frac {3}{11}}$ is approximately. a) 1 b) 0.5223 c) 6.32 d) 2.035 Ans : b) 0.5223 $\\sqrt {\\frac {3}{11}} = \\sqrt {\\frac {3\\times 11}{11 \\times 11}} = \\frac", "very much! i see the way you got the unkown terms on one side, very nice Thanks again, Max 4. Originally Posted by darksupernova Hey there, Getting very stuck on solving these, ive tried using substitution but i dont know how to rearrange it... can you help? $137.9 = x + \\frac{y}{\\sqrt{0.05}}$ and $275.8 = x + \\frac{y}{\\sqrt{0.007}}$ Since 275.8 is double 137.9, then you can go this simpler way: x + y/sqrt(.007) = 2x + 2y/sqrt(.05) 5. Seeing \"x\" alone in both equations, the first thing I would think of is subtracting one equation from the other: $275.8 = x + \\frac{y}{\\sqrt{0.007}}$ $137.9 = x + \\frac{y}{\\sqrt{0.05}}$ $137.9= \\frac{y}{\\sqrt{0.007}}- \\frac{y}{\\sqrt{0.05}}= y\\left(\\frac{1}{\\sqrt{0.007}}- \\frac{1}{\\sqrt{0.05}}\\right)$ $137.9= y\\left(\\frac{\\sqrt{0.05}- \\sqrt{0.007}}{\\sqrt{(0.05)(.007)}}\\right)$ $y= \\frac{137.9\\sqrt{0.00035}}{\\sqrt{0.05}- \\sqrt{0.007}}$", "– – – – (III) Substituting the value of x in equation (II), we get 12+10y93+y2=136$\\frac{\\frac{-12+10y}{9}}{3}+\\frac{y}{2}=\\frac{13}{6}$ 12+10y27+y2=136$\\frac{-12+10y}{27}+\\frac{y}{2}=\\frac{13}{6}$ 24+20y+27y54=136$\\frac{-24+20y+27y}{54}=\\frac{13}{6}$ 47y = 117 + 24 47y = 141 y = 3 – – – – – – – – (IV) Substituting the value of y in equation (III), we get x=12+1039=189=2$x=\\frac{-12+10*3}{9}=\\frac{18}{9}=2$ Hence, x = 2, y = 3 vi) 2x+3y=0$\\sqrt{2}x + \\sqrt{3}y = 0$ – – – – – – – – (I) 3x8y=0$\\sqrt{3}x – \\sqrt{8}y = 0$ – – – – – – – – (II) From equation (I), we get x=3y2$x=\\frac{-\\sqrt{3}y}{\\sqrt{2}}$ – – – – – – – – (III) Substituting the value of x in equation (II), we get 3(3y2)8y=0$\\sqrt{3}\\left ( -\\frac{\\sqrt{3}y}{\\sqrt{2}} \\right )-\\sqrt{8}y=0$ 3y222y=0$-\\frac{3y}{\\sqrt{2}}-2\\sqrt{2}y=0$ y(3222)=0$y\\left ( -\\frac{3}{\\sqrt{2}}-2\\sqrt{2} \\right )=0$ y = 0- – – – – – – – (IV) Substituting the value of y in equation (III), we get x = 0 ∴ x = 0, y = 0 Question 2: Solve 2a + 3b = 11 and 2a – 4b = -24 and calculate the value of m in b = ma + 3. Solution: 2a + 3b = 11 (I) 2a – 4b = -24 (II) From equation (II), we get a=113b2$a=\\frac{11-3b}{2}$ (III) Substituting the value of a in equation (II), we get 2(113b2)4b=24$2\\left(\\frac{11-3b}{2}\\right)-4b=-24$ 11 – 3b – 4b = -24 -7b =", "24 \\setminus \\pm \\sqrt{576 - 12}}{6}$ $=$ $\\setminus \\frac{- 24 \\setminus \\pm \\sqrt{564}}{6}$ Since $564 = 36 \\cdot \\frac{47}{3}$, we can simplfy it out the square root, obtaining $\\setminus \\frac{- 24 \\setminus \\pm 6 \\sqrt{\\frac{47}{3}}}{6}$ and finally we can simplify the whole expression: $\\setminus \\frac{- \\cancel{6} \\cdot 4 \\setminus \\pm \\cancel{6} \\sqrt{\\frac{47}{3}}}{\\cancel{6}}$ into $- 4 \\setminus \\pm \\sqrt{\\frac{47}{3}}$", "# How do you simplify sqrt(1414)? Jun 3, 2017 #### Answer: $\\sqrt{1414}$ is already in simplest form. #### Explanation: The prime factorisation of $1414$ is: $1414 = 2 \\cdot 7 \\cdot 101$ This contains no square factors, so the square root is already in simplest form. $\\textcolor{w h i t e}{}$ Notes Should the $1414$ in the question have been $1444$? If it was, then we would find: $1444 = 2 \\cdot 2 \\cdot 19 \\cdot 19 = {38}^{2}$ So: $\\sqrt{1444} = 38$ We also find that: ${37}^{2} = 1369 < 1414 < 1444 = {38}^{2}$ So $\\sqrt{1414}$ is an irrational number somewhere between $37$ and $38$. If you would like a rational approximation, we can start by linearly interpolating between $37$ and $38$, finding: $\\sqrt{1414} \\approx 37 + \\frac{1414 - 1369}{1444 - 1369} = 37 + \\frac{45}{75} = 37 + \\frac{3}{5} = \\frac{188}{5}$ This will be slightly less than $\\sqrt{1414}$. We find: $\\left(\\frac{188}{5} ^ 2\\right) = \\frac{35344}{25} = \\frac{35350}{25} - \\frac{6}{25} = 1414 - \\frac{6}{25}$ That's not bad, but if we want greater accuracy, we can use a generalised continued fraction based on this approximation. In general we have: $\\sqrt{{a}^{2} + b} = a + \\frac{b}{2 a + \\frac{b}{2 a + \\frac{b}{2 a + \\frac{b}{2 a + \\ldots}}}}$ Putting $a = \\frac{188}{5}$ and $b = \\frac{6}{25}$ we", "= \\frac{23}{\\sqrt{24}}$ and $AE = 10\\sqrt{6}$. Since $\\sin \\theta = \\frac15, GE = \\frac{115}{\\sqrt{24}}$, and $AG = AE - GE = 10\\sqrt{6} - \\frac{115}{\\sqrt{24}} = \\frac{5}{\\sqrt{24}}$. Note that $\\triangle{EB'G} \\sim \\triangle{C'AG}$, so $\\frac{EG}{B'G}=\\frac{C'G}{AG} \\implies C'G = \\frac{25}{\\sqrt{24}}$. Now we have that $AB = AE + EB = 10\\sqrt{6} + 23$, and $B'C' = BC = B'G + C'G = \\frac{23}{\\sqrt{24}} + \\frac{25}{\\sqrt{24}} = \\frac{48}{\\sqrt{24}}=4\\sqrt{6}$. Thus, the area of $ABCD$ is $(10\\sqrt{6} + 23)(4\\sqrt{6}) = 92\\sqrt{6} + 240$, and our final answer is $92 + 6 + 240 = \\boxed{338}$.", "+0 0 35 1 Simplify the following expression to a simplified fraction: $\\frac{\\frac{5}{\\sqrt{80}}}{\\sqrt{5}}.$ Sep 7, 2021 #1 +12224 +1 Simplify the following expression to a simplified fraction:$$\\dfrac{\\dfrac{5}{\\sqrt{80}}}{\\sqrt{5}}$$ Hallo Guest! \\frac{\\frac{5}{\\sqrt{80} }}{\\sqrt{5}} As it is in LaTeX, the denominator of the fraction is $$\\sqrt{5}.$$ So $$\\dfrac{5}{\\sqrt{80}}:\\sqrt{5}=\\dfrac{5}{\\sqrt{5\\cdot 80}}=\\dfrac{\\sqrt{5}\\cdot \\sqrt{5}}{\\sqrt{5\\cdot 80}}=\\sqrt{\\dfrac{5}{80}}=\\sqrt{\\dfrac{1}{16}}=\\color{blue}\\dfrac{1}{4}$$ ! Sep 7, 2021", "# How do you simplify sqrt4/100? $\\frac{\\sqrt{4}}{100} = \\frac{1}{50}$ or $0.02$ $\\sqrt{4} = 2$ $\\frac{\\sqrt{4}}{100} = \\frac{2}{100} = \\frac{1}{50} = 0.02$" ]

[ "# Difference between revisions of \"2005 AMC 8 Problems/Problem 13\" ## Problem The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$? $[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label(\"A\", a, NW); label(\"B\", b, NE); label(\"C\", c, SE); label(\"D\", d, SW); label(\"E\", e, SW); label(\"F\", f, SW); label(\"5\", (0,6.5), W); label(\"8\", (4,9), N); label(\"9\", (8, 4.5), E); [/asy]$ $\\textbf{(A)}\\ 7\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 11$ ## Solution Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\\text{Area} = 52 = 8 \\cdot 9- EF \\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\\boxed{\\textbf{(C)}\\ 9}$.", "(7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw] That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.", "of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ ## Problem 5 $[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label(\"A\",(2.5,-.5),S); label(\"B\",(4.5,-.5),S); label(\"C\",(6.5,-.5),S); label(\"D\",(8.5,-.5),S); label(\"F\",(10.5,-.5),S); label(\"Grade\",(15,-.5),S); label(\"1\",(-1,3),W); label(\"2\",(-1,6),W); label(\"3\",(-1,9),W); label(\"4\",(-1,12),W); label(\"5\",(-1,15),W); [/asy]$ The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory? $\\text{(A)}\\ \\frac{1}{2} \\qquad \\text{(B)}\\ \\frac{2}{3} \\qquad \\text{(C)}\\ \\frac{3}{4} \\qquad \\text{(D)}\\ \\frac{4}{5} \\qquad \\text{(E)}\\ \\frac{9}{10}$ ## Problem 6 A stack of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high? $\\text{(A)}\\ 250 \\qquad \\text{(B)}\\ 550 \\qquad \\text{(C)}\\ 667 \\qquad \\text{(D)}\\ 750 \\qquad \\text{(E)}\\ 1250$ ## Problem 7 A \"stair-step\" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\\text{th}$ row is $[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]$ $\\text{(A)}\\ 34 \\qquad \\text{(B)}\\ 35 \\qquad", "are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face? $[asy] unitsize(2 cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*\"R\", (x + z)/2); label(rotate(-5)*slant(0.5)*\"B\", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*\"G\", ((x + z) + (x + y))/2); label(rotate(-3)*\"W\", (x + z)/2 + trans); label(rotate(50)*slant(-1)*\"B\", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*\"R\", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*\"P\", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*\"R\", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*\"G\", ((x + z) + (x + y))/2 + 2*trans); [/asy]$ $\\textbf{(A) }\\text{red}\\qquad\\textbf{(B) }\\text{white}\\qquad\\textbf{(C) }\\text{green}\\qquad\\textbf{(D) }\\text{brown}\\qquad\\textbf{(E) }\\text{purple}$ ## Problem 13 A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "(A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); /* labels */ label(\"x\",(A+C)/2,(1.2,-1.2)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_3$, or a regular, space, given a point $x$ and a closed set $A$ in a topological space $X$ that are disjoint, there exists open sets $U,V$ such that $x \\in U, A \\subset V$ and $U,V$ are disjoint. An example of a Hausdorff space that is not regular is the space $\\mathbb{R}_k$, the k-topology (or in more generality, a subspace of $\\mathbb{R}$ consisting of $\\mathbb{R}$ missing a countable number of elements). ## Normal $[asy] defaultpen(linewidth(1) + linetype(\"6 6\")); pair A=(0,0),B=(1.6,0),C=(1.6,1),D=(0,1),shiftfactor=(2.5,0.2); /* draw an \"open set\" using Bezier\" */ picture neighborhood(){ picture pic; path p = A{(0.2,-0.6)}..(2*A+B)/3{(0.4,-0.1)}..(A+2*B)/3{(0.4,-0.1)}..B{(0.3,0.7)}..(B+C)/2{(-0.5,0.7)}..C{(-0.1,0.6)}..(2*C+D)/3{(-0.7,0.5)}..(C+2*D)/3{(-0.5,0.5)}..D{(-0.1,-0.5)}..(A+D)/2{(0.6,-0.7)}..cycle; fill(pic,p,rgb(0.9,0.9,0.9)); draw(pic,p); return pic; } path oSC = (A+C)/2+(-0.25,-0.15)--(A+C)/2+(-0.25,0.15)--(A+C)/2+(0.25,0.15)--(A+C)/2+(0.25,-0.15)--cycle; pair SC = (A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); fill(oSC,rgb(0.7,0.7,0.7)); draw(oSC,linewidth(1)+linetype(\"6 4\")); label(\"A\",(A+C)/2,(2,-1)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_4$, or a normal, space, given any two disjoint closed sets $A,B$ in a topological space $X$, there exists open sets $U,V$ such that $A \\subset U, B \\subset V$ and $U,V$ are disjoint. An example of a regular space that is not normal is the Sorgenfrey plane.", "can be tiled without restriction. $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); filldraw(shift((3,0))*unitsquare,mediumgray); filldraw(shift((2,1))*unitrect,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(-1,1)); filldraw(shift((11,1))*unitsquare,mediumgray); filldraw(shift((10,0))*unitrect,mediumgray); filldraw(shift((9,1))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(8-1,1)); filldraw(shift((19,1))*unitsquare,mediumgray); filldraw(shift((18,1))*unitsquare,mediumgray); filldraw(shift((18,0))*unitrect,mediumgray); draw(shift((16,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(16-1,1)); [/asy]$ In the first, second, and third cases, we see that we can tile the rest of the board in $s_{n-1}$, $s_{n-2}$, and $t_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that $$t_n=s_{n-1}+s_{n-2}+t_{n-2}\\tag{3}.$$ For $u_n$, note that a tiling can end in one of the two following ways such that the rest of the board can be tiled without restriction. $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); filldraw(shift((3,1))*unitsquare,mediumgray); filldraw(shift((2,0))*unitrect,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(-1,1)); filldraw(shift((11,0))*unitsquare,mediumgray); filldraw(shift((10,1))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(8-1,1)); [/asy]$ In either case, we are left with a board with a corner removed, hence we can tile the rest of the board in $t_{n-1}$ ways in each case. Hence for $n\\ge 4$, we see that $$u_n=2t_{n-1}\\tag{4}.$$ Subtracting (3) from (2), we find that $$s_n-t_n=s_{n-1}-t_{n-1}.$$ Therefore, if $s_{n-1}=t_{n-1}$, then $s_n=t_n$. Since $s_3=t_3$ and $s_4=t_4$, we see that $s_n=t_n$ for all $n\\ge 3$. Therefore, (2) can be rewritten as $$s_n=s_{n-1}+2s_{n-2}\\tag{5},$$ and (4) can be rewritten as $$u_n=2s_{n-1}\\tag{6}.$$ Now by (1), we know that $$a_{n}+a_{n-1}=u_n+s_n+u_{n-1}+s_{n-1}.\\tag{7}$$ In particular, $a_4+a_3=6+5+2+3=2^4$, so the statement is true for $n=4$. Then by (7), and then substituting (5) and (6) (where these are valid for $n\\ge 5$), we find \\begin{align*} a_n+a_{n-1}&=u_n+s_n+u_{n-1}+s_{n-1}\\\\ &=2s_{n-1}+(s_{n-1}+2s_{n-2})+u_{n-1}+s_{n-1}\\\\ &=4s_{n-1}+2s_{n-2}+u_{n-1}.\\tag{8} \\end{align*} But then by (5), and then", "Difference between revisions of \"1985 AJHSME Problems/Problem 4\" Problem The area of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ Solution Solution 1 $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"G\",(6,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of. If we continue segment $\\overline{FE}$ until it reaches the right side at $G$, we create two rectangles - one on the top and one on the bottom. We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\\times5 = 30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other. Note that $GC+GB=9$, and $GB=AF=5$, so we must have $$GC+5=9\\Rightarrow GC=4$$ The area of the bottom rectangle is then $$(DC)(GC)=4\\times 4=16$$ Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$. $\\boxed{\\text{C}}$ Solution 2 $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE);", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "2020 AMC 8 Problems/Problem 22 Problem When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below. $[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(\"N\",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(\"if N is even\",(9.25,1.25),N); label(\"if N is odd\",(9.25,-1.25),N); label(\"\\frac N2\",(12,1.25)); label(\"3N+1\",(12.6,-1.25)); [/asy]$ For example, starting with an input of $N=7,$ the machine will output $3 \\cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$$$7 \\to 22 \\to 11 \\to 34 \\to 17 \\to 52 \\to 26$$When the same $6$-step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$$$N \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to 1$$ $\\textbf{(A) }73 \\qquad \\textbf{(B) }74 \\qquad \\textbf{(C) }75 \\qquad \\textbf{(D) }82 \\qquad \\textbf{(E) }83$ Solution 1 We start with the final output of $1$ and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage: $$\\{1\\}\\leftarrow\\{2\\}\\leftarrow\\{4\\}\\leftarrow\\{1,8\\}\\leftarrow\\{2,16\\}\\leftarrow\\{4,5,32\\}\\leftarrow\\{1,8,10,64\\}$$ where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come", "cm. A rock with volume $1000 \\text{ cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise? $\\textbf{(A)}\\ 0.25\\qquad\\textbf{(B)}\\ 0.5\\qquad\\textbf{(C)}\\ 1\\qquad\\textbf{(D)}\\ 1.25\\qquad\\textbf{(E)}\\ 2.5$ Problem 22 Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell? $[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label(\"+\", (1.5,1.5)); label(\"+\", (3.5,1.5)); label(\"+\", (2.5,4.5));[/asy]$ $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 24\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 35$ Problem 23 A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? $\\textbf{(A)}\\ 0\\qquad\\textbf{(B)}\\ 1\\qquad\\textbf{(C)}\\ 2\\qquad\\textbf{(D)}\\ 3\\qquad\\textbf{(E)}\\ 5$ Problem 24 In the multiplication problem below $A$, $B$, $C$, $D$ and are different digits.", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "adjacent side vectors $\\overrightarrow{(a,b)}, \\overrightarrow{(c,d)}$ is given by $\\overrightarrow{(a,b)} \\times \\overrightarrow{(c,d)} = ad-bc$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) { return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); } /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i) dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype(\"2 2\")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label(\"R\",shiftL+(r/2,0),S); /* right diagram */ for(int i = 0; i < 9; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); if (i % 3 == 1) { filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); } } for(int i = 9; i < 12; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype(\"2 2\")); } [/asy]$ The area of a dodecagon is $3R^2$, where $R$ is the circumradius. $[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype(\"2 4\"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP(\"(a,b)\",A,W); MP(\"(c,d)\",B,E); MP(\"(c,-d)\",B2,E); [/asy]$ The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given", "figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ /* dots and labels */ label(\"f\",(-6.12,-0.12),NE * labelscalefactor); label(\"a = 2\",(1.46,0.24),NE * labelscalefactor,zzttqq); dot((-3.14,0),dotstyle); label(\"A\",(-3.06,0.12),NE * labelscalefactor); dot((2.2,2.04),dotstyle); dot((4.5,2.08),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ Here is the Asymptote code produced: (with some spacing added) /* Geogebra to Asymptote conversion, documentation at userscripts.org/scripts/show/72997 */ import graph; size(14.26cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.22,xmax = 8.04,ymin = -4.18,ymax = 4.76; /* image dimensions */ pen fueaev = rgb(0.96,0.92,0.9); pen zzttqq = rgb(0.6,0.2,0); real f2 (real x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); filldraw((-3.14,0)--(0,0)--(0,3.14)--(-3.14,pi )--cycle, fueaev, zzttqq); Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "what he had eaten. How many more dollars did Dave pay than Doug? $\\mathrm{(A) \\ } 1\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 3\\qquad \\mathrm{(D) \\ } 4\\qquad \\mathrm{(E) \\ } 5$ ## Problem 6 The $8\\times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$? $[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label(\"A\",(0,4),NW); label(\"B\",(18,4),NE); label(\"C\",(18,-4),SE); label(\"D\",(0,-4),SW); label(\"y\",(3,4),S); label(\"y\",(15,-4),N); label(\"18\",(9,4),N); label(\"18\",(9,-4),S); label(\"8\",(0,0),W); label(\"8\",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$ $\\mathrm{(A) \\ } 6\\qquad \\mathrm{(B) \\ } 7\\qquad \\mathrm{(C) \\ } 8\\qquad \\mathrm{(D) \\ } 9\\qquad \\mathrm{(E) \\ } 10$ ## Problem 7 Mary is $20\\%$ older than Sally, and Sally is $40\\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday? $\\mathrm{(A) \\ } 7\\qquad \\mathrm{(B) \\ } 8\\qquad \\mathrm{(C) \\ } 9\\qquad \\mathrm{(D) \\ } 10\\qquad \\mathrm{(E) \\ } 11$ ## Problem 8 How many sets of two or more consecutive positive integers have a sum of $15$? $\\mathrm{(A) \\ } 1\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 3\\qquad \\mathrm{(D) \\ } 4\\qquad \\mathrm{(E) \\ } 5$ ## Problem 9 Oscar buys $13$ pencils and $3$ erasers for $\\textdollar 1.00$. A pencil costs more than", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "we see that $$a_n=u_n+s_n\\tag{1}.$$ For the sake of convenience in determining recurrence relations, we define another type of board with two $1\\times n$ boards where a specific corner is removed (without loss of generality, we place this in the lower left hand corner). Let $t_n$ be the number of ways to tile such a board without placeing two of the same type of tile atop each other. Once again, we compute that $t_3=3$ and $t_4=5$ as shown below. $[asy] unitsize(10); draw((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1),linewidth(2)); draw((1,1)--(1,2)^^(2,0)--(2,1)); draw(shift((0,-2.5))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((0,-2.5))*((1,1)--(1,2)^^(2,1)--(2,2))); draw(shift((4,0))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((4,0))*((2,1)--(2,2))); draw(shift((11,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,0))*((1,1)--(1,2)^^(2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((11,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,-2.5))*((1,1)--(1,2)^^(2,0)--(2,1)^^(3,1)--(3,2))); draw(shift((16,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,2)^^(3,1)--(3,2))); draw(shift((16,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,-2.5))*((2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((21,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((21,0))*((2,0)--(2,2)^^(3,0)--(3,1))); [/asy]$ We can determine reccurence relations for $s_n$, $t_n$, and $u_n$ in terms of each other. For $s_n$, note that a tiling can end in one of the three following ways such that the rest of the board can be tiled without restriction (the placed tiles are shaded). $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); fill(shift((3,0))*unitsquare,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((0,0))*((3,0)--(3,1))); label(\"\\dots\",(-1,1)); fill(shift((10,1))*unitsquare,mediumgray); fill(shift((10,0))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((8,0))*((2,0)--(2,2))); label(\"\\dots\",(8-1,1)); fill(shift((18,0))*unitrect,mediumgray); fill(shift((17,1))*unitrect,mediumgray); draw(shift((16,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,1)^^(1,1)--(1,2))); label(\"\\dots\",(16-1,1)); [/asy]$ In the first, second, and third cases, we see that we can tile the rest of the board in $t_{n-1}$, $t_{n-2}$, and $s_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that $$s_n=t_{n-1}+t_{n-2}+s_{n-2}\\tag{2}.$$ For $t_n$, note that a tiling can end in one of the three following ways such that the rest of the board", "reduce the number of tiles in the set to one? $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 11 \\qquad \\text{(C)}\\ 18 \\qquad \\text{(D)}\\ 19 \\qquad \\text{(E)}\\ 20$ Problem 24 Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina? $\\text{(A)}\\ 2/5 \\qquad \\text{(B)}\\ 9/20 \\qquad \\text{(C)}\\ 1/2 \\qquad \\text{(D)}\\ 11/20 \\qquad \\text{(E)}\\ 24/25$ Problem 25 $[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,N); label(\"52\",(A+B)/2,S); label(\"39\",(C+D)/2,N); label(\"12\",(B+C)/2,E); label(\"5\",(D+A)/2,W); [/asy]$ In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$ (diagram not to scale). The area of $ABCD$ is $\\text{(A)}\\ 182 \\qquad \\text{(B)}\\ 195 \\qquad \\text{(C)}\\ 210 \\qquad \\text{(D)}\\ 234 \\qquad \\text{(E)}\\ 260$" ]

[ "# Difference between revisions of \"2005 AMC 8 Problems/Problem 13\" ## Problem The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$? $[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label(\"A\", a, NW); label(\"B\", b, NE); label(\"C\", c, SE); label(\"D\", d, SW); label(\"E\", e, SW); label(\"F\", f, SW); label(\"5\", (0,6.5), W); label(\"8\", (4,9), N); label(\"9\", (8, 4.5), E); [/asy]$ $\\textbf{(A)}\\ 7\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 11$ ## Solution Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\\text{Area} = 52 = 8 \\cdot 9- EF \\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\\boxed{\\textbf{(C)}\\ 9}$.", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < 2; ++i) /* left */ fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */ fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label(\"\\frac 12\",(s*3/4,s/2),sm); label(\"\\cdots\",(s*1/4,s/2),sm); label(\"\\frac 12\",shiftR+(s*3/4,s/2),sm); label(\"\\cdots\",shiftR+(s*1/4,s/2),sm); label(\"\\frac 14\",shiftR+(s*1/4,s*3/4),sm); label(\"\\frac 18\",shiftR+(s*3/8,s/4),sm); htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label(\"1\",(s/2,-1),S,sm); label(\"1\",shiftR+(s/2,-1),S,sm); [/asy]$ The infinite geometric series $\\frac 12 + \\frac {1}{2^2} + \\frac {1}{2^3} + \\cdots = 1$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){ for(int i = 0; i < 4; ++i){ draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype(\"2 2\")); draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype(\"2 2\")); } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); } for(int i = 0; i < n; ++i) { fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]); fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]); fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]); fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]); drawTriGrid(h/3^i); } [/asy]$ The infinite geometric series $\\frac 13 + \\frac {1}{3^2} + \\frac {1}{3^3} + \\cdots = \\frac 12$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0); void drawEquilaterals(pair A, real s){ filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]); filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]); filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); } for(int i = 0; i < n; ++i) drawEquilaterals(shiftR + (0,h-h/(2^i)", "cycle); /* manual arrowhead? avoid resizing */ /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype(\"4 4\")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < n+1; ++j) filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i)); /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label(\"n\",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label(\"1\",shiftR+(n,-n),S,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is $n(n+1)/2$. $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; real r = 0.35, h = 3/4; /* radius size and horizontal spacing */ pair shiftR = (h*(n+1)+r, 0); pen colors(int i){ /* shading */ if(i == n) return red; return rgb(5/n,0.4+5/(2n),1-5/n); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } pair getCenter(int i, int j){ return ((2*j-i)*h,-i);} /* triangle */ for(int i = 0; i < n+1; ++i){ draw((-i*h,-i)--(i*h,-i)); /* horizontal lining */ for(int j = 0; j <= i; ++j) filldraw(circle(getCenter(i,j),r), colors(i)); } /* fill in circle in row 4, column 3 */ filldraw(circle(getCenter(3,2),r),blue); draw(getCenter(n,2)-- getCenter(3,2)-- getCenter(n,n+2-3)); makeshiftarrow(getCenter(n,2),pi/4,0.5); makeshiftarrow(getCenter(n,n+2-3),3*pi/4,0.5); htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label(\"n\",shiftR+(-1,(-n+1)/2),E,fontsize(10)); [/asy]$ The sum of the first $n$ positive integers is ${n+1 \\choose 2}$.[1] $[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35;", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)", "[/asy]$ $\\textbf{(D)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,0)*cat); dot((0,1.5)); [/asy]$ $\\textbf{(E)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,1)*cat); dot((1.5,0)); [/asy]$ ## Problem 24 A ship travels from point A to point B along a semicircular path, centered at Island X. Then it travels along a straight path from B to C. Which of these graphs best shows the ship's distance from Island X as it moves along its course? $[asy]size(150); pair X=origin, A=(-5,0), B=(5,0), C=(0,5); draw(Arc(X, 5, 180, 360)^^B--C); dot(X); label(\"X\", X, NE); label(\"C\", C, N); label(\"B\", B, E); label(\"A\", A, W);[/asy]$ $\\textbf{(A)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((4,10), 4, 0, 180)^^(8,10)--(16,12)); [/asy]$ $\\textbf{(B)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((12,10), 4, 180, 360)^^(0,10)--(8,10)); [/asy]$ $\\textbf{(C)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw((0,8)--(10,10)--(16,8)); [/asy]$ $\\textbf{(D)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((12,10), 4, 0, 180)^^(0,10)--(8,10)); [/asy]$ $\\textbf{(E)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw((0,6)--(6,6)--(16,10)); [/asy]$ ## Problem 25 In the figure, the area of square WXYZ is $25 \\text{cm}^2$. The four smaller squares have sides 1 cm long,", "+ dashed); } draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6)); draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6)); label(\"time\", (5,-0.5) + i*transx + j*transy); label(rotate(90)*\"distance\", (-0.5,2.5) + i*transx + j*transy); } }} draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp)); draw((0,0)--(10,5),linewidth(1.5*bp)); draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp)); draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp)); draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp)); draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp)); draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp)); draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp)); draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp)); draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp)); label(\"(A)\", (-1,6)); label(\"(B)\", (-1,6) + transx); label(\"(C)\", (-1,6) + 2*transx); label(\"(D)\", (-1,6) + transy); label(\"(E)\", (-1,6) + transx + transy); [/asy]$ ## Problem 6 There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other $80$ points, what is the probability that the line $PQ$ is a line of symmetry for the square? $[asy] draw((0,0)--(0,8)); draw((0,8)--(8,8)); draw((8,8)--(8,0)); draw((8,0)--(0,0)); dot((0,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((0,6)); dot((0,7)); dot((0,8)); dot((1,0)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((1,6)); dot((1,7)); dot((1,8)); dot((2,0)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((2,6)); dot((2,7)); dot((2,8)); dot((3,0)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((3,6)); dot((3,7)); dot((3,8)); dot((4,0)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); dot((4,6)); dot((4,7)); dot((4,8)); dot((5,0)); dot((5,1)); dot((5,2)); dot((5,3)); dot((5,4)); dot((5,5)); dot((5,6)); dot((5,7)); dot((5,8)); dot((6,0)); dot((6,1)); dot((6,2)); dot((6,3)); dot((6,4)); dot((6,5)); dot((6,6)); dot((6,7)); dot((6,8)); dot((7,0)); dot((7,1)); dot((7,2)); dot((7,3)); dot((7,4)); dot((7,5)); dot((7,6)); dot((7,7)); dot((7,8)); dot((8,0)); dot((8,1)); dot((8,2)); dot((8,3)); dot((8,4)); dot((8,5)); dot((8,6)); dot((8,7)); dot((8,8)); label(\"P\",(4,4),NE); [/asy]$ $\\textbf{(A) }\\frac{1}{5}\\qquad\\textbf{(B) }\\frac{1}{4} \\qquad\\textbf{(C) }\\frac{2}{5} \\qquad\\textbf{(D) }\\frac{9}{20} \\qquad\\textbf{(E) }\\frac{1}{2}$ ## Problem", "such a problem does not occur with simple line pattern. Here a workaround. It is sufficient to define a new pen like dotted pen but with adding an offset : pen mdotted=linetype(new real[] {0,4},offset=.5); import math; import patterns; unitsize(1cm); defaultpen(1); pair C1=(-3,0),C2=(2.5,0); real r=4; path EllipseA=scale(.75,.55)*circle(C1,r); path EllipseB=scale(.5,.35)*circle(C2,r); pen mdotted=linetype(new real[] {0,4},offset=.5); fill(EllipseA,pattern(\"fillA\")); fill(EllipseB,pattern(\"fillB\")); picture AinterB; fill(AinterB,EllipseA,white); fill(AinterB,EllipseA,pattern(\"AinterB\")); clip(AinterB,EllipseB); draw(EllipseA^^EllipseB); label(\"$C$\",C1,dir(180),UnFill); label(\"$D$\",C2,UnFill); label(\"$E$\",(0,0),UnFill); pair A=point(EllipseA,1.25),B=point(EllipseB,.5); label(\"$A$\",A,6N); label(\"$B$\",shift(.2,0)*B,6N); label(\"$G$\",(0,2),9N); // dot((0,1)); draw(A--shift(0,.6)*A); draw(B--shift(.2,.7)*B); draw(shift(-2.5,.75)*(0,1)--shift(-.5,2)*(0,1)); draw(shift(.2,0)*(0,1)--shift(0,2)*(0,1)); draw(shift(1.5,0)*(0,1)--shift(.5,2)*(0,1)); shipout(bbox(5mm,white)); and the result • While waiting for ..., your answer is accepted. Asymptote is my main problem. Thanks. – user197952 Dec 12 '19 at 12:55 • @justonly : thanks for finding this bug – O.G. Dec 12 '19 at 14:42 To use a dotted pen in a PostScript fill pattern, one needs to disable Asymptote's automatic scaling to the pen width and its line length adjustment: import patterns; unitsize(1cm); It looks like a problem with your viewer. Do you view the dvi output? • Yes, it looks like a problem with my TexStudio. I run it by TexStudio (latex-dvips-ps2pdf -dNOSAFER <name>.ps <name>.pdf) and the output of it as displayed. However, the pdf file is no problem. Thanks. – user197952 Dec 10 '19 at 3:19 • For Windows you should use Sumatra PDF for the external viewer. It is much more faster than", "# Difference between revisions of \"1994 AHSME Problems/Problem 2\" ## Problem A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? $[asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label(\"6\", (1.5,6)); label(\"?\", (1.5,2.5)); label(\"14\", (6.5,6)); label(\"35\", (6.5,2.5)); [/asy]$ $\\textbf{(A)}\\ 10 \\qquad\\textbf{(B)}\\ 15 \\qquad\\textbf{(C)}\\ 20 \\qquad\\textbf{(D)}\\ 21 \\qquad\\textbf{(E)}\\ 25$ ## Solution [asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label(\"7\",position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red); draw(BF,L=Label(\"5\",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(FC,L=Label(\"2\",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(DH,L=Label(\"3\",position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red); label(\"6\", (1.5,6)); label(\"15\", (1.5,2.5),blue); label(\"14\", (6.5,6)); label(\"35\", (6.5,2.5)); [/asy] We can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\\times 5=\\boxed{\\textbf{(B) }15}$. ## Solution 2 [asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path CH=C--H; path BF=B--F; path FC=F--C; path DH=D--H; draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(CH,L=Label(\"b\",position=MidPoint,align=(0,1))); draw(BF,L=Label(\"y\",position=MidPoint,align=(1,0))); draw(FC,L=Label(\"x\",position=MidPoint,align=(1,0))); draw(DH,L=Label(\"a\",position=MidPoint,align=(0,1))); label(\"6\", (1.5,6)); label(\"14\", (6.5,6)); label(\"35\", (6.5,2.5)); [/asy] In case its not immediately obvious from inspection what the dimensions of the small rectangles are, we can work it out. We know $ax$ and $bx$ and $by$ and we want to know $ay$. We can compute it as follows $ay = \\frac{(ax)(by)}{bx} = \\frac{ 6\\cdot 35 }{14} = 15$ and the answer is $\\fbox{B}$", "parallelogram draw(pic,T); draw(pic,(Bi+Di)/2--shift(2.5*dir(125))*(((Di+(D-A)/n)+(Bi+(D-A)/n))/2),EndArrow); arrow from it to its copy draw(pic,shift(2.5*dir(125))*T); its copy on top label(pic,\"Area$($\",shift(2.5*dir(125))*(Bi+(D-A)/2/n),W); text to the left of the copy of left parallelogram label(pic,\"$)=$\",shift(2.5*dir(125))*(Di+(D-A)/2/n),E); text to the right of the copy of left parallelogram T=Di--Ci--(Ci+(D-A)/n)--(Di+(D-A)/n)--cycle; right parallelogram draw(pic,T); draw(pic,(Ci+Di)/2--shift(1.7*dir(100))*shift(1.5*E)*(((Di+(D-A)/n)+(Ci+(D-A)/n))/2),EndArrow); arrow from it to its copy on the top draw(pic,shift(1.7*dir(100))*shift(1.5*E)*T); its copy above label(pic,\"$=\\frac{BD}{DC}\\cdot$Area$($\",shift(1.7*dir(100))*shift(1.5*E)*(Di+(D-A)/2/n),W); text to the left of the copy label(pic,\"$)$\",shift(1.7*dir(100))*shift(1.5*E)*(Ci+(D-A)/2/n),E); text to the right of the copy T=Ci--(Ci+(C-A)/n)--(Ci+(D-A)/n)--cycle; small triangle to the right draw(pic,T); draw(pic,(Ci+(C-A)/2/n+(D-A)/2/n)--shift(1.7*dir(-45))*shift(Bi-Ci)*(Ci+(C-A)/2/n),EndArrow); draw(pic,shift(1.7*dir(-45))*shift(Bi-Ci)*T); its copy on the bottom label(pic,scale(0.8)*minipage(\"Error of order \\smallskip $\\frac{1}{n^2}\\cdot$Area$(ABC)$\",2.5cm),shift(1.7*dir(-45))*shift(Bi-Ci)*(Ci+(C-A)/2/n+(D-A)/2/n),S); //text below its copy on the bottom Bi=(3*A+(n-3)*B)/n; Ci=(3*A+(n-3)*C)/n; Di=(3*A+(n-3)*D)/n; label(\"$B_{i}$\",Bi,WNW); label(\"$D_{i}$\",Di,dir(32)); label(\"$C_{i}$\",Ci,ENE); label(pic,\"$B_{i}$\",Bi,W); label(pic,\"$D_{i}$\",Di,S); label(pic,\"$C_{i}$\",Ci,E); \\end{asy} \\label{fig:ratios} \\end{figure} By summing up $n$ such relations we get that $\\area(BDA)=\\frac{BD}{DC}\\cdot\\area(DCA)$ up to an error of order $n\\cdot\\frac{1}{n^2}\\cdot \\area(ABC)=\\frac{1}{n}\\cdot\\area(ABC)$. Hence, as $n$ grows arbitrarily large the equality $\\area(BDA)=\\frac{BD}{DC}\\cdot\\area(DCA)$ becomes arbitrarily precise: It must hold exactly. \\end{proof} \\begin{remark} We could of course prove this lemma by using the formula that expresses the area of a triangle in terms of its base and height, noticing that the triangles $BDA$ and $DCA$ share a common height from vertex $A$; and indeed such a proof looks simpler. However we wanted our arguments to rely only on ratios of lengths and ratios of areas. These notions are intrinsic to the affine geometry of the plane, whereas", "sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\"a\",a/2*expi(rot1),SE,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\"c\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\"2 2\")); draw(I--E, linetype(\"2 2\")); draw(I--F, linetype(\"2 2\")); label(\"a\",(A+B)/2,S); label(\"b\",(B+C)/2,NE); label(\"c\",(A+C)/2,NW); label(\"r\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "(A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); /* labels */ label(\"x\",(A+C)/2,(1.2,-1.2)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_3$, or a regular, space, given a point $x$ and a closed set $A$ in a topological space $X$ that are disjoint, there exists open sets $U,V$ such that $x \\in U, A \\subset V$ and $U,V$ are disjoint. An example of a Hausdorff space that is not regular is the space $\\mathbb{R}_k$, the k-topology (or in more generality, a subspace of $\\mathbb{R}$ consisting of $\\mathbb{R}$ missing a countable number of elements). ## Normal $[asy] defaultpen(linewidth(1) + linetype(\"6 6\")); pair A=(0,0),B=(1.6,0),C=(1.6,1),D=(0,1),shiftfactor=(2.5,0.2); /* draw an \"open set\" using Bezier\" */ picture neighborhood(){ picture pic; path p = A{(0.2,-0.6)}..(2*A+B)/3{(0.4,-0.1)}..(A+2*B)/3{(0.4,-0.1)}..B{(0.3,0.7)}..(B+C)/2{(-0.5,0.7)}..C{(-0.1,0.6)}..(2*C+D)/3{(-0.7,0.5)}..(C+2*D)/3{(-0.5,0.5)}..D{(-0.1,-0.5)}..(A+D)/2{(0.6,-0.7)}..cycle; fill(pic,p,rgb(0.9,0.9,0.9)); draw(pic,p); return pic; } path oSC = (A+C)/2+(-0.25,-0.15)--(A+C)/2+(-0.25,0.15)--(A+C)/2+(0.25,0.15)--(A+C)/2+(0.25,-0.15)--cycle; pair SC = (A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); fill(oSC,rgb(0.7,0.7,0.7)); draw(oSC,linewidth(1)+linetype(\"6 4\")); label(\"A\",(A+C)/2,(2,-1)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_4$, or a normal, space, given any two disjoint closed sets $A,B$ in a topological space $X$, there exists open sets $U,V$ such that $A \\subset U, B \\subset V$ and $U,V$ are disjoint. An example of a regular space that is not normal is the Sorgenfrey plane.", "n = floor(n2*(n2+1)/2); // number of colors, number of layers real h = 0.6; // scale factor of diagram pair shiftR1 = (n*h+1,0), // middle diagram shift offset shiftR2 = shiftR1 + (n*h+1,0); // right diagram shift offset int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */ /* draw tick line with label, segment between A and B */ void htick(pair A, pair B,pair ticklength = (0.15,0)) { draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } /* gradient triangle */ for(int i = 0; i < n; ++i){ for(int j = 0; j < 2*i+1; ++j){ filldraw(shift(shiftR1)*scale(h)*shift((j-i,-i))*unitsquare,colors(i)); /* if(j % lvl(i) == 0 && j != lvl(i)^2) draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((0,0)--(0,1)--(1,1)), heavy); if(j == 2*i) // right border draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((1,0)--(1,1)--(0,1)), heavy); */ } draw(shift(shiftR1)*scale(h)*shift((-i,-i))*((0,0)--(0,1)--(1,1)), heavy); draw(shift(shiftR1)*scale(h)*shift(( i,-i))*((1,0)--(1,1)--(0,1)), heavy); } // return kth triangular number (actually, 1+2+...+k) int tri(int k) { return ((int) (k*(k+1)/2)); } for(int i = 0; i < n2; ++i) { draw(shift(shiftR1)*scale(h)*shift((0-tri(i),0-tri(i)))*((0,1)--(2*tri(i)+1,1)), heavy); if(i % 2 == 0) { // vertical heavy lines for odd layers draw(shift(shiftR1)*scale(h)*((-i/2,1-tri(i))--(-i/2,-i-tri(i))), heavy); draw(shift(shiftR1)*scale(h)*((1+i/2,1-tri(i))--(1+i/2,-i-tri(i))), heavy); } else { // jagged heavy lines for even layers pair jag1 = (-(i+1)/2,-(i-1)/2-tri(i)), jag2 = (1+(i+1)/2,-(i-1)/2-tri(i)); draw(shift(shiftR1)*scale(h)*(jag1+(0,1+(i-1)/2) -- jag1 -- jag1+(1,0) -- jag1+( 1,-(i+1)/2)), heavy); draw(shift(shiftR1)*scale(h)*(jag2+(0,1+(i-1)/2) -- jag2 -- jag2-(1,0) -- jag2+(-1,-(i+1)/2)), heavy); } } draw(shift(shiftR1)*scale(h)*shift((-n2*(n2+1)/2,-n2*(n2+1)/2))*((1,1)--(2*n2*(n2+1)/2,1)),", "filldraw(shift(shiftR+B)*(A--reflectI--reflectC--cycle), c3, heavy); filldraw(shift(shiftR)*(I--B--B+reflectI--cycle), c2, heavy); // dashed perpendiculars; arbitrary coding here draw(shift(shiftR)*(I--F), linetype(\"2 2\")); draw(shift(shiftR)*(B--B+(0,inradius(A,B,C))), linetype(\"2 2\")); draw(shift(shiftR+B)*(reflectI--foot(reflectI,A,B)), linetype(\"2 2\")); draw(shift(shiftR)*rightanglemark(I,F,A)); draw(shift(shiftR)*rightanglemark(B,B+(0,inradius(A,B,C)),I)); draw(shift(shiftR+B)*rightanglemark(reflectI,foot(reflectI,A,B),A)); label(\"a\",shiftR+(A+B)/2,S); label(\"b\",shiftR+(I+B+reflectI)/2,N); label(\"c\",shiftR+B+(A+reflectC)/2,S); label(\"r\",shiftR+(I+F)/2,W); [/asy]$ The area of a triangle is given by $A = \\frac{1}{2} \\cdot r \\cdot (a+b+c) = rs$, where $r$ is the inradius and $s$ is the semiperimeter.[10] (Comment: we do not need to re-arrange the triangles to a trapezoid to see this, but this re-arrangement works due to alternate interior angles/angle bisector properties of the incenter.) $[asy]unitsize(15); defaultpen(linewidth(0.7) + fontsize(10)); pen heavy = linewidth(1); real a = 4.5, b = 2.5, c = 2, d = 4; pair A = (a,b), B = (a+c,b+d), C = (c,d), D = IP(B--B+2*(A-B), (0,0)--(a,0)), F = IP(B--B+2*(C-B), (0,0)--(0,d)), G = IP(B--D,(0,d)--(a+c,d)), H = IP(B--F,(a,0)--(a,b+d)), shiftR = (a+c+1,0), shiftR2 = 2*shiftR; // left diagram filldraw((0,0)--(a,0)--(a,d)--(0,d)--cycle, rgb(0.5, 1, 0.5)); draw((0,0)--A--B--C--cycle); draw(shift((a,d))*xscale(c)*yscale(b)*unitsquare); draw(A--D ^^ C--F, linetype(\"2 2\")); draw((0,0)--(a,0)--(a,b)--cycle, heavy); draw(shift(C)*((0,0)--(a,0)--(a,b)--cycle), heavy); label(\"a\",(a/2,0),S); label(\"d\",(0,d/2),W); label(\"b\",B-(0,b/2),E); label(\"c\",B-(c/2,0),N); label(\"(a,b)\",(a,b),SE,fontsize(8)); label(\"(c,d)\",(c,d),NW,fontsize(8)); // middle diagram filldraw(shift(shiftR)*((0,0)--A--(a,d)--G--B--C--(0,d)--cycle), rgb(0.5, 1, 0.5)); filldraw(shift(shiftR)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR+(a,d))*xscale(c)*yscale(b)*unitsquare); draw(shift(shiftR)*(A--D ^^ C--F), linetype(\"2 2\")); draw(shift(shiftR)*((0,0)--(0,d)--(c,d)--cycle), heavy); draw(shift(shiftR+A)*((0,0)--(0,d)--(c,d)--cycle), heavy); label(\"a\",shiftR+(a/2,0),S); label(\"d\",shiftR+(0,d/2),W); label(\"b\",shiftR+B-(0,b/2),E); label(\"c\",shiftR+B-(c/2,0),N); // right diagram filldraw(shift(shiftR2)*((0,0)--A--G--(a,d)--H--C--cycle), rgb(0.5, 1, 0.5)); filldraw(shift(shiftR2)*(G--(a,d)--H--B--cycle), rgb(0.3, 0.9, 0.3)); filldraw(shift(shiftR2)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); filldraw(shift(shiftR2)*(H--(a,b+d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR2)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR2)*(A--D ^^ C--F), linetype(\"2 2\")); label(\"a\",shiftR2+(a/2,0),S); label(\"d\",shiftR2+(0,d/2),W); label(\"b\",shiftR2+B-(0,b/2),E); label(\"c\",shiftR2+B-(c/2,0),N); label(\"\\overrightarrow{(a,b)}\",shiftR2+(a,b),SE,fontsize(8)); label(\"\\overrightarrow{(c,d)}\",shiftR2+(c,d),NW,fontsize(8)); [/asy]$ The area of a parallelogram with", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$ ## Solution 1 (No trig) Let's redraw the diagram, but extend some helpful lines. $[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label(\"A\", A, SW); B=(6,15); label(\"B\", B, NW); C=(30,15); label(\"C\", C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"O\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label(\"3\", midpoint(A--P), S); label(\"9\", midpoint(P--Q), S); label(\"16\", midpoint(Q--C), S); label(\"x\", (5.5,13.75), W); label(\"20\", (20.25,15), N); label(\"6\", (5.25,0), S); label(\"6\", (1.5,3.75), W); label(\"x\", (8.25,15),N); label(\"14+x\", (17.25,0), S); label(\"6-x\", (27,15), N); label(\"6+x\", (27,7.5), W); label(\"6\\sqrt{3}\", (30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and" ]

[ "# Difference between revisions of \"2005 AMC 8 Problems/Problem 13\" ## Problem The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$? $[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label(\"A\", a, NW); label(\"B\", b, NE); label(\"C\", c, SE); label(\"D\", d, SW); label(\"E\", e, SW); label(\"F\", f, SW); label(\"5\", (0,6.5), W); label(\"8\", (4,9), N); label(\"9\", (8, 4.5), E); [/asy]$ $\\textbf{(A)}\\ 7\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 11$ ## Solution Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\\text{Area} = 52 = 8 \\cdot 9- EF \\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\\boxed{\\textbf{(C)}\\ 9}$.", "of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ ## Problem 5 $[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label(\"A\",(2.5,-.5),S); label(\"B\",(4.5,-.5),S); label(\"C\",(6.5,-.5),S); label(\"D\",(8.5,-.5),S); label(\"F\",(10.5,-.5),S); label(\"Grade\",(15,-.5),S); label(\"1\",(-1,3),W); label(\"2\",(-1,6),W); label(\"3\",(-1,9),W); label(\"4\",(-1,12),W); label(\"5\",(-1,15),W); [/asy]$ The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory? $\\text{(A)}\\ \\frac{1}{2} \\qquad \\text{(B)}\\ \\frac{2}{3} \\qquad \\text{(C)}\\ \\frac{3}{4} \\qquad \\text{(D)}\\ \\frac{4}{5} \\qquad \\text{(E)}\\ \\frac{9}{10}$ ## Problem 6 A stack of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high? $\\text{(A)}\\ 250 \\qquad \\text{(B)}\\ 550 \\qquad \\text{(C)}\\ 667 \\qquad \\text{(D)}\\ 750 \\qquad \\text{(E)}\\ 1250$ ## Problem 7 A \"stair-step\" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\\text{th}$ row is $[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]$ $\\text{(A)}\\ 34 \\qquad \\text{(B)}\\ 35 \\qquad", "(7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw] That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face? $[asy] unitsize(2 cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*\"R\", (x + z)/2); label(rotate(-5)*slant(0.5)*\"B\", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*\"G\", ((x + z) + (x + y))/2); label(rotate(-3)*\"W\", (x + z)/2 + trans); label(rotate(50)*slant(-1)*\"B\", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*\"R\", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*\"P\", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*\"R\", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*\"G\", ((x + z) + (x + y))/2 + 2*trans); [/asy]$ $\\textbf{(A) }\\text{red}\\qquad\\textbf{(B) }\\text{white}\\qquad\\textbf{(C) }\\text{green}\\qquad\\textbf{(D) }\\text{brown}\\qquad\\textbf{(E) }\\text{purple}$ ## Problem 13 A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "(A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); /* labels */ label(\"x\",(A+C)/2,(1.2,-1.2)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_3$, or a regular, space, given a point $x$ and a closed set $A$ in a topological space $X$ that are disjoint, there exists open sets $U,V$ such that $x \\in U, A \\subset V$ and $U,V$ are disjoint. An example of a Hausdorff space that is not regular is the space $\\mathbb{R}_k$, the k-topology (or in more generality, a subspace of $\\mathbb{R}$ consisting of $\\mathbb{R}$ missing a countable number of elements). ## Normal $[asy] defaultpen(linewidth(1) + linetype(\"6 6\")); pair A=(0,0),B=(1.6,0),C=(1.6,1),D=(0,1),shiftfactor=(2.5,0.2); /* draw an \"open set\" using Bezier\" */ picture neighborhood(){ picture pic; path p = A{(0.2,-0.6)}..(2*A+B)/3{(0.4,-0.1)}..(A+2*B)/3{(0.4,-0.1)}..B{(0.3,0.7)}..(B+C)/2{(-0.5,0.7)}..C{(-0.1,0.6)}..(2*C+D)/3{(-0.7,0.5)}..(C+2*D)/3{(-0.5,0.5)}..D{(-0.1,-0.5)}..(A+D)/2{(0.6,-0.7)}..cycle; fill(pic,p,rgb(0.9,0.9,0.9)); draw(pic,p); return pic; } path oSC = (A+C)/2+(-0.25,-0.15)--(A+C)/2+(-0.25,0.15)--(A+C)/2+(0.25,0.15)--(A+C)/2+(0.25,-0.15)--cycle; pair SC = (A+C)/2+shiftfactor; path pSC = SC+(-0.25,-0.15)--SC+(-0.25,0.15)--SC+(0.25,0.15)--SC+(0.25,-0.15)--cycle; add(yscale(1.05)*neighborhood()); add(shift(shiftfactor)*neighborhood()); dot((A+C)/2); fill(pSC,rgb(0.7,0.7,0.7)); draw(pSC,linewidth(1)+linetype(\"6 4\")); fill(oSC,rgb(0.7,0.7,0.7)); draw(oSC,linewidth(1)+linetype(\"6 4\")); label(\"A\",(A+C)/2,(2,-1)); label(\"B\",(A+C)/2 + shiftfactor,(2,-1)); label(\"U\",B,(1,-1)); label(\"V\",B+shiftfactor,(1,-1)); [/asy]$ In a $T_4$, or a normal, space, given any two disjoint closed sets $A,B$ in a topological space $X$, there exists open sets $U,V$ such that $A \\subset U, B \\subset V$ and $U,V$ are disjoint. An example of a regular space that is not normal is the Sorgenfrey plane.", "Difference between revisions of \"1985 AJHSME Problems/Problem 4\" Problem The area of polygon $ABCDEF$, in square units, is $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 46 \\qquad \\text{(D)}\\ 66 \\qquad \\text{(E)}\\ 74$ $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ Solution Solution 1 $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE); label(\"D\",(2,0),SW); label(\"E\",(2,4),NE); label(\"F\",(0,4),SW); label(\"G\",(6,4),SW); label(\"6\",(3,9),N); label(\"9\",(6,4.5),E); label(\"4\",(4,0),S); label(\"5\",(0,6.5),W); [/asy]$ Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of. If we continue segment $\\overline{FE}$ until it reaches the right side at $G$, we create two rectangles - one on the top and one on the bottom. We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\\times5 = 30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other. Note that $GC+GB=9$, and $GB=AF=5$, so we must have $$GC+5=9\\Rightarrow GC=4$$ The area of the bottom rectangle is then $$(DC)(GC)=4\\times 4=16$$ Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$. $\\boxed{\\text{C}}$ Solution 2 $[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label(\"A\",(0,9),NW); label(\"B\",(6,9),NE); label(\"C\",(6,0),SE);", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "can be tiled without restriction. $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); filldraw(shift((3,0))*unitsquare,mediumgray); filldraw(shift((2,1))*unitrect,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(-1,1)); filldraw(shift((11,1))*unitsquare,mediumgray); filldraw(shift((10,0))*unitrect,mediumgray); filldraw(shift((9,1))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(8-1,1)); filldraw(shift((19,1))*unitsquare,mediumgray); filldraw(shift((18,1))*unitsquare,mediumgray); filldraw(shift((18,0))*unitrect,mediumgray); draw(shift((16,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(16-1,1)); [/asy]$ In the first, second, and third cases, we see that we can tile the rest of the board in $s_{n-1}$, $s_{n-2}$, and $t_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that $$t_n=s_{n-1}+s_{n-2}+t_{n-2}\\tag{3}.$$ For $u_n$, note that a tiling can end in one of the two following ways such that the rest of the board can be tiled without restriction. $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); filldraw(shift((3,1))*unitsquare,mediumgray); filldraw(shift((2,0))*unitrect,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(-1,1)); filldraw(shift((11,0))*unitsquare,mediumgray); filldraw(shift((10,1))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(4,1)--(4,2)--(0,2)),linewidth(2)); label(\"\\dots\",(8-1,1)); [/asy]$ In either case, we are left with a board with a corner removed, hence we can tile the rest of the board in $t_{n-1}$ ways in each case. Hence for $n\\ge 4$, we see that $$u_n=2t_{n-1}\\tag{4}.$$ Subtracting (3) from (2), we find that $$s_n-t_n=s_{n-1}-t_{n-1}.$$ Therefore, if $s_{n-1}=t_{n-1}$, then $s_n=t_n$. Since $s_3=t_3$ and $s_4=t_4$, we see that $s_n=t_n$ for all $n\\ge 3$. Therefore, (2) can be rewritten as $$s_n=s_{n-1}+2s_{n-2}\\tag{5},$$ and (4) can be rewritten as $$u_n=2s_{n-1}\\tag{6}.$$ Now by (1), we know that $$a_{n}+a_{n-1}=u_n+s_n+u_{n-1}+s_{n-1}.\\tag{7}$$ In particular, $a_4+a_3=6+5+2+3=2^4$, so the statement is true for $n=4$. Then by (7), and then substituting (5) and (6) (where these are valid for $n\\ge 5$), we find \\begin{align*} a_n+a_{n-1}&=u_n+s_n+u_{n-1}+s_{n-1}\\\\ &=2s_{n-1}+(s_{n-1}+2s_{n-2})+u_{n-1}+s_{n-1}\\\\ &=4s_{n-1}+2s_{n-2}+u_{n-1}.\\tag{8} \\end{align*} But then by (5), and then", "2020 AMC 8 Problems/Problem 22 Problem When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below. $[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(\"N\",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(\"if N is even\",(9.25,1.25),N); label(\"if N is odd\",(9.25,-1.25),N); label(\"\\frac N2\",(12,1.25)); label(\"3N+1\",(12.6,-1.25)); [/asy]$ For example, starting with an input of $N=7,$ the machine will output $3 \\cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$$$7 \\to 22 \\to 11 \\to 34 \\to 17 \\to 52 \\to 26$$When the same $6$-step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$$$N \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to \\rule{0.5cm}{0.15mm} \\to 1$$ $\\textbf{(A) }73 \\qquad \\textbf{(B) }74 \\qquad \\textbf{(C) }75 \\qquad \\textbf{(D) }82 \\qquad \\textbf{(E) }83$ Solution 1 We start with the final output of $1$ and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage: $$\\{1\\}\\leftarrow\\{2\\}\\leftarrow\\{4\\}\\leftarrow\\{1,8\\}\\leftarrow\\{2,16\\}\\leftarrow\\{4,5,32\\}\\leftarrow\\{1,8,10,64\\}$$ where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "reduce the number of tiles in the set to one? $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 11 \\qquad \\text{(C)}\\ 18 \\qquad \\text{(D)}\\ 19 \\qquad \\text{(E)}\\ 20$ Problem 24 Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina? $\\text{(A)}\\ 2/5 \\qquad \\text{(B)}\\ 9/20 \\qquad \\text{(C)}\\ 1/2 \\qquad \\text{(D)}\\ 11/20 \\qquad \\text{(E)}\\ 24/25$ Problem 25 $[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,N); label(\"52\",(A+B)/2,S); label(\"39\",(C+D)/2,N); label(\"12\",(B+C)/2,E); label(\"5\",(D+A)/2,W); [/asy]$ In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$ (diagram not to scale). The area of $ABCD$ is $\\text{(A)}\\ 182 \\qquad \\text{(B)}\\ 195 \\qquad \\text{(C)}\\ 210 \\qquad \\text{(D)}\\ 234 \\qquad \\text{(E)}\\ 260$", "what he had eaten. How many more dollars did Dave pay than Doug? $\\mathrm{(A) \\ } 1\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 3\\qquad \\mathrm{(D) \\ } 4\\qquad \\mathrm{(E) \\ } 5$ ## Problem 6 The $8\\times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$? $[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label(\"A\",(0,4),NW); label(\"B\",(18,4),NE); label(\"C\",(18,-4),SE); label(\"D\",(0,-4),SW); label(\"y\",(3,4),S); label(\"y\",(15,-4),N); label(\"18\",(9,4),N); label(\"18\",(9,-4),S); label(\"8\",(0,0),W); label(\"8\",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$ $\\mathrm{(A) \\ } 6\\qquad \\mathrm{(B) \\ } 7\\qquad \\mathrm{(C) \\ } 8\\qquad \\mathrm{(D) \\ } 9\\qquad \\mathrm{(E) \\ } 10$ ## Problem 7 Mary is $20\\%$ older than Sally, and Sally is $40\\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday? $\\mathrm{(A) \\ } 7\\qquad \\mathrm{(B) \\ } 8\\qquad \\mathrm{(C) \\ } 9\\qquad \\mathrm{(D) \\ } 10\\qquad \\mathrm{(E) \\ } 11$ ## Problem 8 How many sets of two or more consecutive positive integers have a sum of $15$? $\\mathrm{(A) \\ } 1\\qquad \\mathrm{(B) \\ } 2\\qquad \\mathrm{(C) \\ } 3\\qquad \\mathrm{(D) \\ } 4\\qquad \\mathrm{(E) \\ } 5$ ## Problem 9 Oscar buys $13$ pencils and $3$ erasers for $\\textdollar 1.00$. A pencil costs more than", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "adjacent side vectors $\\overrightarrow{(a,b)}, \\overrightarrow{(c,d)}$ is given by $\\overrightarrow{(a,b)} \\times \\overrightarrow{(c,d)} = ad-bc$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) { return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); } /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i) dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype(\"2 2\")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label(\"R\",shiftL+(r/2,0),S); /* right diagram */ for(int i = 0; i < 9; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); if (i % 3 == 1) { filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); } } for(int i = 9; i < 12; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype(\"2 2\")); } [/asy]$ The area of a dodecagon is $3R^2$, where $R$ is the circumradius. $[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype(\"2 4\"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP(\"(a,b)\",A,W); MP(\"(c,d)\",B,E); MP(\"(c,-d)\",B2,E); [/asy]$ The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given", "drawn to scale. What is the length in $X$, in centimeters? $[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label(\"X\",(3.4,1.5)); label(\"6\",(7.6,1.5)); label(\"1\",(7.6,3.5)); label(\"1\",(8.4,4.6)); label(\"2\",(9.4,4.6)); label(\"2\",(10.4,6)); label(\"3\",(8.4,7.4)); label(\"1\",(7.5,7.8)); label(\"2\",(6,8.5)); label(\"1\",(4.7,7.8)); label(\"1\",(4.3,7.5)); label(\"1\",(3.5,6.5)); label(\"4\",(1.8,6.5)); label(\"1\",(-0.5,5.5)); label(\"2\",(0.8,4.5)); label(\"2\",(1.5,3.8)); label(\"2\",(2.8,2.6));[/asy]$ $\\textbf{(A)}\\hspace{.05in}1\\qquad\\textbf{(B)}\\hspace{.05in}2\\qquad\\textbf{(C)}\\hspace{.05in}3\\qquad\\textbf{(D)}\\hspace{.05in}4\\qquad\\textbf{(E)}\\hspace{.05in}5$ ## Problem 6 A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches? $\\textbf{(A)}\\hspace{.05in}36\\qquad\\textbf{(B)}\\hspace{.05in}40\\qquad\\textbf{(C)}\\hspace{.05in}64\\qquad\\textbf{(D)}\\hspace{.05in}72\\qquad\\textbf{(E)}\\hspace{.05in}88$ ## Problem 7 Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test? $\\textbf{(A)}\\hspace{.05in}90\\qquad\\textbf{(B)}\\hspace{.05in}92\\qquad\\textbf{(C)}\\hspace{.05in}95\\qquad\\textbf{(D)}\\hspace{.05in}96\\qquad\\textbf{(E)}\\hspace{.05in}97$ ## Problem 8 A shop advertises everything is \"half price in today's sale.\" In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price? $\\textbf{(A)}\\hspace{.05in}10\\qquad\\textbf{(B)}\\hspace{.05in}33\\qquad\\textbf{(C)}\\hspace{.05in}40\\qquad\\textbf{(D)}\\hspace{.05in}60\\qquad\\textbf{(E)}\\hspace{.05in}70$ ## Problem 9 a number of two-legged birds and a number of four-legged mammals. On one visit to", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "cm. A rock with volume $1000 \\text{ cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise? $\\textbf{(A)}\\ 0.25\\qquad\\textbf{(B)}\\ 0.5\\qquad\\textbf{(C)}\\ 1\\qquad\\textbf{(D)}\\ 1.25\\qquad\\textbf{(E)}\\ 2.5$ Problem 22 Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell? $[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label(\"+\", (1.5,1.5)); label(\"+\", (3.5,1.5)); label(\"+\", (2.5,4.5));[/asy]$ $\\textbf{(A)}\\ 16\\qquad\\textbf{(B)}\\ 24\\qquad\\textbf{(C)}\\ 25\\qquad\\textbf{(D)}\\ 26\\qquad\\textbf{(E)}\\ 35$ Problem 23 A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? $\\textbf{(A)}\\ 0\\qquad\\textbf{(B)}\\ 1\\qquad\\textbf{(C)}\\ 2\\qquad\\textbf{(D)}\\ 3\\qquad\\textbf{(E)}\\ 5$ Problem 24 In the multiplication problem below $A$, $B$, $C$, $D$ and are different digits." ]

[ "+0 # Pls help 0 92 1 Compute:$$0.\\overline{7}-0.\\overline{4}+0.\\overline{2}$$ . Express your answer as a common fraction. Sep 22, 2020", "+0 Decimals 0 49 1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ May 15, 2022 #1 +13731 +1 Express as a common fraction the reciprocal of $$0.\\overline{81}$$ Hello Guest! $$x=0.\\overline{81}\\\\ 100x=81.\\overline{81}\\\\ 99x=81\\\\ x=\\dfrac{9}{11}$$ $$\\color{blue}0.\\overline{81}=\\dfrac{9}{11}$$ ! May 15, 2022", "\"common fraction\" • fraction", "3}}$? Feb 6: What common fraction is equal to $20.1\\bar 8$? Feb 7: If ${1\\over { {1\\over{{1\\over n} +{1\\over 3} }}+{1\\over{{1\\over n} +{1\\over 3} }}} } = {5 \\over 12}$, what is the value of n? Feb 8: If ${ {2x} \\over {x-3}} - 2 = {4\\over{x+2}}$, what is the value of x? Feb 9: What fraction of $3\\over 8$ is $9\\over 16$? Feb 10: What is coordinate of the point that is $1\\over 3$ distance from $(3,0)$ to $(0,-6)$?", "+0 0 582 2 +46 What is $0.4\\overline8 + 0.\\overline{37}$? Express your answer as a common fraction in lowest terms. Sep 5, 2020 #1 +35059 +1 $$0.4\\overline8 + 0.\\overline{37}?$$ .488888.... = 44/90 .373737.... = 37/99 I will let you take it from here.... find the common denominator and convert the fractions, add them then simplify Sep 5, 2020 #2 0 22/45+37/99 Guest Sep 7, 2020", "+0 # Fractions 0 82 1 Express as a fraction in lowest terms: $0.\\overline{12} + 0.\\overline{001}$ May 25, 2022", "you'll get $\\overline{9} = - 1$.", "+0 # Express $4.\\overline{054}$ as a common fraction in lowest terms. 0 126 1 Express $4.\\overline{054}$ as a common fraction in lowest terms. Aug 31, 2020 #1 0 If we expand it out, we get 4.054054054..... Setting this equal to x, we have: 1000x = 4054.054054054... x = 4.054054054... Subtracting the equations, we get: 999x = 4050 x = 4050/999 x = 450/111 x = 150/37 Jan 14, 2021", "a fraction! #### Examples ##### Question 1 Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. ##### Question 2 Starting with $x=0.\\overline{216}$x=0.216, express $x$x as a fraction in simplest form. ##### Question 3 Starting with $x=0.6\\overline{87}$x=0.687, express $x$x as a fraction in simplest form.", "+0 # Decimals 0 35 5 Express as a common fraction the reciprocal of $0.\\overline{72}$. Jun 28, 2022 #1 +1 Express as a common fraction the reciprocal of $0.\\overline{72}$. I can give you the answer. The decimal equals 8 / 11 so its reciprocal would be 11 / 8. I can't tell you how to calculate it though. I guessed and got lucky. I guessed 8/9 ... that was too big, so I guessed 8/11 and that was it. . Jun 28, 2022 #2 +2275 +1 Let $$x = 0. \\overline{72}$$ This means $$100x = 72.\\overline{72}$$ Subtracting the 2 gives $$99x = 72$$, meaning $$x = {72 \\over 99} ={ 8 \\over 11}$$. The reciprocal of this, as Guest found, is $$\\color{brown}\\boxed{11 \\over 8}$$ Jun 28, 2022 #3 +69 +2 Just a fun fact, if it repeats every one digit, then the fraction version is [digit] / 9. If it repeats every two digits, then it is [digit] / 99. You can see the pattern here. ⚡⚡⚡ ShockFish Jun 28, 2022 edited by ShockFish Jun 28, 2022 #4 +2275 +1 Hi Shockfish, nice to see you here BuilderBoi Jun 28, 2022 #5 +69 +2 Heyyy just wondering if you were here too ShockFish Jun 28, 2022", "+0 # Help me 0 69 1 When the repeating decimal $$0.\\overline{12}$$ is expressed as a common fraction in lowest terms, what is the sum of its numerator and denominator? Sep 22, 2020 ### 1+0 Answers #1 0 4 0.12121212 • • • = —— 33 so the numerator & denominator added together total 37 . Sep 22, 2020", "# Convert recurring decimals to fractions ## Interactive practice questions Starting with $x=0.\\overline{1}$x=0.1, express $x$x as a fraction in simplest form. Easy Approx a minute Starting with $x=2.\\overline{3}$x=2.3, express $x$x as a fraction in simplest form. Starting with $x=1.\\overline{7}$x=1.7, express $x$x as a fraction in simplest form. Starting with $x=3.\\overline{6}$x=3.6, express $x$x as a fraction in simplest form.", "a fraction, $0.\\overline{72}$, is equal to $\\dfrac{8}{11}$. 10. Which of the following shows the conversion of $0.48$ to fraction? 11. Which of the following shows the conversion of $5.25$ to fraction?", "+0 # PLZ Help i need an explanation on how to do this!! 0 58 1 Given$$\\frac{1}{3x+3}=3$$ express x as a common fraction. Plz give an explanation! Jun 18, 2020 #1 +902 +1 1 = 9x + 9 9x = 1-9 = -8 x = -8/9 Jun 18, 2020", "# Common Fraction Written by Jerry Ratzlaff. Posted in Mathematics Both top and bottom are integers. • $$\\; 1/8\\;$$ and $$\\;3/4 \\;$$", "# 1.3 (overline) is equal to 1.3 (overline) is expressed as 1.333333333….. 1.3 (overline) is equivalent to the fraction form 4 / 3.", "# Common Fraction Written by Jerry Ratzlaff on . Posted in Mathematics Both top and bottom are integers. • $$\\; 1/8\\;$$ and $$\\;3/4 \\;$$", "+0 # decimals 0 75 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jun 19, 2021 #1 0 0.4888... = 22/45 0.3777... = 17/45 22/45 + 17/45 = 39/45 Jun 20, 2021", "* 1$Fraction(Integer).", "#### Problem 54E Express the number as a ratio of integers. $10.1 \\overline{35}=10.135353535 \\ldots$" ]

[ "# Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. <br/> <br/>(i) $0 . \\overline{6}$<br/> <br/> (ii) $0.4 \\overline{7}$ <br/> <br/>(iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$", "# Why is ⅓ = 0.3333....? • ## Why is it true that $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ ? I wish I had a very simple and clever way to show this, but all the simple and clever ways usually use some idea that is already a step more advanced. For example, you could say that $$0.\\overline{999}\\ldots = 1$$ and then divide both sides by $$3$$ (but knowing the fact that $$0.\\overline{999}\\ldots = 1$$ is even harder to prove!). Or you could start from the fact that $$0.\\overline{111}\\ldots = \\frac{1}{9}$$ and multiply both sides by $$3,$$ but wait, isn't that just the reverse of what we did to get to this place? We were originally trying to show $$\\frac{1}{9} = 0.\\overline{111}\\ldots$$ in the first place, assuming that we knew the decimal representation of $$\\frac{1}{3}!$$ (Round and round we go!) I think the best way to show why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is just to show the long division. Then you'll find that $$1$$ divided by $$3$$ equals $$0.\\overline{333}\\ldots.$$ . This isn't hard, but there are some basic concepts to lay down first. : place value and the idea of division as a reverse-multiplication operation. ## Place value I have some affectionate names that I personally like to use when referring to place value digits. You can make up your own names,", "the confusion the following representation is adapted. The number is given as $0.3\\overline{74}$$0.3 \\overline{74}$ The line over $74$$74$ represents that $74$$74$ is repeated. Convert $\\frac{1}{3}$$\\frac{1}{3}$ into a fraction. The answer is $0.\\overline{3}$$0. \\overline{3}$ Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. convert back to fraction How do we convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction? To convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction, the following steps are used $x=0.\\overline{4}$$x = 0. \\overline{4}$ $10x=4.\\overline{4}$$10 x = 4. \\overline{4}$ Subtracting the two equations $9x=4$$9 x = 4$ $x=\\frac{4}{9}$$x = \\frac{4}{9}$ examples Converting $2.\\overline{4}$$2. \\overline{4}$ into a fraction. $x=2.\\overline{4}$$x = 2. \\overline{4}$ $10x=24.\\overline{4}$$10 x = 24. \\overline{4}$ subtracting the two above $9x=22$$9 x = 22$ $x=\\frac{22}{9}$$x = \\frac{22}{9}$ $x=2\\frac{4}{9}$$x = 2 \\frac{4}{9}$ To convert $2.23\\overline{43}$$2.23 \\overline{43}$ into a fraction: $x=2.23\\overline{43}$$x = 2.23 \\overline{43}$ $100x=223.43\\overline{43}$$100 x = 223.43 \\overline{43}$ Subtracting the above, $99x=221.20$$99 x = 221.20$ $990x=2212$$990 x = 2212$ $x=\\frac{2212}{990}$$x = \\frac{2212}{990}$. Another method is as follows. $100x=223+0.\\overline{43}$$100 x = 223 + 0. \\overline{43}$ $10000x=223×100+43.\\overline{43}$$10000 x = 223 \\times 100 + 43. \\overline{43}$ Subtracting the two equations $9900x=223×\\left(100-1\\right)+43$$9900 x = 223 \\times \\left(100 - 1\\right) + 43$ $x=\\frac{223}{100}+\\frac{43}{9900}$$x = \\frac{223}{100} + \\frac{43}{9900}$ summary Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. Conversion of Repeating Decimals to equivalent", "# Question #0426c Jan 5, 2017 $0.8 \\overline{7} = \\frac{79}{90}$ #### Explanation: Let $x = 0.8 \\overline{7}$ $\\implies 10 x = 8. \\overline{7}$ $\\implies 10 x - x = 8. \\overline{7} - 0.8 \\overline{7}$ $\\implies 9 x = 8.7 - 0.8$ $\\implies 9 x = 7.9$ $\\implies 9 x = \\frac{79}{10}$ $\\implies x = \\frac{79}{10} \\cdot \\frac{1}{9}$ $\\therefore x = \\frac{79}{90}$ This technique of multiplying by $10$ (or a power of $10$) and subtracting can be used to find the fractional representation of any number with infinitely repeating decimal digits. This answer contains a detailed explanation.", "Express the following in the form $\\frac{p}{q}$, where p and q are integers and q ≠ 0. (i) $0 . \\overline{6}$ (ii) $0.4 \\overline{7}$ (iii) $0 . \\overline{001}$ Solution: (i) $0 . \\overline{6}=0.666 \\ldots$ Let $x=0.666 \\ldots$ $10 x=6.666 \\ldots$ $10 x=6+x$ $9 x=6$ $x=\\frac{2}{3}$ (ii) $0 . \\overline{47}=0.4777 \\ldots . .$ $=\\frac{4}{10}+\\frac{0.777}{10}$ $10 x=7.777 \\ldots$ $10 x=7+x$ $x=\\frac{7}{9}$ $\\frac{4}{10}+\\frac{0.777 \\ldots}{10}=\\frac{4}{10}+\\frac{7}{90}$ $=\\frac{36+7}{90}=\\frac{43}{90}$ (iii) $0 . \\overline{001}=0.001001 \\ldots$ Let $x=0.001001 \\ldots$ $1000 x=1.001001 \\ldots$ $1000 x=1+x$ $999 x=1$ $x=\\frac{1}{999}$ Editor", "the value of your number is equal to $$x=0.\\overline1 -\\sum_{n=1}^\\infty 10^{1-\\frac{(n+1)(n+2)}{2}}$$ Since $0.\\overline1=\\frac{1}{9}$, then we have $$x=\\frac{1}{9}-\\sum_{n=1}^\\infty 10^{1-\\frac{(n+1)(n+2)}{2}}$$ Which can be simplified to $$x=\\frac{1}{9}-\\sum_{n=1}^\\infty (\\sqrt{10})^{-n^2-3n}$$ As far as I know, this is as much as it can be simplified without bringing in other non-elementary functions. However, you might be able to use this function to simplify it further.", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "you have to multiply your number by 10³. Now you have $\\boxed{\\begin{array}{lcr}1000x&=&2345.345345345... \\\\ x&=&2.345345345...\\end{array}}$ Subtract and you have $999x = 2343~\\iff~x = \\frac{2343}{999}$", "+0 # pls help me 0 85 3 What is $\\frac{0.\\overline{72}}{0.\\overline{27}}$? Express your answer as a common fraction in lowest terms. Jul 27, 2022 #1 -1 The expression reduces to 2/9. Jul 27, 2022 #2 +2448 0 Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1... BuilderBoi Jul 28, 2022 #3 +2448 0 I'll do the numerator, and you can do the denominator. For the numerator, let $$x = 0.{ \\overline{72}}$$. This means $$100x = {72. \\overline{72}}$$ (move the decimal point 2 places to the right) Subtracting the 2 yields us $$99x = 72$$ (the repeating parts cancel out) Dividing by 99, we find that $$x = {72 \\over 99}$$ Jul 28, 2022", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "\\frac{9}{10^9} + \\frac{0}{10^{10}} + \\frac{1}{10^{11}}+\\frac{2}{10^{12}}+\\cdots \\right], \\end{array}$ which is the announced formula. The general result can be stated as follows. Proposition 1. Let ${b \\geq 3}$ be a fixed basis. Then, when everything is written in basis ${b}$, for any ${n \\geq 1}$, one has $\\displaystyle \\begin{array}{rcl} \\frac{1}{\\overline{b-1} \\, \\overline{b-1}\\cdots\\overline{b-1} \\, \\overline{b-1}^{2}} = 0,(&&00\\cdots00 \\\\ &&00\\cdots01 \\\\ &&00\\cdots02 \\\\ && \\qquad \\vdots\\\\ &&\\overline{b-1} \\, \\overline{b-1}\\cdots\\overline{b-1} \\, \\overline{b-4} \\\\ &&\\overline{b-1} \\, \\overline{b-1}\\cdots\\overline{b-1} \\, \\overline{b-3} \\\\ &&\\overline{b-1} \\, \\overline{b-1}\\cdots\\overline{b-1} \\, \\overline{b-1}), \\end{array}$ where each group containing ${\\cdots}$ in the expression above has ${n}$ digits and ${\\overline{x}}$ denotes the digit ${0 \\leq x in basis ${b}$. Proof: We shall start from the RHS of the equation and work our way towards the LHS. Recall that a periodic decimal expansion in base ${b}$ can be written as a fraction as follows: $\\displaystyle 0,(\\overline{x_{k-1}} \\, \\overline{x_{k-2}} \\, \\overline{x_{k-3}} \\cdots \\overline{x_{1}} \\, \\overline{x_0} ) = \\frac{\\sum_{i=0}^{k-1} \\overline{x_i} b^i}{ (b-1) \\sum_{i=0}^{k-1} b^i}.$ First, let us focus on the case ${n=1}$ and treat the general case at the end of the proof. In this particular case, the RHS of the equation in the statement is equal to $\\displaystyle 0,(012\\cdots \\overline{b-3} \\, \\overline{b-1}) = \\frac{1+\\sum_{i=0}^{b-2} (b-2-i)b^i}{(b-1)\\sum_{i=0}^{b-2}b^i},$ where the 1 in front of the sum stands for the fact that ${\\overline{b-2}}$ is replaced by ${\\overline{b-1}}$ as the last", "Home/Class 9/Maths/ Show that $$1.272727... =1.\\overline {27}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ Speed 00:00 03:43 QuestionMathsClass 9 Show that $$1.272727... =1.\\overline {27}$$ can be expressed in the form $$\\frac {p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\\neq 0.$$ Let $$x=1.272727 \\ldots$$ Since two digits are repeating, we multiply $$x$$ by $$100$$ to get $$100 x=127.2727 \\ldots$$ So, $$100 x=126+1.272727 \\ldots=126+x$$ Therefore, $$100 x-x=126,$$ i.e., $$99 x=126$$ i.e., $$x=\\frac{126}{99}=\\frac{14}{11}$$ You can check the reverse that $$\\frac{14}{11}=1 . \\overline{27}.$$", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", "# How do you express .33 as a fraction in simplest form? Sep 14, 2016 $\\frac{33}{100}$ Sep 14, 2016 As you wrote the question $\\frac{33}{100}$ If you meant $0.333333 . . \\to 0.33 \\overline{3} \\text{ }$ then $\\text{ } \\frac{1}{3}$ #### Explanation: It all depends on what you mean by $0.33$ If you meant it to be exactly as written then we have $\\frac{3}{10} + \\frac{3}{100}$ Write as $\\left(\\frac{3}{10} \\times 1\\right) + \\frac{3}{100}$ But 1 may be written as $\\frac{10}{10}$ $\\left(\\frac{3}{10} \\times \\frac{10}{10}\\right) + \\frac{3}{100}$ $\\frac{30}{100} + \\frac{3}{100} = \\frac{33}{100}$ $\\textcolor{red}{\\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$ If you meant what you wrote to mean $0.33 \\leftarrow$ 3 going on for ever $0.333333333 \\ldots . .$ written as $0.33 \\overline{3}$ $\\textcolor{b r o w n}{\\text{Then this is very different:}}$ Let $x = 0.33 \\overline{3}$ thus $10 x = 3.33 \\overline{3}$ $10 x - x = 3$ $9 x = 3$ $x = \\frac{3}{9} = \\frac{1}{3}$", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "once. Otherwise, you're adding rather than subtracting. • He effectively subtracted twice, but psychologically he had a different problem (or he was just blindly performing the algorithm). – PJTraill Oct 1 '16 at 10:55 The standard subtraction algorithm only works when the larger-magnitude number is on top (in this case, the $8.2$). The most straightforward calculation is to note that the result here will be obviously negative (because the number being subtracted is of larger magnitude), and then perform $8.2 - 4.5$ for the magnitude of the result. Failing to pay attention to which magnitude is larger results in exactly this kind of subtraction error. Expanding on @John's answer https://math.stackexchange.com/a/1948174/169560. You can borrow from \"nothing\" you just need to remember to give it back: $$\\frac{ \\begin{array}[b]{r} 4.5 \\\\ - 8.2 \\end{array} }{}= \\frac{ \\begin{array}[b]{r} \\ ^{-1}14.5 \\\\ - 8.2 \\end{array} }{}= \\frac{ \\begin{array}[b]{r} \\ ^{-1}14.5 \\\\ - 8.2 \\end{array} }{-10+6.3}=3.7$$ This way, if the minuend is smaller than the subtrahend, you will end up with something like $-100 \\ldots 000 + YY \\ldots YY.yy \\ldots yy$ where $x$, $Y$ and $y$ stand for any digits, and the number of zeroes is the same as the number of $Y$s. It should be easy to do this last step, as you just need to pad to a potence of 10", "values of $$n$$: $$\\begin{array}{c|c|} & \\text{n\\in \\mathbb{N}}& \\text{(1- \\frac1{3^n})}\\\\ \\hline a & 1& \\frac23=0.\\overline{6}\\\\ \\hline b & 2& \\frac89=0.\\overline{8}\\\\ \\hline c & 3& \\frac{26}{27}=0.\\overline{962}\\\\ \\hline d & 4& \\frac{80}{81}=0.\\overline{987654320}\\\\ \\hline \\vdots\\\\ \\hline \\infty & n\\rightarrow \\infty& \\approx 1\\\\ \\hline \\end{array}$$ $$s\\not\\in \\{(1- \\frac1{3^n}\\,\\,: n \\in N \\})$$, as $$s$$ does not lie in the set. $$(1- \\frac1{3^n}) \\in [\\frac23, 1)$$. $$\\epsilon \\in\\mathbb{R}, \\epsilon\\gt 0\\implies \\forall \\epsilon \\gt 0: n\\lt \\infty$$, as $$s - \\epsilon \\lt 1$$. For interval $$(s-\\epsilon,s]$$, the lower bound implies for $$\\epsilon\\gt 0$$ that $$s-\\epsilon \\lt 1\\implies n \\lt \\infty$$, & the upper bound implies the max. value is $$s=1$$ at $$n= \\infty$$. All possible values of $$n$$ are covered in the interval $$(s-\\epsilon,s]$$ with $$n=\\infty$$ at the upper bound. 1. Identify the supremum of the set given below, & use the \"new\" definition of supremum to prove your claim: $$\\{(-\\frac12)^n\\,\\,: n \\in N \\}$$ $$\\sup(\\{(-\\frac12)^n\\,\\,: n \\in N \\})= 1.$$ $$s=\\frac14$$, for $$n=2$$. Below is table for set members for few values of $$n$$: $$\\begin{array}{c|c|} & \\text{n\\in \\mathbb{N}}& \\text{(-\\frac12)^n}\\\\ \\hline a & 1& -\\frac12=-0.5\\\\ \\hline b & 2& \\frac14=0.25\\\\ \\hline c & 3& -\\frac 18=-0.125\\\\ \\hline d & 4& \\frac{1}{16}=0.0625\\\\ \\hline \\vdots\\\\ \\hline \\infty & n\\rightarrow \\infty& \\approx 0\\\\ \\hline \\end{array}$$ As $$-(\\frac12)^n\\in \\{-\\frac12, \\cdots, \\frac14\\}$$, & as $$n \\rightarrow \\infty, -(\\frac12)^n \\approx 0$$.", "larger. (Caution:This is true only for proper fractions, that is, fractions between 0 and 1.) Example: ${\\left(\\frac{1}{3}\\right)}^{2}$= $\\frac{1}{9}$ and $\\frac{1}{9}$ is less than $\\frac{1}{3}$ Also and $\\frac{1}{2}$is greater than $\\frac{1}{4}$ For example, the c’s in the expression cannot be canceled. However, the c’s in the expression can be canceled as follows: $\\frac{cx+c}{c}=\\frac{\\overline{)c}\\left(x+1\\right)}{\\overline{)c}}=x+1$", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =" ]

[ "# Question #0426c Jan 5, 2017 $0.8 \\overline{7} = \\frac{79}{90}$ #### Explanation: Let $x = 0.8 \\overline{7}$ $\\implies 10 x = 8. \\overline{7}$ $\\implies 10 x - x = 8. \\overline{7} - 0.8 \\overline{7}$ $\\implies 9 x = 8.7 - 0.8$ $\\implies 9 x = 7.9$ $\\implies 9 x = \\frac{79}{10}$ $\\implies x = \\frac{79}{10} \\cdot \\frac{1}{9}$ $\\therefore x = \\frac{79}{90}$ This technique of multiplying by $10$ (or a power of $10$) and subtracting can be used to find the fractional representation of any number with infinitely repeating decimal digits. This answer contains a detailed explanation.", "the confusion the following representation is adapted. The number is given as $0.3\\overline{74}$$0.3 \\overline{74}$ The line over $74$$74$ represents that $74$$74$ is repeated. Convert $\\frac{1}{3}$$\\frac{1}{3}$ into a fraction. The answer is $0.\\overline{3}$$0. \\overline{3}$ Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. convert back to fraction How do we convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction? To convert $0.\\overline{4}$$0. \\overline{4}$ into a fraction, the following steps are used $x=0.\\overline{4}$$x = 0. \\overline{4}$ $10x=4.\\overline{4}$$10 x = 4. \\overline{4}$ Subtracting the two equations $9x=4$$9 x = 4$ $x=\\frac{4}{9}$$x = \\frac{4}{9}$ examples Converting $2.\\overline{4}$$2. \\overline{4}$ into a fraction. $x=2.\\overline{4}$$x = 2. \\overline{4}$ $10x=24.\\overline{4}$$10 x = 24. \\overline{4}$ subtracting the two above $9x=22$$9 x = 22$ $x=\\frac{22}{9}$$x = \\frac{22}{9}$ $x=2\\frac{4}{9}$$x = 2 \\frac{4}{9}$ To convert $2.23\\overline{43}$$2.23 \\overline{43}$ into a fraction: $x=2.23\\overline{43}$$x = 2.23 \\overline{43}$ $100x=223.43\\overline{43}$$100 x = 223.43 \\overline{43}$ Subtracting the above, $99x=221.20$$99 x = 221.20$ $990x=2212$$990 x = 2212$ $x=\\frac{2212}{990}$$x = \\frac{2212}{990}$. Another method is as follows. $100x=223+0.\\overline{43}$$100 x = 223 + 0. \\overline{43}$ $10000x=223×100+43.\\overline{43}$$10000 x = 223 \\times 100 + 43. \\overline{43}$ Subtracting the two equations $9900x=223×\\left(100-1\\right)+43$$9900 x = 223 \\times \\left(100 - 1\\right) + 43$ $x=\\frac{223}{100}+\\frac{43}{9900}$$x = \\frac{223}{100} + \\frac{43}{9900}$ summary Representation of Repeating Decimals : The repetitive pattern in decimal digits is represented with an over-line. Conversion of Repeating Decimals to equivalent", "# Express Question: Express $0 . \\overline{4}$ as a rational number in simplest form. Solution: Let $x$ be $0 . \\overline{4}$ $x=0 . \\overline{4}$ Multiplying both sides by 10, we get $10 x=4 . \\overline{4}$ Subtracting (1) from (2), we get $10 x-x=4 . \\overline{4}-0 . \\overline{4}$ $\\Rightarrow 9 x=4$ $\\Rightarrow x=\\frac{4}{9}$ Thus, simplest form of $0 . \\overline{4}$ as a rational number is $\\frac{4}{9}$.", "# RD Sharma Solutions Class 9 Number System Exercise 1.2 ## RD Sharma Solutions Class 9 Chapter 1 Ex 1.2 Q1. Express the following rational numbers as decimals: (i) $\\frac{42}{100}$ (ii) $\\frac{327}{500}$ (iii) $\\frac{15}{4}$ Solution: (i) By long division method 100) $\\overline{42}$ ( 0.42 400 $\\overline{200}$ 200 $\\overline{0}$ Therefore, $\\frac{42}{100}$ = 0.42 (ii) By long division method 500) $\\overline{327.000}$ ( 0.654 3000 $\\overline{2700}$ 2500 $\\overline{2000}$ 2000 $\\overline{0}$ Therefore, $\\frac{327}{500}$ = 0.654 (iii) By long division method 4) $\\overline{15.00}$ ( 3.75 12 $\\overline{30}$ 28 $\\overline{20}$ 20 $\\overline{0}$ Therefore, $\\frac{15}{4}$ = 3.75 Q2. Express the following rational numbers as decimals: (i) $\\frac{2}{3}$ (ii) –$\\frac{4}{9}$ (iii) –$\\frac{2}{15}$ (iv) –$\\frac{22}{13}$ (v) $\\frac{437}{999}$ Solution: (i) By long division method 3) $\\overline{2.0000}$ ( 0.66 18 $\\overline{20}$ 18 $\\overline{2}$ Therefore, $\\frac{2}{3}$ = 0.66 (ii) By long division method 9) $\\overline{4.000}$ ( 0.444 3600 $\\overline{4000}$ 3600 $\\overline{4000}$ 3600 $\\overline{400}$ Therefore, – $\\frac{4}{9}$ = – 0.444 (iii) By long division method 15) $\\overline{2.00}$ ( 1.333 15 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{50}$ 45 $\\overline{5}$ Therefore, $\\frac{2}{15}$ = -1.333 (iv) By long division method 13) $\\overline{22.000}$ ( 1.69230769 13 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{30}$ 26 $\\overline{40}$ 39 $\\overline{100}$ 91 $\\overline{90}$ 78 $\\overline{120}$ 117 $\\overline{3}$ Therefore, – $\\frac{22}{13}$ = – 1.69230769 (v) By long division method 999) $\\overline{437.0000}$ ( 0.43743 3996 $\\overline{3740}$ 2997 $\\overline{7430}$ 6993 $\\overline{4370}$ 3996 $\\overline{3740}$ 2997 $\\overline{743}$", "@funnyorangutan-1 Thanks for your very polite and curious responses. 🙂 The first question about why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ is quite long, so if you don't mind, I'll put that in it's own post, here. https://forum.poshenloh.com/topic/545/why-is-0-3333 ⬅ ⬅ ⬅ Answer to why $$\\frac{1}{3} = 0.\\overline{333}\\ldots$$ For your other question, about why $$0.\\overline{222}\\ldots = \\frac{2}{9} \\text{ and } 0.\\overline{444}\\ldots = \\frac{4}{9}$$ and so on, maybe this diagram will help a bit: M1-forum-fractions-with-denominator-9.png You can treat $$0.\\overline{111}\\ldots$$ as a starting point from which you can multiply by any factor in order to get any fraction of the form $$\\frac{n}{9},$$ where $$n$$ is an integer. \"Oh, but didn't you leave out the $$\\frac{3}{9} \\text{ and } \\frac{6}{9}?$$\" Very perceptive question! 🙂 I left those out because $$\\frac{3}{9}$$ and $$\\frac{6}{9}$$ aren't in simplified form. You wouldn't put \"$$\\frac{3}{9}$$\" on a test as an answer. Instead, you would put $$\\frac{1}{3}.$$ And you wouldn't write \"$$\\frac{6}{9}$$\" on a test as an answer either -- you would write $$\\frac{2}{3}.$$ So even though the pattern holds for multiplying $$0.\\overline{111}\\ldots$$ by both $$3$$ and $$6,$$ I left it out, but it is very nice to know that everyone holds across math, and that everything works as it should, because $$0.\\overline{333}\\ldots = \\frac{3}{9} = \\frac{1}{3}$$ Great, math is consistent! 🙂 When the world seems like it's failing you and", "# 0.999... $0.999\\ldots$ (or $0.\\overline{9}$) is an equivalent representation of the real number $1$. It is often mistaken that $0.999\\ldots \\neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\\frac 1{\\infty}$ term left over at the end, etc.). ## Proofs ### Fractions Since $\\frac 13 = 0.\\overline{3} = 0.333\\ldots$, multiplying both sides by $3$ yields $1 = 0.999\\ldots$ Alternatively, $\\frac 19 = 0.\\overline{1} = 0.111\\ldots$, and then multiply both sides by $9$. ### Manipulation Let $x = 0.999\\ldots$ Then \\begin{align*} 10x &= 9.999\\ldots\\\\ x &= 0.999\\ldots \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) Subtracting, \\begin{align*} 9x &= 9\\\\ x &= 1 \\end{align*} (Error compiling LaTeX. ! Package amsmath Error: \\begin{align*} allowed only in paragraph mode.) ### Infinite series $0.999\\ldots = 0.9 + 0.09 + 0.009 + \\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\ldots$ This is an infinite geometric series, so $0.999\\ldots = \\frac{\\frac{9}{10}}{1 - \\frac{1}{10}} = 1$", "+0 # pls help me 0 85 3 What is $\\frac{0.\\overline{72}}{0.\\overline{27}}$? Express your answer as a common fraction in lowest terms. Jul 27, 2022 #1 -1 The expression reduces to 2/9. Jul 27, 2022 #2 +2448 0 Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1... BuilderBoi Jul 28, 2022 #3 +2448 0 I'll do the numerator, and you can do the denominator. For the numerator, let $$x = 0.{ \\overline{72}}$$. This means $$100x = {72. \\overline{72}}$$ (move the decimal point 2 places to the right) Subtracting the 2 yields us $$99x = 72$$ (the repeating parts cancel out) Dividing by 99, we find that $$x = {72 \\over 99}$$ Jul 28, 2022", "\\overline{62)} & \\overline{124} & (2 \\\\ & & & & 124 & \\\\ & & & & \\overline{0} & \\end{array}$ Therefore $\\frac{682}{868}$ $=$ $\\frac{682 \\div 62}{868 \\div 62}$ $=$ $\\frac{11}{14}$ ii) $\\frac{777}{1147}$ $\\begin{array}{r r r r r } 777) & \\overline{1147} & (1 & & \\\\ & 777 & & & \\\\ & \\overline{370)} & \\overline{777} & (2 & \\\\ & & 740 & & \\\\ & & \\overline{37)} & \\overline{370} & (10 \\\\ & & & 370 & \\\\ & & & \\overline{0} & \\end{array}$ Therefore $\\frac{777}{1147}$ $=$ $\\frac{777 \\div 37}{1147 \\div 37}$ $=$ $\\frac{21}{31}$ iii) $\\frac{1095}{1168}$ $\\begin{array}{r r r r } 1095) & \\overline{1168} & (1 & \\\\ & 1095 & & \\\\ & \\overline{73)} & \\overline{1095} & (15 \\\\ & & 1095 & \\\\ & & \\overline{0} & \\end{array}$ Therefore $\\frac{1095}{1168}$ $=$ $\\frac{1095 \\div 73}{1068 \\div 73}$ $=$ $\\frac{15}{16}$ $\\\\$ Question 4. Which of the following numbers are co-primes? i) $18, 25$ ii) $62, 81$ iii) $69, 92$ Note: The natural numbers are said to be co-primes if the HCF of the numbers is $1$. You can calculate the HCF by using any of the two methods…prime factorization or long division method. i) $18, 25$ HCF of $18, 25 = 1$ Hence, $18$ and $25$ are co-primes. ii) $62, 81$ HCF of $62, 81 =", "at 9:23 @MarcvanLeeuwen: any of this without knowledge of limits would be gibberish. This is often where the confusion lies. – robjohn Mar 20 '13 at 9:42 The trick in this is that 10 x 0.3... is not 3.3... it is 3.3... with a 0 on the end ;) – JamesRyan Mar 20 '13 at 17:06 You can find the sum of $\\frac{3}{10} + \\frac{3}{100} + \\frac{3}{1000} + \\cdots$ using the formula of sum of infinite geometric progression. $$a_1 = \\frac{3}{10}$$ $$r=\\frac{1}{10}$$ $$\\sum =\\frac{a_1}{1-r}=\\frac{3}{10}\\times\\frac{10}{9} =\\frac{1}{3}$$ - The problematic part of the question is \"no matter how many ones you add, 0.111... will never equal precisely 1/9.\" In this (imprecise) context $0.111\\ldots$ is an infinite sequence of ones; the sequence of ones does not terminate, so there is no place at which to add another one; each one is already followed by another one. Thus, $10\\times0.111\\ldots=1.111\\ldots$ is precise. Therefore, $9\\times0.111\\ldots=1.000\\ldots=1$ is precise, and $0.111\\ldots=1/9$. I say \"imprecise\" because we also say $\\pi=3.14159\\dots$ where ... there means an unspecified sequence of digits following. A more precise way of writing what, in the context of this question, we mean by $0.111\\dots$ is $0.\\overline{1}$ where the group of digits under the bar is to be repeated without end. In this question, $0.333\\ldots=0.\\overline{3}$, and just as above, $10\\times0.\\overline{3}=3.\\overline{3}$, and therefore, $9\\times0.\\overline{3}=3.\\overline{0}=3$, which means", "add the one from 7/7, giving us $\\frac{207}{7} =29.\\overline{571428}$. ### Division by 9 The fraction 1/9 and its integer multiples are fairly straightforward - they are simply equal to a decimal point followed by the one-digit the numerator repeating to infinity: $\\frac{1}{9} = 0.\\overline{11}$ $\\frac{2}{9} = 0.\\overline{22}$ $\\frac{3}{9} = 0.\\overline{33}$ $\\frac{4}{9} = 0.\\overline{44}$ $\\frac{5}{9} = 0.\\overline{55}$ $\\frac{6}{9} = 0.\\overline{66}$ $\\frac{7}{9} = 0.\\overline{77}$ $\\frac{8}{9} = 0.\\overline{88}$ To solve a problem such as 367/9, we reduce it to $\\frac{367}{9} = \\frac{300}{9} + \\frac{60}{9} + \\frac{7}{9}$ $\\frac{367}{9} = 33.\\overline{33} + 6.\\overline{66} + 0.\\overline{77}$ First add $33.\\overline{33} + 6.\\overline{66} = 40.0$. Then add $40.0 + 0.\\overline{77} = 40.\\overline{77}$. ### Division by 11 The fraction 1/11 and its integer multiples are fairly straightforward - they are simply equal to a decimal point followed by the product of nine and the the integer multiple repeating to infinity: $\\frac{1}{11} = 0.\\overline{09}$ $\\frac{2}{11} = 0.\\overline{18}$ $\\frac{3}{11} = 0.\\overline{27}$ $\\frac{4}{11} = 0.\\overline{36}$ $\\frac{5}{11} = 0.\\overline{45}$ $\\frac{6}{11} = 0.\\overline{54}$ $\\frac{7}{11} = 0.\\overline{63}$ $\\frac{8}{11} = 0.\\overline{72}$ $\\frac{9}{11} = 0.\\overline{81}$ $\\frac{10}{11} = 0.\\overline{90}$ ## Estimation The best way to make estimation quickly in mental math is to round to one or two significant digits (that is, round it to the nearest place of the highest order(s) of magnitude), and then proceed with typical operations. Thus, 1242 * 15645 is approximately", "# How do you express .33 as a fraction in simplest form? Sep 14, 2016 $\\frac{33}{100}$ Sep 14, 2016 As you wrote the question $\\frac{33}{100}$ If you meant $0.333333 . . \\to 0.33 \\overline{3} \\text{ }$ then $\\text{ } \\frac{1}{3}$ #### Explanation: It all depends on what you mean by $0.33$ If you meant it to be exactly as written then we have $\\frac{3}{10} + \\frac{3}{100}$ Write as $\\left(\\frac{3}{10} \\times 1\\right) + \\frac{3}{100}$ But 1 may be written as $\\frac{10}{10}$ $\\left(\\frac{3}{10} \\times \\frac{10}{10}\\right) + \\frac{3}{100}$ $\\frac{30}{100} + \\frac{3}{100} = \\frac{33}{100}$ $\\textcolor{red}{\\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$ If you meant what you wrote to mean $0.33 \\leftarrow$ 3 going on for ever $0.333333333 \\ldots . .$ written as $0.33 \\overline{3}$ $\\textcolor{b r o w n}{\\text{Then this is very different:}}$ Let $x = 0.33 \\overline{3}$ thus $10 x = 3.33 \\overline{3}$ $10 x - x = 3$ $9 x = 3$ $x = \\frac{3}{9} = \\frac{1}{3}$", "+0 # Decimals 0 159 1 What is $0.4\\overline{8} + 0.3\\overline{7}$? Express your answer as a common fraction in lowest terms. Jan 12, 2022 #1 +13900 +1 $$What\\ is\\ 0.4\\overline{8} + 0.3\\overline{7}?$$ Hello Guest! $$0.4\\overline{8} + 0.3\\overline{7}=0.8\\overline 6$$ $$0.86666\\overline 6$$ = x -$$\\underline{0.66666\\overline 6}$$ = 2/3 $$0.2$$ = 1/5 $$x-\\frac{2}{3}=\\frac{1}{5}\\\\ x=\\frac{1}{5}+\\frac{2}{3}=\\frac{3+2\\cdot 5}{15}\\\\ x=\\frac{13}{15}$$ $$0.4\\overline{8} + 0.3\\overline{7}=\\dfrac{13}{15}$$ ! Jan 12, 2022", "(i) from (ii), we get: 9x = 12 = x = $\\frac{4}{3}$ $1.\\overline{)3}=\\frac{4}{3}$ (iii) $0.\\overline{)34}$ Let x$0.\\overline{)34}$ x = 0.3434... ...(i) 100x = 34.3434... ...(ii) On subtracting (i) from (ii), we get: 99x = 34 = x = $\\frac{34}{99}$ $0.\\overline{)34}=\\frac{34}{99}$ (iv) $3.\\overline{)14}$ Let x = $3.\\overline{)14}$ x = 3.1414... ...(i) 100x = 314.1414... ...(ii) On subtracting (i) from (ii), we get: 99x = 311 = x = $\\frac{311}{99}$ $3.\\overline{)14}$ = $\\frac{311}{99}$ (v) $0.\\overline{)324}$ Let x = $0.\\overline{)324}$ x = 0.324324... ...(i) 1000x = 324.324324... ...(ii) On subtracting (i) from (ii), we get: 999x = 324 = x = $\\frac{324}{999}$=$\\frac{12}{37}$ $0.\\overline{)324}$= $\\frac{12}{37}$ (vi) $0.1\\overline{)7}$ Let x = $0.1\\overline{)7}$ x = 0.1777... 10x = 1.777... ...(i) 100x = 17.777.... ...(ii) On subtracting (i) from (ii). we get: 90x = 16 = x = $\\frac{8}{45}$ $0.1\\overline{)7}$ = $\\frac{8}{45}$ (vii) $0.5\\overline{)4}$ Let x = $0.5\\overline{)4}$ x = 0.5444... 10x = 5.4444... ...(i) 100x = 54.4444... ...(ii) On subtracting (i) from (ii), we get: 90x = 49 = x = $\\frac{49}{90}$ $0.5\\overline{)4}$= $\\frac{49}{90}$ (viii) $0.1\\overline{)63}$ Let x$0.1\\overline{)63}$ x = 0.16363... 10x = 1.6363... ...(i) 1000x = 163.6363... ...(ii) On subtracting (i) from (ii), we get: 990x = 162 = x = $\\frac{162}{990}=\\frac{9}{55}$ $0.1\\overline{)63}$ = $\\frac{9}{55}$ #### Question 4: Write, whether the given statement is true or false. Give reasons. (i) Every natural number", "$\\overline{)\\text{}S\\infty =\\frac{a}{1-r}\\text{}}$ Example (Change recurring decimal to fraction) Express each of the following recurring decimals as a fraction in its lowest terms. (a) 0.8888 ... (b) 0.171717... (c) 0.513513513 …. Solution: (a) 0.8888 = 0.8 + 0.08 + 0.008 +0.0008 + ….. (recurring decimal) $\\begin{array}{l}GP,\\text{}a=0.8,\\text{}r=\\frac{0.08}{0.8}=0.1\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.8}{1-0.1}\\\\ {S}_{\\infty }=\\frac{0.8}{0.9}\\\\ {S}_{\\infty }=\\frac{8}{9}\\to \\overline{)\\begin{array}{l}\\text{check using calculator}\\\\ \\text{}\\frac{8}{9}=0.888888....\\end{array}}\\end{array}$ (b) 0.17171717 ….. = 0.17 + 0.0017 + 0.000017 + 0.00000017 + ….. $\\begin{array}{l}GP,\\text{}a=0.17,\\text{}r=\\frac{0.0017}{0.17}=0.01\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.17}{1-0.01}=\\frac{0.17}{0.99}=\\frac{17}{99}\\to \\overline{)\\begin{array}{l}\\text{remember to check the}\\\\ \\text{answer using calculator}\\end{array}}\\end{array}$ (c) 0.513513513….. = 0.513 + 0.000513 + 0.000000513 + ….. $\\begin{array}{l}GP,\\text{}a=0.513,\\text{}r=\\frac{0.00513}{0.513}=0.001\\\\ {S}_{\\infty }=\\frac{a}{1-r}\\\\ {S}_{\\infty }=\\frac{0.513}{1-0.001}=\\frac{0.513}{0.999}=\\frac{513}{999}=\\frac{19}{37}\\end{array}$ # 3. The nth Term of Geometric Progressions (Part 2) (D) The Number of Term of a Geometric Progression Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term Example 2: Find the number of terms for each of the following geometric progressions. (a) 2, 4, 8, ….., 8192 (b) $\\frac{1}{4},\\text{}\\frac{1}{6},\\text{}\\frac{1}{9},\\text{}.....,\\text{}\\frac{16}{729}$ (c) $-\\frac{1}{2},\\text{1},\\text{}-\\text{2},.....,\\text{64}$ Solution: (a) $\\begin{array}{l}2,\\text{}4,\\text{}8,.....,\\text{}8192←\\left(\\text{Last term is given)}\\\\ a=2\\\\ r=\\frac{{T}_{2}}{{T}_{1}}=\\frac{4}{2}=2\\\\ {T}_{n}=8192\\\\ a{r}^{n-1}=8192←\\left(Then\\text{th term of GP,}{T}_{n}=a{r}^{n-1}\\right)\\\\ \\left(2\\right){\\left(2\\right)}^{n-1}=8192\\\\ {2}^{n-1}=4096\\\\ {2}^{n-1}={2}^{12}\\\\ n-1=12\\\\ n=13\\end{array}$ (b) $\\begin{array}{l}\\frac{1}{4},\\text{}\\frac{1}{6},\\text{}\\frac{1}{9},\\text{}.....,\\text{}\\frac{16}{729}\\\\ a=\\frac{1}{4},r=\\frac{\\frac{1}{6}}{\\frac{1}{4}}=\\frac{2}{3}\\\\ {T}_{n}=\\frac{16}{729}\\\\ a{r}^{n-1}=\\frac{16}{729}\\\\ \\left(\\frac{1}{4}\\right){\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{16}{729}\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{16}{729}×4\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}=\\frac{64}{729}\\\\ {\\left(\\frac{2}{3}\\right)}^{n-1}={\\left(\\frac{2}{3}\\right)}^{6}\\\\ \\therefore n-1=6\\\\ n=7\\end{array}$ (c) $\\begin{array}{l}-\\frac{1}{2},\\text{1},\\text{}-\\text{2},.....,\\text{64}\\\\ a=-\\frac{1}{2},\\text{}r=\\frac{-2}{1}=-2\\\\ {T}_{n}=64\\\\ a{r}^{n-1}=64\\\\ \\left(-\\frac{1}{2}\\right){\\left(-2\\right)}^{n-1}=64\\\\ {\\left(-2\\right)}^{n-1}=64×-2\\\\ {\\left(-2\\right)}^{n-1}=-128\\\\ {\\left(-2\\right)}^{n-1}={\\left(-2\\right)}^{7}\\\\ n-1=7\\\\ n=8\\end{array}$ (E) Three consecutive terms of a geometric progression If e, f and g are 3 consecutive terms of", "+0 # Can I get help with 1 question? 0 107 1 +37 Express $$\\frac{0.\\overline{12}}{0.\\overline{321}}$$as a common fraction. Thanks for the help Jun 19, 2021 #1 +2250 +1 12 with line = 12/99 321 with line = 321/999 (12/99)/(321/999) Can you take it from here? =^._.^= Jun 19, 2021", "# Question #a1e8f Jan 6, 2017 $0.6 \\overline{12} = \\frac{101}{165}$ #### Explanation: Let $x = 0.6 \\overline{12}$ Note that the repeating portion of the decimal is $2$ digits long, so we multiply $x$ by ${10}^{2} = 100$. $100 x = 61.2 \\overline{12}$ $\\implies 100 x - x = 61.2 \\overline{12} - 0.6 \\overline{12}$ $\\implies 99 x = 61.2 - 0.6$ $\\implies 99 x = 60.6$ $\\implies 99 x = \\frac{606}{10}$ $\\implies x = \\frac{606}{10} \\cdot \\frac{1}{99}$ $\\therefore x = \\frac{101}{165}$ For a detailed approach to general questions of this type, see this question. Jan 6, 2017 $\\frac{101}{165}$ #### Explanation: Note that $0.6 \\overline{12}$ is the same as $0.612121212 \\ldots$ Let $x = 0.6 \\overline{12}$ Then $10 x = 6.121212 \\ldots$ and $1000 x = 612.121212 \\ldots$ So $1000 x - 10 x = 612.121212 \\ldots$ $\\text{ \"ul(color(white)(61)6.121212...) larr\" subtract}$ $\\text{ } 1000 x - 10 x = 606.000000 \\ldots . .$ $\\implies x \\left(1000 - 10\\right) = 990 x = 606$ $\\implies x = \\frac{606}{990} = \\frac{101}{165}$", "#10 jvang veganjoy \"Joey\" Nov 2015 Middle of Nowhere,AR 2·229 Posts Quote: Originally Posted by Uncwilly Random Synthetic Division is playing on the Who Stage at Bonaroo in 2019. We touched on piecewise functions today (somehow I'm the only student who has used these before?), and someone asked some question that gave me a good excuse to bring up whether $$0.999... = 1$$ (alternatively $$0.\\overline{9} = 1$$). Proofs that I've seen that support the equality include:$\\frac{1}{3} = 0.\\overline{3}, \\frac{1}{3} * 3 = 1, 0.\\overline{3} * 3 = 1$ and$x = 0.\\overline{9}, 10x = 9.\\overline{9}$Subtract $$x$$ from both sides to get$9x = 9, x = 1$. My teacher had a refutation for both; for the first, she said that we are unable to express $$\\frac{1}{3}$$ as a decimal quantity, so $$\\frac{1}{3} \\neq 0.\\overline{3}$$ and the rest of that proof is unfounded. I can go along with that, but I wasn't sure about the second one. She said that, since $$0.\\overline{9}$$ consists of infinitely many decimal places, we are unable to accurately use it in operations, such as multiplication and subtraction. Are these refutations mathematically and/or logically sound? Edit: I briefly mentioned one of the few other arguments against the equality back in the Learning to Learn thread here. However, that argument is based upon the hyperreal number set, which", "≠ 0. (p, q ϵ Z, q ≠ 0). (i) $$0 . \\overline{6}$$ (ii) $$0 . \\overline{47}$$ (iii) $$0 . \\overline{001}$$ (i) $$0 . \\overline{6}$$ = 0.6666………. Let x = 0.6666…………. ∴ x = 0.6666……….. Multiplying both sides by 10, 10x = 6.666….. 10x = 6 + 0.666 10x = 6 + x 10x – x = 6 9x = 6 $$x=\\frac{6}{9}=\\text { form of } \\frac{\\mathrm{p}}{\\mathrm{q}}$$ ∴ $$0 . \\overline{6}=\\frac{2}{3}$$ (ii) $$0 . \\overline{47}$$ = 0.4777….. Let x = $$0 . \\overline{47}$$ then x = 0.4777…….. Here digit 4 is not repeating but 7 is repeated. Let x = 0.4777 Multiplying both sides by 10. 10x = 4.3 + x 10x -x = 4.3 9x = 4.3 Multiplying both sides by 10. 90x = 43 ∴ $$0.4 \\overline{7}=\\frac{43}{90}$$ (iii) $$0 . \\overline{001}$$ x = 0.001001………. Multiplying both sides by 1000 1000x = 1.001001 1000x = 1 + x 1000x – x = 1 999x = 1 $$x=\\frac{1}{999}$$ ∴ $$0 . \\overline{001}=\\frac{1}{999}$$ Question 4. Express 0.99999…….. in the form $$\\frac{p}{q}$$ , Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense. Express 0.99999 in the form of $$\\frac{p}{q}$$, Let x = 0.99999 … Multiplying by 10. 10x = 9.9999…… 10x = 9 + 0.9999……. 10x = 9 + x 10x", "$= \\dot{4}$ This is a pure recurring decimal. $\\\\$ Question 5: Convert each of the following decimals into a fraction : (i) $0.007 (ii)$latex 0.0875 (iii) $9.005 (iv)$latex 4.096 (i) $0.007 =$ $\\frac{7}{1000}$ (ii) $0.0875 =$ $\\frac{875}{10000}$ $=$ $\\frac{175}{2000} = \\frac{7}{40}$ (iii) $9.005 =$ $\\frac{9005}{1000}$ $=$ $\\frac{1801}{200}$ (iv) $4.096 =$ $\\frac{4096}{1000}$ $=$ $\\frac{2048}{500}$ $=$ $\\frac{1024}{250}$ $=$ $\\frac{512}{125}$ $\\\\$ Question 6: Express each of the following decimals as a rational number: (i) $0. \\dot{7}$ (ii) $0.\\overline{36}$ (iii) $4.\\overline{26}$ (iv) $0. 6\\dot{3}$ (v) $0.22\\overline{73}$ (vi) $2.1\\overline{36}$ (i) $0. \\dot{7}$ Let $x=0.7777777 \\ldots$ $\\Rightarrow 10x=7.777777 \\ldots$ $\\Rightarrow 9x = 7 \\ or \\ x =$ $\\frac{7}{9}$ (ii) $0.\\overline{36}$ Let $x=0.363636 \\ldots$ $\\Rightarrow 100x=36.363636 \\ldots$ $\\Rightarrow 99x=36$ $\\ or \\ x=$ $\\frac{4}{11}$ (iii) $4.\\overline{26}$ Let $x=4.26262626 \\ldots$ $\\Rightarrow 100x=426.262626 \\ldots$ $\\Rightarrow 99x=422$ $\\ or \\ x=$ $\\frac{422}{99}$ (iv) $0. 6\\dot{3}$ Let $x=0.63333333 \\ldots$ $\\Rightarrow 10x=6.3333333 \\ldots$ $\\Rightarrow 100x=63.333333 \\ldots$ $\\Rightarrow 90x=57$ $\\ or \\ x=$ $\\frac{19}{30}$ (v) $0.22\\overline{73}$ Let $x=0.227373737373 \\ldots$ $\\Rightarrow 100x=22.73737373737 \\ldots$ $\\Rightarrow 10000x=2273.7373737373 \\ldots$ $\\Rightarrow 9900x=2251$ $\\ or \\ x=$ $\\frac{2251}{9900}$ (vi) $2.1\\overline{36}$ Let $x=2.13636363636 \\ldots$ $\\Rightarrow 10x=21.3636363636 \\ldots$ $\\Rightarrow 1000x=2136.36363636 \\ldots$ $\\Rightarrow 990 x = 2115$ $\\ or \\ x =$ $\\frac{2115}{990}$ $=$ $\\frac{235}{110}$ $latex \\\\$ Question 7: Write the following fractions in descending order by converting them into decimals: (i) $\\frac{8}{25}$ $,$ $\\frac{4}{15}$ $,$ $\\frac{9}{20}$ $,$ $\\frac{3}{8}$", "to base ten: $$\\frac 1 {11} = \\frac 9 {99}= \\frac 9 {10^2-1}$$, so $$\\frac 1 {11} = .\\overline{09}$$" ]

[ "perimeter of a right triangle with the legs 14 cm and 21 cm long?", "25 comprise a Pythagorean triple? 3. What is the perimeter of a right triangle with a hypotenuse of 30 inches and a leg of 18 inches? 4. What is the area of a right triangle with a hypotenuse of 30 inches and a leg of 18 inches?", "Free Version Easy # Calculating the Area of a Right Triangle Given the Legs GEOM-BN1SYL Triangle $MAT$ is a right triangle with leg lengths of 8 and 6 inches. Find the area of triangle $MAT$ in square inches. A $7$ B $24$ C $30$ D $40$ E $48$", "hypotenuse of 15 centimeters. How high is it? 5. A triangular sign has an area of 105 square feet and a height of 15 feet. How long it its base?", "+0 # Help! 0 103 2 +26 The hypotenuse of a right triangle measures $6\\sqrt{2}$ inches and one angle is $45^{\\circ}$. What is the number of square inches in the area of the triangle? Jul 15, 2020 #1 +26 0 Oh! Got it!! If one acute angle of a right triangle is, then the other is, so the triangle is a 45-45-90 triangle. The hypotenuse is times the length of each leg, so each leg has 6. Therefore, the area of the triangle is. Jul 15, 2020 #2 +2428 +1 Nice one! I think you're right :D Jul 15, 2020", "# I need more! Geometry Level 1 A right triangle has a leg of length $7\\text{ m}$ and an area of $84\\text{ m}^2$. What is its perimeter? ×", "# What is the area of a right triangle that has legs 7cm and 4cm long? Jun 28, 2015 It's 14cm. #### Explanation: A = $\\frac{a . b}{2}$ Jun 28, 2015 I found $14 c {m}^{2}$. Have a look:", "leg of an isosceles right triangle [#permalink] 12 Apr 2018, 09:54 Display posts from previous: Sort by", "# Triangle? Geometry Level 3 Find the area of a right triangle whose hypotenuse length is 10 and whose altitude to the hypotenuse is 6. × Problem Loading... Note Loading... Set Loading...", "? Free Version Difficult # Area: Right Triangle/Hypotenuse Given GEOM-EA3FXH One of the legs in right triangle NEA is twice as long as the other leg. The hypotenuse is $6\\sqrt{5}$ feet. Find the area of the triangle in square feet. A $6$ B $24$ C $36$ D $72$ E $144$", "# Square In Right Triangle Geometry Level 2 Find the area (in $$\\text{cm}^2$$) of the shaded green square in the blue isosceles right triangle, whose legs have legnth 6 cm. ×", "1 pt 22. What is the area of the triangle shown below?", "# Find the Area of the Right-angled Triangle with Hypotenuse 40 Cm and One of the Other Two Sides 24 Cm. - Mathematics Sum Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm. #### Solution In right angled triangle ABC Hypotenuse AC = 40 cm One side AB = 24 cm ∴ BC = sqrt(\"AC\"^2 - \"AB\"^2) = sqrt(40^2 - 24^2) = sqrt(1600 - 576) = sqrt(1024) = 32 cm ∴ Area = 1/2 \"AB\" xx \"BC\" = 1/2 xx 24 xx 32 \"cm\"^2 = 384 \"cm\"^2 Concept: Perimeter of Triangles Is there an error in this question or solution? #### APPEARS IN Selina Concise Mathematics Class 8 ICSE Chapter 20 Area of a Trapezium and a Polygon Exercise 20 (A) | Q 13 | Page 224", "Browse Questions # Prove that the area of right-angled traingle of a given hypotenuse is maximum when the triangle is isosceles. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "# The triangle is 24ft tall and the hypotenuse is 26ft, how do you find the area? Jun 9, 2015 $\\cos A = \\frac{24}{26} = 0.923 - \\to \\sin A = \\sin 22.62 = 0.38$ Base CB = $a = 26 \\left(\\sin A\\right) = 26 \\left(0.38\\right) = 10$ $A r e a : \\frac{10 \\left(24\\right)}{2} = 120 s q f t$", "# Square In Right Triangle Geometry Level 2 Find the area (in $$\\text{cm}^2$$) of the shaded green square in the blue isosceles right triangle, whose legs have length 6 cm. ×", "sqrt(7^2 + 24^2) = sqrt(49 + 576) = sqrt625 = 25 cm Let the given inscribed circle touches the sides of the given triangle at points A, B and C respectively. Find the angles of the triangle View solution. Calculate the radius of a inscribed circle of a right triangle if given legs and hypotenuse ( r ) : radius of a circle inscribed in a right triangle : = Digit 2 1 2 4 6 10 F. =. And if someone were to say what is the inradius of this triangle right over here? © Last Updated: 18 July 2019. , - legs of a right triangle. Let the angles be 2x, 3x and 4x. The radius of the incircle of a right triangle can be expressed in terms of legs and the hypotenuse of the right triangle. 3 Diagnosis; 4 Treatment of joint disease ... radius of incircle of right angle triangle Palindromic rheumatism is characterized by sudden and recurrent attacks of painful swelling of one or more joints. Hence the area of the incircle will be PI * ((P + B – H) / … Switch; Flag; Bookmark; 114. For right triangles In the case of a ... where the diameter subtends a right angle to any point on a circle's circumference. We can also", "12.5 units 7. One leg of a right triangle is 18 cm long. The other leg is 4 cm less than the hypotenuse. What are the lengths of this leg and hypotenuse? a. 34.5 cm and 38.5 cm b. 40.5 cm and 44.5 cm c. 38.5 cm and 42.5 cm d. 43.5 cm and 39.5 cm", "a right triangle if leg a = 6 cm and a section of the hypotenuse, which is located adjacent the second leg b is 5cm. • Body diagonal The cuboid has a volume of 32 cm3. Its side surface area is double as one of the square bases. What is the length of the body diagonal?", "What is the area of the triangle?" ]

[ "is the long leg which is the short leg times the square root of 3. $\\displaystyle AB=15\\sqrt{2}\\cdot \\sqrt{3}=15\\sqrt{6}$", "+0 # DUDE HELP ME OUT 0 130 4 In a certain isosceles triangle, the base is $1\\frac 12$ times as long as each leg and the perimeter is $63.$ How long is the base? Mar 8, 2020 #1 +21725 +3 isosceles triangle means two legs are the same length and base is 1/12 as long as leg x = leg length x + x + x/12 =63 solve fo 'x' the leg length then base = x/12 Mar 8, 2020 edited by ElectricPavlov Mar 8, 2020 #2 -3 Mar 8, 2020 #3 -3 its wrong Mar 8, 2020 #4 +4 +2 Let's try setting x equal to the length of each leg. Then, we have two sides of length x and one of length $$\\dfrac 32\\cdot x,$$ so the perimeter is $$x+x+\\dfrac 32\\cdot x,$$which is $$\\dfrac 72\\cdot x.$$We also know the perimeter is 63 so we set our two expressions for the perimeter equal to each other: $$\\frac 72\\cdot x = 63.$$ Multiplying both sides by $$\\frac{2}{7}$$ gives The base is $$1\\frac 12\\cdot 18 = \\boxed{27}$$ Hope it helps! Mar 10, 2020 edited by TechnoWizard Mar 10, 2020 edited by TechnoWizard Mar 17, 2020", "be the sides of a right triangle? If so, is it a special right triangle? 1. Each leg is $\\frac{16}{\\sqrt{2}} = \\frac{16}{\\sqrt{2}} \\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} = \\frac{16\\sqrt{2}}{2} = 8\\sqrt{2}$. 2. Short leg is $\\frac{\\sqrt{6}}{\\sqrt{3}} = \\sqrt{\\frac{6}{3}} = \\sqrt{2}$ and hypotenuse is $2\\sqrt{2}$. 3. Short leg is $\\frac{12}{\\sqrt{3}} = \\frac{12}{\\sqrt{3}} \\cdot \\frac{\\sqrt{3}}{\\sqrt{3}} = \\frac{12\\sqrt{3}}{3} = 4\\sqrt{3}$ and hypotenuse is $8\\sqrt{3}$. 4. The hypotenuse is $4\\sqrt{10} \\cdot \\sqrt{2} = 4\\sqrt{20} = 8\\sqrt{5}$. 5. Each leg is $\\frac{5\\sqrt{2}}{\\sqrt{2}} = 5$. 6. The short leg is $\\frac{15}{2}$ and the long leg is $\\frac{15\\sqrt{3}}{2}$. 7. If the diagonal of a square is 6 ft, then each side of the square is $\\frac{6}{\\sqrt{2}}$ or $3\\sqrt{2} \\approx 4.24 \\ ft$. 8. These are not dimensions for a special right triangle, so to find the diagonal (both are the same length) do the Pythagorean Theorem: $10^2 + 20^2 & = d^2\\\\100 + 400 & = d^2\\\\\\sqrt{500} & = d\\\\10\\sqrt{5} & = d$ So, if each diagonal is $10\\sqrt{5}$, two diagonals would be $20\\sqrt{5} \\approx 45 \\ ft.$ Pablo needs 45 ft of lights for his yard. 9. $2:2:2\\sqrt{3}$ does not fit into either ratio, so it is not a special right triangle. To see if it is a right triangle, plug these values into the Pythagorean Theorem: $2^2 + 2^2 & = \\left ( 2\\sqrt{3} \\right )^2\\\\4", "Free Version Moderate # Area: Isosceles Triangle/ Given the Legs and the Base GEOM-LHRJET The legs of isosceles triangle $RST$ are 10 inches, and the base of the triangle is 12 inches. Find the area of triangle $RST$ in square inches. A $24$ B $48$ C $60$ D $96$ E $120$", "legs of a right triangle are in the ratio $\\frac{\\sqrt{5}}{2}.$ If the hypotenuse is 9 units long, find the area (in square units) of the triangle. 50.) In $\\bigtriangleup ABC, \\angle C = 90^\\circ$ and $cos A = \\frac{3}{5}.$ Find cos B. 26.) x = 6, 9 27.) $140^\\circ$ 28.) $18^\\circ$ 29.) 10 30.) 52 31.) 3 32.) 20 33.) $105^\\circ$ 34.) 5 35.) 6.5 36.) 4 37.) 15 38.) 17.5 or $\\frac{35}{2}$ 39.) 9 40.) 8 cm 41.) 16 : 25 or $\\frac{16}{25}$ 42.) (13, 3) 43.) $\\bigtriangleup BDC$ 44.) $3 \\sqrt{2}$ 45.) $6 \\sqrt{6}$ 46.) $\\sqrt{2}$ 47.) 15 cm 48.) 13 m 49.) $9 \\sqrt{5}$ 50.) $\\dfrac{4}{5}$ or 0.8 View Part I here: 2017 Grade 9 Math Challenge – Elimination Round with answer key – Part I This entry was posted in Grade 9-10 and tagged , , , , . Bookmark the permalink.", "## anonymous one year ago The legs of a right triangle measure 5√2 cm and 4√2 cm. Find the area. 1. anonymous |dw:1436967717958:dw| 2. rishavraj $area = \\frac{ 1 }{ 2 } \\times a \\times b$ 3. anonymous does that mean 40 divided by 2 which in that case would be 20? 4. rishavraj the legs measurement are $5\\sqrt{2}~~~ and ~~~~ 4\\sqrt{2}$ so area = $\\frac{ 1 }{ 2 } \\times 5\\sqrt{2} \\times 4\\sqrt{2}$ 5. rishavraj yupif u mean tht ur answer is 20 then thts correct 6. anonymous okay thank you!", "$4\\sqrt{3}$ is the longer leg because it is opposite the $60^\\circ$. So, in the $x:x\\sqrt{3}:2x$ ratio, $4\\sqrt{3} = x\\sqrt{3}$, therefore $x = 4$ and $2x = 8$. The short leg is 4 and the hypotenuse is 8. b. 17 is the hypotenuse because it is opposite the right angle. In the $x:x\\sqrt{3}:2x$ ratio, $17 = 2x$ and so the short leg is $\\frac{17}{2}$ and the long leg is $\\frac{17\\sqrt{3}}{2}$. c. 15 is the long leg because it is opposite the $60^\\circ$. Even though 15 does not have a radical after it, we can still set it equal to $x\\sqrt{3}$. $x\\sqrt{3} & = 15\\\\x & = \\frac{15}{\\sqrt{3}} \\cdot \\frac{\\sqrt{3}}{\\sqrt{3}} = \\frac{15\\sqrt{3}}{3} = 5\\sqrt{3} \\quad \\text{So, the short leg is} \\ 5\\sqrt{3}.$ Multiplying $5\\sqrt{3}$ by 2, we get the hypotenuse length, which is $10\\sqrt{3}$. d. $24\\sqrt{21}$ is the length of the hypotenuse because it is opposite the right angle. Set it equal to $2x$ and solve for $x$ to get the length of the short leg. $2x & = 24\\sqrt{21}\\\\x &= 12\\sqrt{21}$ To find the length of the longer leg, we need to multiply $12\\sqrt{21}$ by $\\sqrt{3}$. $12\\sqrt{21} \\cdot \\sqrt{3} = 12\\sqrt{3 \\cdot 3 \\cdot 7} = 36\\sqrt{7}$ The length of the longer leg is $36\\sqrt{7}$. Be careful when doing these problems. You can always check your answers by finding", "of the shorter leg is $6$ ft. The length of the longer leg is $6+2$ or $8$ ft and the hypotenuse is $6+4$ or $10$ ft.", "of the leg. $\\mathrm {Therefore\\;(L_S/L_N)^2 = 4 \\\\ L_S/L_N = 2}$", "otherwise, so we actually care about the time when $d = 300$. So we have a triangle with hypotenuse $300$ and one leg $200$, so the other leg is of length $sqrt{300^2 – 200^2}=100 sqrt 5=x$. Now we put it all together. From $d(t)^2 = 200^2 + x(t)^2$, we differentiate to get $2d(t)d'(t) = 2x(t)x'(t)$. We are looking to find $d'(t)$ when $d(t) = 300$, and we found that at this time $x(t) = 100sqrt{5}$. We also know that $x'(t) = 20$ at all times. Plugging these in, we get that $2 cdot 300 cdot d'(t) = 2 cdot 100sqrt 5 cdot 20$, or that $d'(t)=dfrac{200 sqrt 5 cdot 20}{600} = dfrac{20 sqrt 5}{3}$ feet per second. The class did really, really well on this quiz. I was impressed. The vase majority of the class made at least an 8. The only mistake on average was to mistake which leg was implied by “distance from the kite to the girl.” This entry was posted in Brown University, Math 90, Mathematics and tagged , , , , , , , . Bookmark the permalink.", "the negative value. Considering Equation (3), ${c}_{1} / {c}_{2} = \\frac{{b}_{1} + {b}_{2}}{{a}_{1} + {a}_{2}}$ $\\frac{36}{40.34} = \\frac{40 + 32}{a}$ where $a = \\left({a}_{1} + {a}_{2}\\right)$ $a = \\frac{72 \\cdot 40.34}{36} = 80.68$ Considering Eqn (2), $\\frac{a - {a}_{2}}{a} _ 2 = \\frac{{c}_{1} + {c}_{2}}{{b}_{1} + {b}_{2}} = \\frac{76.34}{72}$ #80.68 - a_2 = (76.34/72) * a_2 $148.34 {a}_{2} = 80.68 \\cdot 72$ ${a}_{2} = \\frac{80.68 \\cdot 72}{148.34} = 39.16$ ${a}_{1} = 80.68 - {a}_{2} = 80.68 - 39.16 = 41.52$ Perimeter of the triangle ABC $P = \\left(a + b + c\\right) = 80.68 + 72 + 76.34 \\textcolor{b l u e}{= 229.02}$ Similarly, we can find P for Question (b) ## A right triangle has a hypotenuse of 30 inches. Both of its legs are the same length. How long is a leg of this right triangle? sankarankalyanam Featured 1 week ago Length of a leg $\\vec{A C} = \\vec{B C} = \\textcolor{g r e e n}{21.21}$ #### Explanation: In triangle ABC in the above figure' hypotenuse vec(AB) = 30\" $\\hat{B} , \\hat{A}$ are equal and $\\hat{C} = {90}^{0}$ $\\therefore \\hat{B} = \\hat{A} = \\frac{180 - 90}{2} = {45}^{0}$ Using trigonometric functions, #vec(AC) = vec(AB) * sin B = 30 * sin(45) = 30 * (1/sqrt2) = 21.21\"# #vec(BC) = vec(AB) = 21.21\"# ## Find the equation", "The longer leg of a right triangle is 3 inches more than 3 times the length of the shorter leg. The area of the triangle is 84 square inches. How do you find the perimeter of a right triangle? P = 56 square inches. Explanation: See figure below for better understanding. $c = 3 b + 3$ $\\frac{b . c}{2} = 84$ $\\frac{b . \\left(3 b + 3\\right)}{2} = 84$ $3 {b}^{2} + 3 b = 84 \\times 2$ $3 {b}^{2} + 3 b - 168 = 0$ ${b}_{1} = 7$ ${b}_{2} = - 8$ (impossible) So, $b = 7$ $c = 3 \\times 7 + 3 = 24$ ${a}^{2} = {7}^{2} + {24}^{2}$ ${a}^{2} = 625$ $a = \\sqrt{625} = 25$ $P = 7 + 24 + 25 = 56$ square inches", "taken together is base times height 23. misty1212 ok i see $\\frac{6}{\\sqrt{3}}=2\\sqrt3$ 24. misty1212 so total area is $6\\times 2\\sqrt3 = 12\\sqrt 3$ 25. anonymous fml i've been doin it all wrong this whole time 26. misty1212 you got the height right away then i think you confused yourself and turned the height in to something else first you said it was 6 (correct) then you said it was $$4\\sqrt3$$ 27. anonymous So |dw:1441315715658:dw| 28. misty1212 yeah that looks good to me 29. anonymous i dont get why it's 2 root 3 is the problem like why does it convert that way and the process 30. anonymous i thought in a 30-60-90 triangle the root three would replace the 6 31. misty1212 k lets go slow you said you had a 30 - 60 -90 triangle right? 32. anonymous yea 33. misty1212 the ratios are $1:\\sqrt3:2$ for short leg ; long leg ; hypotenuse 34. anonymous yea 35. misty1212 the long leg here is 6 36. misty1212 so the short leg is $\\frac{6}{\\sqrt3}=2\\sqrt3$ 37. anonymous how? 38. misty1212 you multiply the short leg by root 3 to get the long leg right? 39. misty1212 so to go the other way, from the long leg to the short leg, you divide 40. anonymous yea but like why is it", "$x : x : x \\sqrt{2}$ ratio. $AB = 9 \\sqrt{2}$ because it is equal to $AC$ . So, $BC = 9 \\sqrt{2} \\cdot \\sqrt{2} = 9 \\cdot 2 = 18$ . 2. Use the $x : x : x \\sqrt{2}$ ratio. We need to solve for $x$ in the ratio. $12 \\sqrt{2} &= x \\sqrt{2}\\\\12 &= x$ 3. $x=5\\sqrt{3}\\cdot \\sqrt{2}=5\\sqrt{6}$ 1. Find the length of the missing sides. 2. Find the value of $x$ and $y$ . 3. $x$ is the hypotenuse of a 30-60-90 triangle and $y$ is the longer leg of the same triangle. The shorter leg has a length of $6$ . 1. We are given the hypotenuse. $2x = 20$ , so the shorter leg, $f = 10$ , and the longer leg, $g = 10 \\sqrt{3}$ . 2. We are given the hypotenuse. $2x &= 15 \\sqrt{6}\\\\x &= \\frac{15 \\sqrt{6}}{2}\\\\$ The, the longer leg would be $y = \\left ( \\frac{15 \\sqrt{6}}{2} \\right ) \\cdot \\sqrt{3} = \\frac{15 \\sqrt{18}}{2} = \\frac{45 \\sqrt{2}}{2}$ 3. We are given the shorter leg. $& x=2(6)\\\\& x = 12\\\\& \\text{The longer leg is}\\\\& y= 6 \\cdot \\sqrt{3} = 6 \\sqrt{3}$ ### Practice 1. In a 30-60-90 triangle, if the shorter leg is 5, then the longer leg is __________ and the hypotenuse is ___________. 2. In a", "is the length of the bottom of the leg, we know the leg is $$x*\\sqrt{3}=bottom~~leg$$ 25. anonymous 3.5 sqrt 3 26. Zale101 Yes :) 27. Zale101 Hypotenuse=7 right side leg= 3.5 bottom leg=3.5* sqrt(3) 28. Zale101 How do you solve the perimeter now, what do you do? 29. anonymous 30. anonymous @Zale101 31. Zale101 yep, you got it now 32. anonymous i still dont understand how that turns out to be one of the answer choices though @zale101 33. Zale101 $$\\LARGE \\frac{7}{2}+7+\\frac{7}{2}\\sqrt{3}$$ this is your perimeter But if we add $$\\large\\frac{7}{2}+7$$ $$\\Large\\frac{7}{2}+\\frac{7}{1}=\\frac{7+14}{2}=\\frac{21}{2}$$ So, we will end up with $$\\LARGE \\frac{21}{2}+\\frac{7}{2}\\sqrt{3}$$ as your simplified perimeter 34. Zale101 a.7/2+21/2 sqrt 3 b. 21+7 sqrt 3 c. 7+21 sqrt 3 d.21/2 + 7/2 sqrt 3 <-------- Your Perimeter was hidden here (but simplified )", "BC = 4 Cos 60 = BC/AB AB = BC/Cos 60 = BC/(1/2) = 2BC = 8 Option E Manager Joined: 07 May 2015 Posts: 175 Location: India GMAT 1: 660 Q48 V31 GPA: 3 Re: If the area of ΔABC is , what is the length of AB? [#permalink] ### Show Tags 02 Jan 2016, 07:43 This is an 30- 60 -90 triangle so the lengths of the sides will be i nratio: x(sqrt(3)) : x: 2x Area given is 8(sqrt(3)) 1/2 * x(sqrt(3)) * x = 8(sqrt(3)) thus x = 4 since BC = 2x => 8 Ans:E SC Moderator Joined: 22 May 2016 Posts: 1827 If the area of ΔABC is , what is the length of AB? [#permalink] ### Show Tags 12 Aug 2017, 15:25 Bunuel wrote: If the area of ΔABC is $$8\\sqrt{3}$$, what is the length of AB? A. 4 B. 5 C. 6 D. 7 E. 8 Attachment: 2015-12-27_2136.png The missing angle A must be 30° I have a hard time tracking on side names, so I use terms such as \"short leg.\" 30-60-90 right triangles have sides in ratio short leg: long leg: hypotenuse $$x: x\\sqrt{3}: 2x$$ Let short leg BC = base = x Let long leg AC = height=$$x\\sqrt{3}$$ Area Δ = $$\\frac{b*h}{2}$$, given as $$8\\sqrt{3}$$ $$\\frac{x *", "leg equals $4\\cdot 4/3=8\\cdot 2/3$ so that the vertical leg is $2/3$ of the side of the square. The hypotenuse of that triangle equals $4\\cdot 5/3=8\\cdot 5/6,$ i.e. $5/6$ of the side of the square. Thus the long leg of the small protruding triangle at the left edge of the square is $1/6$ of the side of the square, making the short side $1/8$ and the hypotenuse $5/24.$ It follows that the second fold marks exactly $8\\cdot 1/8=1$ unit up from the bottom edge of the square. ### References 1. Kazuo Haga. et al, Origamics: Mathematical Explorations Through Paper Folding, World Scientific Publishing Company, 2008 Paper Folding Geometry", "then x can be any value from 1 to 2 if y = 1991 then x = 1 if y = 1992 then x = 1. So the total of all possibilities is $2 \\times \\sum_{i=1} ^{996} i = 2 \\times \\frac {996 \\times 997} 2 = 993,012$. However, given the strange wording of the question perhaps OP means that no single leg may exceed length of 1994. If that's what he meant then there are 1994 choices for length of leg x and 1994 choices for leg y, so the total number of possibilities is 1994 x 1994 = 3,976,036. I followed your last sentence and erred my counts should have been 1993 and 1993 but aren't they additive.You are notallowed to make an equilateral triangle for example", "+0 # something 0 26 2 The longer leg of a right triangle is three times as long as the shorter leg. The hypotenuse is sqrt(5) What is the area of this triangle? Oct 13, 2021 #1 +23 0 We know that shortleg^2 + longleg^2 = $$\\sqrt{5}^2$$ or 5 If the shorter leg is x then the longer leg would be 3x Therefore we have: $$x^2 + 9x^2 = 5$$ simplify it to $$10x^2 = 5$$ divide 4 each side $$x^2 = \\frac{5}{10}$$ square root both side $$x = \\sqrt{\\frac{5}{10}}$$ So the shorter leg is $$\\sqrt{\\frac{5}{10}}$$ and the longer leg is $$3 \\sqrt{\\frac{5}{10}}$$ Now we need to find the area Multiply them: $$3\\sqrt{\\frac{5}{10}}\\cdot \\sqrt{\\frac{5}{10}} = 3 \\cdot \\frac{5}{10} = \\frac{15}{10} = \\frac{3}{2}$$ And don't forget we also have to divide by 2 $$\\frac{3}{2} \\div 2 = \\frac{3}{4}$$ Oct 14, 2021 #2 0 Try this: x2 + (3x)2 = 5 Guest Oct 14, 2021", "of the remaining parts of the two legs are respectively $$8-a$$ and $$4-a$$. So the total area which is $$\\frac{1}{2}.a.(8-a)+a^{2}+\\frac{1}{2}.a.(4-a)$$ will be equal to $$16$$ (the area of the whole triangle) Solving the equation you get $$x=\\frac{8}{3}$$ and hence the area is $$\\frac{64}{9}$$ · 2 years, 4 months ago See this problem. Staff · 2 years, 4 months ago Could you elaborate his solution @Calvin Lin · 2 years, 4 months ago You should comment on his solution and ask him to clarify. Staff · 2 years, 4 months ago Comment deleted Jan 08, 2015 Comment deleted Jan 08, 2015 Sorry i misread the question · 2 years, 4 months ago" ]

[ "# I need more! Geometry Level 1 A right triangle has a leg of length $7\\text{ m}$ and an area of $84\\text{ m}^2$. What is its perimeter? ×", "? Free Version Difficult # Area: Right Triangle/Hypotenuse Given GEOM-EA3FXH One of the legs in right triangle NEA is twice as long as the other leg. The hypotenuse is $6\\sqrt{5}$ feet. Find the area of the triangle in square feet. A $6$ B $24$ C $36$ D $72$ E $144$", "Rectangles Diagonal Calculation I was having a problem with the following question, and could use some help: If a rectangle with a perimeter of 48 inches is equal in area to a right triangle with legs of 12 inches and 24 inches, what is the rectangle's diagonal? The answer to the above question is $12\\sqrt{2}$. Frankly I am a bit confused by a part of the question stating in area to a right triangle with legs of 12 inches and 24 inches Is the area of rectangle equal to the area of the triangle? Frankly, the phrase \"equal to triangle\" doesn't say much. And 12 and 24 the lengths of which sides of a triangle? Am I missing something here or are my concerns valid? Edit: After reading the suggestions posted here, here is what I did, and I am getting a square instead of a rectangle: $$2x + 2y = 48 \\qquad (A)$$ $$xy = 144 \\qquad (B)$$ so $x=144/y$, inserting in $(A)$ I get $x=12$. So is this actually a square? - The phrase \"with a perimeter of 48 inches\" describes the rectangle; the phrase \"with legs of 12 inches and 24 inches\" describes the right triangle. If you hide these phrases, then you can read the question like this: \"If this rectangle is equal in", "Free Version Moderate # Area: Isosceles Triangle/ Given the Legs and the Base GEOM-LHRJET The legs of isosceles triangle $RST$ are 10 inches, and the base of the triangle is 12 inches. Find the area of triangle $RST$ in square inches. A $24$ B $48$ C $60$ D $96$ E $120$", "is the area of the path, in square feet? (A) $$25\\pi$$ (B) $$38\\pi$$ (C) $$55\\pi$$ (D) $$57\\pi$$ (E) $$64\\pi$$ In solving this problem we first must recognize that the flower bed is the right triangle with sides of y yards, x yards, and z yards. We are given that the area of the bed (which is the right triangle) is 24 square yards. Since we know that area of a triangle is ½ Base x Height, we can say: 24 = ½(xy) 48 = xy We also know that x = y + 2, so substituting in y + 2 for x in the area equation we have: 48 = (y+2)y 48 = y^2 + 2y y^2 + 2y – 48 = 0 (y + 8)(y – 6) = 0 y = -8 or y = 6 Since we cannot have a negative length, y = 6. We can use the value for y to calculate the value of x. x = y + 2 x = 6 + 2 x = 8 We can see that 6 and 8 represent two legs of the right triangle, and now we need to determine the length of z, which is the hypotenuse. Knowing that the length of one leg is 6 and the other leg is 8, we know", "# A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/2, the angle between sides B and C is pi/12, and the length of B is 4, what is the area of the triangle? Dec 21, 2016 The area of the triangle is: $S = 8 \\cdot \\left(2 - \\sqrt{3}\\right)$ #### Explanation: This is a right triangle, where $A$ and $B$ are the legs, and $C$ the hypotenuse. In a right triangle the ratio between the legs equals the tangent of the angle opposite the leg at the numerator, so that we have: $\\frac{A}{B} = \\tan \\left(\\frac{\\pi}{12}\\right)$ or: $A = B \\tan \\left(\\frac{\\pi}{12}\\right) = 4 \\left(2 - \\sqrt{3}\\right)$ And the area is: $S = \\frac{1}{2} A \\cdot B = 8 \\cdot \\left(2 - \\sqrt{3}\\right)$", "+0 # Help! 0 103 2 +26 The hypotenuse of a right triangle measures $6\\sqrt{2}$ inches and one angle is $45^{\\circ}$. What is the number of square inches in the area of the triangle? Jul 15, 2020 #1 +26 0 Oh! Got it!! If one acute angle of a right triangle is, then the other is, so the triangle is a 45-45-90 triangle. The hypotenuse is times the length of each leg, so each leg has 6. Therefore, the area of the triangle is. Jul 15, 2020 #2 +2428 +1 Nice one! I think you're right :D Jul 15, 2020", "# sin of what is equal to sqrt3/2? May 3, 2018 $\\sin 60$ degrees or $\\frac{\\pi}{3}$ radians #### Explanation: In a 30-60-90 triangle, the sides are in the ratio $x : x \\sqrt{3} : 2 x$ (smallest leg:longest leg:hypotenuse). $\\sin$ is opposite side over hypotenuse The opposite side for the $90$ degree angle is the hypotenuse, so $\\sin 90$ is $1$ The opposite side for the $30$ degree angle is the smallest leg ($x$). The opposite side for the $60$ degree angle is the longest leg ($x \\sqrt{3}$). $\\frac{x \\sqrt{3}}{2 x} = \\frac{\\sqrt{3}}{2}$ May 3, 2018 $\\rightarrow x = n \\pi + {\\left(- 1\\right)}^{n} \\cdot \\left(\\frac{\\pi}{3}\\right)$ where $n \\rightarrow Z$ #### Explanation: Let $\\sin x = \\frac{\\sqrt{3}}{2}$ $\\rightarrow \\sin x = \\sin \\left(\\frac{\\pi}{3}\\right)$ $\\rightarrow x = n \\pi + {\\left(- 1\\right)}^{n} \\cdot \\left(\\frac{\\pi}{3}\\right)$ where $n \\rightarrow Z$", "+0 # DUDE HELP ME OUT 0 130 4 In a certain isosceles triangle, the base is $1\\frac 12$ times as long as each leg and the perimeter is $63.$ How long is the base? Mar 8, 2020 #1 +21725 +3 isosceles triangle means two legs are the same length and base is 1/12 as long as leg x = leg length x + x + x/12 =63 solve fo 'x' the leg length then base = x/12 Mar 8, 2020 edited by ElectricPavlov Mar 8, 2020 #2 -3 Mar 8, 2020 #3 -3 its wrong Mar 8, 2020 #4 +4 +2 Let's try setting x equal to the length of each leg. Then, we have two sides of length x and one of length $$\\dfrac 32\\cdot x,$$ so the perimeter is $$x+x+\\dfrac 32\\cdot x,$$which is $$\\dfrac 72\\cdot x.$$We also know the perimeter is 63 so we set our two expressions for the perimeter equal to each other: $$\\frac 72\\cdot x = 63.$$ Multiplying both sides by $$\\frac{2}{7}$$ gives The base is $$1\\frac 12\\cdot 18 = \\boxed{27}$$ Hope it helps! Mar 10, 2020 edited by TechnoWizard Mar 10, 2020 edited by TechnoWizard Mar 17, 2020", "The longer leg of a right triangle is 3 inches more than 3 times the length of the shorter leg. The area of the triangle is 84 square inches. How do you find the perimeter of a right triangle? P = 56 square inches. Explanation: See figure below for better understanding. $c = 3 b + 3$ $\\frac{b . c}{2} = 84$ $\\frac{b . \\left(3 b + 3\\right)}{2} = 84$ $3 {b}^{2} + 3 b = 84 \\times 2$ $3 {b}^{2} + 3 b - 168 = 0$ ${b}_{1} = 7$ ${b}_{2} = - 8$ (impossible) So, $b = 7$ $c = 3 \\times 7 + 3 = 24$ ${a}^{2} = {7}^{2} + {24}^{2}$ ${a}^{2} = 625$ $a = \\sqrt{625} = 25$ $P = 7 + 24 + 25 = 56$ square inches", "$$p ^ *$$. I don't usually deal with the measured and negative Sobolev spaces, so I don't know much about them. Help would be great and I definitely need it. And any additional reference besides the two above would be nice. Thanks in advance. ## dnd 3.5e – three-dimensional measurements As usual in D & D 3.5e, I refer to \"squares\" when I really mean \"cubes\". Just take \"square\" as game jargon, which it is in this case. Anyway, the \"5-foot, 10-foot.\" is an approximation of the cost of diagonals $$1.5 times$$ the distance, which itself approximates the cost $$sqrt {2} times approx.1,414 times$$ (Pythagoras' theorem says a right angle $$a$$ because the legs have a hypotenuse of $$sqrt {2} times a$$). A \"double diagonal\" is the hypotenuse of a right angle with legs of $$a$$ and $$sqrt {2} times a$$, this is how the hypotenuse will be $$sqrt {3} times a$$, so we need an approximation of $$sqrt {3} times approx.1,732 times$$. If we round it up $$1.75 times$$we need \"5 feet, 10 feet, 10 feet, 10 feet\". (So ​​moving four squares costs 35 feet of motion –$$1.75 times$$ the 20 ft. it would normally take. Obviously \"5 feet, 10 feet, 10 feet, 10 feet\". is a pain, and it is also much more questionable to start", "know $f = 2\\cdot e$ You have 4 congruent right triangles with the legs $\\frac12 f$ and $\\frac12 e$ and the hypotenuse of 15 cm: $\\left(\\frac12 f \\right)^2 + \\left(\\frac12 e \\right)^2 = 15^2$ Substitute f = 2e: $\\left(e \\right)^2 + \\left(\\frac12 e \\right)^2 = 15^2$ $\\frac54 e^2 = 225~\\implies~e^2=180~\\implies~e=6\\sqrt{5}$ Consequently $f = 12 \\sqrt{5}$ The area of a rhombus is calculated by: $a=\\dfrac12 \\cdot e \\cdot f$ With your values you get: $a = \\dfrac12 \\cdot 6\\sqrt{5} \\cdot 12 \\sqrt{5} = 180$", "+0 Pythagoras 0 95 1 What is the perimeter of a right triangle with a 34 inch hypotenuse and a 1 foot leg? Write your answer in inches in simplest form. Mar 30, 2021 #1 +544 +1 1 foot = 12 inches. This means the other leg is: $$\\sqrt{34^2-12^2} = \\sqrt{1012} = 2\\sqrt{253}$$ Which means the perimeter of this triangle is $$2\\sqrt{253} + 12 + 34 = 46 + 2\\sqrt{253} ≈ 77.81$$ Mar 30, 2021 #1 +544 +1 1 foot = 12 inches. This means the other leg is: $$\\sqrt{34^2-12^2} = \\sqrt{1012} = 2\\sqrt{253}$$ Which means the perimeter of this triangle is $$2\\sqrt{253} + 12 + 34 = 46 + 2\\sqrt{253} ≈ 77.81$$", "view it Senior SC Moderator Joined: 22 May 2016 Posts: 2642 A flat triangular cornfield has the dimensions shown in the figure abo [#permalink] ### Show Tags 15 May 2017, 09:07 1 phangureh wrote: Attachment: gmat190507-1-735112.jpg A flat triangular cornfield has the dimensions shown in the figure above. If y^2 = 2, what is the area of the field in square miles? (A) 1/4 (B) √3/4 (C) 1/2 (D) √3/2 (E) 1 If a right triangle's hypotenuse is twice the length of the shorter leg (here, hypotenuse = y and shorter leg = $$\\frac{y}{2}$$), it is a 30-60-90 triangle with side ratio x : $$\\sqrt{3}$$x : 2x We also know that $$y^2$$=2 Take the square root of both sides to get y = $$\\sqrt{2}$$ Short leg is $$\\frac{y}{2}$$ = $$\\frac{√2}{2}$$ Longer leg is $$\\sqrt{3}$$*$$\\frac{√2}{2}$$ = $$\\frac{√6}{2}$$ Area = (b*h)/2 $$\\frac{√6}{2}$$ * $$\\frac{√2}{2}$$ divided by 2 $$\\frac{√12}{4}$$ = $$\\frac{(2√3)}{4}$$ divided by 2 $$\\frac{(2√3)}{4}$$ ÷ 2 = $$\\frac{(2√3)}{8}$$= $$\\frac{√3}{4}$$ _________________ Listen, are you breathing just a little, and calling it a life? -- Mary Oliver For practice SC questions with official explanations that were posted and moderated by the SC Team, go to SC Butler here: https://gmatclub.com/forum/project-sc-butler-get-2-sc-questions-everyday-281043.html Non-Human User Joined: 09 Sep 2013 Posts: 10564 Re: A flat triangular cornfield has the dimensions shown in the figure abo [#permalink] ###", "# The diagonal of a square is sqrt 84. What is the area of the square. The diagonal of a square is the hypotenuse of a 45-45-90 triangle. In a 45-45-90 triangle the hypotenuse is the leg multiplied by `sqrt(2).` Therefore, if given the hypotenuse, divide by `sqrt(2)` to find the length of the leg. `sqrt(84)/sqrt(2) =sqrt(42)` Therefore each side of the square is `sqrt(42).` To find the... Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime. The diagonal of a square is the hypotenuse of a 45-45-90 triangle. In a 45-45-90 triangle the hypotenuse is the leg multiplied by `sqrt(2).` Therefore, if given the hypotenuse, divide by `sqrt(2)` to find the length of the leg. `sqrt(84)/sqrt(2) =sqrt(42)` Therefore each side of the square is `sqrt(42).` To find the area of the square: `A=s^2.` `A = (sqrt(42))^2` Therefore the area of the square is: `42 u^2` Approved by eNotes Editorial Team The length of the diagonal of a square with side S is equal to `sqrt(S^2 + S^2)` = `S*sqrt 2` . If the diagonal of a square is D, the length of the side is `D/sqrt 2` A square has a diagonal of length `sqrt 84` . The area of the square is `(sqrt 84)^2/2 = 84/2 =", "# Volume of Sold with a Known Crosssection 1. Apr 7, 2007 ### mmg0789 the base is bounded by x^2 + y^2 = 4 the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure i just need to know what area formula i would start out with, is it just bh/2 ? Last edited: Apr 7, 2007 2. Apr 7, 2007 ### HallsofIvy Staff Emeritus An \"isosceles right triangle perpendicular to the x axis\" has hypotenuse y. Since it is isosceles, a2+ a2= 2a2= y2. That is, each leg has length $a= y\\sqrt{2}$ and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y. 3. Apr 9, 2007 ### mmg0789 hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be (1/4)y^2. Right? 4. Apr 10, 2007 ### MalayInd yes, you are right.", "# The hypotenuse of an isosceles right triangle is 12 inches. What is the length of each leg? Jan 8, 2017 Length of each leg is $8.4852$ inches. #### Explanation: In a right triangle, square on the hypotenuse is equal to sum of the squares on the other two sides. Here, other two sides as it is an isosceles right triangle. Let these sides be each $x$ inches. Hence ${x}^{2} + {x}^{2} = {12}^{2}$ or $2 {x}^{2} = 144$ or ${x}^{2} = 72$ i.e. $x = \\sqrt{72} = 6 \\sqrt{2} = 6 \\times 1.4142 = 8.4852$ Hence, length of each leg is $8.4852$ inches.", "\\sqrt{2} = 9 \\cdot 2 = 18$ . 2. Use the $x : x : x \\sqrt{2}$ ratio. We need to solve for $x$ in the ratio. $12 \\sqrt{2} &= x \\sqrt{2}\\\\12 &= x$ 3. $x=5\\sqrt{3}\\cdot \\sqrt{2}=5\\sqrt{6}$ ### Practice 1. In an isosceles right triangle, if a leg is $x$ , then the hypotenuse is __________. 2. In an isosceles right triangle, if the hypotenuse is $x$ , then each leg is __________. 3. A square has sides of length 15. What is the length of the diagonal? 4. A square’s diagonal is 22. What is the length of each side? 5. A square has sides of length $6\\sqrt{2}$ . What is the length of the diagonal? 6. A square has sides of length $4 \\sqrt{3}$ . What is the length of the diagonal? 7. A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal). 8. Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is $s$ , what are the lengths of the legs of the isosceles triangles? Find the lengths of the missing sides. Simplify all radicals.", "Free Version Easy # Calculating the Area of a Right Triangle Given the Legs GEOM-BN1SYL Triangle $MAT$ is a right triangle with leg lengths of 8 and 6 inches. Find the area of triangle $MAT$ in square inches. A $7$ B $24$ C $30$ D $40$ E $48$", "$\\textcolor{w h i t e}{\\text{XXX}} t = r \\cdot \\cos \\left({60}^{\\circ}\\right)$ $\\textcolor{w h i t e}{\\text{XXXx}} = 7 \\cdot \\frac{\\sqrt{3}}{2}$ and $\\textcolor{w h i t e}{\\text{XXX}} s = 2 t = 7 \\cdot \\sqrt{3}$ $\\textcolor{w h i t e}{\\text{XXXx}} = 12.11$ (since we are told to use $\\sqrt{3} = 1.73$) Perimeter $= 3 s$ $\\textcolor{w h i t e}{\\text{XXXXXX}} = 3 \\times 12.11 = 36.33$ ## The base of an isosceles triangle is 5, the height lowered from one end of the base is 4. What is the surface area? Roella W. 6.383561643835616 years ago $10$ #### Explanation: The area of any triangle is given by $\\frac{1}{2} \\cdot b \\cdot h$ where $b$ is the base and $h$ is the hieght. Therefore the area here is $\\frac{1}{2} \\cdot 5 \\cdot 4 = 10$ ## The length of a leg of an isosceles right triangle is 5sqrt2. How do you find the length of the hypotenuse? Don't Memorise 6.380821917808219 years ago The hypotenuse $A B = 10 c m$ #### Explanation: The above triangle is a right angled isosceles triangle , with $B C = A C$ The length of the leg given $= 5 \\sqrt{2} c m$ (assuming units to be in cm) So, $B C = A C = 5 \\sqrt{2} c m$ The value of the" ]

[ "777 How many 3-digit numbers are a multiple of 7? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "7?...", "i<7 and 7", "digit of $3^3$ which is 7..", "7 total.", "# Least common multiple ### Problem What is the least common multiple of 7 and 5? Another way to say this is:", "200 is not a multiple of 7. There just isn't enough information.", "# Multiples of 7 and 13 How many 3 digit positive integers $$N$$ are there, such that $$N$$ is a multiple of both 7 and 13? ×", "is the least common multiple of the $n_i$. -", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "+0 # Math Help 0 82 1 +169 How many positive multiples of 7 that are less than 1000 end with the digit 3? matth99 Aug 16, 2018 #1 +20153 +1 How many positive multiples of 7 that are less than 1000 end with the digit 3? $$\\begin{array}{|r|r|r|} \\hline & n & 7n \\\\ \\hline 1. &9& 63\\\\ 2. &19& 133\\\\ 3. &29& 203\\\\ 4.&39& 273\\\\ 5.&49& 343\\\\ 6.&59& 413\\\\ 7.&69& 483\\\\ 8.&79& 553\\\\ 9.&89& 623\\\\ 10.&99& 693\\\\ 11.&109& 763\\\\ 12.&119& 833\\\\ 13.&129& 903\\\\ 14.&139& 973\\\\ \\hline \\end{array}$$ 14 positive multiples of 7 that are less than 1000 end with the digit 3. heureka Aug 16, 2018", "Are you ready to take on this challenging Number Systems problem? # Multiples of 7 and 13 How many 3 digit positive integers $N$ are there, such that $N$ is a multiple of both 7 and 13? 3 tries left ·", "multiples of 8 and 12 but 24 is the lowest.", "7; hence the difference of any pair is a multiple of 7. -", "# Multiples of 5 and 7 Number Theory Level 3 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# \"Double\" Divisors If a three-digit number $$\\overline{abc}$$ has 28 positive divisors, how many positive divisors does the six-digit number $$\\overline{abcabc}$$ have? ×", "positive 12 minus 7 which is 5. Yeah. Good. OK. Good. There, (a) and (d) work.", "# Heaven seven Number Theory Level 5 How many positive divisors of $$20020^{12}$$ have 7 as their units digit? ×", "the 49th negative multiple of 7, we will simply multiply 7 by -49. Hence, $$7\\;\\times\\;-49\\;=\\;-343$$ . Therefore, the 49th negative multiple of 7 is –343. The table below shows the result if we do skip counting and multiplication. Multiplication is the simplest way to find the multiples of 7. Skip counting, on the other hand, is a fun way to determine multiples. ## List of First 30 Multiples of 7 The table below shows the first 30 positive multiples of 7. We cannot list all the possible multiples of 7 as there are infinitely many natural numbers. Thus, there are also infinitely many multiples of 7. ## Did you know that… The number 7 is significantly common in nature than most of us realize? There are seven oceans, seven continents, and seven colors in the rainbow. The Earth was also created in seven days, including the rest day. This explains why if people are asked their favorite number from 1 to 10, they usually answer 7. Seven is indeed a remarkable number. Can you think of other things that make seven an interesting number? ## Problems Involving Multiples of 7 Multiples of 7 are just not a list of numbers that we need to memorize – they can also help us solve real-life problems. Let’s try to solve", "positive number." ]

[ "777 How many 3-digit numbers are a multiple of 7? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "The total number of combinations are $7735+70+15+6=\\boxed{7826}.$", "counting number. So a multiple of $$3$$ would be the product of a counting number and $$3$$. Below are the first six multiples of $$3$$. $\\begin{split} 1 \\cdot 3 & = 3 \\\\ 2 \\cdot 3 & = 6 \\\\ 3 \\cdot 3 & = 9 \\\\ 4 \\cdot 3 & = 12 \\\\ 5 \\cdot 3 & = 15 \\\\ 6 \\cdot 3 &= 18 \\end{split} \\nonumber$ We can find the multiples of any number by continuing this process. Table $$\\PageIndex{1}$$ shows the multiples of $$2$$ through $$9$$ for the first twelve counting numbers. Table $$\\PageIndex{1}$$ Counting Number 1 2 3 4 5 6 7 8 9 10 11 12 Multiples of 2 2 4 6 8 10 12 14 16 18 20 22 24 Multiples of 3 3 6 9 12 15 18 21 24 27 30 33 36 Multiples of 4 4 8 12 16 20 24 28 32 36 40 44 48 Multiples of 5 5 10 15 20 25 30 35 40 45 50 55 60 Multiples of 6 6 12 18 24 30 36 42 48 54 60 66 72 Multiples of 7 7 14 21 28 35 42 49 56 63 70 77 84 Multiples of 8 8 16 24 32 40 48 56 64 72 80 88 96 Multiples of", "3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, … Then, the product of 3 and 7, 3 x 7, is found by skip-counting by three until the 7th multiple of 3; 3, 6, 9, 12, 15, 18, 21 Each time the student voices a multiple of 3, she counts the multiple. I preferred to teach my students to keep count by pressing a finger onto their desk with each multiple counted. I call this the student’s ‘bookkeeping’ method. Some of the skip-count patterns are more easily learned than others, and when teaching this strategy to my students, I employed skip-count patterns for 2, 3, 5, and 10. Once a student masters the 2 skip-count pattern, and then the 3 skip-count pattern, I would teach the ‘double skip-count’ heuristic. The 2, 5, and 10 skip-count patterns come quite naturally to most elementary school age students. It is necessary, however, to teach the 2 skip-count pattern to 40 rather than 20 (20 = 2 x 10). This facilitates using the 2 skip-count pattern to effect the 4 skip-count pattern. The 4 skip-count pattern is nothing more than the 2 skip-count pattern with each second multiple voiced. It look like; 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28,", "7 to the numbers starting from 7. So, to show the first ten multiples of 7 by skip counting, we have, Start: 7 7 + 7 = 14 14 + 7 = 21 21 + 7 = 28 28 + 7 = 35 35 + 7 = 42 42 + 7 = 49 49 + 7 = 56 56 + 7 = 63 63 + 7 = 70 ### Multiples of 7 by Multiplication We find the multiples of 7 by multiplying seven by the counting numbers. Hence, we will multiply 7 by 1, 7 by 2, 7 by 3, and so on. The table below shows the first ten multiples of 7 by multiplying 7 by the first ten counting numbers. ### Multiples of 7 by Division Numbers that can be divided by seven precisely without any remainder are said to be multiples of seven. Let us have the following examples: ( a ) 56 ÷ 7 = 8; since 56 can be divided by 7 without a remainder, it is a multiple of 7. ( b ) 77 ÷ 7 = 11; since 77 can be divided by 7 without a remainder, it is a multiple of 7. ( c ) 98 ÷ 7 = 14; since 98 can be divided by 7 without a remainder,", "skip count by 2, starting at 0, the next number will be $0 + 2 = 2$, then, $2 + 2 = 4$, then $4 + 2 = 6$, then $6 + 2 = 8$, and then, 10, 12, 14, 16, 18 and so on. Similarly, skip counting by 3 will be done by adding 3 to every subsequent number. The resulting series will be as follows: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60 and so on. Did you notice how the result of skip counting yielded the same answers as the multiplication table of 3? Skip counting is majorly applied in multiplication tables where you would skip counts to get the answer to a particular multiple. For example, while reading out the multiplication table of 5, if you had to find the multiple of 30, you would skip count by 5 until you reach 30 to determine the multiple, which will be 6. ## Types of Skip Counting ### Forward Skip Counting When we perform forward skip count, we count or add numbers in the forward direction to a number. So, if we were supposed to skip count by 10, we would follow the pattern below: • $0 + 10 = 10$ • $10", "# What is the least common multiple of 3, 4, and 7? Least Common Multiple of $3$, $4$ and $7$ is $84$. As $3$ and $7$ are prime and the only factor of $4$ (other than $1$ and itself) is $2$, another prime, with no common factor between three numbers. Hence, Least Common Multiple of $3$, $4$ and $7$ is $3 \\times 4 \\times 7 = 84$.", "35, 42, .... 98 } as {x : x is a multiple of 7, 10 < x < 100}.", "# What are the multiples of 3? ## Skip counting by 3’s Skip counting is the method of skipping certain numbers in the number sequence. To skip count by 3’s start with the number 3 and mark every 3rd number from the given number. Let us skip count by 3 for the numbers up to 100. The skip counting here also gives the multiples of 3 up to 100. ## Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 … The multiples of 3 are all the values that we get by multiplying any number by 3 The multiples of 3 are the same as the values we get by skip counting by 3. Example: The first five multiples of three are obtained by multiplying the numbers up to 5 by 3. $$1 \\times 3 = 3$$ $$2 \\times 3 = 6$$ $$3 \\times 3 = 9$$ $$4 \\times 3 = 12$$ $$5 \\times 3 = 15$$ We can find any number of multiples in a similar manner. ## The nth multiple of 3 We know that we can find any number of multiples of 3 by multiplying the given number by 3. The value of the multiple of any nth value is given by 3n. Example: Find the 16th multiple of", "they are not, namely $(1,1,x)$ and $(3,3,x)$. Thus, the actual answer is $730 - 2 = \\boxed{728}$.", "# Sequence of Powers of 7 Jump to navigation Jump to search ## Integer Sequence The sequence of powers of $7$ begins: $1, 7, 49, 343, 2401, 16 \\, 807, 117 \\, 649, 823 \\, 543, 5 \\, 764 \\, 801, 40 \\, 353 \\, 607, \\ldots$", "the 49th negative multiple of 7, we will simply multiply 7 by -49. Hence, $$7\\;\\times\\;-49\\;=\\;-343$$ . Therefore, the 49th negative multiple of 7 is –343. The table below shows the result if we do skip counting and multiplication. Multiplication is the simplest way to find the multiples of 7. Skip counting, on the other hand, is a fun way to determine multiples. ## List of First 30 Multiples of 7 The table below shows the first 30 positive multiples of 7. We cannot list all the possible multiples of 7 as there are infinitely many natural numbers. Thus, there are also infinitely many multiples of 7. ## Did you know that… The number 7 is significantly common in nature than most of us realize? There are seven oceans, seven continents, and seven colors in the rainbow. The Earth was also created in seven days, including the rest day. This explains why if people are asked their favorite number from 1 to 10, they usually answer 7. Seven is indeed a remarkable number. Can you think of other things that make seven an interesting number? ## Problems Involving Multiples of 7 Multiples of 7 are just not a list of numbers that we need to memorize – they can also help us solve real-life problems. Let’s try to solve", "that and you get a multiple of $$7$$.", "# 92nd Problem 2016 How many multiples of 13 are included between 12548 and 19987 both inclusive? ×", "are $7+42+42+42$ \"good\" combinations, for an answer of $133 + 780 = \\boxed{913}$. -jackshi2006", "# What are the multiples of 79 [SOLVED] Multiple Calculator ## The multiples of 79 Answer : 79,158,237,316,395,474,553,632,711,790,869,948,1027,1106,1185,1264,1343,1422,1501,1580,1659,1738,1817,1896,1975,2054,2133,2212,2291,2370,2449,2528,2607,2686,2765,2844,2923,3002,3081,3160,3239,3318,3397,3476,3555,3634,3713,3792,3871,", "× # What is the least common multiple of 2, 3, and 7? Apr 13, 2016 Least Common Multiple is $42$. #### Explanation: Multiples of $2$ are $\\left\\{2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 , 28 , 30 , 32 , 34 , 36 , 38 , 40 , 42 , 44 , 46 , 48 , 50 , \\ldots\\right\\}$ Multiples of $3$ are $\\left\\{3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48 , 51 , . .\\right\\}$ Multiples of $7$ are $\\left\\{7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , \\ldots\\right\\}$ Common multiples are $\\left\\{42 , \\ldots\\right\\}$ Least Common Multiple is $42$.", "× 6 = 48 or Forty Eight. Skip counting: 8, 16, 24, 32, 40, 48. Question 2. 8 × 8 = ______ 8 × 8 = _(2 × 8) + (2 × 8) + (2 × 8) + (2 × 8)_= 16 + 16 + 16 + 16_= 64 or Sixty Four___. Explanation: 8 × 8 = (2 × 8) + (2 × 8) + (2 × 8) + (2 × 8) 8 × 8 = 16 + 16 + 16 + 16 8 × 8 = 32 + 16 + 16 8 × 8 = 48 + 16 8 × 8 = 64 or Sixty Four. Skip counting: 8, 16, 24, 32, 40, 48, 56, 64. Question 3. 8 × 7 = _____ 8 × 7 = (2 × 7) + (2 × 7) + (2 × 7) + (2 × 7)_= 14 + 14 + 14 + 14_= 56 or Fifty Six___. Explanation: 8 × 7 = (2 × 7) + (2 × 7) + (2 × 7) + (2 × 7) 8 × 7 = 14 + 14 + 14 + 14 8 × 7 = 28 + 14 + 14 8 × 7 = 42 + 14 8 × 7 = 56 or Fifty Six. Skip counting: 8, 16, 24, 32, 40, 48, 56.", "the sets. Let's list out the possible sets. Since they are in lists of 4, we need multiples of 6 or 7. Let's list them out. 11, 12, 13, 14 12, 13, 14, 15 But if a student has 11 or 15, they would know. So that wouldn't work. The second set would be this: $\\{34,35,36,37\\}$ $\\{47,48,49,50\\}$ $\\{76,77,78,79\\}$ $\\{89,90,91,92\\}$ They couldn't know since the multiples of 7 is 35, and if you list it out, they wouldn't know about it since there are other sets such as 35, 36, 37, 38, and 33, 34, 35, 36. The people with these sets would say no, so the second set would WORK. The other sets are only 2 numbers, and since 2 numbers are sufficiently smaller than our 2nd set, we have concluded the answer is 37+50+79+92 = $\\boxed{258}$. ~Arcticturn ## Solution 2 (Illustrations) Note that $\\text{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$ In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S.$ By observations,", "+0 # Math Help 0 82 1 +169 How many positive multiples of 7 that are less than 1000 end with the digit 3? matth99 Aug 16, 2018 #1 +20153 +1 How many positive multiples of 7 that are less than 1000 end with the digit 3? $$\\begin{array}{|r|r|r|} \\hline & n & 7n \\\\ \\hline 1. &9& 63\\\\ 2. &19& 133\\\\ 3. &29& 203\\\\ 4.&39& 273\\\\ 5.&49& 343\\\\ 6.&59& 413\\\\ 7.&69& 483\\\\ 8.&79& 553\\\\ 9.&89& 623\\\\ 10.&99& 693\\\\ 11.&109& 763\\\\ 12.&119& 833\\\\ 13.&129& 903\\\\ 14.&139& 973\\\\ \\hline \\end{array}$$ 14 positive multiples of 7 that are less than 1000 end with the digit 3. heureka Aug 16, 2018" ]

[ "+0 # Math Help 0 82 1 +169 How many positive multiples of 7 that are less than 1000 end with the digit 3? matth99 Aug 16, 2018 #1 +20153 +1 How many positive multiples of 7 that are less than 1000 end with the digit 3? $$\\begin{array}{|r|r|r|} \\hline & n & 7n \\\\ \\hline 1. &9& 63\\\\ 2. &19& 133\\\\ 3. &29& 203\\\\ 4.&39& 273\\\\ 5.&49& 343\\\\ 6.&59& 413\\\\ 7.&69& 483\\\\ 8.&79& 553\\\\ 9.&89& 623\\\\ 10.&99& 693\\\\ 11.&109& 763\\\\ 12.&119& 833\\\\ 13.&129& 903\\\\ 14.&139& 973\\\\ \\hline \\end{array}$$ 14 positive multiples of 7 that are less than 1000 end with the digit 3. heureka Aug 16, 2018", "777 How many 3-digit numbers are a multiple of 7? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "+0 # Math Help 0 403 1 +169 How many positive multiples of 7 that are less than 1000 end with the digit 3? Aug 16, 2018 #1 +24983 +1 How many positive multiples of 7 that are less than 1000 end with the digit 3? $$\\begin{array}{|r|r|r|} \\hline & n & 7n \\\\ \\hline 1. &9& 63\\\\ 2. &19& 133\\\\ 3. &29& 203\\\\ 4.&39& 273\\\\ 5.&49& 343\\\\ 6.&59& 413\\\\ 7.&69& 483\\\\ 8.&79& 553\\\\ 9.&89& 623\\\\ 10.&99& 693\\\\ 11.&109& 763\\\\ 12.&119& 833\\\\ 13.&129& 903\\\\ 14.&139& 973\\\\ \\hline \\end{array}$$ 14 positive multiples of 7 that are less than 1000 end with the digit 3. Aug 16, 2018", "## anonymous 5 years ago What is the sum of the first 55 positive multiples of 3? Please show your work. 1. anonymous $3+6+9+\\cdots+162+165$ $=3(1)+3(2)+ \\cdots + 3(54)+3(55)$ $=3(1+2+3+ \\cdots +54+55) = 3 \\cdot \\frac{55(56)}{2}=4620$ 2. anonymous Nice! I should've thought of that one. Thanks Find more explanations on OpenStudy", "the 49th negative multiple of 7, we will simply multiply 7 by -49. Hence, $$7\\;\\times\\;-49\\;=\\;-343$$ . Therefore, the 49th negative multiple of 7 is –343. The table below shows the result if we do skip counting and multiplication. Multiplication is the simplest way to find the multiples of 7. Skip counting, on the other hand, is a fun way to determine multiples. ## List of First 30 Multiples of 7 The table below shows the first 30 positive multiples of 7. We cannot list all the possible multiples of 7 as there are infinitely many natural numbers. Thus, there are also infinitely many multiples of 7. ## Did you know that… The number 7 is significantly common in nature than most of us realize? There are seven oceans, seven continents, and seven colors in the rainbow. The Earth was also created in seven days, including the rest day. This explains why if people are asked their favorite number from 1 to 10, they usually answer 7. Seven is indeed a remarkable number. Can you think of other things that make seven an interesting number? ## Problems Involving Multiples of 7 Multiples of 7 are just not a list of numbers that we need to memorize – they can also help us solve real-life problems. Let’s try to solve", "+0 # Let $m$ and $n$ denote the greatest and least positive three-digit multiples of 7, respectively. What is the value of $m + n$ ? 0 381 1 +402 Let $m$ and $n$ denote the greatest and least positive three-digit multiples of 7, respectively. What is the value of $m + n$ ? RektTheNoob Aug 13, 2017 #1 +20037 0 Let $$m$$ and $$n$$ denote the greatest and least positive three-digit multiples of 7, respectively. What is the value of $$m + n$$ ? $$\\begin{array}{|rcll|} \\hline m &=& 7\\cdot \\left \\lfloor{ \\frac{999}{7} }\\right \\rfloor{} \\\\ &=& 7\\cdot 142 \\\\ &=& 994 \\\\\\\\ n &=& 7\\cdot \\left \\lceil{ \\frac{100}{7} }\\right \\rceil{} \\\\ &=& 7\\cdot 15 \\\\ &=& 105 \\\\\\\\ m+n &=& 994 + 105 \\\\ &=& 1099 \\\\ \\hline \\end{array}$$ heureka Aug 14, 2017", "number; 77 = 7 × 11; 77 is a composite number; * Positive integers that are only dividing by themselves and 1 are ...", "# Multiples of 7 and 13 How many 3 digit positive integers $$N$$ are there, such that $$N$$ is a multiple of both 7 and 13? ×", "# How many multiple? Number Theory Level 1 How many multiples of 7 are between 345 and 563 inclusive? ×", "+0 # counting +1 74 1 How many multiples of 3 are between 62 and 275? Apr 3, 2021 ### 1+0 Answers #1 +484 0 The next multiple of 3 after 62 is 63...and 273 is the nearest multiple of 3 that is less than 275... the number of multiplyes is given by: $\\frac{273-63}{3}+1=\\frac{210}{3}+1=70+1=\\boxed{71}$ *Credit, CPhill's answer to this problem: https://web2.0calc.com/questions/how-many-multiples-of-3-are-between-62-and-215* Apr 3, 2021", "7×14=98 7×15=105 7×16=112 7×17=119 7×18=126 7×19=133 7×20=140 So the first 20 multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133 and 140. ## FAQs on Multiples of 7 Q: How do we find multiples of 7? Ans: Multiplying 7 with natural numbers we get the multiples of 7. Q. What are the first 5 multiples of 7? Ans: The first 5 multiples of 7 are 7, 14, 21, 28, and 35. Q. What are the first 10 multiples of 7? Ans: The first 10 multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, and 70 Q. Are factors of 7 and multiples of 7 the same? Ans: Yes, the factors of 7 and multiples of 7 are the same. Q. What is the least common multiple of 5 and 7 Ans: The least common multiple of 5 and 7 is 35.", "gives us the largest three digit multiple of 125, $$\\boxed{875}$$. • Why can't we have $N=376$ (even and $125$ divides $N-1$) as a solution to $N(N-1)(N+1)\\equiv0\\pmod{1000}$? – J. W. Tanner Mar 13 '19 at 14:22", "710 \\qquad 147$$ We know that 1470 and 1047 are greater than the three three-digit numbers, since 1470 and 1047 are both greater than one thousand, and the three-digit numbers are less than one thousand. 1470 is greater than 1047 because 1470 contains one thousand and four hundreds, while 1047 contains one thousand and less than one hundred. We know that 847 is greater than 710 because 847 contains eight hundreds, and 710 contains less than eight hundreds. 710 is greater than 147 because 710 contains seven hundreds, and 147 contains less than two hundreds.", "+0 Math 0 105 1 +182 Given that a and b are real numbers such that -3 ≤ a ≤ 1 and -2 ≤ b ≤ 4, and values for a and b are chosen at random, what is the probability that product a * B is positive? wiskdls Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero. This is because zero is neither positive nor negative, and 0 * n = 0. The first step is to find the total number of combination: (a,b). This can be done by: $$(1-(-3)+1)\\cdot(4-(-2)+1) =5\\cdot7=35.$$ There are a total of 35 different combinations. Out of those combinations there are: ( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 ) ( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 ) 10 that work. The probabilty of this happening is: $$\\boxed{\\frac{10}{35}=\\frac{2}{7}}$$ I hope this helped, Gavin. GYanggg Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive,", "# What are the Multiples of 7? Seven is considered a lucky number in many cultures around the world. For some, the number seven represents completeness and perfection. More so, in Pythagorean numerology, 7 represents spirituality. This is because it comprises 4 and 3, which are physical and spiritual numbers, respectively. Not only does the number seven signify the jackpot on slot machines, but it is also significant for countless stories and folklore, religion, and philosophy. It is no surprise why most people are drawn into the number 7. Are you getting excited to see the remarkable side of 7 and its multiples? ## Multiples of are 77, 14, 21, 28, 35, 42, 49 … A multiple of 7 is a numerical value formed by multiplying one natural number by another natural number or counting number. Do you notice something on the numbers above? If you observe the numbers 7, 14, 21, 28, 32…, the difference between two consecutive numbers is always 7. Say $$21\\;-\\;14\\;=\\;7$$ , and $$35\\;-\\;28\\;=\\;7$$ . Amazing, right? They are called multiples of 7 because every time we divide a number by 7, it will give us a remainder of zero. Thus, to say that a number is a multiple of 7, it should be in the form of 7n where n is any natural", "Are you ready to take on this challenging Number Systems problem? # Multiples of 7 and 13 How many 3 digit positive integers $N$ are there, such that $N$ is a multiple of both 7 and 13? 3 tries left ·", "digit of $3^3$ which is 7..", "there are two cases possible Case 1: When 7 is at the left extreme In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9) So total ways 3(8)(7)= 168 Case 2: When 7 is not at the extremes Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left) Hence in total 3(7)(7)=147 ways Total ways 168+147=315 ways $ab\\ =\\ 4^{2017}=2^{4034}$ The total number of factors = 4035. out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017. And since the given number is a perfect square we have one set of two equal factors. .’. many pairs(a, b) of positive integers are there such that $a\\leq b$ and $ab=4^{2017}$ = 2018. The number of multiples of 2 between 1 and 120 = 60 The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12 The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7 Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60", "seven positive multiples of three is closest to which of the following powers of 10? (A) 10^9 (B) 10^8 (C) 10^7 (D) 10^6 (E) 105 Senior Manager Joined: 20 Mar 2018 Posts: 465 Location: Ghana Concentration: Finance, Real Estate Re: The product of the first seven positive multiples of three is closest [#permalink] ### Show Tags 23 Jan 2020, 02:31 1 The product of the first seven positive multiples of three is closest to which of the following powers of 10? Multiple of 3 is in the form 3x where x= any positive integer First seven = 3,6,9,12,15,18,21 Product = 3•(2•3)•(3^2)•(2^2•3)•(3•5)•(2•3^2)•(3•7) = 2^4•3^9•5•7 = 2^3•3^9•7•10= (2^3•70)•3^9 = 560•19,683= 11,022,480 10,000,000 < 11,022,480 < 12,000,000 10^7 < 11,022,480 < 12•10^6 ~ 10^8 10^7 < 11,022,480 < 10^8 .: 11,022,480 is close to 10^7 since it difference is smaller than to 10^8 Hit that C Posted from my mobile device CrackVerbal Quant Expert Joined: 12 Apr 2019 Posts: 385 Re: The product of the first seven positive multiples of three is closest [#permalink] ### Show Tags 23 Jan 2020, 03:40 1 This is an excellent question which tests you on your ability to do quick approximations. Of course, you also need to know some specific numbers here if you want to be good with your approximation. The first seven positive", "568 to 7 Since the number 7 is in the name of the blog (for some reason that I don’t remember), a limited amount of associated numerology seems appropriate, if only as a filler. 568 is the smallest positive integer whose seventh power is the sum of seven seventh powers. $\\displaystyle 568^7=525^7+439^7+430^7+413^7+266^7+258^7+127^7$ According to Wolfram MathWorld, it is not known whether the seventh power of a positive integer can be written as the sum of 3, 4, 5, or 6 seventh powers. More numerology: • The fouriest transform of 568 is in base seven: 568=14417 • The digit 7 is conspicuously missing from the digits of 568" ]

[ "Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know? 6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow? 7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? 11. Alexandra, who is 5 ft 8 in tall", "5. Priya and Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know? 6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow? 7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? 11. Alexandra, who is 5 ft", "## anonymous one year ago A monument casts a 40-foot shadow. Adam is 6 feet tall and casts a 5-foot shadow. How tall is the monument? A. 33 ft B. 30 ft C. 40 ft D. 48 ft 1. Nnesha |dw:1440599315146:dw| 2. anonymous is it 48 3. anonymous o6 30 4. Nnesha so both triangles are similar you can set the proportion$\\huge\\rm \\frac{ height }{ base }=\\frac{ height }{ base }$ 5. Nnesha looks good.", "is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Thomas, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building?", "# The shadow cast by a one-foot ruler is 8 inches long. At the same time, the shadow cast by a pine tree is 24 feet long. What is the height, in feet, of the pine tree? I found: $36 \\text{ft}$", "your $$6’3″$$ friend casts a $$2’$$ whereas a tree casts a $$9’$$ shadow. How tall is the tree?", "# The sun casts a shadow of a pine tree on the ground. The tree is 65 feet tall. The angle of depression is 53.4°. What is the length of the shadow? Dec 19, 2016 $48.3$ feet #### Explanation: Here height of the tree, its shadow and ground form a right angled triangle. Angle of depression = angle of elevation = 53.4° Therefore $\\tan 53.4 = \\text{height\"/\"base\" = \"opposite\"/\"adjacent}$. or, shadow = $\\frac{65}{\\tan 53.4} = \\frac{65}{1.3465} = 48.27$ feet So, length of the shadow = $48.3$ feet.", "klepkowy7c 2022-07-23 The person who is 70 inches tall cast a shadow is 54 inches long. If a tree casts a shadow that is 28 feet long at te same time, then how tall is the tree to the nearest foot? Bianca Chung Expert Convert inches into feet 5.84 feet height of tree having 4.5 geet shadow 28feet shadow of tree having height of $\\frac{28×4.5}{5.84}=21.57534247$ height of tree Do you have a similar question?", "ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ryan is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Thomas, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? ## Vocabulary Term Definition AA Similarity Postulate If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar. Proportion A proportion is an equation that shows two equivalent ratios. Interactive Element Video: Scale and Indirect Measurements Activities: Indirect Measurement Discussion Questions Study Aids: Polygon Similarity Study Guide Practice: Indirect Measurement Applications Real World: Mighty Measurements", "the wire into 3 pieces. The second piece was 6 inches longer than the first piece. The third piece is 9 inches longer than the second. What was the length of each piece of wire? The answers I got w... ##### MEASUREMENT: A woman is 5 ft tall and her shadow is 4ft long a nearby tree has a shadow 30 ft long MEASUREMENT:A woman is 5 ft tall and her shadow is 4ft longa nearby tree has a shadow 30 ft long. how tall is the tree? my answer i got was; 37.5 ft is that right?a man is standing near a totem pole. the man is 74 in tall. his shadow is 4ft long. the shadow of the totem pole is 6ft long. how tall is... ##### Chris has twice as many 1 dollar bills as pennies he has twice as many dimes as he has 10 dollar bills he has twice as many pennies as he has dimes how much money does chris have chris has twice as many 1 dollar bills as pennies he has twice as many dimes as he has 10 dollar bills he has twice as many pennies as he has dimes how much money does chris have?... ##### A carpenter wants to put four shelves on an 8-foot wall so", "distance across the Grand Canyon). 2. Find \\begin{align*}OE\\end{align*}. 3. Find \\begin{align*}EA\\end{align*}. 4. Mark is 6 ft tall and casts a 15 ft shadow. A the same time of day, a nearby building casts a 30 ft shadow. How tall is the building? 5. Karen and Jeff are standing next to each other. Karen casts a 10 ft shadow and Jeff casts an 8 ft shadow. Who is taller? How do you know? 6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow? 7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ryan is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is", "know how to figure out the height of the statue. ### Guidance We can use the properties of similar figures to measure things that are challenging to measure directly. We call this type of measurement indirect measurement. Jamie’s Dad is six feet tall. Standing outside, his shadow is eight feet long. A tree is next to him. The tree has a shadow that is sixteen feet long. Given these dimensions, how tall is the tree? That is a good question! Think about a tree, it makes a shadow and the line from the end of the shadow to the top of the tree creates a triangle. It sounds confusing, but here is a diagram to help you. Here is a palm tree. You can see from the picture that the tree itself is one side of the triangle, that the shadow length is another side of the triangle and that the diagonal from the top of the tree to the top of the shadow forms the hypotenuse (the longest side) of the triangle. How would this work with the shadow of a person? There is a triangle here too. Just because it is on an angle don’t let that fool you. It is still a triangle. Alright, now let’s go back to the problem again. Jamie’s dad is", "building 300 feet high cast a shadow 50 feet long. What was the angle of elevation of the Sun?...", "building's shadow but still comes up way short. this is the problem (btw in the real life photo the bridge is 1 cm, and ... Geography (w/ geometry) So I'm trying to calculate the latitude of where I am via a few factors. My lesson says I need to know the following: The date of my measurements: 4/3/17 the time: 12:30 my height: 5' 10\" my shadows length: 4' 6\" It then says first, compute the angle of the sun when the photo ... You've been banned from posting here for posting what you claim are answers to a geometry test. Geometry Enlarge 6 in x 10 in picture to 3 feet x 4 feet. What is the scale factor and how much of the space will be used? geometry A vertical mast is on top of building and is positioned 6' from edge of building. The mast casts a 90' shadow from front of building. Another building 24' tall casts a shadow 64'. What is the height of the mast? Geometry A tree cast a shadow on the ground and forms an angle of elevation of 22°. If the shadow is 15 feet long, how tall is the tree? Geometry Given: ABCD is a parallelogram m∠A = 60º ; BK ⊥AD AK = KD; Perimeter of ABCD", ". Jamie’s Dad is six feet tall. Standing outside, his shadow is eight feet long. A tree is next to him. The tree has a shadow that is sixteen feet long. Given these dimensions, how tall is the tree? That is a good question! Think about a tree, it makes a shadow and the line from the end of the shadow to the top of the tree creates a triangle. It sounds confusing, but here is a diagram to help you. Here is a palm tree. You can see from the picture that the tree itself is one side of the triangle, that the shadow length is another side of the triangle and that the diagonal from the top of the tree to the top of the shadow forms the hypotenuse (the longest side) of the triangle. How would this work with the shadow of a person? There is a triangle here too. Just because it is on an angle don’t let that fool you. It is still a triangle. Alright, now let’s go back to the problem again. Jamie’s dad is six feet tall. Standing outside, his shadow is eight feet long. A tree is next to him. The tree has a shadow that is sixteen feet long. Given these dimensions, how tall is the tree? To", "mirror. She is 18 in from the mirror and her eyes are 5 ft 3 in above the ground. The angle formed by her line of sight and the ground is congruent to the angle formed by the reflection of the building and the ground. You may wish to draw a diagram to illustrate this problem. How tall is the building? 2. Sebastian is curious to know how tall the announcer’s box is on his school’s football field. On a sunny day he measures the shadow of the box to be 45 ft and his own shadow is 9 ft. Sebastian is 5 ft 10 in tall. Find the height of the box. 3. Juanita wonders how tall the mast of a ship she spots in the harbor is. The deck of the ship is the same height as the pier on which she is standing. The shadow of the mast is on the pier and she measures it to be 18 ft long. Juanita is 5 ft 4 in tall and her shadow is 4 ft long. How tall is the ship’s mast? 4. Evan is 6 ft tall and casts a 15 ft shadow. A the same time of day, a nearby building casts a 30 ft shadow. How tall is the building? 5. Priya and", "# Measurment #### DOOKIE A 70-foot tree has a 35-foot shadow. If the building next to the tree has an 80-foot shadow, how tall is the building? #### skeeter MHF Helper A 70-foot tree has a 35-foot shadow. If the building next to the tree has an 80-foot shadow, how tall is the building? set up a ratio ... $$\\displaystyle \\frac{70}{35} = \\frac{h}{80}$$ DOOKIE", "is . In the meantime, Andrea has paced off the length of the tree’s shadow and finds that it is feet long. ### 1. How might Andrea and Bonita use this information, along with their knowledge of trigonometric ratios, to calculate the height of the tree? (Andrea and Bonita know they can find the value of any trigonometric ratio they might need for any acute angle using a calculator.) Part 2: What’s your angle? After their rest, Andrea and Bonita are going for a walk straight up the side of the hill. Andrea decided to stretch before heading up the hill while Bonita thought this would be a good time to get a head start. Once Bonita was away from Andrea, she stopped to take a break and looked at her GPS device that told her that she had walked and had already increased her elevation by . With a bit of time to waste, Bonita wrote down the trigonometric ratios for and for . ### 2. Name the trigonometric ratios for and for that involve the given sides. When Andrea caught up, she said, “What about the unknown angle measures? When I was at the bottom and looked up to see you, I was thinking about the ‘upward’ angle measure from me to you. Based on your", "## anonymous one year ago A building casts a 50-foot shadow. Alex is 5 feet tall and casts a 6-foot shadow. Approximately how tall is the building? A. 60 ft B. 42 ft C. 24 ft D. 50 ft 1. anonymous its 50 feet 2. Nnesha |dw:1440424485130:dw| 3. Nnesha so angle angle similarity there sides should be in proportion $\\huge\\rm \\color{red}{\\frac{ height }{ base }} =\\frac{ height }{ base }$ red part is height/base of big triangle 4. Nnesha what is the height and base of big trinagle ? 5. Nnesha let me know what you get..", "angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the height of the tree. #### The length of the shadow of a tower standing on level plane is found to be 2x meters The length of the shadow of a tower standing on level plane is found to be 2x meters longer when the sun’s attitude is 30° than when it was 30°. Prove that the height of tower is $$x(\\sqrt{3}+1)$$ meters." ]

[ "☐C. $$35\\%$$ ☐D. $$40\\%$$ 4- The width of a box is one third of its length. The height of the box is one third of its width. If the length of the box is 36 cm, what is the volume of the box? ☐A. $$81 cm^3$$ ☐B. $$162cm^3$$ ☐C. $$243cm^3$$ ☐D. $$1,728cm^3$$ 5- A tree 32 feet tall casts a shadow 12 feet long. Jack is 6 feet tall. How long is Jack’s shadow? ☐A. 2.25 feet ☐B. 4 feet ☐C. 4.25 feet ☐D. 8 feet 6- When a number is subtracted from 28 and the difference is divided by that number, the result is 3. What is the value of the number? ☐A. 2 ☐B. 4 ☐C. 7 ☐D. 12 7- An angle is equal to one ninth of its supplement. What is the measure of that angle? ☐A. 9 ☐B. 18 ☐C. 25 ☐D. 60 8- John traveled 150 km in 6 hours and Alice traveled 140 km in 4 hours. What is the ratio of the average speed of John to average speed of Alice? ☐A. 3∶2 ☐B. 2∶3 ☐C. 5∶7 ☐D. 5∶6 9-If A={2,5,11,15}, B={1,2,3,4,5,6}, and C={5,7,9,11,13}, then which of the following set is $$(A∪B)∩C$$? ☐A. {1,2,3,4,5,6,11,15} ☐B.{1,2,3,4,5,6,7,11,13,15} ☐C. {5,11,13,15} ☐D. {5,11} 10- If $$4n-3≥1$$, what is the least possible value of $$4n+3$$? ☐A.", "Let’s think about how Henry and Carmen could figure out the height of the statue. We know that Henry is five feet tall and that his shadow is half as long as he is tall. Now we can write a ratio to compare Henry’s height to his shadow’s length. $\\frac{Henry' s \\ height}{Shadow' s \\ length} = \\frac{5 \\ feet}{2.5 \\ feet}$ Next, we figure out the height of the statue. Henry and Carmen figure out very quickly that they need to learn the length of the shadow of the statue to figure out the height of the statue. Once they know the length of the shadow, they can use proportional reasoning and indirect measurement to figure out the statue’s height. Approximating 1 foot using a length a little longer than Henry’s sneaker, they measure $32 \\frac{1}{2}$ feet. It is not an exact measure, but they feel that it is very close. Now they write the following proportion. $\\frac{5 \\ ft}{2.5 \\ ft} = \\frac{x}{32.5 \\ ft}$ Taking out a notebook, Carmen cross multiplies to solve the proportion. $5(32.5) &= 2.5x\\\\162.5 &= 2.5x\\\\x &= 65$ The sculpture is approximately 65 feet tall. After completing their work, Henry and Carmen check out their answer with the curator of the museum. The statue is actually 65.6 feet tall. Their work", "# The shadow cast by a one-foot ruler is 8 inches long. At the same time, the shadow cast by a pine tree is 24 feet long. What is the height, in feet, of the pine tree? I found: $36 \\text{ft}$", "Only one positive value 🙂 10. jc says: I guess the theory is that a portrait which is more than phi in its height:width ratio would look “too tall,” while one less than phi would look “too wide.” In other words, why paintings are generally neither squares nor long bars, but a certain shape. 1. David Starner says: That’s the theory, but what are the ratios of portraits in practice? To one digit, The Mona Lisa is 1.5, The Girl with a Pearl Earring is 1.2, Van Gogh’s Starry Night was 1.3. The Persistence of Memory is 1.4. Leonardo da Vinci’s Last Supper is 1.9. In fact, not a single painting I looked at had a ratio of phi. I finally found Bottecelli’s Birth of Venus, which does have such a ratio, but it seems to be a rarity. Movies are neither squares or (with rare exceptions, long bars), but they range from 4:3 (1.33), 16:9 (1.78), 1.85:1, all the way to 2.39:1. 1.66:1 is the closest standard movie ratios come to phi, and that doesn’t seem particularly common or close. Paper sizes are 8.5 x 11 (1.29), An (n=4 for normal paper) (sqrt(2) ~ 1.414), or legal at 8.5 x 14 (1.65). Legal is sort of phi ratio, and yet no one seems to be demanding to", "# Britney is 5 feet tall and casts a 3 1/2 -foot shadow at 10:00 A.M. At that time, a nearby tree casts a 17-foot shadow. Two hours later, Britney's shadow is 2 feet long. What is the length of the shadow of the tree at this time? May 11, 2018 The tree is (approximately) 9.71 feet tall at noon. #### Explanation: Because the length of an object's shadow is proportional to its height (at a given time of day), this kind of question is solved by using the proportion relation: \"$a$ is to $b$ as $c$ is to $d$\". Mathematically, this is written as: $\\frac{a}{b} = \\frac{c}{d}$ They key to solving this question is finding the pieces of information that (a) give us 3 of these 4 values and (b) also let us use this relation to find the 4th. What we have: $\\left.\\begin{matrix}\\null & \\text{Britney\" & \"tree\" \\\\ \"height\" & 5 & - \\\\ \"10:00 shadow\" & \"3.5 ft\" & \"17 ft\" \\\\ \"12:00 shadow\" & \"2 ft} & \\square\\end{matrix}\\right.$ The box marks what we want to solve for. We need to use the \"12:00 shadow\" info, because it has the value we want to solve for. We can not use the height info, because it also has missing data. (We can only solve for the", "## Algebra 1 This is a proportion problem. You already know that a 6-feet tall object creates a 9-feet tall shadow. The variable is the height of a tree with a 48 ft shadow. You would use the equation shown below: $\\frac{6}{9}=\\frac{x}{48}$ Then, cross multiply and solve. $6\\times48=x\\times9$ $9x=288$ $(9x)\\div9=(288)\\div9$ $x=32$ The tree is 32 ft tall.", "# The ratio of the height of a tower and the length of its shadow on the ground is $\\sqrt{3}:1$. What is the angle of elevation of the sun? • 0 What are you looking for?", "the max height after $1.75\\text{ seconds}$ and the maximum height is $49\\text{ feet}$. Hence, the maximum height is $49\\text{ feet}$ and we attain it after $1.75\\text{ seconds}$", "feet tall? Set up a proportion to represent the problem: $\\frac{4\\;\\text{ft}}{9\\;\\text{ft}} = \\frac{35\\;\\text{ft}}{\\text{x ft}}$ It is important to notice that the ratios in any proportion must match in arrangement. In the proportion above, the height of each object is placed in the numerator, and the length of each shadow is placed in the denominator. The problem can still be solved if each shadow length is in the numerator and each object height is in the denominator, but the problem cannot be solved if one ratio shows the object’s height in the numerator and the other ratio shows the object’s height in the denominator. Cross multiplication yields a shadow length of $$78.75\\;\\text{ft}$$. Proportions are useful for finding the unknown side lengths of similar objects, particularly triangles. Consider the following example involving similar triangles. Solve for $$x$$: A proportion can be established between the two triangles (as they share two sides): $\\frac{x}{6} = \\frac{x + 8}{18}$ which becomes $$18x = 6x + 48$$, so $$x = 4$$ #### Determining Rate A rate expresses the relationship between two quantities of different units. Miles per hour, dollars per minute, and degrees per second are all examples of rates. Unit rates express a rate in which the denominator is reduced to one. For example, if Sally drives 200 miles in 3.5 hours, the", "If the width to height ratio is to be $$3:2$$, then to how many pixels should he set the height? 28. If a video monitor is produced in the width to height ratio of $$16:9$$ and the width of the monitor is $$40$$ inches, then what is the height? 1. $$960$$ people 3. $$276$$ students 5. $$5,000$$ male students 7. $$4,245,672$$ births 9. $$8$$ cups of lemon juice 11. $$24$$ minutes 13. $$56$$ miles 15. \\$$$19.50$$ 17. $$66$$ ears 19. $$275$$ calories 21. $$24.9$$ pounds 23. $$60$$ times 25. $$285$$ peanuts 27. $$480$$ pixels Exercise $$\\PageIndex{7}$$ Similar Triangles If triangle $$ABC$$ is similar to triangle $$RST$$, find the remaining two sides given the information. 1. $$a=6, b=8, c=10,$$ and $$s=16$$ 2. $$b=36, c=48, r=20,$$ and $$t=32$$ 3. $$b=2, c=4, r=6,$$ and $$s=4$$ 4. $$b=3, c=2, r=10,$$ and $$t=12$$ 5. $$a=40, c=50, s=3,$$ and $$t=10$$ 6. $$c=2, r=7, s=9,$$ and $$t=4$$ 7. At the same time of day, a tree casts a $$12$$-foot shadow while a $$6$$-foot man casts a $$3$$-foot shadow. Estimate the height of the tree. 8. At the same time of day, a father and son, standing side by side, cast a $$4$$-foot and $$2$$-foot shadow, respectively. If the father is $$6$$ feet tall, then how tall is his son? 9. If the $$6-8-10$$ right triangle", "### Home > INT3 > Chapter 11 > Lesson 11.2.1 > Problem11-61 11-61. While trying to measure the height of a tree, Julie notices that a $3.5$-foot post has a $4.25$-foot shadow. If the tree’s shadow is $100$ feet long, how tall is the tree? Draw and label a picture. The ratio between the height of the object and the length of the shadow will be the same for both the tree and the post.", "so it likely tests how well we know Ratios and Proportions. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question. ## What Do We Know? Let’s carefully read through the question and make a list of the things that we know. 1. We have one pole that is \\$12\\$ feet tall casting a shadow \\$8\\$ feet long 2. A nearby pole casts a shadow that is \\$10\\$ feet long 3. The lengths of the shadow are proportional to the height of their corresponding poles ## Develop a Plan The question tells us that the pole height and shadow lengths are proportional. When two quantities are proportional, their ratio will remain constant. In other words, we can set the ratio of the pole height to the shadow length for the first pole equal to the ratio of the pole height to the shadow length for the second pole. \\$\\${\\Height_1}/{\\Shadow_1}={\\Height_2}/{\\Shadow_2}\\$\\$ It doesn’t matter which pole we label \\$1\\$ or \\$2\\$. Let’s go ahead and label the pole we have all the information about as pole \\$2\\$. So for pole \\$2\\$, we know that its height is \\$12\\$ feet and its shadow length is \\$8\\$ feet. We also know that the shadow length of the other pole, pole \\$1\\$, is \\$10\\$", "## 'Weekly Problem 15 - 2006' printed from http://nrich.maths.org/ The ratio width : height of television screens is changing from the traditional $4:3$ to the widescreen $16:9$.", "## Algebra 1 Mitchell is 5.25 feet tall and casts a .5 foot shadow. The tree has a height of $x$ feet and casts a 2.5 foot shadow. $5.25/.5 = x/2.5$ (compares height to shadow) $10.5 = x/2.5$ $10.5*2.5 = x*2.5/2.5$ $26.25 = x$ $.25*12= 3$ inches", "be $155\\times 2+110=420$ so height to length ratio should be $0.73$, you got $0.72$ which is pretty close. • I've edited your answer to make it match the question and removed the reference to my (now) deleted comment. (I also change \\cdot to \\times, because sometimes that looks like a decimal delimiter on some devices.) [Flag this comment obsolete when read ;)] – Martin - マーチン Jul 24 '17 at 10:53 • Thanks. Could you please compare d/2r to make direct connection to the question? – Pygmalion Jul 24 '17 at 12:18", "### Home > CCA2 > Chapter 7 > Lesson 7.1.3 > Problem7-38 7-38. While trying to measure the height of a tree, Julie noticed that a $3.5$-foot post had a $4.25$-foot shadow. If the tree’s shadow is $100$ feet long, how tall is the tree? Draw and label a picture. The ratio between the height of the object and the length of the shadow will be the same for both the tree and the post.", "image ratios and scale proportionally This is a simple calculator to help you work out the aspect ratio of an image, and the size of that image when it's resized, keeping the same proportions. PosterBurner's image aspect ratio calculator is a tool that takes your image and finds the best matching print sizes for your image. You can use our aspect ratio calculator to convert the size of an image from original one to another sizes keeping the shape of the object and make him look natural without stretches. In, for example, a group of images that all have an aspect ratio of 16:9, one image might be 16 inches wide and 9 inches high, another 16 centimeters wide and 9 centimeters high, and a third might be 8 yards wide and 4.5 yards high. The value of the golden ratio, which is the limit of the ratio of consecutive Fibonacci numbers, has a value of approximately 1.618 . In the printing situation, the existing image is usually a different shape than the paper we want to print it on. Click OK button your own image, or finds the missing value in a set of ratios most common ratios... The also chenge the pass word ( width to height 103 1 1 gold badge 1..., but your blog", "no matter how big or small, will always have the width of 16 and the height of 9 units of the same length. Therefore, 16 centimeters wide and 9 centimeters high image as well as 32 centimeters wide Aspect Ratio Calculator (ARC) - Andrew Hedges ## May 2, 2019 How the Lighthouse image aspect ratio audit fails #. Lighthouse flags any image with a rendered aspect ratio 5 percent or more different from its Formula. Say you have a photo that is 1600 x 1200 pixels, but your blog only has space for a photo 400 pixels wide. To find the Calculate the Aspect Ratio (ARC) here by entering your in pixel or ratio ✅. Change the image aspect ratio via this Ratio Calculator ✅. The pixel aspect calculator There is a simple formula for calculating aspect ratios: aspectRatio = ( oldWidth / oldHeight ) . For instance if you want to know the new height of an object, you can The corresponding formula for the adjusted width is adjusted width = * original width / original height , or if you want to do it your way, Feb 24, 2020 Each image has a unique shape- this is called the aspect ratio. The aspect ratio of an image is the relation of its' width to height. For", "# To find the height of a building, Oscar stuck the end of a $$4 \\frac {1}{2}$$ -foot stick 6 inches straight into the ground. He measured the shadow of the stick and found it was 1 foot long. Oscar then measured the building's shadow and found it was 16 feet long. How many feet tall is the building? 64 feet Explanation The part of the stick that is above the ground is 4 ft 6 in – 6 in, or 4 ft high. It casts a shadow 1 ft long. The shadow of the building is 16 ft long. This creates two similar right triangles, so the ratio of the two sides of the small triangle is equal to the ratio of the corresponding sides of the large triangle. This fact lets you write the following proportion: 1:4 = 16:? Since the shadow of the building is 16 times the shadow of the stick, the height of the building is 16 times the height of the stick: $$4 \\times 16=64$$", "$H$ is the height of your object and $D$ is the object distance. Therefore, you have $$h = f \\frac{H}{D}$$ Plugging in the numbers, your image will have a height of 0.9375mm, so to have that cover 10 pixels, (remember the height is from the center of the image, not the bottom, to the top) you will need pixels with a spacing of $\\frac{0.9375}{10} = 0.09375 \\text{ mm}$, or $93.75\\ \\mu m$" ]

[ "Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know? 6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow? 7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? 11. Alexandra, who is 5 ft 8 in tall", "# The sun casts a shadow of a pine tree on the ground. The tree is 65 feet tall. The angle of depression is 53.4°. What is the length of the shadow? Dec 19, 2016 $48.3$ feet #### Explanation: Here height of the tree, its shadow and ground form a right angled triangle. Angle of depression = angle of elevation = 53.4° Therefore $\\tan 53.4 = \\text{height\"/\"base\" = \"opposite\"/\"adjacent}$. or, shadow = $\\frac{65}{\\tan 53.4} = \\frac{65}{1.3465} = 48.27$ feet So, length of the shadow = $48.3$ feet.", "5. Priya and Meera are standing next to each other. Priya casts a 10 ft shadow and Meera casts an 8 ft shadow. Who is taller? How do you know? 6. Billy is 5 ft 9 inches tall and Bobby is 6 ft tall. Bobby's shadow is 13 feet long. How long is Billy's shadow? 7. Sally and her little brother are walking to school. Sally is 4 ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ray is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Jason, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? 11. Alexandra, who is 5 ft", "## anonymous one year ago A monument casts a 40-foot shadow. Adam is 6 feet tall and casts a 5-foot shadow. How tall is the monument? A. 33 ft B. 30 ft C. 40 ft D. 48 ft 1. Nnesha |dw:1440599315146:dw| 2. anonymous is it 48 3. anonymous o6 30 4. Nnesha so both triangles are similar you can set the proportion$\\huge\\rm \\frac{ height }{ base }=\\frac{ height }{ base }$ 5. Nnesha looks good.", "# The shadow cast by a one-foot ruler is 8 inches long. At the same time, the shadow cast by a pine tree is 24 feet long. What is the height, in feet, of the pine tree? I found: $36 \\text{ft}$", "Let’s think about how Henry and Carmen could figure out the height of the statue. We know that Henry is five feet tall and that his shadow is half as long as he is tall. Now we can write a ratio to compare Henry’s height to his shadow’s length. $\\frac{Henry' s \\ height}{Shadow' s \\ length} = \\frac{5 \\ feet}{2.5 \\ feet}$ Next, we figure out the height of the statue. Henry and Carmen figure out very quickly that they need to learn the length of the shadow of the statue to figure out the height of the statue. Once they know the length of the shadow, they can use proportional reasoning and indirect measurement to figure out the statue’s height. Approximating 1 foot using a length a little longer than Henry’s sneaker, they measure $32 \\frac{1}{2}$ feet. It is not an exact measure, but they feel that it is very close. Now they write the following proportion. $\\frac{5 \\ ft}{2.5 \\ ft} = \\frac{x}{32.5 \\ ft}$ Taking out a notebook, Carmen cross multiplies to solve the proportion. $5(32.5) &= 2.5x\\\\162.5 &= 2.5x\\\\x &= 65$ The sculpture is approximately 65 feet tall. After completing their work, Henry and Carmen check out their answer with the curator of the museum. The statue is actually 65.6 feet tall. Their work", "# Britney is 5 feet tall and casts a 3 1/2 -foot shadow at 10:00 A.M. At that time, a nearby tree casts a 17-foot shadow. Two hours later, Britney's shadow is 2 feet long. What is the length of the shadow of the tree at this time? May 11, 2018 The tree is (approximately) 9.71 feet tall at noon. #### Explanation: Because the length of an object's shadow is proportional to its height (at a given time of day), this kind of question is solved by using the proportion relation: \"$a$ is to $b$ as $c$ is to $d$\". Mathematically, this is written as: $\\frac{a}{b} = \\frac{c}{d}$ They key to solving this question is finding the pieces of information that (a) give us 3 of these 4 values and (b) also let us use this relation to find the 4th. What we have: $\\left.\\begin{matrix}\\null & \\text{Britney\" & \"tree\" \\\\ \"height\" & 5 & - \\\\ \"10:00 shadow\" & \"3.5 ft\" & \"17 ft\" \\\\ \"12:00 shadow\" & \"2 ft} & \\square\\end{matrix}\\right.$ The box marks what we want to solve for. We need to use the \"12:00 shadow\" info, because it has the value we want to solve for. We can not use the height info, because it also has missing data. (We can only solve for the", "the wire into 3 pieces. The second piece was 6 inches longer than the first piece. The third piece is 9 inches longer than the second. What was the length of each piece of wire? The answers I got w... ##### MEASUREMENT: A woman is 5 ft tall and her shadow is 4ft long a nearby tree has a shadow 30 ft long MEASUREMENT:A woman is 5 ft tall and her shadow is 4ft longa nearby tree has a shadow 30 ft long. how tall is the tree? my answer i got was; 37.5 ft is that right?a man is standing near a totem pole. the man is 74 in tall. his shadow is 4ft long. the shadow of the totem pole is 6ft long. how tall is... ##### Chris has twice as many 1 dollar bills as pennies he has twice as many dimes as he has 10 dollar bills he has twice as many pennies as he has dimes how much money does chris have chris has twice as many 1 dollar bills as pennies he has twice as many dimes as he has 10 dollar bills he has twice as many pennies as he has dimes how much money does chris have?... ##### A carpenter wants to put four shelves on an 8-foot wall so", "## Algebra 1 Mitchell is 5.25 feet tall and casts a .5 foot shadow. The tree has a height of $x$ feet and casts a 2.5 foot shadow. $5.25/.5 = x/2.5$ (compares height to shadow) $10.5 = x/2.5$ $10.5*2.5 = x*2.5/2.5$ $26.25 = x$ $.25*12= 3$ inches", "If the width to height ratio is to be $$3:2$$, then to how many pixels should he set the height? 28. If a video monitor is produced in the width to height ratio of $$16:9$$ and the width of the monitor is $$40$$ inches, then what is the height? 1. $$960$$ people 3. $$276$$ students 5. $$5,000$$ male students 7. $$4,245,672$$ births 9. $$8$$ cups of lemon juice 11. $$24$$ minutes 13. $$56$$ miles 15. \\$$$19.50$$ 17. $$66$$ ears 19. $$275$$ calories 21. $$24.9$$ pounds 23. $$60$$ times 25. $$285$$ peanuts 27. $$480$$ pixels Exercise $$\\PageIndex{7}$$ Similar Triangles If triangle $$ABC$$ is similar to triangle $$RST$$, find the remaining two sides given the information. 1. $$a=6, b=8, c=10,$$ and $$s=16$$ 2. $$b=36, c=48, r=20,$$ and $$t=32$$ 3. $$b=2, c=4, r=6,$$ and $$s=4$$ 4. $$b=3, c=2, r=10,$$ and $$t=12$$ 5. $$a=40, c=50, s=3,$$ and $$t=10$$ 6. $$c=2, r=7, s=9,$$ and $$t=4$$ 7. At the same time of day, a tree casts a $$12$$-foot shadow while a $$6$$-foot man casts a $$3$$-foot shadow. Estimate the height of the tree. 8. At the same time of day, a father and son, standing side by side, cast a $$4$$-foot and $$2$$-foot shadow, respectively. If the father is $$6$$ feet tall, then how tall is his son? 9. If the $$6-8-10$$ right triangle", "klepkowy7c 2022-07-23 The person who is 70 inches tall cast a shadow is 54 inches long. If a tree casts a shadow that is 28 feet long at te same time, then how tall is the tree to the nearest foot? Bianca Chung Expert Convert inches into feet 5.84 feet height of tree having 4.5 geet shadow 28feet shadow of tree having height of $\\frac{28×4.5}{5.84}=21.57534247$ height of tree Do you have a similar question?", "ft tall and has a shadow that is 3 ft long. Her little brother's shadow is 2 ft long. How tall is her little brother? 8. Ryan is outside playing basketball. He is 5 ft tall and at this time of day is casting a 12 ft shadow. The basketball hoop is 10 ft tall. How long is the basketball hoop's shadow? 9. Jack is standing next to a very tall tree and wonders just how tall it is. He knows that he is 6 ft tall and at this moment his shadow is 8 ft long. He measures the shadow of the tree and finds it is 90 ft. How tall is the tree? 10. Thomas, who is 4 ft 9 inches tall is casting a 6 ft shadow. A nearby building is casting a 42 ft shadow. How tall is the building? ## Vocabulary Term Definition AA Similarity Postulate If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar. Proportion A proportion is an equation that shows two equivalent ratios. Interactive Element Video: Scale and Indirect Measurements Activities: Indirect Measurement Discussion Questions Study Aids: Polygon Similarity Study Guide Practice: Indirect Measurement Applications Real World: Mighty Measurements", "### Home > INT3 > Chapter 11 > Lesson 11.2.1 > Problem11-61 11-61. While trying to measure the height of a tree, Julie notices that a $3.5$-foot post has a $4.25$-foot shadow. If the tree’s shadow is $100$ feet long, how tall is the tree? Draw and label a picture. The ratio between the height of the object and the length of the shadow will be the same for both the tree and the post.", "your $$6’3″$$ friend casts a $$2’$$ whereas a tree casts a $$9’$$ shadow. How tall is the tree?", "is . In the meantime, Andrea has paced off the length of the tree’s shadow and finds that it is feet long. ### 1. How might Andrea and Bonita use this information, along with their knowledge of trigonometric ratios, to calculate the height of the tree? (Andrea and Bonita know they can find the value of any trigonometric ratio they might need for any acute angle using a calculator.) Part 2: What’s your angle? After their rest, Andrea and Bonita are going for a walk straight up the side of the hill. Andrea decided to stretch before heading up the hill while Bonita thought this would be a good time to get a head start. Once Bonita was away from Andrea, she stopped to take a break and looked at her GPS device that told her that she had walked and had already increased her elevation by . With a bit of time to waste, Bonita wrote down the trigonometric ratios for and for . ### 2. Name the trigonometric ratios for and for that involve the given sides. When Andrea caught up, she said, “What about the unknown angle measures? When I was at the bottom and looked up to see you, I was thinking about the ‘upward’ angle measure from me to you. Based on your", "is?” he whispered. “You can figure that out quite easily with math,” Mrs. Gilson said overhearing the conversation. “How can I do that?” Henry smirked. “How tall are you?” Henry looked at Carmen and then back at Mrs. Gilson. “I have an idea,” he said smiling. Let’s think about how Henry and Carmen could figure out the height of the statue. We know that Henry is five feet tall and that his shadow is half as long as he is tall. Now we can write a ratio to compare Henry’s height to his shadow’s length. \\begin{align*}\\frac{Henry' s \\ height}{Shadow' s \\ length} = \\frac{5 \\ feet}{2.5 \\ feet}\\end{align*} Next, we figure out the height of the statue. Henry and Carmen figure out very quickly that they need to figure out the length of the shadow of the statue to figure out the height of the statue. Once they know the length of the shadow, they can use proportional reasoning and indirect measurement to figure out the statue’s height. Approximating 1 foot using a length a little longer than Henry’s sneaker, they measure \\begin{align*}32 \\frac{1}{2}\\end{align*} feet. It is not an exact measure, but they feel that it is very close. Now they wrote the following proportion. \\begin{align*}\\frac{5 \\ ft}{2.5 \\ ft} = \\frac{x}{32.5 \\ ft}\\end{align*} Taking out a notebook,", "Andrea's room is 9 feet. Now, you need to figure out the perimeter of Andrea's room so Andrea will know how many feet of lights she needs. You know the room is a square with a side length of 9 feet. Substitute s=9 into the square perimeter formula. The answer is that the perimeter of her room is 36 feet, so she needs 36 feet of lights. Example $$\\PageIndex{2}$$ The Hegazzi family is designing a summer garden based on the model shown below. The bottom of the plot is a square and the top of the plot is a rectangle. If the total area of both plots in the garden is 139 square feet, what is the length of one side of the square plot (indicated with a question mark below)? Solution First, use the given information to find the area of the rectangle plot. You will then use this information to help you to find the area of the square plot, which will allow you to find the side length of the square plot. The rectangle has a length of 15 feet and a width of 6 feet. The formula to find the area of a rectangle is $$A=lw$$. Substitute the values for the length and the width of the rectangle into the formula and evaluate to", "perpendicular. Also according to his equipment, he is 193 meters from the other entrance of the tunnel, at an ang... ### Maura found that a redwood tree that is 32 yards tall casts a shadow 60 yards long. ThenMaura noticed Maura found that a redwood tree that is 32 yards tall casts a shadow 60 yards long. Then Maura noticed that a cell phone tower nearby casts a shadow 45 yards long. How tall is the cell phone tower?... ### Describe how masters of the Italian renaissance redefined art through realism Describe how masters of the Italian renaissance redefined art through realism... ### Carly is making 3 1/2 batches of biscuit. if one batch calls for2 1/3 cups of flour how much flour will she need Carly is making 3 1/2 batches of biscuit. if one batch calls for2 1/3 cups of flour how much flour will she need... ### Coordinate plane with triangles efg and ghi with e at negative 6 comma 2, f at negative 2 comma 6, g Coordinate plane with triangles efg and ghi with e at negative 6 comma 2, f at negative 2 comma 6, g at negative 2 comma 2, h at negative 2 comma 4, and i at 0 comma 2 which set of transformations would prove δefg ~ δghi?", "##### GED Test One or two questions on the GED Math test will ask you to find an unknown quantity of something, using a proportional relationship. The following practice questions ask you to do this by setting up a proportion where you are given three related values and you have to find the missing fourth value. ## Practice Questions 1. On a sunny day, a tree casts a shadow that is 33 feet long. At the same time, a nearby 4-foot-tall mailbox casts a shadow that is 5.5 feet long. What is the height of the tree? Round your answer to the nearest foot. A. 24 feet B. 33 feet C. 38 feet D. 45 feet 2. Meg can swim 4 laps in 11 minutes. How long should she plan to practice if she needs to swim 10 laps, assuming her speed is constant? A. 4.4 B. 15.7 C. 22 D. 27.5 1. The correct answer is A. Set up a proportion for the mailbox with the height as the numerator of the fraction and the shadow length as the denominator: Substitute the given shadow length of the tree into the proportion and let x equal the unknown height of the tree: Multiply both sides by 33 to solve for x: Alternatively, you can solve the proportion by", "know how to figure out the height of the statue. ### Guidance We can use the properties of similar figures to measure things that are challenging to measure directly. We call this type of measurement indirect measurement. Jamie’s Dad is six feet tall. Standing outside, his shadow is eight feet long. A tree is next to him. The tree has a shadow that is sixteen feet long. Given these dimensions, how tall is the tree? That is a good question! Think about a tree, it makes a shadow and the line from the end of the shadow to the top of the tree creates a triangle. It sounds confusing, but here is a diagram to help you. Here is a palm tree. You can see from the picture that the tree itself is one side of the triangle, that the shadow length is another side of the triangle and that the diagonal from the top of the tree to the top of the shadow forms the hypotenuse (the longest side) of the triangle. How would this work with the shadow of a person? There is a triangle here too. Just because it is on an angle don’t let that fool you. It is still a triangle. Alright, now let’s go back to the problem again. Jamie’s dad is" ]

[ "# Arithmetic Mean • November 8th 2006, 06:37 AM Dragon Arithmetic Mean The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? • November 8th 2006, 07:01 AM earboth Quote: Originally Posted by Dragon The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? Hello, Dragon, the sum of the first nine numbers is: 9 * 54 = 486. $\\frac{9*54+u+v}{11}=66$", "Math Help - Arithmetic Mean 1. Arithmetic Mean The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? 2. Originally Posted by Dragon The arithmetic mean of nine numbers is 54. Of two number u and v are added to the list the mean of the eleven numer list becomes 66. what are the values of u and v? Hello, Dragon, the sum of the first nine numbers is: 9 * 54 = 486. So your problem becomes: $\\frac{9*54+u+v}{11}=66$ That means: u + v = 240 or: u = 240 - v So you can construct a pair of numbers with this property. EB", "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is : [A] 4 [B] 5 [C] 14 [D] 18", "# If the mean of nine numbers is 105, what is the sum of the numbers? \"(Average)\"={\"(Sum)\"}/{9}=105 $\\text{(Sum)} = 105 \\setminus \\times 9 = 945$", "is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 _________________ Math Expert Joined: 02 Sep 2009 Posts: 59561 In a numerical table with 10 rows and 10 columns, each entry is either [#permalink] ### Show Tags 25 Jul 2017, 23:10 24 18 carcass wrote: In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 We have 10*10 = 100 entries. \"The number of 9s in the nth row is n – 1 for each n from 1 to 10\" means that: In the 10th row the number of 9s is 10 - 1 = 9; In the 9th row the number of 9s is 9 - 1 = 8; In the 8th row the number of 9s is 8 - 1 = 7; ... In the 1st row the number of 9s is 1 - 1 = 0. Thus, the total number of 9s is 9 + 8 + 7 + 6 +", "is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 _________________ Math Expert Joined: 02 Sep 2009 Posts: 57022 In a numerical table with 10 rows and 10 columns, each entry is either [#permalink] ### Show Tags 25 Jul 2017, 23:10 18 16 carcass wrote: In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 We have 10*10 = 100 entries. \"The number of 9s in the nth row is n – 1 for each n from 1 to 10\" means that: In the 10th row the number of 9s is 10 - 1 = 9; In the 9th row the number of 9s is 9 - 1 = 8; In the 8th row the number of 9s is 8 - 1 = 7; ... In the 1st row the number of 9s is 1 - 1 = 0. Thus, the total number of 9s is 9 + 8 + 7 + 6 +", "## Question Between two numbers, whose sum is 13/6, an even number of arithmetic means are inserted, the sum of these means exceed their number by unity. How many means are there? Joshi sir comment", "## List $$K$$ consists of seven numbers. Is the average (arithmetic mean) of the seven numbers negative? ##### This topic has expert replies Moderator Posts: 2058 Joined: 29 Oct 2017 Thanked: 1 times Followed by:5 members ### List $$K$$ consists of seven numbers. Is the average (arithmetic mean) of the seven numbers negative? by M7MBA » Sat Sep 11, 2021 2:06 pm 00:00 A B C D E ## Global Stats List $$K$$ consists of seven numbers. Is the average (arithmetic mean) of the seven numbers negative? (1) Four of the seven numbers in list $$K$$ are negative. (2) The sum of the seven numbers in list $$K$$ is negative.", "## Arithmetic Mean The arithmetic mean of a set of nine different positive integers is 123456789. Each number in the set contains a different number of digits with the greatest value being a nine-digit number. Find the value of each of the nine numbers. Source: mathcontest.olemiss.edu 1/26/2009 SOLUTION Given $\\left \\{a_1,a_2,\\cdots,a_9\\right \\}$ a set of nine different positive integers, the arithmetic mean of the set is $\\frac{1}{9}\\sum_{i=1}^{9} a_i$ Define the set of as follows: $a_1=1$ $a_2=11$ $a_3=111$ $a_4=1111$ $a_5=11111$ $a_6=111111$ $a_7=1111111$ $a_8=11111111$ $a_9=111111111$ Then, $\\sum_{i=1}^9 a_i=123456789$ Since we want the arithmetic mean to be 123456789, all we have to do is multiply each $a_i$ by 9. Answer: 9; 99; 999; 9999; 99999; 999999; 9999999; 99999999; 999999999.", "# Just 9 around Algebra Level pending If the arithmetic mean of nine numbers in the given set $$\\{9,99,999,9999,.....,999999999\\}$$ is a nine digit number $$N$$, then $$N$$ doesn't contain ×", "66 arithmetic questions. 17 A partition of 1000 into nine parts 1 answers, 451 views number-theory graph-theory arithmetic The sum of nine whole numbers is 1000. If those numbers are placed on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (i.e. ... 13 My Three Children The sum of the ages of my three children is 40. Though the ages of my two daughters are relatively prime (i. e. they have no common divisor), the age of each of them does have a common divisor greater ... 20 A Tour Around a Triangle 1 answers, 666 views number-theory graph-theory arithmetic Place the 18 even integers between 2 and 36 in the empty nodes of this triangular graph in such a way that if a path is drawn by coloring in red all the edges joining any two nodes whose numbers add ... 5 Six sisters on the ski lift 1 answers, 309 views number-theory arithmetic The sum of the ages of six sisters known to me is 92. Though there is no single whole number greater than 1 that simultaneously divides the ages of any three of them, I did notice this morning, while ... 20 The 10,958 Problem Here", "list divided by the number of items in the list. ... Again, changing the order of the three members of the list does not change the result: A = (8 + 11 + 2)/3 = 7, and that 7 is between 2 and 11. This summation method is easily generalized for lists with any number of elements. However, the mean of a list of integers is not necessarily an integer. \"The average family has 1.7 children\" is a jarring way of making a statement that is more appropriately expressed by \"the average number of children in the collection of families examined is 1.7\". ### Geometric mean With geometric mean, instead of adding numbers, the numbers are multiplied. Thus, the geometric mean of 2 and 8 is obtained by solving for G in the following equation: 2 * 8 = G * G. Thus, the geometric mean of 2 and 8 is G = sqrt(2 * 8) = 4. And again it is seen that changing the order of the members of the list to be averaged does not change the result: G = sqrt(8 * 2) = 4. In order to make sense of the requirement that the mean must be at least as big as the smallest member of the list and no bigger than the largest,", "arithmetic mean of the first ten prime numbers 6. Find the arithmetic mean of the first eleven even numbers 7. The number of members in 10 families of a society are 2, 4, 3, 4, 2, 2, 3, 5, 1, 4. Find the mean number of members per family. 8. The following are the number of maths book issued in a school library during a week: 222, 120, 201, 322, 167, 273, 406 and 345 Find the average number of maths book issued per day. 9. If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x. 10. The daily minimum temperature recorded (in degree F) at a place during a week was as under: Saturday Monday Tuesday Wednesday Thursday Friday 34.5 35.8 38.3 37.8 35.1 33.9 Find the mean temperature?. 11. The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean? 12. The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean? 13. The mean of 15 numbers is 27. If each number is multiplied by 5, what will be the mean of the new numbers? 14. The mean of 15 numbers is 63. If each number is", "In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the ~$n$~ throw is ~$n$~-1 for each n from 1 to 10, what is the average(arithmetic mean) of all the numbers in the table? 9.45 9.50 9.55 9.65 9.70", "### The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number. A. 25 B. 26 C. 27 D. 28 $\\begin{array}{l}\\text{Excluded number}\\\\ \\text{= (18 × 5) – (16 × 4)}\\\\ \\text{= 90 – 64}\\\\ \\text{= 26}\\end{array}$ $$\\begin{array}{l}\\text{Excluded number}\\\\ \\text{= (18 × 5) – (16 × 4)}\\\\ \\text{= 90 – 64}\\\\ \\text{= 26}\\end{array}$$", "# A set of 12 numbers has an arithmetic mean of 20. if the numbers 10 and 14 are removed from the set, what is the arithmetic mean of the 10 numbers remaining in the set? A set of 12 numbers has an arithmetic mean of 20. if the numbers 10 and 14 are removed from the set, what is the arithmetic mean of the 10 numbers remaining in the set? You can still ask an expert for help ## Want to know more about Polynomial arithmetic? • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Asma Vang For a set of positive numbers , the arithmetic mean is given as Now, set of 12 numbers which includes 10 and 14 has an arithmetic mean of 20. Let x be the sum of the remaining 10 numbers. Now, we find the arithmetic mean of the remaining 10 numbers whose sum is 216.", "from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 [Reveal] Spoiler: OA _________________ Math Expert Joined: 02 Sep 2009 Posts: 43831 In a numerical table with 10 rows and 10 columns, each entry is either [#permalink] ### Show Tags 25 Jul 2017, 22:10 5 KUDOS Expert's post 5 This post was BOOKMARKED carcass wrote: In a numerical table with 10 rows and 10 columns, each entry is either a 9 or a 10. If the number of 9s in the nth row is n – 1 for each n from 1 to 10, what is the average (arithmetic mean) of all the numbers in the table? A. 9.45 B. 9.50 C. 9.55 D. 9.65 E. 9.70 We have 10*10 = 100 entries. \"The number of 9s in the nth row is n – 1 for each n from 1 to 10\" means that: In the 10th row the number of 9s is 10 - 1 = 9; In the 9th row the number of 9s is 9 - 1 = 8; In the 8th row the number of 9s is 8 - 1 = 7; ... In the 1st row the number of 9s is 1 - 1 = 0.", "Question # Out of 20 numbers , mean of 8 numbers is 40 and the mean of the remaining numbers is 50. Find the mean of 20 numbers. A 60 B 46 C 40 D 45 Solution ## The correct option is B 46Given the mean of 8 numbers = 40 Hence the total of 8 numbers = 40 x 8 = 320 Given the mean of remaining (20 - 8 = 12) numbers = 50 Total of 12 numbers = 50 x 12 = 600 Total of 20 observation = 600 + 320 = 920 Hence the mean of 20 observation =92020=46Mathematics Suggest Corrections 0 Similar questions View More", "# hm if there are two groups with 75 and 65 as harmonic means and containing 15 and 13 observation then the combined harmonic mean is" ]

[ "# Ratio 6 numbers are in the ratio 1:5:1:5:5:5. Their sum is 242. What are the numbers? Correct result: a = 11 b = 55 c = 11 d = 55 e = 55 f = 55 #### Solution: $a=\\frac{1\\cdot 242}{1+5+1+5+5+5}=11$ $b=\\frac{5\\cdot 242}{1+5+1+5+5+5}=55$ $c=\\frac{1\\cdot 242}{1+5+1+5+5+5}=11$ $d=\\frac{5\\cdot 242}{1+5+1+5+5+5}=55$ $e=\\frac{5\\cdot 242}{1+5+1+5+5+5}=55$ $f=\\frac{5\\cdot 242}{1+5+1+5+5+5}=55$ We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Check out our ratio calculator. Do you have a system of equations and looking for calculator system of linear equations? ## Next similar math problems: • Ratio three numbers Three numbers SUV are in the ratio 1:2:3. Their sum is 24. Find this numbers and write their add and sum. • Moneys in triple ratio Milan, John and Lili have a total 344 euros. Their amounts are in the ratio 1:2:5. Determine how much each of them has? • Two squares Two squares whose sides are in the ratio 5:2 have a sum of its perimeters 73 cm. Calculate the sum of the area of these two squares. • Building At the building, we divided 240 boards into two piles in a 5: 3 ratio. How many were fewer boards in the lower pile? • Report 2 A", "I did not really do here anyway) could also be done for some $U_4,V_4,U_6,V_6.$ The other repeats up to $10^{12}$ are $41=V_3(4)=U_2(6),\\ 89=V_3(5)=U_2(9),\\ 1189=V_3(11)=U_2(34)$ along with $3191=U_5(5)=U_2(56)$ and $13201=V_5(7)=V_3(24).$ Note: to check up to $10^{12}$ we can generate the $U_3(m)$ and $V_3(m)$ up to $m=10^4$ along with any $U_i(m) \\lt 10^{12}$ and $V_i(m) \\lt 10^{12}$ for $i \\gt 3.$ In all this is about $43000$ vaues. We could also generate $U_2(m)=m^2+m-1$ up to $m=10^6$ but $m^2+m-1=v$ for $v=\\frac{-1+\\sqrt{5+4v}}{2}$ so it is better to just check which of the other values make the expression under the radical a square. However this does make it harder to check for the smallest gaps. It could still be done but I did not. My feeling is that there are a handful of repeated terms for coincidental reasons and that it is reasonable on random grounds to expect that there are only finitely many. Quite possibly just the $10$ I mentioned. There does not seem to be any underlying meaning for the coincidences. For example $239=U_3(6) \\approx \\frac{239+169\\sqrt{2}}{2}\\approx 239.001046$ and also $239=U_2(15)\\approx \\frac{239}{2}+\\frac{209\\sqrt{221}}{26} \\approx 239.0003219.$ I do not see anything deep here. However the fact that the rational and irrational parts are nearly equal is not a coincidence. Other thoughts: In a sense, $U(m)$ and $V(m)$ are just scaled versions of the powers", "## Return to Answer 2 added 386 characters in body Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$. If there are $7$ or more children, then the product is at least $8!>7056$, so there are at most $6$ children. Furthermore, if there are $6$ children, the sum is at most $84-8-7=69$, so the product is at most $69^2=4761$, but the product is at least $7!=5040$, so there cannot be $6$ children. Let $S$ denote the sum and $P$ the product. By the AM-GM inequality, we have $\\frac{S}{n} \\ge \\sqrt[n]{P} = \\sqrt[n]{S^2}$, so $\\frac{S^n}{n^n} \\ge S^2$, or $S^{n-2} \\ge n^n$, where $n$ is the number of children. This means that $n \\ge 3$, since $n^n \\ge 1$, which would contradict $n=2$. I'm not sure there's much else you can do without getting into some messy casework. To rule out $3$ and $5$, you can divide into cases like \"Suppose at least two children are older than 11,\" etc, and use similar arguments regarding sums and products as above. To then find the result given that $n=4$, you'll need to use some divisibility arguments and, yes, a little bit of casework. 1 Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$. If there", "# The sum of two numbers is 48. One number is 3 times as large as the other. What are the numbers? ## I just do not understand how to solve it. Apr 22, 2018 $12$ and $36$. #### Explanation: Let one number be $n$. Then the other number will be $3 n$. As per question $n + 3 n = 48$ $\\implies 4 n = 48$ $n = \\frac{48}{4} = 12$ Therefore, one number is $12$ and the other number is $3 \\cdot 12 = 36$", "Thread: Summing Series -2 questions. 1. Summing Series -2 questions. Hey, I've got 2 unsolved problems that need help! 1. Find the sum of the multiples of 7 which are less than 10000. My answer is 714264285 but the book says 7142142? 2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n. A step by step guide would be really appreciated for this one. Thanks in advance! 2. Hello, Originally Posted by Nyoxis Hey, I've got 2 unsolved problems that need help! 1. Find the sum of the multiples of 7 which are less than 10000. My answer is 714264285 but the book says 7142142? There are 1428 multiples of 7 between 1 and 10000, because $\\frac{10000}{7} \\approx 1428.571428571429$. Therefore, you can write the sum this way : $S=7 \\cdot 1+7 \\cdot 2+7 \\cdot 3+\\dots+7 \\cdot 1427+7 \\cdot 1428$ $S=7 \\left(1+2+\\dots+1427+1428\\right)$ What's in brackets is the sum of the 1428 first integers. We know that $1+\\dots+n=\\frac{n(n+1)}{2}$ So here, $1+2+\\dots+1428=\\frac{1428 \\cdot 1429}{2}=\\boxed{1020306}$ Hence $S=7 \\cdot 1020306=\\boxed{7142142}$ How did you find 714264285 ? 3. Originally Posted by Nyoxis 2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n. A step by step guide would be really appreciated for this one. Write it this way : $S=\\sum_{k=0}^{n-1} (k+1)(n-k)$ $S=\\sum_{k=0}^{n-1} \\left[nk+n-k-k^2\\right]=\\sum_{k=0}^{n-1} (n-1)k+\\sum_{k=0}^{n-1} n-\\sum_{k=0}^{n-1} k^2$", "sum is 17157 #### Question 11: Find the sum of all multiples of 9 lying between 300 and 700. All the multiples of 9 lying between 300 and 700 are 306, 315, 324, 333, ..., 693. This is an AP in which a = 306, d = (315 - 306) = 9 and l = 693. Let the number of terms be n. Then Tn = 693 a + (n - 1)d = 693 ⇒ 306 + (n - 1​) ⨯​ 9 = 693 ⇒ 9n = 396 ⇒ n = 44 ∴ Required sum = $\\frac{n}{2}\\left(a+l\\right)$ = Hence, the required sum is 21978. #### Question 12: Find the sum of all three-digit natural numbers which are divisible by 13. All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,..., 988. This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988 Let the number of terms be n. Then Tn = 988 a + (n - 1)d = 988 ⇒ 104 + (n -1​) ⨯​ 13 = 988 ⇒ 13n = 897 ⇒ n = 69 ∴ Required sum = $\\frac{n}{2}\\left(a+l\\right)$ = Hence, the required sum is 37674. #### Question 13: Find the sum of first 15 multiples of 8. The first 15 multiples of", "# Sum of digits raised to a power Let $S$ equal the sum of the digits of $2014^{2014}$. Let $T$ equal the sum of the digits of $S$. Let $U$ equal the sum of the digits of $T$. What is $U$? Because $10^n\\text{ mod }9=1$, the sum of digits $\\text{mod }9$ is the number itself $\\text{mod }9$. $$2014^{2014}\\equiv 7^{2014}\\equiv 7\\cdot(7^3)^{671}\\equiv7\\cdot1^{671}\\equiv7\\quad\\text{(mod }9)$$ What remains is to check that $U$ can't be bigger than 7. The least such number would be $16$. Clearly, $2014^{2014}<10^{4\\cdot2014}$, so $S$ is at most $9\\cdot8056<10^5$. Similarly $T$ is at most $9\\cdot5=45$, so $U$ is at most $4+9=13$. Therefore the answer is $7$. I don't know what $U$ is, but I can give an upper bound for it. $$2014^{2014} < 3000^{3000} = 27000000000^{1000} = (2.7*10^{10})^{1000} < (10^{11})^{1000} = 10^{11000}$$ So $2014^{2014}$ is at most $11000$ digits long, which means $S$ is at most $99000$, which means $T$ is at most $9+8+9+9+9=44$, which means $U$ is at most $3+9=12$.", "+0 # help 0 93 2 What is the sum of the series 7 + 10 + 17 + 24 + ... + 479? May 3, 2020 #1 0 7 + sum_(n=1)^68(7 n + 3) = 16,633 May 3, 2020 #2 +732 +2 Hi guest! The formula for a series is $$\\frac{a_1+a_n}{2}\\cdot n$$ n = how many terms there are So, we know $$a_1=10$$ since it's the first term (in this case let's save the 7 for the end, since it doesn't follow the pattern) $$a_n=479$$ since it's the last term There are 68 terms because there are 479-10=469, 469/7 = 67 terms, but you have to add 1 since we didn't count 10, so there are 68 terms. Now, all we have to do is plug the equation in! $$\\frac{10+479}{2} \\cdot 68=16626$$. We have to add that first 7 too, so $$16626+7=\\boxed{16633}$$. I hope this helped you, guest! :) May 3, 2020", "these latter five are the numbers we want to add. $\\frac1{19}+ \\frac2{19}+ \\frac3{19}+ \\frac 1{17}+ \\frac2{17}= \\frac6{19} + \\frac 3{17} = \\frac{6\\cdot17 + 3\\cdot19}{17\\cdot19} = \\frac{159}{323}$ and so the answer is $159 + 323 = \\boxed{482}$.", "sum of it's digits is V (and V is also a divisor of y). Then N is simply U written $\\frac{y}{V}$ times. For example, with x=y=2002 The divisors of 2002 are: 1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,20 02. We consider some multiples of 2002 and the sum of their digits: 2002 4 4004 8 6006 12 8008 16 10010 2 *Opalg's example 12012 6 14014 10 16016 14 14 divides 2002, and $\\frac{2002}{14}=143$. So, we just have to write 16016 143 times (16016160161601616016..........1601616016). That's another number that's divisible by 2002 and whose digits sum to 2002. Also, since $\\frac{16016}{2002}=8$, I believe dividing this number (1601616016...) by 8 will yield a number 143 digits long with 8 for all it's digits (i.e. 888888...8888 where 8 is written 143 times). Some interesting things to think about are: 1A) if there is a case in which no multiple of x has digits that sum up to a divisor of y. 1B) If the above case exist, does it necessarily mean that there does not exist a number N that satisfies the condition? (Look at number 3 below) 2) if x=y, will there always exist a number N? 3) The method we used takes advantage of repeating the number, but what about cases with non-repeating numbers?", "combinations for this case. If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\\cdot 4!=48$ combinations for this case. Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\\binom{4}{2}$ to choose the numbers and $\\frac{4!}{2!2!}=6$ ways to arrange them, so $6\\cdot 6=36$. If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case. Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\\frac{200}{1296}=\\frac{25}{162}$, yielding an answer of $25+162=\\boxed{187}$. -Stormersyle ## Solution 4 (Casework) Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$, and how many ways they can be ordered $1^2$- $1,1,1,1$- $1$ way. $2^2$- $4,1,1,1$ or $2,2,1,1$- $\\binom{4}{2}+4=10$ ways. $3^2$- $3,3,1,1$- $\\binom{4}{2}=6$ ways. $4^2$- $4,4,1,1$, $2,2,2,2$, or $2,2,4,1$- $\\binom{4}{2}+1+12=19$ ways. $5^2$- $5,5,1,1$- $\\binom{4}{2}=6$ ways. $6^2$- $6,6,1,1$, $1,2,3,6$, $2,3,2,3$, $3,3,4,1$- $2*\\binom{4}{2}+4!+12=48$ ways. $7^2$- Since there is a prime greater than 6 in its prime", "since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ ## Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ ## Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 ~ pi_is_3.14", "of members in the first sum, so: $n = \\frac{168}{4} = 42$ Finally: $\\sum = 3 \\cdot 42 \\cdot \\left(2 \\cdot 42 + 3\\right) = 126 \\cdot 87 = 10962$", ". . $\\vdots$ $\\text{ten 9's}\\begin{Bmatrix}9&0 \\\\ 9&1 \\\\ 9& 2 \\\\ \\vdots \\\\ 9&9 \\end{Bmatrix}\\;\\;\\text{0 to 9 = 45}$ In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times. Sum of the units digits: . $9 \\times 45 \\:=\\:405$ In the tens-column, we have: . $\\text{ten 1's + ten 2's + ten 3's + . . . + ten 9's}$ . . $\\;=\\;10(1) + 10(2) + 10(3) + \\cdots + 10(9) \\;=\\;10(1 + 2 + 3 +\\cdots + 9) \\;=\\;10(45)$ Sum of the tens digits: . $450$ Therefore, the sum of the digits is: . $405 + 450 \\;=\\;{\\color{blue}855}$ 12. Originally Posted by Chris L T521 $S_n=\\frac{n}{2}(a+z)$", "# Thread: How many ways can the number 50 be witten as the sum of 10 non-negative, even number? 1. ## How many ways can the number 50 be witten as the sum of 10 non-negative, even number? Hi, Please help me to find out: in how many ways the number 50 can be witten as the sum of 10 non-negative, even numbers? Thanks alot. 2. Hello, gravity2910! In how many ways the number 50 can be witten as the sum of 10 nonnegative, even numbers? I hope the order of the numbers is important. .** For example: $\\{1,1,1,1,1,1,1,1,1,41\\}$ is different from $\\{41,1,1,1,1,1,1,1,1,1\\}$ Consider fifty 1's in a row. Group them in pairs .... and we have twenty-five 2's in a row. . . . . . $\\underbrace{\\boxed{2}\\;\\;\\boxed{2}\\;\\;\\boxed{2}\\;\\ ;\\boxed{2}\\;\\;\\hdots\\;\\;\\boxed{2}}_{\\text{t\\!went\\ !y-five pairs}}$ There are 24 spaces between these pairs. We will insert 9 dividers into those spaces. For example: . $\\boxed{2}\\;\\boxed{2}\\;\\bigg|\\;\\boxed{2}\\;\\bigg| \\; \\boxed{2}\\;\\boxed{2}\\; \\boxed{2}\\;\\boxed{2}\\;\\bigg| \\;\\hdots \\; \\bigg|\\;\\boxed{2}\\;\\boxed{2}\\;\\boxed{2}$ . . . . . . . . . . represents the sum: . $4 + 2 + 8 + \\hdots + 6$ Therefore, there are: . $_{24}C_9 \\;=\\;{24\\choose9} \\;=\\;\\frac{24!}{9!\\,15!} \\;=\\;\\boxed{1,\\!307,\\!504}$ ways. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~", "3:55 To start, you should work out the sum of all the cubes, preferably using the formula $$1^3+2^3+\\cdots+12^3=\\frac{12^2\\times13^2}{4}=6084\\ .$$ So you need to find six of your twelve cubes which add up to $3042$. I should think it's trial and error from here, but would be glad to see if anyone has a smarter solution. • Thanks for your suggestion David :) I'll give it a go and see where it gets me. – A is for Ambition Mar 24 '15 at 8:08 First, you've noticed that each side of the equation has to sum up to $3042$. Since $11^3+12^3=3059>3042$, each one of them has to be on a different group. Since $3042$ is even, each group must contain an even number of odd numbers. So each group must contain $0$ or $2$ or $4$ or $6$ odd numbers. With $0$ odd numbers on one group, it will definitely sum up to a larger value than the other group. With $6$ odd numbers on one group, it will definitely sum up to a smaller value than the other group. So the remaining options are: • \"$12$\" with $3$ out of $5$ other even numbers and $2$ out of $5$ odd numbers other than \"$11$\" • \"$12$\" with $1$ out of $5$ other even numbers and $4$ out of", "easy to work towards a sum of 27. The children could be possibly: 1, 6, 6, 6 and 8. because $1+6+6+6+8=27=3^3$ and $1\\cdot 6\\cdot 6\\cdot 6\\cdot 8=1728=12^3$ One important thing to notice is that, since the product of the ages must be a cube, the prime factors of all ages can only appear 3 times, 6 times, 9 times, etc. I first tried with the product $2^3\\cdot3^3=216$, but couldn't manage to distribute the factors such that the sum would also be a cube. Then I experimented with $2^6+3^3$ and I had more luck. • This is right, but having triplets is very rare. – Jamal Senjaya Oct 7 '16 at 9:16 • +1 because that was a really speedy answer, and technically OP asked for reasonable, not probable. Kudos to Lynn! – Xenocacia Oct 7 '16 at 9:20 I think $4, 16, 35, 35, 35$ works as well - even though it is not a common, but still a possible solution: sum is $125=5^3$, product is $2744000=140^3$ Other - not yet mentioned solutions are: $2, 4, 7, 7, 7$ with $\\sum=3^3, \\prod=14^3$ $1, 1, 2, 2, 2$ with $\\sum=2^3, \\prod=2^3$ $0, 2, 2, 2, 2$ with $\\sum=2^3, \\prod=0^3$ $0, 2, 5, 9, 11$ with $\\sum=3^3, \\prod=0^3$ • My solution was the second one 1,1,2,2,2. Although this means once", "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 Solution 6 We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _", "126. So 630 is the LCD for our calculation. 90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\\frac{29}{90} = \\frac{7 \\cdot 29}{7 \\cdot 90} = \\frac{203}{630}$ . 126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\\frac{13}{126} = \\frac{5 \\cdot 13}{5 \\cdot 126} = \\frac{65}{630}$ . Now we complete the problem: $\\frac{29}{90} - \\frac{13}{126} = \\frac{203}{630} - \\frac{65}{630} = \\frac{138}{630}$ . This fraction simplifies . To be sure of finding the simplest form for $\\frac{138}{630}$ , we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23. $\\frac{138}{630} = \\frac{2 \\cdot 3 \\cdot 23}{2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7}$ ; one factor of 2 and one factor of 3 cancels out, leaving $\\frac{23}{3 \\cdot 5 \\cdot 7}$ or $\\frac{23}{105}$ as our answer. Identify and Apply Properties of Addition Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties . Commutative property: When two numbers are added, the sum is the same even if the order of the items being added changes. Example: $3 + 2 = 2", "so these latter five are the numbers we want to add. $\\frac1{19}+ \\frac2{19}+ \\frac3{19}+ \\frac 1{17}+ \\frac2{17}= \\frac6{19} + \\frac 3{17} = \\frac{6\\cdot17 + 3\\cdot19}{17\\cdot19} = \\frac{159}{323}$ and so the answer is $159 + 323 = \\bf 482$." ]

[ "### The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is. A. 414 B. 420 C. 620 D. 621 $\\begin{array}{l}\\text{Sum of numbers}\\\\ \\text{= (41.4 × 15)}\\\\ \\text{= 621}\\end{array}$", "# IMAT 2015 Q57 [Arithmetic Mean] The arithmetic mean of the three numbers a, b, c is 8. Find the arithmetic mean of the four numbers: a + 1, b + 2, c + 6, 3. A. 7 B. 11 C. 5 D. 9 E. 27 The arithmetic mean A is defined by the formula: A = \\frac{1}{n}\\sum_{i=1}^{n}a_i = \\frac{a_1 + a_2 + \\ldots + a_n}{n} So the arithmetic mean of three numbers a, b, c is A_1 =\\frac{a+b+c}{3} = 8. The arithmetic mean for four numbers a+1, b+2, c+6, 3 is A_2 =\\frac{(a+1)+(b+2)+(c+6)+3}{4} = \\frac{a+b+c+9+3}{4}= \\frac{a+b+c+12}{4} = \\frac{a+b+c}{4} +\\frac{12}{4} = \\frac{a+b+c}{4} + 3. From the first arithmetic mean, we can express what a+b+c is and use it in the second mean in order to get the value needed. A_1=\\frac{a+b+c}{3} = 8 \\to a+b+c = 8 \\cdot 3\\to a+b+c = 24. Using this information in the A_2 we get: A_2 = \\frac{a+b+c}{4} + 3 = \\frac{24}{4} + 3 = 6 + 3 = 9. Which gives us the correct solution (D) 9.", "# The sum of 40 numbers in an arithmetic sequence is 8000. If the first number is the sequence is 83, what is the last number? Mar 8, 2016 $317$ #### Explanation: In an arithmetic sequence the difference between consecutive members is constant. Therefore the sequence is $83 , 83 + d , 83 + 2 d , \\ldots . .83 + 39 d$ The $i$th member is $83 + \\left(i - 1\\right) d$ As shown on Arithmetic Sequences the formula for the sum of members is ${\\sum}_{0}^{n - 1} \\left(a + k d\\right) = \\frac{n}{2} \\cdot \\left(2 a + \\left(n - 1\\right) d\\right)$ In this case ${\\sum}_{0}^{39} = \\frac{40}{2} \\cdot \\left(2 \\cdot 83 + 39 \\cdot d\\right) = 8000$ $166 + 39 d = 400$ $39 d = 234$ $d = \\frac{234}{39} = 6$ The $40$th number is therefore $83 + 39 \\cdot 6 = 317$", "arithmetic means of the sets: 1. 261, 286, 257, 284, 258, 281, 258, 285, 267, 275, 258, 284, 260, 285, 258, 284 2. 43, 44, 45, 45, 46, 46, 47, 47, 47, 48, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 50, 50, 50, 51 3. 14, 28, 42, 56, 23, 67, 12, 45, 93, 52 Solution: For each set, find the sum of the values in the set, and divide by the number of values: a) The sum of the values is 4,341. There are 16 values. $\\frac{4,341}{16}=271.31$ b) The sum of the values is 1,198. There are 25 values. $\\frac{1,198}{25}=47.92$ c) The sum of the values is 432. There are 10 values. $\\frac{432}{10}=43.2$ Example C Using the data from the chart, find: a) the mean number of prank calls per five-year interval, and b) the mean number of reported prank calls per year. Solution: a) First, find the sum of the number of reported calls: $188+149+161+172+154+172=996 \\ calls$ The chart indicates that there are six five-year intervals with recorded values. So, dividing the sum by the number of values, we get: $\\frac{996}{6}=166 \\ calls \\ per \\ five \\ year \\ interval$ Note that the value “166” is not the actual count of any of the recorded intervals. The arithmetic mean obviously does", "### If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is : A. 10 B. 12 C. 14 D. 16 $\\begin{array}{l}\\text{Arithmetic mean}\\\\ \\Large\\frac{\\text{ Total sum }}{\\text{ Total number }}\\\\ \\text{According to the question,}\\\\ \\Rightarrow10=\\frac{7+5+13+x+9}{5}\\\\ \\text{⇒ 50 = x + 34}\\\\ \\text{⇒ x = 50 – 34}\\\\ \\text{⇒ x = 16}\\end{array}$", "Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too! # Insert four arithmetic means between 4 and 29 Question: Insert four arithmetic means between 4 and 29 Solution: To find: Four arithmetic means between 4 and 29 Formula used: (i) $d=\\frac{b-a}{n+1}$, where, $d$ is the common difference n is the number of arithmetic means (ii) $A_{n}=a+n d$ We have 4 and 29 Using Formula, $d=\\frac{b-a}{n+1}$ $d=\\frac{29-4}{4+1}$ $d=\\frac{25}{5}$ $d=5$ Using Formula, $A_{n}=a+n d$ First arithmetic mean, $A_{1}=a+d$ = 4 + 5 = 9 Second arithmetic mean, $A_{2}=a+2 d$ $=4+2(5)$ $=4+10$ $=14$ Third arithmetic mean, $\\mathrm{A}_{3}=\\mathrm{a}+3 \\mathrm{~d}$ $=4+3(5)$ $=4+15$ = 19 Fourth arithmetic mean, $\\mathrm{A}_{4}=\\mathrm{a}+4 \\mathrm{~d}$ $=4+4(5)$ $=4+20$ $=24$ Ans) The four arithmetic means between 4 and 29 are 9, 14, 19 and 24", "but we may NOT ADD MEANS! The mean of 3 numbers is 7 . $\\text{There are 3 numbers, the mean is 7,\" rArr \"the total is } 3 \\times 7 = 21$ The mean of 4 numbers is 8 $\\text{There are 4 numbers, the mean is 8,\" rArr \"the total is } 4 \\times 8 = 32$ The mean of 5 number is 11 $\\text{There are 5 numbers, mean is 11\" rArr \"the total is } 5 \\times 11 = 55$ We can add numbers: $3 + 4 + 5 = 12$ We can add totals: $21 + 32 + 55 = 108$ We can find the mean of the 12 numbers: $M = \\frac{108}{12} = 9$", "Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here! MD # Find the mean of each set of values.See Example 1.$$3 \\quad 4 \\quad 7 \\quad 7 \\quad 8 \\quad 11 \\quad 16$$ ## 8 #### Topics Graphs and Statistics ### Discussion You must be signed in to discuss. ### Video Transcript on the mean of a set of values. We need toe fined the sum of old values and divide this by the total number of values. For example, if we're given the base up great. So 77 eight, 11 16 and we want to find the me and of this Davis up, we first need to dio the some of these values three plus for plus seven who's seven plus eight with 11 cough 16 and divide this by the 10th. The number values. As we can see there are 1234567 values, so we would divide this by seven. This dip SOS 56 divided by seven, which equals eight, so the mean MD #### Topics Graphs and Statistics", "of members in the first sum, so: $n = \\frac{168}{4} = 42$ Finally: $\\sum = 3 \\cdot 42 \\cdot \\left(2 \\cdot 42 + 3\\right) = 126 \\cdot 87 = 10962$", "# How do you find the mean of $7420,$3210, $6720,$5780, $1850,$4350? Jul 11, 2018 The formula for mean is $\\frac{\\sum x}{n}$. #### Explanation: We can find the mean of these numbers by using the mean formula. $\\frac{\\sum x}{n}$ $\\sum$ is a symbol meaning 'sum.' In this case, it tells us to take the sum of $x$, which refers to the values we're finding the mean of. $n$ stands for the number of terms in the set. In this problem, there are $6$ numbers. $\\frac{7420 + 3210 + 6720 + 5780 + 1850 + 4350}{6} \\approx 4888.3$ Therefore, the mean is approximately $4888.3$.", "# The mean of 3 numbers is 7. The mean of 4 other numbers is 8. The mean of 5 other numbers is 11. What is the mean of all 12 numbers? Aug 11, 2016 $9$. #### Explanation: Let ${\\overline{x}}_{i}$ be the mean of a group ${G}_{i}$ of ${n}_{i}$ observations, where, $i = 1 , 2 , 3 , \\ldots , m$. Then, the mean $\\overline{x}$ of the combined group $G = {G}_{1} \\cup {G}_{2} \\cup {G}_{3} \\cup \\ldots \\cup {G}_{m}$ is given by, barx=(Sigma_(i=1)^(i=m) n_ibarx_i)/(Sigma_(i=1)^(i=m) n_i. We have, ${n}_{1} = 3 , {n}_{2} = 4 , {n}_{3} = 5 , {\\overline{x}}_{1} = 7 , {\\overline{x}}_{2} = 8 , {\\overline{x}}_{3} = 11$. Hence, $\\overline{x} = \\frac{3 \\cdot 7 + 4 \\cdot 8 + 5 \\cdot 11}{3 + 4 + 5} = \\frac{108}{12} = 9$, as derived by Lyn. ! Enjoy Maths.! Aug 11, 2016 $9$ #### Explanation: To work out the mean (what we understand as the 'average') we do the following; Mean = (\" sum (total) of all the values\")/(\"the number of values\")\" OR \" \" M=T/N As with other formulae in this form, we have 3 ways in which it can be used. $M = \\frac{T}{N} \\text{ \"T = MxxN\" } N = \\frac{T}{M}$ Remember that we may add Totals , we may add Numbers ,", "If the sum of n arithmetic means between 9 and 51 is 270, Question: If the sum of n arithmetic means between 9 and 51 is 270, then the value of n is ___________. Solution: Let a = 9 b = 51 Given sum of arithmetic means between 9 and 51 is 270. Let A1A2, ______ An be arithmetic means between and b. ∴ aA1A2, _____ An , are in A.P Here total number of terms is n + 2 first term is a last term is b. $\\therefore$ Sum of first $n+2$ terms $=\\frac{(n+2)}{2}\\{a+b\\}$ $A_{1}+A_{2}+\\ldots \\ldots \\ldots+A_{n}=\\frac{n(a+b)}{2}$ i. e $270=n\\left(\\frac{9+51}{2}\\right)$ i. e $270=n\\left(\\frac{60}{2}\\right)$ i. e $\\frac{270}{30}=n$ i. e $n=9$ Hence, value of n is 9.", "The sum of sixteen and six times a number t is eighty-two. What is the number? Dec 6, 2016 $t = 11$ Explanation: First, let's write the equation we need to solve one step at a time: \"six times a number $t$\" can be written as: $6 \\cdot t$ Then, \"The sum of sixteen and \" this number can be written as: $\\left(6 \\cdot t\\right) + 16$ Finally, this term \"is eighty-two\" gives us: $\\left(6 \\cdot t\\right) + 16 = 82$ We can now solve for $t$: $\\left(6 \\cdot t\\right) + 16 - 16 = 82 - 16$ $\\left(6 \\cdot t\\right) + 0 = 66$ $6 t = 66$ $\\frac{6 t}{6} = \\frac{66}{6}$ $\\frac{\\cancel{6} t}{\\cancel{6}} = 11$ $t = 11$", "two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean. Sol: Calculated mean of 100 items = 64 Sum of 100 items, as calculated = (100×64)=6400$(100\\times 64)=6400$ Correct sum of these items = [6400 – (wrong items) + (correct items)] = [6400 – (26 + 9) + (36 + 90)] = [6400 – 35 + 126] Therefore, the Correct mean = 6491100=64.91$\\frac{6491}{100}=64.91$ Q.15: The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number. Sol: Mean of 6 numbers = 23 Sum of 6 numbers = 23×6=138$23\\times 6=138$ Again, mean of 5 numbers = 20 Sum of 5 number = 20×5=100$20 \\times 5=100$ The excluded number = (Sum of 6 number) – (sum of 5 numbers) = (138-100) = 38 Therefore, the excluded number = 38 #### Practise This Question When alpha particles are sent through a thin metal foil, only one out of ten thousand rebounded. This observation led to the conclusion that:", "### The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number. A. 25 B. 26 C. 27 D. 28 $\\begin{array}{l}\\text{Excluded number}\\\\ \\text{= (18 × 5) – (16 × 4)}\\\\ \\text{= 90 – 64}\\\\ \\text{= 26}\\end{array}$ $$\\begin{array}{l}\\text{Excluded number}\\\\ \\text{= (18 × 5) – (16 × 4)}\\\\ \\text{= 90 – 64}\\\\ \\text{= 26}\\end{array}$$", "# Thread: How many ways can the number 50 be witten as the sum of 10 non-negative, even number? 1. ## How many ways can the number 50 be witten as the sum of 10 non-negative, even number? Hi, Please help me to find out: in how many ways the number 50 can be witten as the sum of 10 non-negative, even numbers? Thanks alot. 2. Hello, gravity2910! In how many ways the number 50 can be witten as the sum of 10 nonnegative, even numbers? I hope the order of the numbers is important. .** For example: $\\{1,1,1,1,1,1,1,1,1,41\\}$ is different from $\\{41,1,1,1,1,1,1,1,1,1\\}$ Consider fifty 1's in a row. Group them in pairs .... and we have twenty-five 2's in a row. . . . . . $\\underbrace{\\boxed{2}\\;\\;\\boxed{2}\\;\\;\\boxed{2}\\;\\ ;\\boxed{2}\\;\\;\\hdots\\;\\;\\boxed{2}}_{\\text{t\\!went\\ !y-five pairs}}$ There are 24 spaces between these pairs. We will insert 9 dividers into those spaces. For example: . $\\boxed{2}\\;\\boxed{2}\\;\\bigg|\\;\\boxed{2}\\;\\bigg| \\; \\boxed{2}\\;\\boxed{2}\\; \\boxed{2}\\;\\boxed{2}\\;\\bigg| \\;\\hdots \\; \\bigg|\\;\\boxed{2}\\;\\boxed{2}\\;\\boxed{2}$ . . . . . . . . . . represents the sum: . $4 + 2 + 8 + \\hdots + 6$ Therefore, there are: . $_{24}C_9 \\;=\\;{24\\choose9} \\;=\\;\\frac{24!}{9!\\,15!} \\;=\\;\\boxed{1,\\!307,\\!504}$ ways. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~", "01:59 Thanks all for helping me on this Intern Joined: 10 Aug 2014 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink] 08 Sep 2014, 22:09 Bunuel wrote: gmatpapa wrote: If numbers N and K are added to set X{2,8,10,12}, its mean will increase by 25%. What is the value of $$N^2+2NK+K^2$$ ? A. 28 B. 32 C. 64 D. 784 E. 3600 Mean of {2,8,10,12} is $$mean=\\frac{2+8+10+12}{4}=\\frac{32}{4}=8$$ --> new mean thus should equal to $$8*1.25=10$$, so $$\\frac{2+8+10+12+n+k}{6}=10$$ (note that now we have the set of 6 terms not 4) --> $$n+k=60-32=28$$ --> $$n^2+2nk+k^2=(n+k)^2=28^2=784$$. Magician!!! Why do my eyes escape these little details like the a+b whole square formula being expanded in the question :// Re: If numbers N and K are added to set X{2,8,10,12}, its mean w [#permalink] 08 Sep 2014, 22:09 Similar topics Replies Last post Similar Topics: 2 The arithmetic mean of the list of numbers above is 4. If k 3 13 Mar 2014, 01:31 9 A certain set of numbers has an average (arithmetic mean) of 7 21 Oct 2013, 19:13 3 Set S consists of positive numbers. If -1 is added 5 16 Apr 2013, 18:38 1 If the average (arithmetic mean) of the four", "# Given the set: {9,0,6,-10,2,x,3} , for what x would the mean of the set be -2? Jun 10, 2018 $x$ would have to be $- 24$ #### Explanation: Use the formula for mean. $\\frac{9 + 0 + 6 - 10 + 2 + x + 3}{7} = - 2$ $10 + x = - 14$ $x = - 24$ Hopefully this helps!", "Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here! MD # Find the mean of each set of values.See Example 1.$$0 \\quad 0 \\quad 3 \\quad 4 \\quad 7 \\quad 9 \\quad 12$$ ## 5 #### Topics Graphs and Statistics ### Discussion You must be signed in to discuss. ### Video Transcript if we won't find the mean of assessing valleys, we need to find the song of all values and divide this by the total number of values. For example, if we're given the status up zero zero three, uh seven, nine. 12. Find the symbol values we would adult all the numbers and the status up. We need to include those errors as these are important. Well, im within kindle the title number of values. Uh, I mean it cools zero plus zero plus three. It was for with seven plus nine. Who's 12 divided by the total number values. Just 123456 happen. We have this up. We can see that this equals 35 divided by seven means the final answer on the mean of this status that equals five. MD #### Topics Graphs and Statistics", "of six numbers ​$÷$ 6 ⇒ Sum of the six numbers = 25 x 6 = 150. The included number = Sum of the six numbers - Sum of the five numbers ⇒​The included number = 150 - 135 ⇒The included number = 15. #### Question 23: The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean. #### Answer: Let x1, x2, x3, ... x75 be 75 numbers with their mean equal to 35. Then, x1x2 x3 +... + x75 = 35 ×75 x1 x2 x3 +... + x75 2625 The new numbers are 4x1, 4x2, 4x3, ...4x75. Let M be the arithmetic mean of the new numbers. Then, View NCERT Solutions for all chapters of Class 7" ]

[ "store. She bought vegetables for $2, Fruits for$3, Loaf of bread $1, a can of mushrooms$4, and a can of Baked beans $3. How much did she spend in all? 8. Sylvia has gone to a boutique; she buys a stole for$6, a coat for $12, a pair of sandals for$34, and a bag for $10. Calculate the total amount of money she spent. 9. Phoebe buys 3 pencils, 4 erasers, 2 sharpeners, and 1 ruler. How many things has she bought in all? Answers of word problems : 1) 32 2) 64 3) 100 4) 21 5)$70 6) $12 7)$13 8) \\$62 9) 10", "friend 7 cents. Now he has 51 cents in the wallet. How many cents previously had in his wallet? • Gasoline price At 34.4 cents per liter, how much will 42 liters of gasoline cost? • Trickster Figliarko has 13 cents less than Hana. Together they have 57 cents. How many cents has Hana and how many Figliarko? • Parcel 4 Send a parcel by messenger within city limits costs 60 cents for the first pound and 48 cents for each additional pound. What is the cost, in cents, of sending a parcel weighing p=4 pounds? • Birthday party For her youngest son's birthday party, mother bought 6 3/4 kg of hotdog and 5 1/3 dozens bread rolls. Hotdogs cost 160 per kilogram and a dozen bread roll costs 25. How much did she spend in all? • Rolls Mom every day buys 6 rolls. On Friday buys 2 times as much. How many rolls buys on Friday? • Vacation Smiths paid a deposit for a vacation of two-sevenths of the total price of the vacation. Then paid also € 550. How much cost their vacation? • Apples and pears Apples cost 50 cents piece, pears 60 cents piece, bananas cheaper than pears. Grandma bought 5 pieces of fruit. There was only one banana, and I paid 2", "Judy bought a quantity of pens in packages of 5 for $0.80 per package. She sold all of the pens in packages of 3 for$0.60 per package. If Judy’s profit from the pens was \\$8.00, how many pens did she buy and sell? 40 80 100 200 400", "# Cindy Cindy bought four slices of pizza for Php 95.60. Whole pizza has 6 slices. How much does each slice cost? s = 23.9 PHP ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! ## Related math problems and questions: • A pizza 2 A pizza is cut into slices that are each 1/6 of the whole. John is going to eat 1/2 of the whole pizza. How many slices will John eat? • Slices of pizza Maria ate 1/4 of a pizza. If there were 20 slices of pizza, how many slices did Maria eat? • Pizza 5 You have 2/4 of a pizza, and you want to share it equally between 2 people. How much pizza does each person get? • Grayson Grayson bought snacks for his team's practice. He bought a bag of oranges for $3.11 and a 24-pack of juice bottles. The total cost before tax was$50.15. How much was each bottle of juice? • ZOO 2 Valerie bought tickets to the zoo for the family. She bought 3 adult tickets for $9.50 each and 2 children's tickets for$4.50 each. How much did the tickets cost Valerie in all? • Two cakes Two cakes were each cut into 8 slices. Maria ate 1/8 of the", "# Groceries Arturo buys two cans of tomato soup that cost $0.52 each and 12 cans of chicken noodle soup that cost$0.51 each. How much is the total cost of these groceries? x = 7.16 USD ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you!", "did she pay for the box of cereal? 182. Coupon Diana bought a can of coffee that cost $7.99.7.99.$ She had a coupon for $0.750.75$ off, and the store doubled the coupon. How much did she pay for the can of coffee? 183. Diet Leo took part in a diet program. He weighed $190190$ pounds at the start of the program. During the first week, he lost $4.34.3$ pounds. During the second week, he had lost $2.82.8$ pounds. The third week, he gained $0.70.7$ pounds. The fourth week, he lost $1.91.9$ pounds. What did Leo weigh at the end of the fourth week? 184. Snowpack On April $1,1,$ the snowpack at the ski resort was $44$ meters deep, but the next few days were very warm. By April $5,5,$ the snow depth was $1.61.6$ meters less. On April $8,8,$ it snowed and added $2.12.1$ meters of snow. What was the total depth of the snow? 185. Coffee Noriko bought $44$ coffees for herself and her co-workers. Each coffee was $3.75.3.75.$ How much did she pay for all the coffees? 186. Subway Fare Arianna spends $4.504.50$ per day on subway fare. Last week she rode the subway $66$ days. How much did she spend for the subway fares? 187. Income Mayra earns $9.259.25$ per hour. Last week she worked", "Buy 1: $99 - Ultra-Premium Chocolate Ice Cream$98 - Ultra-Premium Vanilla Ice Cream The shaping helped a lot, but now our AI decision resulted in substantial overspending. To fix this issue, we can use a pruning criteria and add a fallback. • #9 Preferring Ice Cream and Gelato, of the items that cost less than $20, pick any random one of most preferred food items or if there are none, leave the store without buying anything. Scenario 7 - Buy 1:$1 - Extra Gummy Bears $1 - Extra Napkins$5 - Chocolate Ice Cream $5 - Vanilla Ice Cream$5 - Strawberry Ice Cream $5 - Cherry Vanilla Ice Cream$6 - Tiramisu Gelato $6 - Limoncello Gelato$50 - Rocky Road Ice Cream $99 - Ultra-Premium Chocolate Ice Cream$98 - Ultra-Premium Vanilla Ice Cream The AI’s final possible decisions from this set are tiered as: Tier 1 - $5 - Chocolate Ice Cream Tier 1 -$5 - Vanilla Ice Cream Tier 1 - $5 - Strawberry Ice Cream Tier 1 -$5 - Cherry Vanilla Ice Cream Tier 1 - $6 - Tiramisu Gelato Tier 1 -$6 - Limoncello Gelato Tier 2 - \\$1 - Extra Gummy Bears This selection set seems fantastic! Any AI picking from that tiered set is making a very reasonable decision. So, we need a combination of", "(Carrots are a standout case: you can go through the hassle of cleaning, peeling, and chopping a sad bowl of fresh carrots which you must eat before they go bad, or you can eat some delicious pre-chopped carrots whenever you feel like pulling them out of your freezer, possibly a year later.) (I wouldn’t take ‘health’ too seriously as a criterion. Diet/nutrition research is one of the worst fields in all medicine. Don’t sacrifice your quality of life now for some small late-late QALYs which may not exist at all.) 2. investigate all local groceries. The average price can differ considerably between stores. In my own feasible shopping area, I have Walmart, Target, Shoppers, Aldi, Giant and some others (BJ’s is the major alternative but I’ve never been convinced I would be able to buy enough to benefit). When I switched from NEX to Walmart, I saved a good 10%; when I switched (most of) my shopping to Aldi, I saved another good 10%. (The cost savings had I started with Whole Foods or Harris Teeter hardly bear thinking on.) There are some disadvantages to shopping at Aldi (more restricted selection, disorganized store, having to remember to bring a quarter for the shopping carts) but saving $20 or$30 is a good salve for the annoyances. It may take", "Ann, Carol, and Judy paid a total of $45 for their dinner at a restaurant. If Ann paid ~$\\frac{2}{5}$~ of the total amount, Carol paid$17, and Judy paid the rest, what fraction of the total amount did Judy pay? ~$\\frac{2}{9}$~ ~$\\frac{14}{45}$~ ~$\\frac{1}{3}$~ ~$\\frac{2}{5}$~ ~$\\frac{7}{15}$~", "# Donna spent a total of $40 at the grocery store. Of this amount, she pent$8 on fruit. What... Donna spent a total of $40 at the grocery store. Of this amount, she pent$8 on fruit. What percentage of the total did she spend on fruit?", "at $0.75 per bottle, 4 pounds of assorted candy at$3.50 per pound, and 16 packages of microwave popcorn costing $0.50 each for her party. What was her total bill? 52. Joe bought four 8-foot 2-by-4 boards for$24.00. How much did he spend per linear foot? 53. Margaret bought two cases of soda at the local discount store for $23.52. If each case contained 24 bottles, how much did she spend per bottle? 54. Billy earns$12.00 per hour and “time and a half” for every hour he works over 40 hours a week. What is his pay for 47 hours of work this week? 55. Audry bought 4 bags of marbles each containing 15 assorted marbles. If she wishes to divide them up evenly between her 3 children, how many will each child receive? 56. Mark and Janet carpooled home from college for the Thanksgiving holiday. They shared the driving, but Mark drove twice as far as Janet. If Janet drove 135 miles, then how many miles was the entire trip? Part B: Order of Operations with Absolute Values Simplify. 57. $3+2|−5|$ 58. $9−4|−3|$ 59. $−(−|2|+|−10|)$ 60. $−(|−6|−|−8|)$ 61. $|−(40−|−22|)|$ 62. $||−5|−|10||$ 63. $−(|−8|−5)2$ 64. $(|−1|−|−2|)2$ 65. $−4+2|22−32|$ 66. $−10−|4−52|$ 67. $−|(−5)2+42÷8|$ 68. 69. $−2[7−(4+|−7|)]$ 70. $3−7 |−2−3|+43$ 71. $7−5|62−52|+(−7)2$ 72. $(−4)2−|−5+(−2)3|−32$ 73. $23−|12−(−43)2|$ 74. $−30|103−12÷15|$ 75. $(−4)3−(2−|−4|)÷|−32+7|$ 76.", "+ Verbal:https://gmatclub.com/forum/gmat-flashcards-108651.html Thursdays with Ron:https://gmatclub.com/forum/manhattan-s-thursdays-with-ron-consolidated-video-index-223612.html#p2138753 Re: A shopkeeper mixes three varieties of rice costing$10, $12,$17 per &nbs [#permalink] 05 Oct 2018, 03:42 Display posts from previous: Sort by", "$7.99.7.99.$ She had a coupon for $0.750.75$ off, and the store doubled the coupon. How much did she pay for the can of coffee? 183. Diet Leo took part in a diet program. He weighed $190190$ pounds at the start of the program. During the first week, he lost $4.34.3$ pounds. During the second week, he had lost $2.82.8$ pounds. The third week, he gained $0.70.7$ pounds. The fourth week, he lost $1.91.9$ pounds. What did Leo weigh at the end of the fourth week? 184. Snowpack On April $1,1,$ the snowpack at the ski resort was $44$ meters deep, but the next few days were very warm. By April $5,5,$ the snow depth was $1.61.6$ meters less. On April $8,8,$ it snowed and added $2.12.1$ meters of snow. What was the total depth of the snow? 185. Coffee Noriko bought $44$ coffees for herself and her co-workers. Each coffee was $3.75.3.75.$ How much did she pay for all the coffees? 186. Subway Fare Arianna spends $4.504.50$ per day on subway fare. Last week she rode the subway $66$ days. How much did she spend for the subway fares? 187. Income Mayra earns $9.259.25$ per hour. Last week she worked $3232$ hours. How much did she earn? 188. Income Peter earns $8.758.75$ per hour. Last week he worked", "$40 from the ATM. She spent$15.11 on a pair of earrings. How much money did she have left? 2. Spending money Marissa found $20 in her pocket. She spent$4.82 on a smoothie. How much of the $20 did she have left? 3. Shopping Adam bought a t-shirt for$18.49 and a book for $8.92 The sales tax was$1.65. How much did Adam spend? 4. Restaurant Roberto’s restaurant bill was $20.45 for the entrée and$3.15 for the drink. He left a $4.40 tip. How much did Roberto spend? 5. Coupon Emily bought a box of cereal that cost$4.29. She had a coupon for $0.55 off, and the store doubled the coupon. How much did she pay for the box of cereal? 6. Coupon Diana bought a can of coffee that cost$7.99. She had a coupon for $0.75 off, and the store doubled the coupon. How much did she pay for the can of coffee? 7. Diet Leo took part in a diet program. He weighed 190 pounds at the start of the program. During the first week, he lost 4.3 pounds. During the second week, he had lost 2.8 pounds. The third week, he gained 0.7 pounds. The fourth week, he lost 1.9 pounds. What did Leo weigh at the end of the fourth week? 8. Snowpack On April 1,", "carrots and oatmeal that would provide sufficient nutrients without giving too much fat. What we want to know is which combination is least expensive. A pound of carrots and a pound of oatmeal both cost about \\$0.50. So, we want to minimize the function$\\$0.5\\cdot C+\\$0.5\\cdot Qwhile still staying inside the blue region. We can write all of these things in a form that operations researchers call a \"linear program.\"2 \\begin{align*}\\text{minimize }\\;\\; & 0.5\\cdot C+0.5\\cdot Q\\\\ \\text{subject to }\\;\\; &172\\cdot C+1742\\cdot Q\\geq 2000\\\\&0.839\\cdot C+25\\cdot Q\\leq 40\\\\&19\\cdot C+15\\cdot Q\\geq 60\\\\&C\\geq 0, \\;Q\\geq 0\\end{align*} It turns out, the optimal solution is that my friend should eat 2.44 pounds of carrots per day and 0.91 pounds of oatmeal. With that combination, he'll get all his nutrients, restrict his fat, and keep his grocery store bill as low as possible (about \\1.73 per day!). Below, I show another graph of the possible carrot-oatmeal combinations. This time, the color of a point represents the cost of that diet. The optimal diet occurs at the black dot, i.e. where the graph is most red. The most expensive diet occurs at the bottom right corner, where the graph is most blue. That diet corresponds to eating nearly 48 pounds of carrots per day! Operations researchers use this sort of linear programming mathematical model to solve all", "the coupon. How much did she pay for the box of cereal? 6. Coupon Diana bought a can of coffee that cost$7.99. She had a coupon for $0.75 off, and the store doubled the coupon. How much did she pay for the can of coffee? 7. Diet Leo took part in a diet program. He weighed 190 pounds at the start of the program. During the first week, he lost 4.3 pounds. During the second week, he had lost 2.8 pounds. The third week, he gained 0.7 pounds. The fourth week, he lost 1.9 pounds. What did Leo weigh at the end of the fourth week? 8. Snowpack On April 1, the snowpack at the ski resort was 4 meters deep, but the next few days were very warm. By April 5, the snow depth was 1.6 meters less. On April 8, it snowed and added 2.1 meters of snow. What was the total depth of the snow? 9. Coffee Noriko bought 4 coffees for herself and her coworkers. Each coffee was$3.75. How much did she pay for all the coffees? 10. Subway Fare Arianna spends $4.50 per day on subway fare. Last week she rode the subway 6 days. How much did she spend for the subway fares? 187. Income Mayra earns$9.25 per hour. Last week she", "+0 # Please answer question 0 102 2 At a mall's food court, Crystal has $\\$7.50\\$ to buy a meal (one entree, one drink and one dessert). The table below lists Crystal's choices and their prices including sales tax. How many distinct possible meals can she afford to buy? $\\begin{array}{|c|c|c|} \\hline \\text{Entrees} & \\text{Drinks} & \\text{Desserts} \\\\ \\hline \\text{Pizza } \\3.50 & \\text{Lemonade } \\1.50 & \\text{Frozen Yogurt } \\3.00 \\\\ \\hline \\text{Corn Dog } \\2.50 & \\text{Soda } \\1.25 & \\text{Cookies } \\2.00 \\\\ \\hline \\text{Fish~\\& Chips } \\3.50 & & \\\\ \\hline \\text{Fried Rice } \\4.75 & & \\\\ \\hline \\end{array}$ Jun 7, 2021 ### 2+0 Answers #1 +208 0 She has four options for Entrees. She has 2 options for drinks She has 2 options for desserts. 4*2*2=16 But some possibilities go over her budget. These are the possibilities: P-L-FY=8 P-S-FY=7.75 F&C-L-FY=8 F&C-S-FY=7.75 FR-S-C=8 FR-S-FY=9 FR-L-C=8.25 FR-L-FY=9.25 these are 8 possibilities. So, she has 16-8=8 distinct possible meals. JP Jun 7, 2021 #2 0 Thank you! Guest Jun 7, 2021", "of microwave popcorn costing $$$0.50$$ each for her party. What was her total bill? 2. Joe bought four $$8$$-foot $$2$$-by-$$4$$ boards for$$$24.00$$. How much did he spend per linear foot? 3. Margaret bought two cases of soda at the local discount store for $$$23.52$$. If each case contained $$24$$ bottles, how much did she spend per bottle? 4. Billy earns$$$12.00$$ per hour and “time and a half” for every hour he works over $$40$$ hours a week. What is his pay for $$47$$ hours of work this week? 5. Audry bought $$4$$ bags of marbles each containing $$15$$ assorted marbles. If she wishes to divide them up evenly between her $$3$$ children, how many will each child receive? 6. Mark and Janet carpooled home from college for the Thanksgiving holiday. They shared the driving, but Mark drove twice as far as Janet. If Janet drove $$135$$ miles, then how many miles was the entire trip? 1. $$32.50$$ 3. $$0.49$$ 5. $$20$$ marbles Exercise $$\\PageIndex{5}$$ Order of Operations with Absolute Values Simplify. 1. $$3+2|−5|$$ 2. $$9−4|−3|$$ 3. $$−(−|2|+|−10|)$$ 4. $$−(|−6|−|−8|)$$ 5. $$|−(40−|−22|)|$$ 6. $$||−5|−|10||$$ 7. $$−(|−8|−5)^{2}$$ 8. $$(|−1|−|−2|)^{2}$$ 9. $$−4+2|2^{2}−3^{2}|$$ 10. $$−10−|4−5^{2}|$$ 11. $$−|(−5)^{2}+4^{2}÷8|$$ 12. $$−(−3−[ 6−|−7|])$$ 13. $$−2[7−(4+|−7|)]$$ 14. $$3−7 |−2−3|+4^{3}$$ 15. $$7−5|6^{2}−5^{2}|+(−7)^{2}$$ 16. $$(−4)^{2}−|−5+(−2)^{3}|−3^{2}$$ 17. $$\\frac{2}{3}−|\\frac{1}{2}−(−\\frac{4}{3})^{2}|$$ 18. $$−30|\\frac{10}{3}−\\frac{1}{2}÷\\frac{1}{5}|$$ 19. $$(−4)^{3}−(2−|−4|)÷|−3^{2}+7|$$ 20. $$[10−3(6−|−8|)] ÷4−5^{2}$$ 1. $$13$$ 3. $$−8$$", "in her piggy bank now? Answer: Given, Cheryl has$25.00 in her piggy bank. She saves $6.00 more and then spends$15.00. 25 + 6 = 31 31 – 15 = 16 Thus Cheryl has \\$16 in her piggy bank now. Scroll to Top", "8/0.04 = 800/4 = 200. Math Expert Joined: 02 Sep 2009 Posts: 53063 Re: Judy bought a quantity of pens in packages of 5 for$0.80 per package [#permalink] ### Show Tags 23 Jul 2017, 23:48 2 carcass wrote: Judy bought a quantity of pens in packages of 5 for $0.80 per package. She sold all of the pens in packages of 3 for$0.60 per package. If Judy's profit from the pens was $8.00, how many pens did she buy and sell? A. 40 B. 80 C. 100 D. 200 E. 400 Cost price: packages of 5 for$0.80 per package = $0.80/5 =$0.16 per pen. Selling price: packages of 3 for $0.60 per package =$0.60/3 = $0.2 per pen. Profit = ¢4 per pen. Total profit = ¢800. The number of pens = 800/4 = 200. Answer: D. _________________ Manager Joined: 22 Nov 2016 Posts: 207 Location: United States Concentration: Leadership, Strategy GPA: 3.4 Re: Judy bought a quantity of pens in packages of 5 for$0.80 per package [#permalink] ### Show Tags 24 Jul 2017, 08:21 1 $$\\frac{x}{5} * (0.8) - \\frac{x}{3}*(0.6)$$ = 8 $$\\frac{2.4x - 3x}{15} = 8$$ x=200 _________________ Kudosity killed the cat but your kudos can save it. Senior Manager Joined: 28 Jun 2015 Posts: 290 Concentration: Finance GPA: 3.5 Re: Judy bought a quantity of" ]

[ "the same number to write the unit rate. $$\\frac{\\FormInput[][math_wrap][]{partial22_wrap}\\FormInput{partial22}}{\\FormInput[][math_wrap][]{partial23_wrap}\\FormInput{partial23}}$$ = $$\\frac{\\FormInput[][math_wrap][]{partial24_wrap}\\FormInput{partial24}}{\\FormInput[][math_wrap][]{partial25_wrap}\\FormInput{partial25}}$$ Think about what your answer means. The price of pineapples was$ per pound.", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> # Proportions ## Multiplication to solve for an unknown given two equal ratios. Estimated8 minsto complete % Progress Practice Proportions Progress Estimated8 minsto complete % Pineapple Frenzy ### Proportions How do you find the price of some quantity given a rate or proportion? Recall that proportions are two ratios set equal to each other. Credit: user: Fæ Source: http://commons.wikimedia.org/wiki/File:Pineapples.jpg [Figure1] You are at the grocery store and need to buy 15 pineapples. The price for pineapples is $12 for 4. You make your way over to the check out and-- Oh no! The cash register is broken! Unfortunately, there is no provided rate for individual pineapples and now you must calculate how much money the total of 15 pineapples will cost. The clerk is becoming frantic and does not know what to do in this situation. Luckily, with the help of proportions, or equal ratios, you can solve for the price of 15 pineapples at the rate of 4 for$12! First we must set up the proportion. Set the two ratios equal to each other, with unknown cost x. [Figure2] Your total cost for 15 pineapples will be \\$45. Yay, problem averted! You pay the clerk and go home happily with your 15 pineapples. ### Creative Applications 1. Your friend Andrew", "price, and so the regular price is twice the sale price. Therefore, the regular price for a full pound of fish is $$2 \\cdot \\ 6 = \\ 12.$$ Thus, D is the correct answer. 3. What is the value of \\begin{align*}4 \\cdot (&-1 + 2 - 3 + 4 - \\\\ &5 + 6 - 7 + \\cdots + 1000)?\\end{align*} $$-10$$ $$0$$ $$1$$ $$500$$ $$2000$$ ###### Solution(s): Within the parentheses, we can group each pair of numbers as follows: \\begin{align*} &(-1 + 2) + (-3 + 4) + \\\\ &\\cdots + (-999 + 1000).\\end{align*} Each pair of grouped terms has a sum of $$1,$$ and as there are $$500$$ such pairs, we have a total value of $$500$$ within the parentheses. Therefore, multiplying this value by $$4$$ yields $$2000.$$ Thus, E is the correct answer. 4. Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $$\\ 2.50$$ to cover her portion of the total bill. What was the total bill? $$\\ 120$$ $$\\ 128$$ $$\\ 140$$ $$\\ 144$$ $$\\ 160$$ ###### Solution(s): Since Judi's portion was fully covered by everyone's contributions, the portion of the bill she was responsible for was equal to $7 \\cdot \\ 2.50 = \\ 17.50.$", "mean in this context? That would mean Michael bought $$2$$ staplers and $$40$$ markers, spending $$19\\cdot2+1.75\\cdot40=108$$ dollars. That is within Michael's budget. In fact, any point on the lower left side of this line represents a total purchase within Michael's budget. The shading in Figure 4.10.5 captures all solutions to $$19x+1.75y\\leq133\\text{.}$$ Some of those solutions have negative $$x$$- and $$y$$-values, which make no sense in context. So in Figure 4.10.6, we restrict the shading to solutions which make physical sense. in-context", "_____ Next, we need to line up the numbers vertically. $2.25 \\\\.10 \\\\.30 \\\\\\underline{+\\ .85} \\\\3.50$ The cost of the ice cream cone is $3.50. Julie took the ten dollar bill and the two quarters from the customer. We can use mental math to figure out the customer’s change. $\\10.50 - 3.50 & = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}\\\\.50 - .50 & = 0\\\\10 - 3 & = 7$ Julie confidently handed the customer$7.00 in change. The customer smiled, thanked Julie and left eating her delicious ice cream cone. ## Vocabulary Here are the vocabulary words in this lesson. Properties the features of specific mathematical situations. Associative Property a property that states that changing the grouping in an addition problem does not change the sum. Commutative Property a property that states that changing the order of the numbers in an addition problem does not change the sum. ## Technology Integration Other Videos: 1. http://www.gamequarium.org/cgi-bin/search/linfo.cgi?id=7544 – Blackboard video on how to add decimals. 2. http://www.gamequarium.org/cgi-bin/search/linfo.cgi?id=7545 – Blackboard video on how to subtract decimals. ## Time to Practice Directions: Add or subtract the following decimals. 1. 4.5 + 6.7 = _____ 2. 3.45 + 2.1 = _____ 3. 6.78 + 2.11 = _____ 4. 5.56 + 3.02 = _____ 5. 7.08 + 11.9 = _____ 6. 1.24 + 6.5 = _____ 7. 3.45 +", "# Question ff115 $52. #### Explanation: There are $4$persons in the Group. To purchase a ticket, one has to pay$7.25. $\\therefore \\text{ cost of 4 tickets } = 4 \\left(7.25\\right) \\ldots . \\left(1\\right)$ To purchase a drink & popcorn, one has to pay $5.75. $\\therefore \\text{ cost of the refreshment for 4 friends} = 4 \\left(5.75\\right) \\ldots . \\left(2\\right)$Clearly, to get the total cost for the group of 4 friends, we have to add (1) & (2), i.e., $\\text{The total cost=} 4 \\left(7.25\\right) + 4 \\left(5.75\\right) \\ldots \\ldots \\ldots \\left(\\star\\right)$Now, here comes the Maths. to save your labour! A Well-known Property of Numbers, called \"The Distributive Property\" says that, $a \\left(b + c\\right) = a \\cdot b + a \\cdot c ,$or, taking the other way round, $a \\cdot b + a \\cdot c = a \\left(b + c\\right)$Observe that, to get the total cost from $\\left(\\star\\right)$, we have to perform multiplication twice, but, by the Distributive Property as shown in the last eqn., we have to multiply only once. $\\text{So, the total cost=} 4 \\left(7.25\\right) + 4 \\left(5.75\\right) = 4 \\left(7.25 + 5.75\\right) = 4 \\left(13.00\\right)$, \"i.e.,\"=$52#.", ", 24.11.2020yoyo0123 # Pineapple works as a natural meat tenderizer true step-by-step explanation: x=14 step-by-step explanation: 3x wow we learn something everyday Step-by-step explanation: wait actually loll Step-by-step explanation: :0 ### Other questions on the subject: Mathematics $$\\begin {array}{c|c}\\underline {\\quad x\\quad}& \\underline{\\quad y\\quad}\\\\ -2& 4\\\\-1& 2\\\\0& 0\\\\1& 2\\\\ 2& 4\\\\\\end{array}$$step-by-step explanation: y...Read More Mathematics, 21.06.2019, seth82 5Step-by-step explanation:First, find the intersections of the parabolas using any method. I choose to use the elimination methodNext, find the distance between the intersected po...Read More (2,20) means that the 2 is 2 hours and 20 is 20$. so there both 20$ for 2 hours only the same price. like it says. so, the dunk tank and the bounce house are the same price (20\\$) f...Read More", "employees their salary however, each pineapple costs . How many pineapples can you order this month beforehand without running over your budget. Write an inequality to represent this situation. Explanation: If we write the number of pineapples ordered as , then is the price of all the pineapples since each one costs . Now, it is alright if you have exactly so it is an inclusive inequality, and you are drawing from an available amount of , so we can write this inequality in words as 6 dollars * (number of pineapples) from 1000 dollars must be greater than or equal to 400 dollars. This can easily be converted to mathematical symbols as ### Example Question #3 : Inequalities Angie and Kirsten are looking to buy a couch that is less than . Kirsten has an individual spending limit of . Which set of inequalities represents a solution for how much each girl will spend on the couch? Explanation: If Kirsten is represented by the variable , and Angie is represented by the variable , then the total amount Kirsten and Angie are spending can be represnted by the sum of . Since Angie and Kirsten only have to spend on a couch, must be less than . Therefore, . Kirsten only has to spend on a couch.", "cart, we calculate the total price, and, later, perform the delivery. Each cart only stores the products' identifiers and quantities (in unspecified units; yes, it's super-simplified). (def cart1 {:banana 10 :pineapple 7 :pears 3}) (def cart2 {:pineapple 3 :mango 9}) Having defined product and cart data, we write a function that, given the products \"database\" and a cart, calculates the total price of the products in the cart. The cart-price function reduces all [product quantity] pairs in the cart, by retrieving the appropriate product map in the product-db, and taking the value associated with its :price key. It multiplies that price with the quantity, and accumulates it in total. (defn cart-price [product-db cart] (reduce (fn [total [product quantity]] (+ total (* (:price (product-db product)) quantity))) 0 cart)) Let's call this function with the available carts, and see it in action. (cart-price products cart1) => 31.699999999999996 (cart-price products cart2) => 23.7 We hopefully have more than one order. Our code can easily process sequences of carts, and compute the total revenue. (reduce + (map (partial cart-price products) [cart1 cart2])) => 55.39999999999999 It's all good; but what does it have to do with linear algebra? ## A more general algorithm In the previous implementation, we entangled the specifics of data storage and the algorithm that computes the total price. In", "unit. We use the unit cost to compare values when the quantities are not the same. To determine the unit cost, divide the cost by the number of units. Example $$\\PageIndex{3}$$ A local supermarket offers a pack of $$12$$ sodas for $$$3.48$$ on sale, and the local discount warehouse offers the soda in a $$36$$-can case for$$$11.52$$. Which is the better value? Solution: Divide the cost by the number of cans to obtain the unit price. Supermarket Discount warehouse $$$3.4812$$ cans$$=$$$$$0.29$$/can $$$11.5236$$ cans$$=$$$$$0.32$$/can Table 2.6.1 The supermarket sale price of $$$3.48$$ for a $$12$$-pack is a better value at$$$0.29$$ per can. A proportion is a statement of equality of two ratios. $\\frac{a}{b}=\\frac{c}{d}$ This proportion is often read “$$a$$ is to $$b$$ as $$c$$ is to $$d$$.” Here is an example of a simple proportion, $$\\frac{1}{2}=\\frac{2}{4}$$ If we clear the fractions by multiplying both sides of the proportion by the product of the denominators, $$8$$, then we obtain the following true statement: \\begin{aligned}\\color{Cerulean}{8\\cdot}\\color{black}{\\frac{1}{2}}&=\\color{Cerulean}{8\\cdot}\\color{black}{\\frac{2}{4}} \\\\ 4\\cdot 1&=2\\cdot 2\\\\ 4&=4 \\end{aligned} Given any nonzero real numbers $$a, b, c,$$ and $$d$$ that satisfy a proportion, multiply both sides by the product of the denominators to obtain the following: \\begin{aligned} \\frac{a}{b}&=\\frac{c}{d} \\\\ \\color{Cerulean}{bd}\\color{black}{\\cdot \\frac{a}{b}}&=\\color{Cerulean}{bd}\\color{black}{\\cdot \\frac{c}{d}} \\\\ ad&=bc \\end{aligned} This shows that cross products are equal, and is commonly referred to as cross", "buying volleyballs. We can form an expression using the above explanation: $$=(100+6\\times x)+68.25$$ $$=(100+6x)+68.25$$ $$=(6x+100)+68.25$$ Using the commutative property of addition $$=6x+(100+68.25)$$ Using the associative property of addition $$=6x+168.25$$ Now, if the value of $$x$$ is 14.5, the total cost will be: $$6(14.5)+168.25$$ $$=87.0+168.25$$ $$=255.25$$", "+ a = x + a) \\\\ & = \\sum (x + a) \\cdot P(X = x) \\\\ & = \\left( \\sum x \\cdot P(X=x) \\right ) + \\left( \\sum a \\cdot P(X=x) \\right) \\\\ & = E[X] + a \\cdot \\left(\\sum P(X=x) \\right) \\\\ & = E[X] + a \\end{aligned} ### (R) Expected value rules Let $$Y = 4X$$ be a new random variable. Is $$E[Y] = 4 \\cdot E[X]$$? ### $$\\text{(R) Let } Y = b X \\quad E[Y] = ?$$ \\begin{aligned} E[Y] & = \\sum y \\cdot P(Y=y) \\\\ & = \\sum (x b ) \\cdot P(X b = x b) \\\\ & = \\sum (x b) \\cdot P(X = x) \\\\ & = b \\cdot \\left( \\sum x \\cdot P(X=x) \\right ) \\\\ & = b \\cdot E[X] \\end{aligned} ### $$\\text{(R) Let } Y = a + b X \\quad E[Y] = ?$$ We can combine these rules: \\begin{aligned} E[Y] & = E[a + b X] \\\\ & = a + E[ b X] \\\\ & = a + b \\cdot E[X] \\end{aligned} ### An example At StatsPie, the number of toppings on the typical pizza is a random variable X having a mean value of 3. The cost of a pizza is $10 plus$1.50 per topping. How much does the average customer spend", "1. $10^2$ 2. $10^3$ 3. $10^4$ 4. $10^5$ Choose < , >, or = to compare. $10^4$ 10,000 1. < 2. > 3. = Ten to the first power is written as which of the following? 1. $100$ 2. $10^1$ 3. $1$ 4. $1000$ Grade 5 Algebraic Expressions CCSS: 5.OA.A.2 Which expression matches the statement, \"add 5 and 3, then subtract from 12\"? 1. $5 + 3 -12$ 2. $12 - 5 +3$ 3. $(5 + 3) -12$ 4. $12 - (5 + 3)$ Use mental math and a pattern to find the product. $(8xx7)xx10^4$ 1. 560 2. 5,600 3. 56,000 4. 560,000 Evaluate. $1^6$ 1. 10 2. 6 3. 1 4. 16 Which of the following shows the number 10,000 written as a power of 10? 1. $10^3$ 2. $10^4$ 3. $10^5$ 4. $10^6$ Use mental math and a pattern to find the product. $(2xx9)xx10^2$ 1. 18 2. 180 3. 1,800 4. 18,000 Jennifer's family is going to a beach resort. Jennifer bought 7 beach towels that cost $13 each to take to the resort. To find the total cost, she added the products of $7 xx 10$ and $7 xx 3$, for a total of$91. What property did Jennifer use? 1. Commutative Property of Multiplication 3. Associative Property of Multiplication 4. Distributive Property Which is another", "the numbers it would no longer be true. If we ask for oranges to pineapples, we want the number of oranges listed first, followed by the number of pineapples second. When we write a ratio, we can simplify in a similar way to when we worked with fractions, with one exception (the exception is covered below). Notice how each part of the ratio is divisible by 3. This means we can write our ratio as: $$1:2$$ $$\\frac{1}{2}$$ $$1\\hspace{.2em}to \\hspace{.2em}2$$ If we rearrange the picture a bit, it becomes completely clear that for each orange, there are two pineapples. So what if we reversed the order of the words in our ratio? We would want to reverse the numbers involved. What is the ratio of pineapples to oranges? $$\\frac{6}{3}=\\frac{2}{1}$$ Now, here is where our exception mentioned before comes into play. In this case, we need to be careful. We know when we work with a fraction and we get something like 2/1, this can just be written as the number 2. When we work with a ratio, we need to keep the 1 since it is showing the other part of our ratio. If we see 2:1 or 2/1, we know it's 2 pineapples for every 1 orange. In some cases, we may see a ratio that compares three", "# 2007 AMC 12A Problems/Problem 1 The following problem is from both the 2007 AMC 12A #1 and 2007 AMC 10A #1, so both problems redirect to this page. ## Problem One ticket to a show costs 20\\$at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?$ (Error compiling LaTeX. ! Missing $inserted.)\\mathrm{(A)}\\ 2\\qquad \\mathrm{(B)}\\ 5\\qquad \\mathrm{(C)}\\ 10\\qquad \\mathrm{(D)}\\ 15\\qquad \\mathrm{(E)}\\ 20$ ## Solution $P$ = the amount Pam spent $S$ = the amount Susan spent • $P=5 \\cdot (20 \\cdot .7) = 70$ • $S=4 \\cdot (20 \\cdot .75) = 60$ Pam pays 10 more dollars than Susan.", "the equation to find x 2x + 3(-1) = 5 2x – 3 + 3 = 5 + 3 2x = 8 x = 8/2 = 4 The soultion is (4, -1) Question 7. $$\\left\\{\\begin{array}{l}2x+5y=16 \\\\-4x+3y=20\\end{array}\\right.$$ (________ , ________ ) Answer: The soultion is (-2, 4) Explanation: $$\\left\\{\\begin{array}{l}2x+5y=16 \\\\-4x+3y=20\\end{array}\\right.$$ To eliminate x terms, multiply the 1st equation by 2 2(2x + 5y = 16) 4x + 10y = 32 Add the equations 4x + 10y = 32 +(-4x + 3y = 20) x is eliminated it has reversed coefficients. Solve for y 10y + 3y = 32 + 20 13y = 52 y = 52/13 = 4 Substituting y in either of the equation to find x 2x + 5(4) = 16 2x + 20 – 20 = 16 – 20 2x = -4 x = -4/2 = -2 The soultion is (-2, 4) Question 8. Bryce spent$5.26 on some apples priced at $0.64 each and some pears priced at$0.45 each. At another store he could have bought the same number of apples at $0.32 each and the same number of pears at$0.39 each, for a total cost of $3.62. How many apples and how many pears did Bryce buy? a. Write equations to represent Bryce’s expenditures at each store First store: _____________ Second store: _____________ Type below:", "# Taxes and Sales Alignments to Content Standards: A-SSE.B Judy is working at a retail store over summer break. A customer buys a \\$50 shirt that is on sale for 20% off. Judy computes the discount, then adds sales tax of 10%, and tells the customer how much he owes. The customer insists that Judy first add the sales tax and then apply the discount. He is convinced that this way he will save more money because the discount amount will be larger. 1. Is the customer right? 2. Does your answer to part (a) depend on the numbers used or would it work for any percentage discount and any sales tax percentage? Find a convincing argument using algebraic expressions and/or diagrams for this more general scenario. ## IM Commentary This task is not about computing the final price of the shirt but about using the structure in the computation to make a general argument. The key underlying idea is that multiplication is commutative, which we often just take for granted and don't feel needs any explanation. In this case, the context of the problem makes it not obvious at all that we can switch the order of the two computations, but it becomes quite obvious after observing that the application of both the discount and the sales", "items? A $\\$131.00$B$\\$64.75$ C $\\$142.00$D$\\$123.50$ Question 20 Explanation: The correct answer is (C). Calculate the amount he spent on each item and then add these totals together: $\\require{cancel} 20 \\cancel{\\text{cones}} \\cdot \\dfrac{\\$1.25}{\\cancel{\\text{cone}}} = \\$25.00$ $\\require{cancel} 3 \\cancel{\\text{balls}} \\cdot \\dfrac{\\$25}{\\cancel{\\text{ball}}} = \\$75.00$ $\\require{cancel} 12 \\cancel{\\text{vests}} \\cdot \\dfrac{\\$3.50}{\\cancel{\\text{vest}}} = \\$42.00$ $\\$25.00 + \\$75.00 + \\$42.00= \\$142.00$ Question 21 ### What is the value of the expression given below? $26 − 7(3 + 5) ÷ 4 + 2$ A $14$ B $16.67$ C $23$ D $40$ Question 21 Explanation: The correct answer is (A). We can use PEMDAS to evaluate the expression. (PEMDAS is a technique for remembering the order of operations. It stands for Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction) According to the rules, we must evaluate the expression in the parentheses first: $26 − 7(3 + 5) ÷ 4 + 2$ $= 26 − 7(8) ÷ 4 + 2$ Then we perform multiplication and division in order from left to right as follows: $26 − 7(8) ÷ 4 + 2$ $= 26 − 56 ÷ 4 + 2$ $= 26 − 14 + 2$ Finally, we perform addition and subtraction in order from left to right as follows: $26 − 14 + 2$ $= 12 + 2$ $= 14$ Question 22 ### $5^3 × 8^2 =$", "$$=5\\text {%}$$ of $$8.00$$ ('of' means multiply) $$=5\\text {%}×8.00$$ ( $$5\\text {%}=0.05$$) $$=0.05×8.00$$ $$\\begin{array} {r} 8.00 \\\\ ×0.05 \\\\ \\hline 0.40 \\\\ \\hline \\end{array}$$ Amount of discount $$=0.40$$ Step 2: Subtract the amount of discount from the original price. $$8.00-0.40=7.60$$ Thus, the amount Jessica will pay for a burger $$=7.60$$ #### Ana goes to a bakery to buy a packet of muffins. The bakery is offering a sale of $$10\\text {%}$$ off on each packet of muffins. She buys one packet. If the original price of a packet is $$6.50$$, how much money will she pay for it? A $$6.25$$ B $$7.25$$ C $$5.85$$ D $$4.65$$ × \"Sale of $$10\\,\\text {%}$$ off on the original price\" means there is a discount of $$10\\,\\text {%}$$. Original price of a packet of muffins $$=6.50$$ The amount of discount $$=10\\,\\text {%}$$ of the original price $$=10\\,\\text {%}$$ of $$6.50$$ ('of' means multiply) $$=10\\,\\text {%}×6.50$$ ( $$10\\,\\text {%}=0.10$$ ) $$=0.10×6.50$$ $$\\begin {array} {r} 6.50 \\\\ ×0.10 \\\\ \\hline 0.65 \\\\ \\hline \\end {array}$$ Discount $$=0.65$$ Ana will pay = Original Price – Amount of discount $$=6.50-0.65$$ $$=5.85$$ Hence, option (C) is correct. ### Ana goes to a bakery to buy a packet of muffins. The bakery is offering a sale of $$10\\text {%}$$ off on each packet of muffins. She buys one", "^4C_2 = 6$$ i.e. the two points are selected from $$4$$ collinear points making the straight line only one. Therefore, total no. of straight lines $$= 45 - 6 + 1 = 40$$ 3. Total no. of selections of $$4$$ points out of $$10 = ^{10}C_4 = 210$$ Number of selections when no quadrilateral is formed $$= ^4C_3.^6C_1 + ^4C_4.6C_0 = 25$$ Thus, total no. of quadrilaterals formed $$= 210 - 25 = 185$$ 39. Case I: When fruits of same type are different. Number of selections $$= (^4C_0 + ^4C_1 + \\ldots + ^4C_4)$$ $$(^5C_0 + ^5C_1 + \\ldots + ^5C_5)(^6C_0 + ^6C_1 + \\ldots + ^6C_6) - 1$$ $$= 2^4.2^5.2^6 - 1 = 2^15 - 1$$ Case I: When fruits of same type are same. Zero or more oranges can be selected in $$5$$ ways. Zero or more apples can be selected in $$6$$ ways. Zero or more mangoes can be selected in $$7$$ ways. Thus, total no. of ways $$= 5.6.7 = 210$$ But in one of these ways no fruit is selected, making answer $$209$$. 40. Following case I of previous problem we have total no. of ways $$= (2^5 - 1).(2^4 - 1).2^3 - 1 = 3720$$ 41. Factors of $$21,600$$ are $$2, 2, 2, 2, 2, 3, 3, 3, 5, 5$$ Thus," ]

[ "Find the total number of chocolates they both ate in all. A $$10$$ B $$14$$ C $$13$$ D $$15$$ × Chocolates bought by Sam $$=13$$ He ate $$\\dfrac{1}{2}$$ of the chocolates $$=13\\times\\dfrac{1}{2}=\\dfrac{13}{2}$$ Chocolates bought by Casey $$=10$$ She ate $$\\dfrac{3}{4}$$ of the chocolates $$=\\dfrac{3}{4}\\times10 = \\dfrac{15}{2}$$ $$\\therefore$$ Total number of chocolates they both ate $$=\\dfrac{13}{2}+\\dfrac{15}{2}$$ $$=\\dfrac{13+15}{2}$$ $$=\\dfrac{28}{2}$$ $$=14$$ Thus , the total number of chocolates they both ate equals $$14$$. Hence, option (B) is correct. ### Sam bought $$13$$ chocolates and ate $$\\dfrac{1}{2}$$ of them. Casey bought $$10$$ chocolates and ate $$\\dfrac{3}{4}$$ of them. Find the total number of chocolates they both ate in all. A $$10$$ . B $$14$$ C $$13$$ D $$15$$ Option B is Correct # Mathematical Modeling on Subtraction and Multiplication Here, we will use the combination of subtraction and multiplication to solve the problem. For example: Maria buys $$\\dfrac{1}{2}$$ kilogram(kg) of strawberries at the rate of $$10$$ per kg and Jacob buys the same amount of strawberries at the rate of $$12$$ per kg. Find who spends more and by how much. Rate at which Maria buys strawberries $$=10$$ per kg Quantity of strawberries she buys $$=\\dfrac{1}{2}$$ kg Total price of $$\\dfrac{1}{2}$$ kilogram of strawberries $$=\\dfrac{1}{2}\\times10$$ $$=5$$ Rate at which Jacob buys strawberries $$=12$$ per kg Quantity of strawberries he buys $$=\\dfrac{1}{2}$$ kg", "_____ Next, we need to line up the numbers vertically. $2.25 \\\\.10 \\\\.30 \\\\\\underline{+\\ .85} \\\\3.50$ The cost of the ice cream cone is $3.50. Julie took the ten dollar bill and the two quarters from the customer. We can use mental math to figure out the customer’s change. $\\10.50 - 3.50 & = \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}\\\\.50 - .50 & = 0\\\\10 - 3 & = 7$ Julie confidently handed the customer$7.00 in change. The customer smiled, thanked Julie and left eating her delicious ice cream cone. ## Vocabulary Here are the vocabulary words in this lesson. Properties the features of specific mathematical situations. Associative Property a property that states that changing the grouping in an addition problem does not change the sum. Commutative Property a property that states that changing the order of the numbers in an addition problem does not change the sum. ## Technology Integration Other Videos: 1. http://www.gamequarium.org/cgi-bin/search/linfo.cgi?id=7544 – Blackboard video on how to add decimals. 2. http://www.gamequarium.org/cgi-bin/search/linfo.cgi?id=7545 – Blackboard video on how to subtract decimals. ## Time to Practice Directions: Add or subtract the following decimals. 1. 4.5 + 6.7 = _____ 2. 3.45 + 2.1 = _____ 3. 6.78 + 2.11 = _____ 4. 5.56 + 3.02 = _____ 5. 7.08 + 11.9 = _____ 6. 1.24 + 6.5 = _____ 7. 3.45 +", "restaur Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52917 ### Show Tags 30 Jun 2017, 02:48 Bunuel wrote: Ann, Carol, and Judy paid a total of $45 for their dinner at a restaurant. If Ann paid (2/5)th of what Judy paid, Carol paid$17 and Judy paid the rest, what fraction of the total amount did Judy pay? (A) 2/9 (B) 14/45 (C) 1/3 (D) 2/5 (E) 4/9 Let Judy paid be $$= x$$ Ann paid $$= \\frac{2}{5}x$$ Carol paid $$= 17$$ Total Bill $$= 45$$ Total bill = Ann paid + Judy paid + Carol paid $$45 = x + \\frac{2}{5}x + 17$$ $$x + \\frac{2}{5}x = 45 - 17 => 28$$ $$\\frac{5x + 2x}{5} = 28$$ $$7x = 28*5$$ $$x = \\frac{140}{7} = 20$$ Therefore Judy paid $$= 20$$ Required fraction = Amount Judy paid / Total Bill $$= \\frac{20}{45} = \\frac{4}{9}$$ Senior Manager Joined: 28 May 2017 Posts: 283 Concentration: Finance, General Management ### Show Tags 30 Jun 2017, 06:02 Bunuel wrote: Ann, Carol, and Judy paid a total of $45 for their dinner at a restaurant. If Ann paid (2/5)th of what Judy paid, Carol paid$17 and Judy paid the rest, what fraction of the total amount did Judy pay? (A) 2/9 (B) 14/45 (C) 1/3 (D) 2/5 (E)", "half of $y$: $y \\div 2$ or $\\frac{1}{2}y$ 6. Seconds in $m$ minutes and $h$ hours: $60m$ and $3,600h$ 7. Kobo in $n$ naira: $100n$ 8. Total cost, for example total cost of $p$ pens and $r$ rulers: $p+r$ 9. The area of a square is given by $A=s^2$. The perimeter of a square is given by $P=4s$. 10. The area of a rectangle is given by $A=l \\times b$. The perimeter of a rectangle is given by $P=2l+2b$. After you have put together an algebraic expression, expand it if possible and give the answer in its simplest form. In your answer, state what the variable stands for, for example, \"where $$₦\\,x$$ is the price of a small sachet of milk\", which is used in the exercise below. ### Exercise 6.8: Use algebraic expressions for practical problems 1. A small sachet of milk costs $$₦\\,x$$ and a small sachet of noodles costs $$₦\\,20$$ more. Work out the cost of 4 sachets of milk and 8 sachets of noodles. A small sachet of noodles costs $&#8358;\\,x+20$. \\begin{align} &\\phantom{000}4\\times x+8\\times (x+20) \\\\ &=4x+8(x+20) \\\\ &=4x+8x+160 \\\\ &=12x+160 \\end{align} The cost is $&#8358;\\,12x+160$, where $&#8358;\\,x$ is the price of a small sachet of milk. 2. A shop owner sells rice at $$₦\\,x$$ per cup. She also sells maize per cup, at", "shown in the following figure. Each brownie costs 50¢, each scoop of chocolate ice cream costs 20¢, each bottle of cola costs 30¢, and each piece of pineapple cheesecake costs 80¢. Search Engine Land is the leading industry source for daily, must-read news and in-depth analysis about search engine technology. 09 return for every$1. , bottom and four sides of equal size and square. Subject: Re: Calculus-Optimization problem From: myoarin-ga on 08 Dec 2005 20:21 PST The largest volume box will be a cube, i. The Scottish Resource Allocation formula is used in the allocation of around 70% of the total NHS Budget between the 14 territorial NHS Boards in Scotland. When expecting a major surgery, prepare with the Pre-Op formula to help give your body the edge it needs in healing. Businesses use a price optimisation formula based on the overall demand for their product, their level of competition, and (in the case of manufacturers) the cost of manufacturing their goods. Other Revionics customers include Sally Beauty. The optimization engine is able to forecast demand and cost for a set of prices to calculate net profit. benefit) by taking the difference of the highest they would pay and the actual price they pay. Your go-to source for the latest F1 news, video highlights, GP results, live timing,", "did she pay for the box of cereal? 182. Coupon Diana bought a can of coffee that cost $7.99.7.99.$ She had a coupon for $0.750.75$ off, and the store doubled the coupon. How much did she pay for the can of coffee? 183. Diet Leo took part in a diet program. He weighed $190190$ pounds at the start of the program. During the first week, he lost $4.34.3$ pounds. During the second week, he had lost $2.82.8$ pounds. The third week, he gained $0.70.7$ pounds. The fourth week, he lost $1.91.9$ pounds. What did Leo weigh at the end of the fourth week? 184. Snowpack On April $1,1,$ the snowpack at the ski resort was $44$ meters deep, but the next few days were very warm. By April $5,5,$ the snow depth was $1.61.6$ meters less. On April $8,8,$ it snowed and added $2.12.1$ meters of snow. What was the total depth of the snow? 185. Coffee Noriko bought $44$ coffees for herself and her co-workers. Each coffee was $3.75.3.75.$ How much did she pay for all the coffees? 186. Subway Fare Arianna spends $4.504.50$ per day on subway fare. Last week she rode the subway $66$ days. How much did she spend for the subway fares? 187. Income Mayra earns $9.259.25$ per hour. Last week she worked", "each of them purchased for the household. At the end of each week, they use this data to balance their account. As an aspiring data scientist, you offer your help. Here’s last week’s data: Name Mon Tue Wed Thu Fri Sat Sun Anna Bread: $2.50 Pasta:$4.50 Pencils: $3.25 Milk:$4.80 Cookies: $4.40 Cake:$12.50 Butter: $2.00 Cream:$3.90 Brian Chips: $3.80 Beer:$11.80 Steak: $16.20 Toilet paper:$4.50 Wine: $8.80 Caro Fruit:$6.30 Batteries: $6.10 Newspaper:$2.90 Honey: $3.20 Detergent:$9.95 1. Which variables and which observations would you define here? Go ahead and enter the data into a tibble acc_1. 2. Use acc_1 to answer the following questions (by using dplyr for creating tables that contain the answer): • How much money was spent this week? • Which percentage of the overall amount was spent by each person? • How many items did each person purchase? • How much did each person pay overall? • Who buys the cheapest/most expensive items (on average)? • How much is being spent on each day of the week (overall and on average)? • What is the order of days sorted by the overall amount spent (from most expensive to least expensive)? 1. Interpret and re-create the following graphs (using ggplot2 and possibly dplyr): 1. Bonus task: What do the following plots show? Try re-creating the plots from the", "makes his calculations on a weekly basis, comparing cost to sale prices. With the pies, pasties and sausage rolls in that order he applies them to the cost and sale price columns : The code for this is: $$\\left[\\begin{array} 350&310&270 \\end{array}\\right] \\ \\left[\\begin{array} \\2.10&\\3.60 \\ \\2.05&\\3.60 \\ \\1.90&\\3.10 \\end{array} \\right] \\ = \\ \\left[\\begin{array} \\735.00&\\1260.00 \\ \\635.50&\\1116.00 \\ \\513.00&\\837.00 \\end{array}\\right]$$", "always make your estimate at the store in such a way that it’s higher than the actual price. That way, you are certain to have enough money. You could do like this, for instance: You round the price of the banana from $\\text{}1.07$ to $\\text{}1.00$. Then, the estimated price of the bananas is $\\text{}1.00\\cdot 3=\\text{}3.00$. You can estimate the price of the milk, which is $\\text{}1.92$, to be $\\text{}2.50$. Your estimate of the price is then $\\text{}2.50+\\text{}3.00=\\text{}5.50.$ If you have$$5.50$, you know that you have enough to pay for the items. Example 2 You have $\\text{}\\text{}5.36\\text{}$, and are about to buy a t-shirt for $\\text{}\\text{}3.79\\text{}$. How much money do you have left after buying the t-shirt? In this problem, we’re dealing with differences, and you should therefore either round both numbers up or down. In this instance, you choose to round up. That means you estimate to have $\\text{}5.50-\\text{}4.00=\\text{}1.50$ The exact difference is $\\text{}5.36-\\text{}3.79=\\text{}1.57,$ so you have \\$$0.07$ more left in your wallet than you estimated. Note! When estimating calculations, it’s important to think through whether you want the estimate to be too high or too low. This is a good rule of thumb: Rule ### Tips If it’s money going out of your wallet, the estimate should be too high. If it’s money going into your wallet,", "8/10 • Now add 1 8/10 + 6 = 7 8/10 • Simplify the fraction to its lowest terms. = 7 4/5 ### Practice Questions 1. A woman distributed a fraction of a pineapple among her $6$ daughters. If each person got $\\dfrac{1}{9}$ of the pineapple. What is the total fraction of the pineapple that the woman distributed? 2. Edwin and Ann bought $15$ kg of sweets at their wedding and distributed $\\dfrac{3}{4}$ of it among the visitors. How many sweets did they distribute? 3. My weight was $60$ kg before I lost $\\dfrac{1}{10}$ of the weight in the past $3$ months. How much weight did I lose? 4. Jason had $\\$3140$in his bank account. He spent$\\dfrac{2}{5}$of it to buy essentials from the grocery. How much money did he spend? 5. Stella had$15$liters of milk in a container. If she consumed$\\dfrac{3}{4}$of the milk. How many liters of milk were consumed? 6. A boy walks$3 \\dfrac{1}{2}$kilometers daily. What is the total distance covered in one week? 7. Ahmed read$\\dfrac{2}{3}$of his storybook having$420$pages. If Mike read$\\dfrac{3}{4}$of the same storybook, find out who read many pages and how many were they? 8. A rectangular school garden is$6 \\dfrac{4}{5}$meters long and$1\\dfrac{3}{8}$meters wide. Which of the following shows the area of the garden? 9. It takes$\\dfrac{5}{6}$yards of wool to manufacture a dress. How", "\\frac{7}{8} + 1\\frac{1}{2}\\end{align*} 14. \\begin{align*}-3 \\frac{1}{3} + \\left (-2 \\frac{3}{4} \\right )\\end{align*} In 17 – 20, which property of addition does each situation involve? 1. Whichever order your groceries are scanned at the store, the total will be the same. 2. Suppose you go buy a DVD for $8.00, another for$29.99, and a third for \\$14.99. You can add \\begin{align*}(8 + 29.99) + 14.99\\end{align*} or you can add \\begin{align*}8 + (29.99 + 14.99)\\end{align*} to obtain the total. 3. Shari’s age minus the negative of Jerry’s age equals the sum of the two ages. 4. Kerri has 16 apples and has added zero additional apples. Her current total is 16 apples. 5. Nadia, Peter, and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They figure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the cost of the ice cream was added as tax?", "solution is (-2, 5, 7). ### Problem Set Use a calculator to find the inverse of each matrix below. 1. $$\\begin{bmatrix} 2 & -3 \\\\ 8 & 2 \\end{bmatrix}$$ 2. $$\\begin{bmatrix} -6 & 8 \\\\ -3 & 2 \\end{bmatrix}$$ 3. $$\\begin{bmatrix} 5 & 2 & -1 \\\\ 3 & 5 & 0 \\\\ -4 & 1 & 6 \\end{bmatrix}$$ Solve the systems using matrices and a calculator. 1. $$2x+y &= -1\\\\ -3x-2y &= -3$$ 2. $$2x+10y &= 6\\\\ x+11y &= -3$$ 3. $$-3x+y &= 17\\\\ 2x-3y &= -23$$ 4. $$10x-5y &= 9\\\\ 2.5y &= 5x +4$$ 5. $$4x-3y &= -12\\\\ 7x &= 2y+5$$ 6. $$2x+5y &= 0\\\\ -4x+10y &= -12$$ 7. $$7x+2y+3z &= -12\\\\ -8x-y + 4z &= -6\\\\ -10x+3y+5z &= 1$$ 8. $$2x-4y+11z &= 18\\\\ -3x+5y-13z &= -25\\\\ 5x+10y+10z &= 5$$ 9. $$8x+y-4z &= 4\\\\ -2z-3y+4z &= -7\\\\ 4 x+5y-8z &= 11$$ For the following word problems, set up a system of linear equations to solve using matrices. 1. A mix of 1 lb almonds and 1.5 lbs cashews sells for $15.00. A mix of 2 lbs almonds and 1 lb cashews sells for$17.00. How much does each nut cost per pound? 2. Maria, Rebecca and Sally are selling baked goods for their math club. Maria sold 15 cookies, 20 brownies and 12 cupcakes and raised $23.50.", "usefulness of these models. The cost of raisins varies directly with the number of kilograms of raisins purchased. If the cost is $\\34.25$ when the number of kilograms purchased is $7.5$, calculate the amount of raisins that can be purchased for $\\10.50$. Solution: Let $C$ denote the cost of the raisins and $N$ the number of kilograms of raisins purchased. The direct variation equation is $C = k \\cdot N$. Substitute $C = 34.25, N = 7.5$. $34.25 & = k \\cdot 7.5\\\\4.6 & \\approx k$ With the constant of proportionality known, calculate $N$ when $C = 10.50$. $C & = 4.6 \\cdot N\\\\10.5 & = 4.6 \\cdot N\\\\2.2 & = N$ So the number of kilograms of raisins that can be purchased for $\\10.50$ is $2.2\\;\\mathrm{kg}$. A ball was thrown from the roof of a high building. The velocity $v$ of the falling ball is directly proportional to the time $t$ of the fall. After $4\\;\\mathrm{seconds}$, the velocity of the ball is $85\\;\\mathrm{meters}$ per second. What will the velocity be after $6\\;\\mathrm{seconds}$, assuming it hasn’t hit the ground yet? Hint: Solve for $k$ in the direct variation equation $v = k \\cdot t$ first. NONE Feb 22, 2012 Aug 22, 2014", "each of the shops. The code is: $$\\left[\\begin{array}{c+50C+25.c.c} 72&95&68 \\ \\hdash 54&61&65 \\ \\hdash 48&51&60 \\end{array}\\right] - \\left[\\begin{array}{c+50C+25.c.c} 11&14&12 \\ \\hdash 16&12&22 \\ \\hdash 14&17&15 \\end{array}\\right] = \\left[\\begin{array}{c+50C+25.c.c} 61&81&56 \\ \\hdash 38&49&43 \\ \\hdash 34&34&48 \\end{array}\\right]$$ This code looks more complex than it really is, it is cluttered by the lines and alignment sequences. ## Multiplication Matrices Different than the addition or subtraction matrices, the multiplication matrix comes in three parts, the row matrix, the column matrix and the answer matrix. This implies it has a different construction methodology. And the code for this is: $$\\begin{array} 10&14&16\\end{array} \\ \\left[\\begin{array} 45 \\\\ 61 \\\\ 19 \\end{array}\\right] \\ = \\ \\begin{array} 450&854&304\\end{array}$$ While different, it is not necessarily more complex. For example a problem like: Bill the baker is selling his product to Con the cafe owner, who wants to make sure his overall prices are profitable for himself. Con needs to make sure that his average price is providing sufficient profit to be able to keep the cafes open. Con makes his calculations on a weekly basis, comparing cost to sale prices. With the pies, pasties and sausage rolls in that order he applies them to the cost and sale price columns : The code for this is: $$\\left[\\begin{array} 350&310&270 \\end{array}\\right] \\ \\left[\\begin{array} \\2.10&\\3.60 \\ \\2.05&\\3.60 \\ \\1.90&\\3.10 \\end{array}", "that no matter what else she may choose, Grace has four possible options for her beverage. Thus, the total number of possible breakfast orders will be four times the number of possible orders of main and side (with optional extras). Then we could have proceeded to analyse the number of possible choices for her main dish and her side dish (together with the extras). Breaking down the choices for her main and side dishes into the same cases as before, we could see that there are $$3 · 2 = 6$$ choices in the first case; $$2 · 2 · 2 = 8$$ choices in the second case; $$3 · 4 = 12$$ choices in the third case; and $$2 · 2 · 4 = 16$$ choices in the fourth case. Thus she has a total of $$6 + 8 + 12 + 16 = 42$$ choices for her main and side dishes. The product rule now tells us that she has $$4 · 42 = 168$$ possible orders for her breakfast. Let’s run through one more (simpler) example of using both the sum and product rules, and work out the answer in two different ways. Example $$\\PageIndex{3}$$ Kathy plans to buy her Dad a shirt for his birthday. The store she goes to has three different colours", "chocolates costs $$\\text{R}\\,\\text{15,95}$$. How much does each chocolate cost? $$\\text{R}\\,\\text{15,95} \\div \\text{6} = \\text{R}\\,\\text{2,668} \\ldots$$ rounded off to $$\\text{R}\\,\\text{2,67}$$ per chocolate. A truck driver travels $$\\text{1 500}$$ $$\\text{km}$$ in $$\\text{18}$$ hours. What is his average speed? $$\\text{1 500}\\text{ km} \\div$$ $$\\text{18}$$ $$\\text{h}$$ = $$\\text{83,33}$$$$\\ldots$$ km/h. Round off to $$\\text{83,33}$$ $$\\text{km/h}$$. Nicola is able to input $$\\text{96}$$ words in $$\\text{2}$$ minutes on her laptop. Karen times her own typing speed as $$\\text{314}$$ words in $$\\text{7}$$ minutes. Work out their speeds to see who is faster. Nicola types $$\\text{96}$$ words $$\\div$$ $$\\text{2}$$ min = $$\\text{48}$$ words/min. Karen types $$\\text{314}$$ words $$\\div$$ $$\\text{7}$$ min = $$\\text{44,857}$$$$\\ldots$$ words/min. Round off to a whole word: $$\\text{45}$$ words/min. Nicola is faster. ### Finding missing numbers in ratios and rates (EMGX) Solving problems often involves using ratios and rates to find unknown values. We use a similar process to find missing numbers in ratios and rates. ## Worked example 14: Finding missing numbers in a ratio Thenji makes a fruit salad for breakfast at a restaurant. She uses pieces of fruit in the following ratio: banana : apple : paw-paw $$\\text{1}$$ : $$\\text{2}$$ : $$\\text{3}$$ 1. If she uses $$\\text{20}$$ pieces of apple, how many pieces of banana and paw-paw should she use? 2. If she uses $$\\text{12}$$ pieces of banana, how many pieces of", "soda [3 * 2 * 1... for all 15 days, though again I wonder if it would be 3 * 2 * 3) or maybe I should do a permutation that combines the two? I don't even know how to begin to think about this question. I added up the permutations above but I'm shy of the stated answer. Any insights would be most welcome. • The \"caveat\" implies that whatever lunch she had today doesn't restrict her choices tomorrow, since it came from a different pizzeria. – Micapps Jun 15 '16 at 14:43 Mary has $777 \\cdot 3=2331$ choices for the first day and $2330$ choices each day after that because this day has to be different from the last. This would give $2331 \\cdot 2330^{14} \\approx 3.24\\cdot 10^{50}$ choices. The point of the other pizzeria is to say that she didn't eat here today and has the $2331$ choices tomorrow. Notice that there are $777\\times 3 = 2331$ possible lunches. So Mary must select 15 of them, and then decide in which order to have them. So there are ${2331 \\choose 15} \\times 15! = \\frac{2331!}{2316!}$ ways to plan the lunches. Plugging this in to a computer, this is exactly: $721554522956870749367491575069940819296530824826880000\\approx 7.2\\cdot 10^{53}$ • This is not correct. She is allowed to have the same lunch", "$$\\PageIndex{76}$$ Baking. Nina is making 4 pans of fudge to serve after a music recital. For each pan, she needs $$\\frac{2}{3}$$ cup of condensed milk. 1. How much condensed milk will Nina need? Show your calculation. 2. Measuring cups usually come in sets of $$\\frac{1}{4}$$, $$\\frac{1}{3}$$, $$\\frac{1}{2}$$, and $$1$$ cup. Draw a diagram to show two different ways that Nina could measure the condensed milk needed for $$4$$ pans of fudge. ##### Exercise $$\\PageIndex{77}$$ Portions Don purchased a bulk package of candy that weighs $$5$$ pounds. He wants to sell the candy in little bags that hold $$\\frac{1}{4}$$ pound. How many little bags of candy can he fill from the bulk package? $$20$$ bags ##### Exercise $$\\PageIndex{78}$$ Portions Kristen has $$\\frac{3}{4}$$ yards of ribbon that she wants to cut into $$6$$ equal parts to make hair ribbons for her daughter’s $$6$$ dolls. How long will each doll’s hair ribbon be? ## Writing Exercises ##### Exercise $$\\PageIndex{79}$$ Rafael wanted to order half a medium pizza at a restaurant. The waiter told him that a medium pizza could be cut into $$6$$ or $$8$$ slices. Would he prefer $$3$$ out of $$6$$ slices or $$4$$ out of $$8$$ slices? Rafael replied that since he wasn’t very hungry, he would prefer $$3$$ out of $$6$$ slices. Explain what is wrong with", "Doreen ate the rest of it the next morning. – The Short One May 10 '18 at 21:16 ## 2 Answers Here, I recommend we use variables. Let $p$ represent the pie that is being eaten (somewhat timidly). Jerry eats $1/5$ of it, so $1-1/5 = 4/5$ of the original pie is left: $4p/5$. The his wife comes and eats $1/6$ of what is left of the pie. Take the remaining pie ($4p/5$), and take away $1/6$ of this new pie stuffs, not the original pie. $(4p/5)(1-1/6) = 4p/5 \\cdot 5/6 = 2p/3$. So two-thirds of the pie remains. 1. You are subtracting two fractions of the entire pie. The question asks you first about the enitre pie, then the remainder. 2. 'Of' implies multiplication: $10\\%\\ \\text{of}\\ 500 = .1 \\cdot 500 = 50$. 3. Multiplication is the only way it will work, because subtraction is the incorrect way to think about the wording of the problem. Cheers! If you eat a fraction $q$ of something then what is left is $1-q$ of that something, that is to say you are left with the earlier amount of that something multiplied by $(1-q)$ In effect you subtract from $1$ and then multiply So • They start off with $1$ pie and Jerry eats $\\dfrac15$ of it, leaving $1 \\times", "subject.... ### Jan went grocery shopping and only bought items which had been marked down. The items she bought, along with their prices, can be seen below.ItemFinal PriceMarkdownChicken$8.4715%Milk$2.1620%Onions$0.8910%Potato chips$1.4512%Oranges$1.3625%Flour$4.3918%Compared to if she had bought no items on sale, by what percent was Jan’s total grocery bill discounted? Round to two decimal places.a.15.85%b.16.50%c.16.65% d.17.50% Jan went grocery shopping and only bought items which had been marked down. The items she bought, along with their prices, can be seen below.ItemFinal PriceMarkdownChicken$8.4715%Milk$2.1620%Onions$0.8910%Potato chips$1.4512%Oranges$1.3625%Flour$4.3918%Compared to if she had bought no items on sale,... ### Video Example EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 − 4xy + 1. SOLUTION We first locate the critical points: Video Example EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 − 4xy + 1. SOLUTION We first locate the critical points:... ### I need the answer ASAP :/ I need the answer ASAP :/... ### James answered 14 items correctly on a 16 item quiz. what percent did he answer correctly James answered 14 items correctly on a 16 item quiz. what percent did he answer correctly... ### If f(x) = (x – 2, then find f(-3).​ If f(x) = (x – 2, then find" ]

[ "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$ ## Solution 1 (No trig) Let's redraw the diagram, but extend some helpful lines. $[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label(\"A\", A, SW); B=(6,15); label(\"B\", B, NW); C=(30,15); label(\"C\", C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"O\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label(\"3\", midpoint(A--P), S); label(\"9\", midpoint(P--Q), S); label(\"16\", midpoint(Q--C), S); label(\"x\", (5.5,13.75), W); label(\"20\", (20.25,15), N); label(\"6\", (5.25,0), S); label(\"6\", (1.5,3.75), W); label(\"x\", (8.25,15),N); label(\"14+x\", (17.25,0), S); label(\"6-x\", (27,15), N); label(\"6+x\", (27,7.5), W); label(\"6\\sqrt{3}\", (30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,- ...th(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth( 18 KB (2,742 words) - 19:46, 29 July 2021 • ...,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6) ...--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); 16 KB (2,345 words) - 23:46, 15 January 2022 • [itex]\\text{(v)}$ Without the use of logarithm tables evaluate $\\frac{1}{\\log_{ 4 KB (540 words) - 17:23, 8 October 2014 • [itex]\\text{(v)}$ Without the use of logarithm tables evaluate $\\frac{1}{\\log_{ [itex]\\text{(v)}$ <center> 2 KB (286 words) - 14:14, 21 July 2012 • ...,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6) ...--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); 3 KB (544 words) - 18:57, 17 August 2021 • ...{v} \\rfloor[/itex] means the greatest integer less than or equal to $v$.) 15 KB (2,247 words) - 12:44, 19 February 2020 • ...th>. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interio 14 KB (2,197 words) - 12:34, 12 August 2020 • ...th>. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interi ...math>v=(x,y)[/itex] will create the translation of $S$, $T(v)$ such that it covers both $(0,0)$ and $(1,0)$. 4 KB (658", "real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label(\"1\", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label(\"2\", (2.91,-0.11), SE*lsf); label(\"2\", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label(\"\\theta\", (0.42,0.77), SE*lsf); label(\"2\\theta\", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label(\"2\\theta\", (2.18,-0.3), SE*lsf); dot((0,0)); label(\"B\", (-0.21,-0.2),NE*lsf); dot((4,0)); label(\"A\", (4.03,0.06),NE*lsf); dot((2,0)); label(\"O\", (2.04,0.06),NE*lsf); dot((0.59,0)); label(\"P\", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label(\"C\", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label(\"D\", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label(\"E\", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]$ Solution 2 Let $AC=b$, $BC=a$ by convention. Also, Let $AP=x$ and $BP=y$. Finally, let $\\angle ACP=\\theta$ and $\\angle APC=2\\theta$. We are then looking for $\\frac{AP}{BP}=\\frac{x}{y}$ Now, by arc interceptions and angle chasing we find that $\\triangle BPD \\sim \\triangle CPA$, and that therefore $BD=yb.$ Then, since $\\angle ABD=\\theta$ (it intercepts the same arc as $\\angle ACD$) and $ADB$ is right, $\\cos\\theta=\\frac{DB}{AB}=\\frac{by}{4}$. Using law of sines on $APC$, we additionally find that $\\frac{b}{\\sin 2\\theta}=\\frac{x}{\\sin\\theta}.$ Simplification by the double angle formula $\\sin 2\\theta=2\\sin \\theta\\cos\\theta$ yields $\\cos \\theta=\\frac{b}{2x}$. We equate these expressions for $\\cos\\theta$ to find that $xy=2$. Since $x+y=AB=4$, we have enough information to solve for $x$ and $y$. We obtain $x,y=2 \\pm \\sqrt{2}$ Since we know $x>y$,", "draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\\boxed{\\textbf{(B) }25}$.", "O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"O\", O, ENE); label(\"A\", A, NW); label(\"B\", B, W); label(\"C\", C, E); label(\"H\", H, E); label(\"H'\", K, NE); label(\"X\", X, W); label(\"Y\", Y, NE); label(\"O'\", P, E); label(\"M\", M, NE); label(\"L\", L, NE); label(\"D\", D, NNE); label(\"2\", X -- H, NW); label(\"3\", H -- A, SW); label(\"6\", H -- Y, NW); label(\"R\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy]$ Diagram not to scale. We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\\angle HBC = 90 - \\angle C = \\angle H'AC = \\angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\\triangle BHC$ is the image of circle $O$ over line", "yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$: $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); label(\"P\",P,dir(235)); [/asy]$ This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "\\frac{s-a}{a}$. $[asy] pathpen = linewidth(0.7); size(300); pen d = linetype(\"4 4\") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP(\"D_1\",foot(O,B,C))),E1 = D(MP(\"E_1\",foot(O,A,C),NE)),E2 = D(MP(\"E_2\",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP(\"D_2\",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP(\"F\",D(foot(OA,B,Fa)),NW), G=MP(\"G\",D(foot(OA,C,Ga)),NE); D(OA); D(MP(\"A\",A,N)--MP(\"B\",B,NW)--MP(\"C\",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP(\"P\",IP(D(A--D2),D(B--E2)),NNE)); D(MP(\"Q\",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); [/asy]$ By Menelaus' Theorem on $\\triangle ACD_2$ with segment $\\overline{BE_2}$, it follows that $\\frac{CE_2}{E_2A} \\cdot \\frac{AP}{PD_2} \\cdot \\frac{BD_2}{BC} = 1 \\Longrightarrow \\frac{AP}{PD_2} = \\frac{(c - (s-a)) \\cdot a}{(a-(s-c)) \\cdot AE_1} = \\frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\\blacksquare$ ### Solution 2 The key observation is the following lemma. Lemma: Segment $D_1Q$ is a diameter of circle $\\omega$. Proof: Let $I$ be the center of circle $\\omega$, i.e., $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\\omega$. Let $l$ be the line tangent to circle $\\omega$ at $Q'$, and let $l$ intersect the segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\\omega$ is an excircle of triangle $AB_1C_1$. Let $\\mathbf{H}_1$ denote the", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. ## See Also 2014 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS", "in Shaddoll's solution, and our answer is $p=13+84+85=\\boxed{182}$. ## Solution 4 $[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"A\", A, SE); label(\"B\", B, NW); label(\"C_0\", C0, SW); label(\"C_1\", C1, NE); label(\"C_2\", C2, W); label(\"C_3\", C3, NE); label(\"C_4\", C4, W); label(\"a\", (B+C0)/2, W); label(\"b\", (A+C0)/2, S); label(\"c\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]$ Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\\frac{a+c}{b}$. The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \\cdots$, shown in red in the figure below. Since each of the triangles $\\triangle C_0 C_1 C_2, \\triangle C_2 C_3 C_4, \\dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "$n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\\underbrace{111\\cdots 111}_{x\\text{ times}}$. $\\Box$ ## Picture 1 $[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype(\"8 8\")); label(\"a\",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype(\"8 8\")); label(\"b\",(2,0),SSW); [/asy]$ $$\\text{Find the probability that }b>a \\text{.}$$ $[asy]unitsize(2inch); import olympiad; path c2 = dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180); path c1 = dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60); path c3 = dir(120)..dir(150)..dir(180); path c4 = dir(0)..dir(30)..dir(60); draw(dir(0)..dir(30)..dir(60)..dir(90)..dir(120)..dir(150)..dir(180)--dir(0)); draw(dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60)..dir(90)-dir(90)..dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180)--dir(90)-dir(0)); draw((-1.2,0.8)--(1.2,0.8)); label(\"l_{-n}\", (1.2,0.8),dir(0)); draw((-1.2,0.5)--(1.2,0.5)); label(\"l_0\", (1.2,0.5),dir(0)); draw((-1.2,0.2)--(1.2,0.2)); label(\"l_n\", (1.2,0.2),dir(0)); label(\"\\vdots\", (1.2, 0.4), dir(0)); label(\"\\vdots\", (0, 0.4)); label(\"\\vdots\", (1.2, 0.65), dir(0)); label(\"\\vdots\", (0, 0.65)); label(\"A_{-n}\", intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"C_{-n}\", intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"D_{-n}\", intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"B_{-n}\", intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"X\", intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5)), dir(150)); label(\"Y\", intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5)), dir(30)); label(\"C_{n}\", intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"A_{n}\", intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); label(\"B_{n}\", intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"D_{n}\", intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); dot(intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c3, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c4, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2))); [/asy]$ Two half-circles are drawn as shown above, with a line $l_0$ throught the two intersections points, $X,Y$ of the half-circles. Lines $l_k$ for $k=-n\\to", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational" ]

[ "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"O\", O, ENE); label(\"A\", A, NW); label(\"B\", B, W); label(\"C\", C, E); label(\"H\", H, E); label(\"H'\", K, NE); label(\"X\", X, W); label(\"Y\", Y, NE); label(\"O'\", P, E); label(\"M\", M, NE); label(\"L\", L, NE); label(\"D\", D, NNE); label(\"2\", X -- H, NW); label(\"3\", H -- A, SW); label(\"6\", H -- Y, NW); label(\"R\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy]$ Diagram not to scale. We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\\angle HBC = 90 - \\angle C = \\angle H'AC = \\angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\\triangle BHC$ is the image of circle $O$ over line", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\\boxed{\\textbf{(B) }25}$.", "C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$ ## Solution 1 (No trig) Let's redraw the diagram, but extend some helpful lines. $[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label(\"A\", A, SW); B=(6,15); label(\"B\", B, NW); C=(30,15); label(\"C\", C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"O\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label(\"3\", midpoint(A--P), S); label(\"9\", midpoint(P--Q), S); label(\"16\", midpoint(Q--C), S); label(\"x\", (5.5,13.75), W); label(\"20\", (20.25,15), N); label(\"6\", (5.25,0), S); label(\"6\", (1.5,3.75), W); label(\"x\", (8.25,15),N); label(\"14+x\", (17.25,0), S); label(\"6-x\", (27,15), N); label(\"6+x\", (27,7.5), W); label(\"6\\sqrt{3}\", (30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ Solution Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\\frac{1^2\\sqrt{3}}{4}\\implies\\frac{1\\sqrt{3}}{4}$, and $\\frac{2^2\\sqrt{3}}{4}\\implies\\frac{4\\sqrt{3}}{4}\\implies\\sqrt{3}$ and the area of the large equilateral triangle is $\\frac{7^2\\sqrt{3}}{4}\\implies\\frac{49\\sqrt{3}}{4}$ so the area of the hexagon would be $\\frac{49\\sqrt{3}}{4}-\\frac{\\sqrt {3}}{4}-\\frac{\\sqrt{3}}{4}-\\sqrt{3}\\implies\\boxed{\\frac{43\\sqrt{3}}{4}}$ Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can", "refer to the following diagram. Note that the area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "*/ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label(\"12\",(6,-1),SE*labelscalefactor); label(\" 2\\sqrt{76}-6 \",(14,5),SE*labelscalefactor); label(\"20\",(8,7),SE*labelscalefactor); draw((12,0)--(12,10)); draw((12,10)--(18,10)); /* dots and labels */ dot((0,0),dotstyle); label(\"A\", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label(\"B\", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label(\"C\", (18,10), NE * labelscalefactor); label(\"120^\\circ\", (11,1), NE * labelscalefactor,qqwuqq); dot((12,10),dotstyle); label(\"H\", (12,10), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Now, note that we have to find the coordinates of $E,F$ and $D$. Assume WLOG that $A$ is $(0,0)$ and $B$ be $(12,0)$. $E$ is obviously $(3,0)$, $F$ is going to be $(12+\\frac{9+\\sqrt{73}-12}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$ and the coordinates of $D$ is going to be $(\\frac{9+\\sqrt{73}}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$. Since $D$ and $F$ have the same $y$ value, we can find $[EDF]$ by using $\\frac12*b*h$. The base is going to be equal to the distance from $D$ to $F$, which is the same thing as $12+\\frac{9+\\sqrt{73}-12}{4}-\\frac{9+\\sqrt{73}}{4}=9=b$, and the height is the change in the $y$ coordinate from $E$ to $D$. This is the same thing as $\\frac{\\sqrt{219}-3\\sqrt3}{4}-0=\\frac{\\sqrt{219}-3\\sqrt3}{4}=h$. Hence, plugging these into $[EDF]=\\frac12*b*h$ give us $\\frac12*9*\\frac{\\sqrt{219}-3\\sqrt3}{4}=\\frac{9\\sqrt{219}-27\\sqrt3}{8}$. The answer is thus $9+219+27+3+8=\\boxed{266}$. ## Solution 2 Let $S$ be the area of the large triangle and let $x$ be the area we are trying to find. We have that $\\Delta ADE$ is similar to $\\Delta ABC$ with a ratio of $1:4$ while $\\Delta CDF$ is similar to $\\Delta", "yields $b = \\frac{4a\\sqrt5}{7}$. We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$: $[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,dir(90)); label(\"F\",F,NE); label(\"G\",G,NE); label(\"H\",H,W); label(\"J\",J,S); label(\"K\",K,SE); label(\"L\",L,SE); label(\"M\",M,dir(90)); label(\"N\",N,dir(180)); label(\"P\",P,dir(235)); [/asy]$ This gives that $AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}$ and $ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}$ Since $AE$ = $AM + ME$, we get $2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a$ $\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a$ $\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a$ $\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}$ $\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a$ $\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}$ So our final answer is $(7\\sqrt{11})^2 = \\boxed{539}$", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "(int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ So the answer is $\\boxed{\\textbf{(A)}28}$ ~yofro (Diagram credits to franzliszt) ## Solution 2 Suppose we \"extend\" the chessboard indefinitely to the right: $[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label(\"P\", (5.5,.5)); label(\"Q\", (6.5,7.5)); label(\"X\", (8.5,3.5)); label(\"Y\", (8.5,5.5)); [/asy]$ The total number of paths from $P$ to $Q$ (including invalid paths which cross over the red line) is $\\binom{7}{3} = 35$. We subtract the number of invalid paths that pass through $X$ or $Y$. The number of paths that pass through $X$ is $\\binom{3}{0}\\binom{4}{3} = 4$ and the number of paths that pass through $Y$ is $\\binom{5}{1}\\binom{2}{2} = 5$. However, we overcounted the", "be the intersection of the segments emanating from $A$ and $C$. By Law of Cosines, $$BP' = \\sqrt{1 + 1 - 2\\cos{120^\\circ}} = \\sqrt{3}.$$ So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines $$s = \\sqrt{4 + 1 - 4\\cos{120^\\circ}} = \\boxed{\\textbf{(B)} \\sqrt{7}}.$$ ~ hnkevin42 ## Solution 3 (Answer Choices) $[asy] unitsize(0.4inch); pen p = fontsize(10pt); draw((0,0)--(4,5.65)--(8,0)--cycle); label(\"A\", (4,5.65), N, p); label(\"C\", (8,0), SE, p); label(\"B\", (0,0), SW, p); label(\"P\", (3.5,3.5), E, p); label(\"E\", (2.8191,3.982), NW, p); label(\"F\", (4.848,4.452), NE, p); label(\"G\", (3.5,0), down, p); draw((0,0)--(3.5,3.5)); label(\"\\sqrt{3}\",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label(\"2\",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label(\"1\",(4,5.65)--(3.5,3.5), E,p); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); [/asy]$ We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\\triangle{APC}$, $\\triangle{APB}$, and $\\triangle{BPC}$. Let $s$ be the side of the equilateral triangle, we use the Heron's formula: $$\\triangle{APC} = \\frac{s\\cdot PF}{2} = \\sqrt{\\frac{s+3}{2}\\left(\\frac{s+3}{2}-s\\right)\\left(\\frac{s+3}{2}-1\\right)\\left(\\frac{s+3}{2}-2\\right)}$$ $$\\implies PF = \\frac{\\sqrt{10s^2-s^4-9}}{2s}$$ Similarly, we obtain: $$PE = \\frac{\\sqrt{8s^2-s^4-4}}{2s}$$ $$PG = \\frac{\\sqrt{14s^2-s^4-1}}{2s}$$ By Viviani's theorem, $$\\frac{\\sqrt{10s^2-s^4-9}}{2s}+\\frac{\\sqrt{8s^2-s^4-4}}{2s}+\\frac{\\sqrt{14s^2-s^4-1}}{2s} = \\frac{\\sqrt{3}}{2}s$$ $$\\sqrt{10s^2-s^4-9}+\\sqrt{8s^2-s^4-4}+\\sqrt{14s^2-s^4-1} = \\sqrt{3}s^2$$ Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with" ]

[ "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"O\", O, ENE); label(\"A\", A, NW); label(\"B\", B, W); label(\"C\", C, E); label(\"H\", H, E); label(\"H'\", K, NE); label(\"X\", X, W); label(\"Y\", Y, NE); label(\"O'\", P, E); label(\"M\", M, NE); label(\"L\", L, NE); label(\"D\", D, NNE); label(\"2\", X -- H, NW); label(\"3\", H -- A, SW); label(\"6\", H -- Y, NW); label(\"R\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy]$ Diagram not to scale. We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\\angle HBC = 90 - \\angle C = \\angle H'AC = \\angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\\triangle BHC$ is the image of circle $O$ over line", "C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$ ## Solution 1 (No trig) Let's redraw the diagram, but extend some helpful lines. $[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label(\"A\", A, SW); B=(6,15); label(\"B\", B, NW); C=(30,15); label(\"C\", C, NE); D=(24,0); label(\"D\", D, SE); P=(5.2,2.6); label(\"P\", (5.8,2.6), N); Q=(18.3,9.1); label(\"Q\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"O\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label(\"3\", midpoint(A--P), S); label(\"9\", midpoint(P--Q), S); label(\"16\", midpoint(Q--C), S); label(\"x\", (5.5,13.75), W); label(\"20\", (20.25,15), N); label(\"6\", (5.25,0), S); label(\"6\", (1.5,3.75), W); label(\"x\", (8.25,15),N); label(\"14+x\", (17.25,0), S); label(\"6-x\", (27,15), N); label(\"6+x\", (27,7.5), W); label(\"6\\sqrt{3}\", (30,7.5),W); label(\"T_1\", (10.5,15), N); label(\"T_2\", (10.5,0), S); label(\"T_3\", (4.5,11.25),W); label(\"E\",E, N); label(\"F\",F, S); [/asy]$ We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "\\frac{s-a}{a}$. $[asy] pathpen = linewidth(0.7); size(300); pen d = linetype(\"4 4\") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP(\"D_1\",foot(O,B,C))),E1 = D(MP(\"E_1\",foot(O,A,C),NE)),E2 = D(MP(\"E_2\",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP(\"D_2\",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP(\"F\",D(foot(OA,B,Fa)),NW), G=MP(\"G\",D(foot(OA,C,Ga)),NE); D(OA); D(MP(\"A\",A,N)--MP(\"B\",B,NW)--MP(\"C\",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP(\"P\",IP(D(A--D2),D(B--E2)),NNE)); D(MP(\"Q\",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); [/asy]$ By Menelaus' Theorem on $\\triangle ACD_2$ with segment $\\overline{BE_2}$, it follows that $\\frac{CE_2}{E_2A} \\cdot \\frac{AP}{PD_2} \\cdot \\frac{BD_2}{BC} = 1 \\Longrightarrow \\frac{AP}{PD_2} = \\frac{(c - (s-a)) \\cdot a}{(a-(s-c)) \\cdot AE_1} = \\frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\\blacksquare$ ### Solution 2 The key observation is the following lemma. Lemma: Segment $D_1Q$ is a diameter of circle $\\omega$. Proof: Let $I$ be the center of circle $\\omega$, i.e., $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\\omega$. Let $l$ be the line tangent to circle $\\omega$ at $Q'$, and let $l$ intersect the segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\\omega$ is an excircle of triangle $AB_1C_1$. Let $\\mathbf{H}_1$ denote the", "draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\\boxed{\\textbf{(B) }25}$.", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ Solution Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "length of the blue segment times $\\frac{a+c}{b}$. In other words, we have that $a\\left(\\frac{a+c}{b}\\right) = 6p$. Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = \\boxed{182}$ satisfies this equation. $[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"A\", A, SE); label(\"B\", B, NW); label(\"C_0\", C0, SW); label(\"a\", (B+C0)/2, W); label(\"b\", (A+C0)/2, S); label(\"c\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); [/asy]$ ## Solution 5 This solution proceeds from $\\frac{\\frac{ab}{c}}{1-\\frac{a}{c}} = \\frac{\\frac{ab}{c}}{\\frac{c-a}{c}} = \\frac{ab}{c-a} = 6(a+b+c)$. Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$. Thus $a = 168n^2, b = 26n^2, c = 170n^2$. Since we know all three side lengths are relatively prime, we must divide each by 2 and", "U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,- ...th(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth( 18 KB (2,742 words) - 19:46, 29 July 2021 • ...,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6) ...--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); 16 KB (2,345 words) - 23:46, 15 January 2022 • [itex]\\text{(v)}$ Without the use of logarithm tables evaluate $\\frac{1}{\\log_{ 4 KB (540 words) - 17:23, 8 October 2014 • [itex]\\text{(v)}$ Without the use of logarithm tables evaluate $\\frac{1}{\\log_{ [itex]\\text{(v)}$ <center> 2 KB (286 words) - 14:14, 21 July 2012 • ...,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6) ...--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); 3 KB (544 words) - 18:57, 17 August 2021 • ...{v} \\rfloor[/itex] means the greatest integer less than or equal to $v$.) 15 KB (2,247 words) - 12:44, 19 February 2020 • ...th>. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interio 14 KB (2,197 words) - 12:34, 12 August 2020 • ...th>. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interi ...math>v=(x,y)[/itex] will create the translation of $S$, $T(v)$ such that it covers both $(0,0)$ and $(1,0)$. 4 KB (658", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational" ]

[ "# Difference between revisions of \"2006 AMC 8 Problems/Problem 25\" ## Problem Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? $[asy] path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle); draw(card, linewidth(1)); draw(shift(2.5,0)*card, linewidth(1)); draw(shift(5,0)*card, linewidth(1)); label(\"44\", (1,1.5)); label(\"59\", shift(2.5,0)*(1,1.5)); label(\"38\", shift(5,0)*(1,1.5));[/asy]$ $\\textbf{(A)}\\ 13\\qquad\\textbf{(B)}\\ 14\\qquad\\textbf{(C)}\\ 15\\qquad\\textbf{(D)}\\ 16\\qquad\\textbf{(E)}\\ 17$ ## Solution Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is $59+2=61$. Thus, the first card's hidden number is $61-44=17$, and the last card's hidden number is $61-38=23$. Since the sum of the hidden primes is $2+17+23=42$, the average of the primes is $\\dfrac{42}{3}=\\boxed{\\textbf{(B)} 14}$.", "Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$", "# A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box. Question: A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box. Find the probability that the number on the drawn card is a prime number. Solution: The numbers 3, 5, 7, 9, ..., 35, 37 are in AP. Here, a = 3 and d = 5 − 3 = 2 Suppose there are n terms in the AP. $\\therefore a_{n}=37$ $\\Rightarrow 3+(n-1) \\times 2=37 \\quad\\left[a_{n}=a+(n-1) d\\right]$ $\\Rightarrow 2 n+1=37$ $\\Rightarrow 2 n=37-1=36$ $\\Rightarrow n=18$ ∴ Total number of outcomes = 18 Let E be the event of drawing a card with prime number on it. Out of the given numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. So, the favourable number of outcomes are 11. $\\therefore$ Required probability $=\\mathrm{P}(\\mathrm{E})=\\frac{\\text { Favourable number of outcomes }}{\\text { Total number of outcomes }}=\\frac{11}{18}$", "# Line of primes Discrete Mathematics Level 4 Hugo has written the 9 smallest primes with 2 digits down on 9 cards. Now he places the 9 cards in a line, so that the difference of the numbers written on neighbored cards is always divisible by $$2^m, m \\in \\mathbb{N} .$$ In how many ways can Hugo arrange his deck, so that his deck suffices the property given above? ×", "a = [1243,11311] The prime factors would be calculated as: a' = [[11, 113], [11311]] = [] Because 11 appears in 11311, and 113 appears in 11311, the result would be [], an empty set. Lets look at another example: a = [270,325,192] = [[2, 3, 3, 3, 5], [5, 5, 13], [2, 2, 2, 2, 2, 2, 3]] = [13] Another example: a = [27072,32585,19296] a' = [[2, 2, 2, 2, 2, 2, 3, 3, 47], [5, 7, 7, 7, 19], [2, 2, 2, 2, 2, 3, 3, 67]] a'' = [47,67] A bigger example: a = [56,3924,10000,1240124] a' = [[2, 2, 2, 7], [2, 2, 3, 3, 109], [2, 2, 2, 2, 5, 5, 5, 5], [2, 2, 31, 73, 137]] a'' = [7,109,31,73,137] This is code golf, lowest byte-count wins. # Play time Hello. It is time to play the most popular portuguese card game: Sueca! So for people who does not know the game, it is needed to print a wall poster to make them know the power and the points each card has. So, the job is to print an ASCII art wall as: 11 10 4 3 2 ——— ——— ——— ——— ——— ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ ╭───────╮ |A A| |7 7| |R R| |V V| |D", "- 1) 5 = 4x x= 5/4 Question 199. There are thirty cards numbered from 1 to 30. If a card is drawn at random find the probability that, the drawn card has a prime number- 1. a. 1/2 2. b. 1/3 3. c. 1/4 4. d. 1/5 Solution: The prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are 10 prime numbers between 1 to 30. Total number of possible chances = 30 Total number of favourable chances = 10 Probability of a card drawn from 1 to 30 having a prime number = 10/30 = ⅓ Question 200. An insect which is climbing on a vertical pole in such a way that one day it climbs a height of 2m on the next day it comes down 1 m. If the height of the pole is 12 m, find the no. of days in which it will reach on the top. 1. a. 11 days 2. b. 12 days 3. c. 21 days 4. d. 22 days Solution:", "side and a letter on the other. How many of the cards must be turned over to prove the correctness of this statement: \"If a card has a vowel on one side, then it has a prime number on the other side\" ? . . $[\\,A\\,]\\quad[\\,B\\,]\\quad[\\,C\\,]\\quad[\\;3\\;]\\quad[\\;4\\;]\\quad[\\;5\\;]$ The cards are to be \"turned over\". Does this mean each card has one side already visible? If so, which sides are visible? . . The letters? .the numbers? .a random mix?", "A box contains cords numbered from 1 to 20. A card is drawn at random from the box. Question: A box contains cords numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) a prime number, (ii) a composite number, (iii) a number divisible by 3. Solution: There are 20 cards in the box. One card can drawn at random from the box in 20 ways. ∴ Total number of outcomes = 20 (i) The prime number cards in the box are 2, 3, 5, 7, 11, 13, 17 and 19. There are 8 prime number cards in the box. So, there are 8 ways to draw a card from the box which is a prime number card. Favourable number of outcomes = 8 $\\therefore P($ Number on the drawn card is a prime number $)=\\frac{\\text { Favourable number of outcomes }}{\\text { Total number of outcomes }}=\\frac{8}{20}=\\frac{2}{5}$ (ii) The composite number cards in the box are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. There are 11 composite number cards in the box. So, there are 11 ways to draw a card from the box which is a composite number card. Favourable number of outcomes = 11", "### Prime Magic Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find? ### Got It Article This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy. ### Clever Carl What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that. # Sort Them Out (2) ##### Stage: 2 Challenge Level: Thank you to those who submitted solutions to this problem. There were plenty of opportunities to practise calculations, and also to share different strategies. Courtenay and Max from Ickford Combined School looked at the set of fifteen cards, as did Ms Bean's Math Whizzes from Sylvan Park. They both noticed that the answers are numbers between $21$ and $35$, and could be arranged in ascending or descending order. Courtenay and Max explained how they sorted their cards: First we sorted them into the five time tables and the two times table. Then we sorted them into ones that have a digit that is a two or three and ones", "Select Page Try this beautiful problem from Algebra based on Prime numbers. ## Algebra based on Number theory – AMC-8, 2009 – Problem 23 Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? • $14$ • $12$ • $16$ ### Key Concepts Algebra Number theory card number But try the problem first… Answer:$14$ Source AMC-8 (2006) Problem 25 Pre College Mathematics ## Try with Hints First hint Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. Can you now finish the problem ………. Second Hint Obtain this even number would be to add another even number to 44 Can you finish the problem…….. Final Step Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38.", "$$7.$$ Thus, A is the correct answer. 24. In the multiplication problem below $$A,$$ $$B,$$ $$C,$$ $$D$$ are different digits. What is $$A+B?$$ $\\begin{array}{cccc}& A & B & A\\\\ \\times & & C & D\\\\ \\hline C & D & C & D\\\\ \\end{array}$ $$1$$ $$2$$ $$3$$ $$4$$ $$9$$ ###### Solution(s): Note that $CDCD = 101 \\cdot CD.$ This forces $$ABA = 101.$$ Then $$A = 1,$$ $$B = 0,$$ and $$A + B = 1.$$ Thus, A is the correct answer. 25. Barry wrote $$6$$ different numbers, one on each side of $$3$$ cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? $$13$$ $$14$$ $$15$$ $$16$$ $$17$$ Note that if we add an odd number to $$59,$$ we will get an even number. To make the other two sums even numbers, the primes on the back of $$44$$ and $$38$$ must both be even. They also have to be different, however. Since there is only one even prime, the number on the back of $$59$$ must be an even prime, namely $$2.$$ This means that the sum of the front and back of", "val primes = sieve(num) for (i in primes) print(\"$i \") println() And the output of a run: \\$ kts sieve.kts 500 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 There are many other applications of sets. For instance, when finding your way in a maze, you could store the positions already seen in a set. Similarly, when implementing a computer game, you could store the game positions already evaluated in a set. Sets", "Number Detective Follow the clues to find the mystery number. Red Even You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters? Prime Magic Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find? Sets of Numbers Age 7 to 11 Challenge Level: How many different sets of numbers with at least four members can you find in the numbers in this box? For example, one set could be multiples of $4$ {$8, 36 ...$}, another could be odd numbers {$3, 13 ...$}.", "card has 8 symbols. Katara was very patient: I guess, I would like to talk about this some more when i get home if that's okay Any time. Anyway this evening I cut up some index cards, and found a bunch of stickers in a drawer, and made Katara a projective plane of order 3. This has 13 cards, each with 4 different stickers, and again, every two cards share exactly one sticker. She was very pleased and wanted to know how to make them herself. Each set of cards has an order, which is a non-negative integer. Then there must be !!n^2 + n + 1!! cards, each with !!n+1!! stickers or symbols. When !!n!! is a prime power, you can use field theory to construct a set of cards from the structure of the (unique) field of order !!n!!. ### Fields to projective planes ### Order 2 I'll describe the procedure using the plane of order !!n=2!!, which is unusually simple. There will be !!2^2+2+1 = 7!! cards, each with !!3!! of the !!7!! symbols. Here is the finite field of order 2, called !!GF(2)!!: + 0 1 0 0 1 1 1 0 × 0 1 0 0 0 1 0 1 • The stickers correspond to ordered triples of elements of !!GF(2)!!, except that", "# A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is prime? - Mathematics and Statistics Sum A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that, number on ticket is prime? #### Solution The box contains 75 tickets numbered 1 to 75. ∴ 1 ticket can be drawn from the box in 75C1 = 75 ways. ∴ n(S) = 75 Let C be the event that the number on the ticket is a prime number ∴ C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73} ∴ n(C) = 21 ∴ P(C) = (\"n\"(\"C\"))/(\"n\"(\"S\") = 21/75 = 7/25 Concept: Elementary Properties of Probability Is there an error in this question or solution?", "The series of peaks at the sequences zoo1: Mounts denied: Don. Always even same line it keeps track of which cards you 've matched and you. Number that has an integer value for any$ k = 3^2 = 9 $what type of are.: there should not be one symbol in common with any other in deck! Understand how you have written, although I am trying to follow the matrix generated by Don Simborg formula! Primes you can swap the commented lines to print letters, though they wo n't work powers. In my head for ages algorithm works when n is 4 or 8 ( meaning 5 9... Including boss ), boss asks for handover of work, boss boss. Pattern from the beginning of the lines for$ q $equal to the fact that we want cards have! Grasp the comments about n being a prime number which lines with a pen & paper it... The generators submitted by Karinka, Urmil Karikh and Uwe are working nicely card = n! Online Encyclopedia of integer sequences each column spells out the symbols are these less ones... How you could design a deck,$ s = 3 $because it each. To litigate against other States ' election results some of the pattern when there are links... Games involve finding which symbol is only", "a prime number card. Favourable number of outcomes = 8 ∴ P(Number on the drawn card is a prime number) = (ii) The composite number cards in the box are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. There are 11 composite number cards in the box. So, there are 11 ways to draw a card from the box which is a composite number card. Favourable number of outcomes = 11 ∴ P(Number on the drawn card is a composite number) = (iii) The cards with number divisible by 3 on them in the box are 3, 6, 9, 12, 15 and 18. There are 6 cards with number divisible by 3 on them in the box. So, there are 6 ways to draw a card from the box with number divisible by 3 on them. Favourable number of outcomes = 6 ∴ P(Number on the drawn card is a number divisible by 3) = #### Question 24: A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one-digit number, (ii) a number divisible by 5, (iii) an odd number less than 30, (iv) a composite number between 50 and 70. ​Given number 6, 7, 8, ....", "the resulting parse tree). – Bubbler Dec 26 '19 at 2:04 # Potentially Prime Punch-Card Patterns code-golfprimesnumber-theory ## Punch-cards Let us define a punch-card to be a set of 'open' and 'closed' holes which can slide over the positive integer number line: _________________________ | ___ ___ ___ | --> | | | | | | | | 1 | | 3 | | 5 | | 7 | | 9 10 11 12 13 14 | |___| |___| |___| | |_________________________| --> When the first open hole is on the integer $$\\ p \\$$, then this punchcard (as it slides across the number line) yields the integers $$\\p, p+2,\\$$ and $$\\p+4\\$$. So, we can represent this card as the set $$\\\\{0, 2, 4\\}\\$$. Given a punch-card we may wonder whether it is possible to slide the card to such a position that only prime numbers are under the 'open' holes. Clearly, for $$\\\\{0, 2, 4\\}\\$$, we can position the punch-card to make 3, 5, and 7 visible. However, this is the only solution. Taking each element modulo 3 yields $$\\\\{0, 2, 1\\}\\$$, which contains every possible remainder under division by 3, so one visible integer must be divisible by 3. But we only want primes, and so 3 itself must be visible; this leaves a finite number of", "cards from another of Dan’s games, the delightful Tiny Polka Dots. I deliberately used cards with different representations of 1, 2, 7 and 9 (with 8 as a back-up). Pairs with a different representation of the same number formed a group, one at each table. I gave them the rest of the cards for their number, and we did a quick ‘Notice and Wonder’ on the different representations. #### Different-sized grids for Prime Climb Rather than show students the Prime Climb hundreds chart arranged in columns of 10 (the image earlier in this post), I wanted to foster noticing and wondering by having them construct the charts themselves with mini-cards of the numbers: physically handling, examining, ordering and organising. The (roughly) six students at each table arranged the cards into charts with the number of columns corresponding to their Tiny Polka Dots card (with 1 and 2 corresponding to 11 and 12, respectively). Click on the images below to make them bigger. What do you notice? What do you wonder? To add impetus to the discussion, I relayed that part of the weekly task (contributing towards their course grade) was to individually write a forum post with at least five things that they noticed and wondered about their charts. It was heartening to see students collaboratively generating lists", "Cards bearing numbers 1, 3, 5, …. , 35 are kept in a bag. A card is drawn at random from the bag. Question: Cards bearing numbers 1, 3, 5, …. , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (i) a prime number less than 15, (ii) a number divisible by 3 and 5. Solution: Given number 1, 3, 5, …. , 35 form an AP with a = 1 and d = 2. Let Tn = 35. Then, 1 + (n − 1)2 = 35 ⇒ 1 + 2n − 2 = 35 ⇒ 2n = 36 ⇒ n = 18 Thus, total number of outcomes = 18. (i) Let E1 be the event of getting a prime number less than 15. Out of these numbers, prime numbers less than 15 are 3, 5, 7, 11 and 13. Number of favourable outcomes = 5. $\\therefore \\mathrm{P}$ (getting a prime number less than 15$)=\\mathrm{P}\\left(\\mathrm{E}_{1}\\right)=\\frac{\\text { Number of outcomes favourable to } \\mathrm{E}_{1}}{\\text { Number of all possible outcomes }}$ $=\\frac{5}{18}$ Thus, the probability of getting a card bearing a prime number less than 15 is 518518. (ii) Let E2 be the event of getting a number divisible by 3 and 5. Out" ]

[ "Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$", "Hints Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. Can you now finish the problem ………. Obtain this even number would be to add another even number to 44 Can you finish the problem…….. Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$ Categories ## Largest and smallest numbers | AMC 8, 2006 | Problem 22 Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006. ## Largest and smallest numbers | AMC-8, 2006|Problem 22 Three different one-digit positive integers are placed", "we have; 1568 $\\times$ 185 – 1568 $\\times$ 85 = 1568 $\\times$ (185 – 85) = 1568 $\\times$ 100 = 156800 Q16: (888 + 777 + 555) = (111 $\\times$ ?) (a) 120 (b) 280 (c) 20 (d) 140 Sol: (c) 20 (888 + 777 + 555) = (111 $\\times$ ?) $\\Rightarrow$ (888 + 777 + 555) = 111 $\\times$ (8 + 7 + 5) [by taking 111 common] = 111 $\\times$ 20 = 2220 Q17: The sum of two odd numbers is (a) an odd number (b) an even number (c) a prime number (d) a multiple of 3 Sol: (b) An even number The sum of two odd numbers is an even number. Example: 5 + 3 = 8 Q18:The product of two odd numbers is (a) an odd number (b) an even number (c) a prime number (d) none of these Sol: (a) an odd number The product of two odd numbers is an odd number. Example: 5 $\\times$ 3 = 15 Q19: If a is a whole number such that a + a = a, then a = ? (a) 1 (b) 2 (c) 3 (d) none of these Sol: (d) None of these Given: a is a whole number such that a + a = a If a = 1, then 1 +", "the line exactly if their number of distinct odd prime factors is even, except $1$ goes above the line. Equivalently, numbers are below the line when the sum of their distinct prime factors is even and positive. So, $29$ would go above the line (odd prime), $52 = 2^2 \\times 13$ would go above, and $138 = 2 \\times 3 \\times 23$ would go below.", "a lot of what your confusion results from. Now to this current problem. In order four prime numbers to sum to an odd number, three of them must be odd and one must be even. There is only one even prime, 2. So the number 2 must be among the four primes which sum to an odd number; the other three must be odd. From the first seventeen primes one is even and sixteen are odd numbers. There are $\\binom{16}{3}$ ways to choose three of those odd primes. So there are that many ways to have three odd primes and one even prime from the first seventeen primes. • January 6th 2011, 05:31 PM Ka9 Hi Plato, Answer to the first question is yes. Here and there confusion with the dominate language. But, I am able to follow. Sorry for the inconvience. • January 7th 2011, 01:03 AM chisigma Selecting randomly 2 primes among the first 17, the probability that their sum is odd is... $\\displaystyle p_{o}= \\frac{1}{17} + \\frac{1}{16}\\ \\frac{16}{17} = \\frac{2}{17}$ (1) Selecting randomly 3 primes among the first 17, the probability that their sum is even is... $\\displaystyle p_{e}= \\frac{1}{17} + \\frac{1}{16}\\ \\frac{16}{17} + \\frac {1}{15}\\ \\frac{15}{16}\\ \\frac{16}{17}= \\frac{3}{17}$ (2) Selecting randomly 4 primes among the first 17, the probability that their sum is odd", "to find sum of even numbers. It can be expressed as either the sum of six consecutive prime numbers: 19 + 23 + 29 + 31 + 37 + 41, or the sum of eight consecutive prime numbers: 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37. These numbers form an A.P. 1-50 1-100 1-500 1-1000 Odd Even List Randomizer Random Numbers Number Converters. Sum of first four odd numbers = 1 + 3 + 5 + 7 = 16 (16 = 4 x 4). Logic to find sum of even numbers. FAQ. Sum of first three odd numbers = 1 + 3 + 5 = 9 (9 = 3 x 3). Answer: The two numbers are 15 and 3, since 15 times 3 is 45, and 15 added to 3 is 18. Sol: 1 + 2 + 3+ 4+ 5+ ———-+50 So Here n = 50 = 50 ( 50+1) / 2 = 25 x 51 = 1275 The difference = 290. Product means you need to multiply the numbers (-5 multiplied by -12 = 60). Do I need to enter plus (+) sign between two numbers? To get the average, notice that the numbers are all equally distributed. If we have 100 numbers (1…100), then we clearly have 100 items.", "test is simple. To make the sum of two numbers odd, one of the numbers must be odd and the other even. There is only one even prime, so that limits you to sums of the form $2+p$. Thus the odd numbers that are the sum of two primes are exactly the ones that are two more than a prime. The first few are $$5, 7, 9, 13, 15, 19, 21, 25, 31, 33, 39, 43\\ldots$$ Also note that it is not known whether every even number is the sum of two primes. Every single even number that has been checked has been verified to be the sum of two primes, but we don't know whether it is always true.", "post 4 This post was BOOKMARKED banksy wrote: 187. Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 We know that other than 2, all prime numbers are odd. When you add two odd numbers, you get an even sum. To get an odd sum, one number must be even and then other odd. So to get 19, 45 and 79, one prime must be 2. Now we just need to subtract 2 out of each of these three options to see whether we get another prime. 79 - 2 = 77 which is not prime. So 79 CANNOT be the sum of two prime numbers. Note that there is a conjecture that every even number greater than 2 can be written as sum of two prime numbers. So we don't even need to check for the even sum options. For more on this, check: http://www.veritasprep.com/blog/2014/08 ... t-part-iv/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for \\$199 Veritas Prep Reviews Current Student Joined: 29 Jul 2014 Posts: 17 Followers: 0 Kudos [?]: 1 [0], given: 22 Re: Which of the following CANNOT be the sum of two prime numbers? [#permalink] Show Tags 04 Sep 2015,", "numbers, 7 numbers are odd and one is even i.e. 2 $$C^7_2$$ = 21 (choosing any two odd numbers from 7 $$C^8_2$$ = 28 ( total number of outcomes) $$\\frac{21}{28}$$ i.e. $$\\frac{3}{4}$$ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8011 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If 2 numbers are selected from the first 8 prime numbers, what is the [#permalink] ### Show Tags 12 Oct 2018, 08:35 => In order for the sum to be even, both primes selected must be odd. As 2 is the only even prime number, the number of selections with an even sum is equal to the number of ways to select 2 numbers from these 7 odd prime numbers, or 7C2. The total number of selections of 2 prime numbers from the first 8 prime numbers is 8C2. Therefore, the probability that the sum of the two numbers selected is even is 7C2 / 8C2 = $${\\frac{(7*6)}{(1*2)}}/{\\frac{(8*7)}{(1*2)}} = \\frac{6}{8} = \\frac{3}{4}.$$ Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$79 for 1 month Online Course\" \"Free Resources-30 day online access & Diagnostic Test\" \"Unlimited Access to over 120 free", "you finish the problem…….. Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$ Categories ## Largest and smallest numbers | AMC 8, 2006 | Problem 22 Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006. ## Largest and smallest numbers | AMC-8, 2006|Problem 22 Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers", "$15+k$ and $40-k$ for $0 \\leq k \\leq 12$ Thus, there are the same amount of odd sets as even sets. Therefore, the answer is $\\frac{1}{2} {10 \\choose 5} = 126$.", "# Even numbers have more factors than odd numbers… This was an exercise to show that, in a sense, the even numbers have more prime factors than the odds, but--if it's right-- I still have a question. As an heuristic calculation, we could take a large interval (1, 2N) on which the average number of prime factors with repetitions is $\\mu$ (for sufficiently large N, $\\mu$ is about $\\ln \\ln 2N$; see answer to this problem). WLOG N is even, then the set $S_2 = \\{N+2, N+4, N+6, ..., 2N \\}$ corresponds to a sequence $S_1 = \\{\\frac{N}{2}+1, \\frac{N}{2}+ 2, ..., N \\}$. The average number of primes in $S_1$ is only slightly less than $\\mu$ for large N$^{(1)}$, and so the average number of primes $\\mu_2$ in $S_2$ is $\\mu+1$ primes. Let $\\mu_o$ be the average number of primes of odd numbers in $[N,2N]$. Since on $[N,2N]$ the average number of primes is also about $\\mu$, we have that $$\\mu =\\frac{( \\mu_2 + \\mu_o)}{2} = \\frac{(\\mu + 1 + \\mu_o )}{2},$$ and so the average number of primes for the odd numbers $^{(2)}$ is $$\\mu_o = \\mu - 1 = \\mu_2 - 2$$ so $$\\mu_2 - \\mu_o \\approx 2.$$ This argument has I think a somewhat complicated generalization. Using the same reasoning for multiples of $3, 5,...,p_k$", "two odd numbers is also always even, but in this case the sum of two primes is odd, which means that one of the primes is $2$. This leaves only one possible value for the other root and one possible value for their product $k$. ## Solution 3 By Vieta's you have $$a+b=63$$ $$ab=k$$ Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\\boxed{1}$ solution. ~ pi_is_3.14", "be a prime number. For example, $$7+9+11+13+15 = \\overset{\\text{average}}{11}\\times\\overset{\\text{count}}5 = 55$$ If we allow negative odd numbers into the sum, then we can arrange for the average to be $1$ and any odd number is possible. We can also make any number divisible by $4$, but not even numbers which are twice an odd number. • thanks for the information. – RAJAT303 Jan 18 '17 at 14:13", "Jane Street Test The questions below appeared on an online test administered by Jane Street Capital Management to assess whether a person is worthy of a phone interview. 🔗 Question 1 Suppose we choose four numbers at random without replacement from the first 20 prime numbers. What is the probability that the sum of these four numbers is odd? Solution By definition, an even number is an integer multiple of 2; that is, even numbers have the form $2n$, for some integer $n$. An odd number is one more than an even number; that is, odd numbers have the form $2n+1$ for some integer $n$. If all four numbers are odd, then the sum is even. This follows from the fact that the sum of two odd numbers is even: $(2n+1) + (2k+1) = 2(n+k) + 2 = 2(n+k+1)$. And the sum of two even numbers is again even: $2n+2k = 2(n+k)$. Therefore, in order for the sum of the four primes chosen to be odd, the only even prime number, 2, must be among the four numbers we selected. Here are two ways to compute the probability that 2 is one of the four numbers selected: First, consider the probability of not selecting 2: $$P(\\text{2 is not among the chosen primes}) = \\frac{19}{20} \\frac{18}{19} \\frac{17}{18} \\frac{16}{17} =", "that the answer choices are spread out nicely. To get the LOWEST possible price (after the two discounts), we must apply the biggest discount (which is a 40% discount) So, AFTER the 40% discount, customers will pay 60% of ...” April 20, 2018 Brent@GMATPrepNow posted a reply to The \"prime sum\" of an integer n greater than 1 is in the Problem Solving forum “This question requires us to find the prime factorization of the answer choices A. 440 = (2)(2)(2)(5)(11). PRIME SUM = 2 + 2 + 2 + 5 + 11 = 22 B. 512 = (2)(2)(2)(2)(2)(2)(2)(2)(2) PRIME SUM = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18 C. 620 = (2)(2)(5)(31) PRIME SUM = 2 + 2 + 5 + 31 = 40 ...” April 20, 2018 Brent@GMATPrepNow posted a reply to If x and y are integers, is x+y an even in the Data Sufficiency forum “Target question: Is x+y an even integer? Statement 1: x+3y is even There are 4 possible cases to consider. Let''s test them all Case a: x is ODD and y is ODD. Notice that x +3y = ODD + (3)(ODD) = ODD + ODD = EVEN. In this case, x + y = ODD + ODD =", "movie. Write your own ending. Kermit the Frog Intern Joined: 01 Jan 2018 Posts: 44 Re: If three prime number are randomly selected from prime numbers less [#permalink] ### Show Tags 15 Jun 2018, 07:13 4 1 Only one prime number is even and all other primes are odd. So for the sum to be even : two primes should be odd and the other one should be even prime. No. of prime number (2,3,5,7,11,13,17,19, 23, 29) = 10 numbers. Even prime can be selected in 1C1 ways = 1 Odd prime can be selected in 9C2 ways = 36 Total ways in which 3 numbers can be selected = 10C3 = 120 Therefore percentage of sum of 3 primes being even = (1*36)/120 = (3/10)*100 = 30% OA - C. ##### General Discussion Intern Status: No Progress without Struggle Joined: 04 Aug 2017 Posts: 43 Location: Armenia GPA: 3.4 Re: If three prime number are randomly selected from prime numbers less [#permalink] ### Show Tags 24 Aug 2018, 08:05 1 To have the sum as an even number one of the numbers should be even. ODD+ODD+EVEN=EVEN ODD+EVEN+ODD=EVEN Logically we can conclude that the probability that the prime number 2 will be selected out of 9 other odd numbers is relatively low. _________________ Seryozha Sargsyan 21 Contact: [email protected]", "even number n >= 4 is a sum of two primes. 6) Every even number n >= 6 is a sum of two odd primes. 14) Every even number n >= 14 is a sum of two odd primes twice. 12 = 5 + 7 70) Every even number n >= 70 is a sum of two odd primes three times. 68 = 7 + 61 = 31 + 37 130) Every even number n >= 130 is a sum of two odd primes four times. 128 = 19 + 109 128 = 31 + 97 128 = 61 + 67 154) Every even number n >= 154 is a sum of two odd primes five times. 152 = 3 + 149 152 = 13 + 139 152 = 43 + 109 152 = 73 + 79 190) Every even number n >= 190 is a sum of two odd primes six times. 188 = 7 + 181 188 = 31 + 157 188 = 37 + 151 188 = 61 + 127 188 = 79 + 109 334) Every even number n >= 334 is a sum of two odd primes seven times. 332 = 19 + 313 332 = 61 + 271 332 = 103 + 229 332 = 109 + 223 332 = 139 +", "2... 5 5 and 3....16 insuff combined.. Difference of TWO prime will be ODD only when one is 2, so q is 2 ODD-EVEN = odd p-2=5...p=7 so ans (7-2)(7+2) = 5*9=45 suff C _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html BANGALORE/- Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 4710 GPA: 3.82 Re: What is the value of (p-q)(p+q)? [#permalink] ### Show Tags 27 Dec 2017, 23:29 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) and 2) together first. Conditions 1) and 2) Since $$p$$ and $$q$$ are prime numbers satisfying $$p – q = 5$$, one of them is an odd prime and the other is an even prime. The unique even prime is $$2$$. Therefore, $$q = 2$$ and $$p = 7$$, and $$(p-q)(p+q) = (7-2)(7+2) = 5*9 = 45.$$ Both conditions 1) and 2) together are sufficient. Normally, in problems which require 2 equations such as those in", "3: $$2+3=5=prime$$, distinct primes of 45 are 3 and 5: $$3+5=8\\neq{prime}$$ and distinct primes of 100 are 2 and 5: $$2+5=7=prime$$. $$x$$ can be 12 or 100. Not sufficient. (1)+(2) $$x$$ can only be 100. Sufficient. _________________ Math Expert Joined: 02 Sep 2009 Posts: 62289 Re: The Discreet Charm of the DS [#permalink] ### Show Tags 05 Feb 2012, 03:59 13 26 3. If a, b and c are integers, is abc an even integer? In order the product of the integers to be even at leas on of them must be even (1) b is halfway between a and c --> on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\\frac{a+c}{2}$$. Now, does that mean that at leas on of them is be even? Not necessarily, consider $$a=1$$, $$b=3$$ and $$c=5$$. Of course it's also possible that $$b=even$$, for example if $$a=1$$ and $$b=7$$. Not sufficient. (2) a = b - c --> $$a+c=b$$. Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient. _________________ Math Expert Joined: 02 Sep 2009 Posts: 62289 Re: The Discreet Charm of the DS [#permalink] ### Show Tags 05 Feb" ]

[ "# Difference between revisions of \"2006 AMC 8 Problems/Problem 25\" ## Problem Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? $[asy] path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle); draw(card, linewidth(1)); draw(shift(2.5,0)*card, linewidth(1)); draw(shift(5,0)*card, linewidth(1)); label(\"44\", (1,1.5)); label(\"59\", shift(2.5,0)*(1,1.5)); label(\"38\", shift(5,0)*(1,1.5));[/asy]$ $\\textbf{(A)}\\ 13\\qquad\\textbf{(B)}\\ 14\\qquad\\textbf{(C)}\\ 15\\qquad\\textbf{(D)}\\ 16\\qquad\\textbf{(E)}\\ 17$ ## Solution Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is $59+2=61$. Thus, the first card's hidden number is $61-44=17$, and the last card's hidden number is $61-38=23$. Since the sum of the hidden primes is $2+17+23=42$, the average of the primes is $\\dfrac{42}{3}=\\boxed{\\textbf{(B)} 14}$.", "Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$", "Hints Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. Can you now finish the problem ………. Obtain this even number would be to add another even number to 44 Can you finish the problem…….. Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$ Categories ## Largest and smallest numbers | AMC 8, 2006 | Problem 22 Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006. ## Largest and smallest numbers | AMC-8, 2006|Problem 22 Three different one-digit positive integers are placed", "Select Page Try this beautiful problem from Algebra based on Prime numbers. ## Algebra based on Number theory – AMC-8, 2009 – Problem 23 Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? • $14$ • $12$ • $16$ ### Key Concepts Algebra Number theory card number But try the problem first… Answer:$14$ Source AMC-8 (2006) Problem 25 Pre College Mathematics ## Try with Hints First hint Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. Can you now finish the problem ………. Second Hint Obtain this even number would be to add another even number to 44 Can you finish the problem…….. Final Step Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38.", "you finish the problem…….. Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23 Since the sum of the hidden primes is 2+17+23=42, the average of the primes is $\\frac{42}{3}=14$ Categories ## Largest and smallest numbers | AMC 8, 2006 | Problem 22 Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006. ## Largest and smallest numbers | AMC-8, 2006|Problem 22 Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers", "1, 3, 5, …., 49, i.e., 25 ∴ P(an odd number) = $\\frac{25}{49}$ (ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9 ∴ P(a multiple of 5) = $\\frac{9}{49}$ (iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7 ∴ P(a perfect square number) = $\\frac{7}{49}=\\frac{1}{7}$ (iv) “An even prime number” is 2, i.e., only one number ∴ P(an even prime number) = $\\frac{1}{49}$ Question 46. A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (2015D) (i) divisible by 2 or 3, (ii) a prime number. Solution: (i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13 Total outcomes = 20 Possible outcomes = 13 ∴ P(divisible by 2 or 3) = $\\frac{13}{20}$ (ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8 Total Outcomes = 20 Possible outcomes = 8 ∴ P(a prime number) = $\\frac{8}{20}=\\frac{2}{5}$ Question 47. A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number", "$$7.$$ Thus, A is the correct answer. 24. In the multiplication problem below $$A,$$ $$B,$$ $$C,$$ $$D$$ are different digits. What is $$A+B?$$ $\\begin{array}{cccc}& A & B & A\\\\ \\times & & C & D\\\\ \\hline C & D & C & D\\\\ \\end{array}$ $$1$$ $$2$$ $$3$$ $$4$$ $$9$$ ###### Solution(s): Note that $CDCD = 101 \\cdot CD.$ This forces $$ABA = 101.$$ Then $$A = 1,$$ $$B = 0,$$ and $$A + B = 1.$$ Thus, A is the correct answer. 25. Barry wrote $$6$$ different numbers, one on each side of $$3$$ cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? $$13$$ $$14$$ $$15$$ $$16$$ $$17$$ Note that if we add an odd number to $$59,$$ we will get an even number. To make the other two sums even numbers, the primes on the back of $$44$$ and $$38$$ must both be even. They also have to be different, however. Since there is only one even prime, the number on the back of $$59$$ must be an even prime, namely $$2.$$ This means that the sum of the front and back of", "3) ? 3. $ \\frac{(2+7-1)!}{7!(2-1)!} + \\frac{(2+5-1)!}{5!(2-1)!} + \\frac{(2+3-1)!}{3!(2-1)!} = 8 +6 +4 = 18 $ I used 1 less case than part 2) , in order to cover the \"None of the boxes are empty\" case 1 = odd number 1 case 2 = odd number 3 case 3 = odd number 5 case 4 = odd number 7 (4 cases) . That means that in part 2) i used 4 cases, since there was no restriction, and for part 3) i used 3 cases (for odd numbers 1/3/5) in order the cover the restriction \"None of the boxes are empty\" Where am i mistaken? thanks in advance =) 4. I count 12 With 1 in the 3rd box, the other 2 box's contents must sum to 7 o|oooooo The red line can be placed in 6 positions. With 3 in the 3rd box, the others must sum to 5 o|oooo The red line can have 4 positions. With 5 in the 3rd box, the others must sum to 3 o|oo the red line has only 2 positions. The possibilities are.... 3rd box contains 1 611 161 521 251 431 341 3rd box contains 3 413 143 323 233 3rd box contains 5 125 215 5. Bah stupid me, bad momentum from the 2) question. True you", "number, or those is an odd // position for card numbers whose length is an odd number. Instead of counting two sums (of odd and even digits) and returning their difference, you could use a single sum value, subtracting a digit's value if it's even and adding if it's odd. Output Format: The odd and even numbers interlaced, each separated by a space. Question 253193: Two odd integers and one even integer are multiplied together. If the right words/tokens are used for simple questions (which this is) then you should get a solution in the first page. the version between that concern and the challenge you want to sparkling up is amazingly small: you want to bypass the a number of numbers. Input = 1479386 Output: Odd Sum = (1+7+3+6) =17 Even Sum = (4+9+8) = 21 [CODE]import java. Goldbach's conjecture – and more: Even numbers n > 4208 are sums of two primes (twins!) Prime twins (or twin primes) are odd prime numbers that come in pairs: {3, 5}, {11, 13}, {17, 19}, {29, 31}. A 4 by 4 magic square is a doubly even magic square, one of the three types of magic square. You can learn more tutorials here and Java interview questions for beginners. Sum of positive and negative numbers. Here is the complete", "8, 10, 12, 14, 16, 18, 20, 22, 24 i.e. number of favourable outcomes = 12 Probability of an even number will be P(E) = $$\\\\ \\frac { 12 }{ 25 }$$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 25 }$$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 25 }$$ Question 26. A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Solution: Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 Number of odd numbers = 8 Probability of odd numbers will be", "box. Jack picked up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd? A. $$\\frac{1}{150*50}$$ B. $$\\frac{2}{150*50}$$ C. $$\\frac{1}{54}$$ D. $$\\frac{1}{27}$$ E. $$\\frac{3}{8}$$ To access all the questions: Question of the Week: Consolidated List Actually if you understand the question, you would know that the denominator has to be a multiple of 4 and with just 2s, that is 2^n Only E has 8.. Let's solve it and why denominator should be in form of 2^n Number of prime digits - 2,3,4,7 so 4 of them So each of the three digit can be filled with any of three.. Total ways one can select - 4*4*4 and similarly second in 4*4*4 ways.. Combined 4*4*4*4*4*4 and this will come in denominator and that is why the denominator should have just 2s Only way the sum could be odd is when one ends with 2 and other with a odd prime number.. First chooses - 4*4*1.......1 because only 2 can be placed Second chooses - 4*4*3.... 3 because 2 is not included Total 4*4*1*4*4*3 *2.... 2", "up one ticket from the box, noted the number, and put the ticket back in the box. Then Rose came and did the same. If both picked tickets where all the digits within the picked number are distinct and prime, then find the probability that the sum of the two ticket numbers is odd? A. $$\\frac{1}{150*50}$$ B. $$\\frac{2}{150*50}$$ C. $$\\frac{1}{54}$$ D. $$\\frac{1}{27}$$ E. $$\\frac{3}{8}$$ To access all the questions: Question of the Week: Consolidated List Given the consecutive #'s from 100 to 999, as the #'s on a ticket. Jack & Rose each pick a # which has distinct primes as its digits. Hence the digits are {2,3,5,7} Total 3 digit #'s with distinct primes as its digits are = 4*3*2 = 24 Now for the sum of the #'s picked to be Odd, we need one # as even & other as odd, hence the numbers can be _ _ 2 & _ _ O One # has units digit as even & other is an # 3 digit #. There are (3*2*1) = 6 such even #'s & (2*3*3) = 18 Now Case 1 consider Jack picks the even # & Rose picks the Odd #, hence the probability = (6/24)*(18/24) = 3/16 Now Case 2 consider Rose picks the even # & Jack picks the", "to find sum of even numbers. It can be expressed as either the sum of six consecutive prime numbers: 19 + 23 + 29 + 31 + 37 + 41, or the sum of eight consecutive prime numbers: 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37. These numbers form an A.P. 1-50 1-100 1-500 1-1000 Odd Even List Randomizer Random Numbers Number Converters. Sum of first four odd numbers = 1 + 3 + 5 + 7 = 16 (16 = 4 x 4). Logic to find sum of even numbers. FAQ. Sum of first three odd numbers = 1 + 3 + 5 = 9 (9 = 3 x 3). Answer: The two numbers are 15 and 3, since 15 times 3 is 45, and 15 added to 3 is 18. Sol: 1 + 2 + 3+ 4+ 5+ ———-+50 So Here n = 50 = 50 ( 50+1) / 2 = 25 x 51 = 1275 The difference = 290. Product means you need to multiply the numbers (-5 multiplied by -12 = 60). Do I need to enter plus (+) sign between two numbers? To get the average, notice that the numbers are all equally distributed. If we have 100 numbers (1…100), then we clearly have 100 items.", "the odd numbers from 1 to 1000 in your notebook. bonus cards : 1) cards with odd numbers like 1,3, 5 etc 2) cards with even numbers like 2,4,6 etc 3) cards with count by twos like 2,4 etc. Odd facts are limitless but we tried our best to compile the 80 most interesting facts that you probably didn't know. Now, to check whether num is even or odd, we calculate its remainder using % operator and check if it is divisible by 2 or not. Question 78 Cube of an odd number is even. The trading volume hit $1 million in the first six. All Factors of Numbers 1-100. 100 is an average of odd numbers between 1 and 200 mentioned in the below table, by substituting the total sum and count of numbers in the below formula. Odd numbers can NOT be divided evenly into groups of two. , wholesale. Explain that if the number of counters can be arranged so each counter is. A box contains 100 balls, numbered from 1 to 100. 2 is an even number. 5 : Find the sum of consecutive odd numbers 51 +53 +55 + ———+ 199. In China, it is customary to regard even numbers as being more auspicious than odd ones. Very challenging. Flowchart to Add two", "19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 25 }$$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 25 }$$ Question 26. A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Solution: Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 Number of odd numbers = 8 Probability of odd numbers will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 8 }{ 15 }$$ (ii) Prime number are 2, 3, 5, 7, 11, 13 Number of primes is 6 Probability of prime number will be", "$$\\frac{1}{8}$$ => P(B) = 1- P($$\\overline{B}$$) = $$\\frac{7}{8}$$ ### Solve problem 9.7 page 86 Math textbook 10 Connecting knowledge volume 2 A box of cards numbered 10; 11; ….; 20. Randomly draw two cards from a box. Calculate the probability of the following events: a) C: “Both draw cards have odd numbers”; b) D: “Both cards drawn have even numbers”. Solution method Calculate $$n(\\Omega )$$ – The two cards drawn are odd numbers, so the two cards drawn belong to set {11; 13; 15; 17; 19} => Number of ways to choose – The two cards drawn have even numbers, so the two cards drawn belong to set {10; twelfth; 14; 16; 18; 20} => Number of ways to choose Detailed explanation Draw two cards from 11 cards with the number of ways: $$C_{11}^{2}=55$$ or $$n(\\Omega )$$ = 55. a) Both cards drawn are odd numbers, so 2 cards drawn belong to set {11; 13; 15; 17; 19}. => The number of ways to choose is: $$C_{5}^{2}=10$$. So P(C) = $$\\frac{10}{55}=\\frac{2}{11}$$. b) Both cards drawn have even numbers, so 2 cards drawn belong to set {10; twelfth; 14; 16; 18; 20} => Number of ways to choose is: $$C_{6}^{2}=15$$. So P(D) = $$\\frac{15}{55}=\\frac{3}{11}$$. ### Solve problem 9.8, page 86, Math textbook 10, Connecting knowledge volume 2 A box", "all the cards is $$59 + 2 = 61.$$ This makes the other two primes $$61 - 44 = 17$$ and $$61- 38 = 23.$$ The average of the primes is therefore $\\dfrac{2 + 17 + 23}{3} = \\dfrac{42}{3} = 14.$", "= 0.75$, $P(D) = 0.3$, $P(C|D) = 0.75$ and $P(C \\text{ AND } D) = 0.225$. 1. Are $C$ and $D$ independent? 2. Are $C$ and $D$ mutually exclusive? 3. What is $P(D|C)$? ### try it A student goes to the library. Let events $B$ = the student checks out a book and $D$ = the student checks out a DVD. Suppose that $P(B) = 0.40$, $P(D) = 0.30$ and $P(B \\text{ AND } D) = 0.20$. 1. Find $P(B|D)$. 2. Find $P(D|B)$. 3. Are $B$ and $D$ independent? 4. Are $B$and $D$ mutually exclusive? ### Example In a box there are three red cards and five blue cards. The red cards are marked with the numbers $1, 2$, and $3$, and the blue cards are marked with the numbers $1, 2, 3, 4$, and $5$. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let $R$ = red card is drawn, $B$ = blue card is drawn, $E$ = even-numbered card is drawn. The sample space $S = R1, R2, R3, B1, B2, B3, B4, B5$. $S$ has eight outcomes. • $P(R) = \\displaystyle\\frac{{3}}{{8}}$.$P(B) = \\displaystyle\\frac{{5}}{{8}}$. $P(R \\text{ AND } B) = 0$. (You cannot draw one card that is both red and blue.) • $P(E) = \\displaystyle\\frac{{3}}{{8}}$.", "(1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11. ∴ n(A) = 6 × 6 + 5 × 5 = 36 + 25 = 61 ∴ P(A) = $$\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{61}{121}$$ Let B be the event that the numbers tickets drawn are odd ∴ n(B) = 6 × 6 = 36 ∴ P(B) = $$\\frac{\\mathrm{n}(\\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{36}{121}$$ Since 6 odd numbers are common between A and B. ∴ n(A ∩ B) = 6 × 6 = 36 ∴ P(A ∩ B) = $$\\frac{\\mathrm{n}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{36}{121}$$ ∴ Probability of both the numbers are odd, given that sum is even, is given by P(B/A) = $$\\frac{\\mathrm{P}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{P}(\\mathrm{A})}=\\frac{\\frac{36}{121}}{\\frac{61}{121}}=\\frac{36}{61}$$ Question 4. A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as A: a club card is drawn. B: an ace card is drawn. Determine whether events A and B are independent or not. Solution: One card can be drawn out of 52 cards in 52C1 ways. ∴ n(S) = 52C1 Let A be the event that a club card is drawn. 1 club card out of 13 club cards can be drawn in 13C1 ways. ∴ n(A) = 13C1 ∴ P(A) = $$\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{{ }^{13} \\mathrm{C}_{1}}{{ }^{52} \\mathrm{C}_{1}}$$ Let B be the event that an ace card is", "multiple of 5 = 8 (5, 10, 15, 20, 25, 30, 35, 40) Probability = $\\\\ \\frac { 8 }{ 40 }$ = $\\\\ \\frac { 1 }{ 5 }$ (a) Question 22 If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is (a) $\\\\ \\frac { 7 }{ 25 }$ (b) $\\\\ \\frac { 9 }{ 25 }$ (c) $\\\\ \\frac { 11 }{ 25 }$ (d) $\\\\ \\frac { 13 }{ 25 }$ There are 25 number bearing numbers 1, 2, 3,…,25 Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9 Probability being a prime number = $\\\\ \\frac { 9 }{ 25 }$ (b) Question 23 A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is (a) $\\\\ \\frac { 1 }{ 10 }$ (b) $\\\\ \\frac { 9 }{ 100 }$ (c) $\\\\ \\frac { 1 }{ 9 }$ (d) $\\\\ \\frac { 1 }{ 100 }$ In a box, there are 90 cards bearing numbers 1 to 90 Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9 Probability" ]

[ "widgets wholesale: there’s also rent, employee wages, buying all that music, and so on. Because of overhead, your per-widget average markup can’t be too small: it needs to add up to enough to cover your overhead costs. After that, all the rest of the overhead is profit, and as we’ve seen, the profit margin can be quite small if you’re selling lots of widgets. But, because of overhead, even if you’re selling hojillions of widgets, there’s always going to be some amount below which your per-widget markup cannot go. For instance, if your overhead is, say, $500 a day and you’re selling 2500 widgets a day, then you need there to be, on average, at least a twenty-cent markup per widget to cover overhead. So, if you’re selling the 2500 widgets for$5.40 each, then only $0.20 per widget will be profit, and you’ll make$500 in profit a day. $500 in profit is still not bad, but what if you wanted to be really sure to make$1000 in profit per day? One thing you could try is counting that $1000 per day as part of your overhead. Then, your daily overhead would be$1500 (that is, the actual overhead of $500, plus the profit you want to be sure to make), and you could set your average price at$5.60. As", "currently charging$5 for a widget and the average total cost of... Suppose a business is currently charging $5 for a widget and the average total cost of producing a widget is$7. From that information we know that O fixed costs must be really large for the business. O the revenues from selling widgets must be negative o the profit from selling widgets can be negat...", "The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $$100$$ workers can produce $$300$$ widgets and $$200$$ whoosits. In two hours, $$60$$ workers can produce $$240$$ widgets and $$300$$ whoosits. In three hours, $$50$$ workers can produce $$150$$ widgets and $$m$$ whoosits. Find $$m$$. (第二十五届AIME2 2007 第4题)", "and widgets sell for $1 each, then how many workers should the firm hire if the market wage is$10/hr and there are no other materials or costs of production? Answer: The key to answering this question is to compare productivity for a given amount of time. With one worker, the firm's productivity is 100 widgets per hour. With two workers, their combined productivity is 133 widgets per hour. With three workers, their combined productivity is 120 widgets per hour. For each hour of operation with two workers, the firm's net profit is $113 from selling 133 widgets at$1 each and paying $20 in wages. With one or three workers, the firm's net profit is only$90 (either $100-$10 or $120-$30). Therefore, profit is maximized by hiring two workers. Hopefully this gives you a taste of how firms use marginal productivity in their decision making with regard to the factors of production. Firms purchase and hire only up to the point where the marginal contribution to revenue of the next resource unit equals the marginal cost. The following are other real-world examples of markets for factors of production. Labor Market – An online application for job listings and available job candidates can serve as a market for employees. Land Market – An auction for industrial or agricultural land is an", "product. Variable cost per widget is $1.55; full cost is$2.80. Comparable widgets sell on the open market for \\$3.30 each. Division A can produce up to 3.00 mil...", "just and fair and all would be good. And this perception is gravely mistaken. To make that clear to you, I offer a little fable. Once upon a time, widgets were made by hand. One person, working for one eight-hour day, could make 100 widgets. Most people were employed making widgets full-time. The wage for making widgets was$1 per widget. Then, an inventor came up with a way to automate the production of widgets. For $100 per day, the same cost to hire a worker to make 100 widgets, the machine could instead make 101 widgets. Because it was 1% more efficient, businesses began adopting the new machine, and now made slightly more widgets than before. But some workers who had previously made widgets were laid off, while others saw their wages fall to only$0.99 per widget. If there were more widgets, but fewer people were getting paid less to make them, where did the extra wealth go? To the inventor, of course, who now owns 10% of all widget production and has billions of dollars. Later, another inventor came up with an even better machine, which could make 102 widgets in a day. And that inventor became a billionare too, while more became unemployed and wages fell to $0.98 per widget. And then there was another inventor,", "$3 million worth of widgets was$2 million. But I have other expenses. I can't just put that in my pocket. I had marketing, sales, general and administrative. And then I was left with $1 million. And that's how much operating profit my widget company has produced. But I'm not done yet. We learned in the last video I didn't own the company outright. I had some debt. In this case,$5 million worth of debt. So I have to pay some interest. There was 10% interest. So you take the interest out. You're left with $500,000 of pre-tax income. And of course you have to pay the government. And we can debate on whether we're paying them too much or too little. But they're providing me defense so that other foreign countries can't come and bomb my factories. And they're providing, hopefully, an educated workforce for me. And nice roads and infrastructure and electricity. So who knows? So that's what I'm paying for, arguably. And so I'm left with net income of$350,000. And when people talk about earnings of a company, this is what they're talking about. Net income is also earnings. So when someone says, Google made $4 billion this past quarter-- and quarter just means a quarter of a year, a three month period. And they normally", "Company hasn't spent very much money overhauling its machinery over the years, it is still using old technology to manufacture its widgets, which makes the process more costly due to constant maintenance costs. In addition, XYZ Company loses market share every year to its more modern competitors. XYZ Company's working ratio gets larger each year since its expenses go up and its income falls. Recently it rose above 1, and analysts predict it will continue to rise.", "40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only $$\\frac{1}{4}^{th}$$ of the work was completed, then how many more workers should be recruited so that the work gets completed on time? A. 32 B. 100 C. 200 D. 240 E. 280 Let the daily rate per worker = 1 widget per day, implying that the daily rate for 40 workers = 40 widgets per day. In 20 days, the number of widgets produced by 40 workers = (daily rate for 40 workers)(number of days) = 40*20 = 800 widgets. Since these 800 widgets constitute 1/4 of the total job, we get: $$800 = \\frac{1}{4}j$$ $$j = 3200$$ Remaining work = (total job) - (widgets produced in the first 20 days) = 3200 - 800 = 2400. For 2400 widgets to be produced in the remaining 10 days, the required rate $$= \\frac{work}{time} = \\frac{2400}{10} = 240$$ widgets per day. Since the daily rate must increase from 40 widgets per day to 240 widgets per day, the number of additional workers that must be recruited = 200. _________________ GMAT and GRE Tutor Over 1800 followers [email protected] New York, NY If you find one of my posts", "4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We’ll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. Try It Now that we have the basic idea of the cost origins and how they are related to production, let’s drill down into the details, by examining average, marginal, fixed, and variable costs. Average and Marginal Costs The cost of producing a firm’s output depends on how much labor and capital the firm uses. A list of the costs involved in producing cars will look very different from the costs involved in producing computer software", "lube content to them. We currently are selling the widgets for 3.00 each. The lube costs .50 cents for each ...", "to maximize their profit? 2. A company can produce a maximum of 25 widgets in a day. If they sell x widgets during the day then their profit, in dollars, is given by, $P\\left( x \\right) = 3000 - 40x + 11{x^2} - \\frac{1}{3}{x^3}$ How many widgets should they try to sell in order to maximize their profit? 3. A management company is going to build a new apartment complex. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. will be, $C\\left( x \\right) = 70,000 + \\frac{{2736}}{5}x - \\frac{{211}}{{50}}{x^2} + \\frac{1}{{150}}{x^3}$ The land they have purchased can hold a complex of at most 400 apartments. How many apartments should the complex have in order to minimize the maintenance costs? 4. The production costs of producing x widgets is given by, $C\\left( x \\right) = 2000 + 4x + \\frac{{90,000}}{x}$ If the company can produce at most 200 widgets how many should they produce to minimize the production costs? 5. The production costs, in dollars, per day of producing x widgets is given by, $C\\left( x \\right) = 400 - 3x + 2{x^2} + 0.002{x^3}$ What is the marginal cost when $$x = 20$$ and $$x = 75$$? What do your answers tell you about the production costs? 6. The", "to produce 1 million widgets is far below \\$1 . . .", "2. Economics 3. 1 use the following information to answer the next three... # Question: 1 use the following information to answer the next three... ###### Question details 1. Use the following information to answer the next three questions: Widgets Inc. can hire workers at $40 per day and can rent a unit of equipment for$10 per day. Currently Widgets Inc. spends $200 per day on input a- Draw an isocost curve above for Widgets Inc. spending$200. (label the intercepts) b- What is the slope of the isocost curve? c- If Widgets Inc. hires 4 workers and rents 4 pieces of equipment, then the marginal product of labor = 20 and the marginal product of capital = 10. Given this information, is Widgets Inc. optimizing? If not, then what should they do? 2. Currently Moe’s hires 2 employees, has 1 grill (capital), and produces 20 meals per day. If Moe’s decides to double both inputs (employees and grills). If Moe’s experiences decreasing returns to scale how many meals do you expect to produce after inputs double?", "### Technological unemployment is hiding in plain sight I've just published a new feature-length article on Medium centered around the chart below. Like my blog? Please subscribe and also consider . . . Posted in: automationmedium October 24, 2017 # Basic Income is JUST Compensation ### Recognizing what we owe each other Let's say the cost to produce a widget is $1. What's the cost to produce 1 million widgets? This may sound like an extremely simple word problem that even some preschoolers could solve. However, if you think the answer is$1 million, you would be entirely wrong. The cost to produce 1 million widgets is far below \\$1 . . .", "# Weighted average of multiple weighted factors Lets say a factory machine runs as follows: 1. Day 1: $3$ widgets per minute $\\times 1000$ minutes $\\times 1$ kg per widget $= 3000$ kg. 2. Day 2: $5$ widgets per minute $\\times 800$ minutes $\\times 1.5$ kg per widget $= 6000$ kg. 3. Day 3: $8$ widgets per minute $\\times 300$ minutes $\\times 0.5$ kg per widget $= 1200$ kg. So, over the $3$ days the factory machine has run for $2100$ minutes. I am assuming that if I multiply the weighted average number of widgets with the weighted average kg per widget and multiply this by the number of minutes over the three days $(2100)$ this should equal $10,200$ kg $(3000+6000+1200)$. My weighting calculation is below for widgets per minute and kg per widget. $$(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5)+(8\\cdot300\\cdot0.5)/(1000\\cdot1)+(800\\cdot1.5)+(300\\cdot0.5) = 4.833$$ widgets per minute. $$(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5)+(8\\cdot300\\cdot0.5)/(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5) = 1.085$$ kg per widget. However $4.833$ widgets $\\times 1.085$ kg per widget $\\times 2100$ mins $= 11,012$ kg over the three days. Not the correct $10,200$kg. Why is this the case? I have tried weighting by total for the day and this still does not produce the correct answer. I have been using Excel and have noticed that if I keep either \"widgets per minute\" or \"kg per widget\" the same across all three days and", "pay its employees \\$ 2 more per hour store... Called net sales that the firm takes in as gross profit is the winner markup in dollars bids vs. costs!", "sorts of maintenance costs and that the more tenants renting apartments the more the maintenance costs will be. With this analysis we can see that, for this complex at least, something probably needs to be done to get the maximum profit more towards full capacity. This kind of analysis can help them determine just what they need to do to move towards that goal whether it be raising rent or finding a way to reduce maintenance costs. Note as well that because most apartment complexes have at least a few units empty after a tenant moves out and the like that it’s possible that they would actually like the maximum profit to fall slightly under full capacity to take this into account. Again, another reason to not just assume that maximum profit will always be at the upper limit of the range. Let’s take a quick look at another problem along these lines. Example 2 A production facility is capable of producing 60,000 widgets in a day and the total daily cost of producing $$x$$ widgets in a day is given by, $C\\left( x \\right) = 250,000 + 0.08x + \\frac{{200,000,000}}{x}$ How many widgets per day should they produce in order to minimize production costs? Problem Statement. Show Solution Here we need to minimize the cost subject to", "Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 05 Sep 2018, 01:00 Bunuel wrote: If w widgets cost c cents, how many widgets can you get for d dollars? (A) $$\\frac{100dw}{c}$$ (B) $$\\frac{dw}{100c}$$ (C) $$100cdw$$ (D) $$\\frac{dw}{c}$$ (E) $$cdw$$ Unit cost of widgets=$$\\frac{c}{w}$$ cents Or, $$\\frac{c}{100w}$$ dollars are required to purchase 1 widget Or, $1 is required to purchase $$\\frac{100w}{c}$$widgets hence$d is required to purchase $$\\frac{100dw}{c}$$ Ans. (A) _________________ Regards, PKN Rise above the storm, you will find the sunshine Senior Manager Joined: 07 Jul 2012 Posts: 370 Location: India Concentration: Finance, Accounting GPA: 3.5 Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 05 Sep 2018, 01:30 For c cents, w widgets can be bought ......1 cent.., $$\\frac{w}{c,}$$ widgets can be bought .......100d cents...., $$\\frac{w}{c}$$*100d widgets can be bought = $$\\frac{100dw}{c}$$ _________________ Kindly hit kudos if my post helps! Senior Manager Joined: 29 Dec 2017 Posts: 382 Location: United States Concentration: Marketing, Technology GMAT 1: 630 Q44 V33 GMAT 2: 690 Q47 V37 GPA: 3.25 WE: Marketing (Telecommunications) If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 15 Sep 2018, 08:40 One of my", "$L=g(Q)$ Table 3. Widgets Produced by Workers Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We'll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. #### TRY IT Consider a firm with the following production schedule: Assume that each worker gets paid $15 per hour and that other cost of materials is$2 per item produced. Estimate the cost of producing 45 units. It is ________. Number of Workers Production (units/hr) 1 10 2 30 3 45" ]

[ "and marginal cost of widgets to the value of what a worker produces and what the firm is willing to pay, imagine that an extra hour of work leads to 10 extra widgets being produced. Then at $2 per widget, the value of the extra production from an extra hour of work is$20: the price times the 10 widgets per hour marginal physical product of labor. But since selling the extra widgets would require cutting the price of widgets or advertising, the firm only nets $1.60 in marginal revenue after cutting prices to sell an extra widget, so it is only willing to pay$16 an hour: the marginal revenue of $1.60 a widget times the 10 widgets per hour marginal physical product of labor. (Actually, although this is usually called the marginal revenue product, when marginal revenue is equal to marginal cost, one can also call this the \"marginal cost product.\" And in the short-run when prices are sticky it is the marginal cost product (the marginal cost times the marginal physical product of labor), not the marginal revenue product (marginal revenue times the marginal physical product of labor) that matters. Indeed, while a price is sticky, marginal revenue doesn't do anything at all. Marginal revenue only matters when a firm is in the process of adjusting its", "and widgets sell for $1 each, then how many workers should the firm hire if the market wage is$10/hr and there are no other materials or costs of production? Answer: The key to answering this question is to compare productivity for a given amount of time. With one worker, the firm's productivity is 100 widgets per hour. With two workers, their combined productivity is 133 widgets per hour. With three workers, their combined productivity is 120 widgets per hour. For each hour of operation with two workers, the firm's net profit is $113 from selling 133 widgets at$1 each and paying $20 in wages. With one or three workers, the firm's net profit is only$90 (either $100-$10 or $120-$30). Therefore, profit is maximized by hiring two workers. Hopefully this gives you a taste of how firms use marginal productivity in their decision making with regard to the factors of production. Firms purchase and hire only up to the point where the marginal contribution to revenue of the next resource unit equals the marginal cost. The following are other real-world examples of markets for factors of production. Labor Market – An online application for job listings and available job candidates can serve as a market for employees. Land Market – An auction for industrial or agricultural land is an", "$300$ hours. How many staff hours will the whole project take? Two different methods will be given. Method 1. $15\\%$ of the project takes $300$ hours, therefore $\\phantom{Method 1.}1\\%$ of the project takes $\\frac{300}{15} = 20$ hours (divide both numbers by $15$), so $\\phantom{Method 1.}100\\%$ of the project takes $20 \\times100 = 2000$ hours (multiply both by $100$). Method 2. Write $n$ for the total number of hours needed by staff for the whole project. That $15\\%$ of the total number of hours staff need is $300$ tells us that $\\frac{15}{100}\\times n = 300$. We must solve for $n$: $\\frac{15n}{100} = 300$, so $15n = 100\\times 300$ and therefore $n = \\frac{100\\times 300}{15}= 2000$ hours. 4. ACME Products pays $\\$62.50$for each widget they buy from the importer. They sell each widget for$\\$80.00$. Finding the percentage profit as a proportion of cost price. The profit for each widget is $\\$80.00 - \\$62.50 = \\$17.50$. The profit as a proportion of cost price is$\\frac{17.50}{62.50}$so the percentage profit as a proportion of cost price is$\\frac{17.50}{62.50}\\times100 = 28\\%\\$.", "4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We’ll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. Try It Now that we have the basic idea of the cost origins and how they are related to production, let’s drill down into the details, by examining average, marginal, fixed, and variable costs. Average and Marginal Costs The cost of producing a firm’s output depends on how much labor and capital the firm uses. A list of the costs involved in producing cars will look very different from the costs involved in producing computer software", "equation we have $N$N $=$= $60\\log_216-50$60log2​16−50 $=$= $60\\times4-50$60×4−50 $=$= $240-50$240−50 $=$= $190$190 So there will still be $190$190 widgets produced each month. • If the average salary is $\\$500$$500 per month, what is the expected number of widgets produced? By substituting S=500S=500 into the equation we have NN == 60\\log_2\\left(500+16\\right)-5060log2​(500+16)−50 == 60\\log_2516-5060log2​516−50 == 491491 (to nearest whole widget) The warehouse will produce around 490490 widgets if the employees are given \\500$$500 per month. • How many more widgets will be produced if the monthly salary is doubled to $\\$1000$$1000? First we can find the number of widgets produced when the salary is \\1000$$1000, then we can subtract from this the $491$491 widgets expected when the salary is only $\\$500$$500. NN == 60\\log_2\\left(1000+16\\right)-5060log2​(1000+16)−50 == 60\\log_21016-5060log2​1016−50 == 549549 (to nearest whole widget) By doubling the salary, the warehouse will produce and extra 549-491=58549491=58 widgets. #### Practice question A major communications company found that the more they spend on advertising, the higher their revenue. Their sales revenue, in thousands of dollars, is given by R=10+20\\log_4\\left(x+1\\right)R=10+20log4(x+1), where xx represents the amount they spend on advertising (in thousands of dollars). 1. Determine their sales revenue if they spend no money on advertising. 2. Determine their sales revenue if they spend \\14000$$14000 on advertising. Give your answer to the nearest thousand $dollars$dollars. 3. Would", "$L=g(Q)$ Table 3. Widgets Produced by Workers Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We'll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. #### TRY IT Consider a firm with the following production schedule: Assume that each worker gets paid $15 per hour and that other cost of materials is$2 per item produced. Estimate the cost of producing 45 units. It is ________. Number of Workers Production (units/hr) 1 10 2 30 3 45", "would do 8 divided by 2 to get radius of 4. Then do 3.14($4^{2}$) and round to get your answer. Yes, question 5 is height times length times width, which would be (13)(5)(6) to get 346 s squared. let met look at the rest Question 8 is v=lwh but you have the base area. So you would do area times height which is (12)(8), which gets you the answer as 96 cm cubed Question 7 you need to find the surface area, surface area is (l)(w)(h), so (8)(2)(2), so 72 should be your answer 4. shmote12 says: 9.375. do 2.5 times 3.75 for the answer 5. kukisbae says: Firm should hire the 4th worker as MR > MC. Explanation: Here, we are comparing the marginal cost of hiring 4th worker with the revenue generated by the 4th worker. Marginal cost of hiring 4th worker: = Total cost with 4 workers - Total cost with 3 workers =$4,600 - $4,000 =$600 Total revenue generated by the 4th worker: = Number of units produced by 4th worker × Price of each unit = 50 × $15 =$750 Therefore, the firm should hire the 4th worker as the marginal revenue of 4th worker is greater than its marginal cost. 6. danam04 says: Message me I’ll help Explanation: 7. tankhill5534 says: Triangle:", "under here is another$10, under this green curve. Does it make sense to hire that 1st person? Sure, you're going to get $25 in revenue, and you're going to have$10 in wages, so you're going to have $15 of benefit. You definitely want to hire one person. Does it make sense to hire that 2nd person? Well, the marginal product revenue is going to be$20 from that 2nd person. Your marginal cost from that 2nd person is $10, so you're going to make$10 on that 2nd person, so you should hire that 2nd person. The 3rd person, you're going to make $15 off them, they're going to cost you$10 dollars, so you're going to, and this is all on a per-hour basis. You're going to make $5, so you should hire that. The 4th person, well, now it's interesting. The 4th person, you're going to make$10 in total off of the 4th person, but you're not, but you're also going to have to pay $10 for that 4th person, so it doesn't make sense for you to hire another total 4th person. Now, if you could, it could make sense for you to hire another half person, maybe someone who shows up every other day, if this still holds, or maybe someone who shows up for half an", "Run, was based on the following production function, which is similar to Table 7.3 except for \"widgets\" instead of trees. Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Table 7.4 We can use the information from the production function to determine production costs. What we need to know is how many workers are required to produce any quantity of output. If we flip the order of the rows, we “invert” the production function so it shows $L=f(Q)L=f(Q)$. Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Table 7.5 Now focus on the whole number quantities of output. We’ll eliminate the fractions from the table: Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Table 7.6 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 Table 7.7 This is same cost function with which we began! (shown in Table 7.3) Now that we have the", "2. Economics 3. 1 use the following information to answer the next three... # Question: 1 use the following information to answer the next three... ###### Question details 1. Use the following information to answer the next three questions: Widgets Inc. can hire workers at $40 per day and can rent a unit of equipment for$10 per day. Currently Widgets Inc. spends $200 per day on input a- Draw an isocost curve above for Widgets Inc. spending$200. (label the intercepts) b- What is the slope of the isocost curve? c- If Widgets Inc. hires 4 workers and rents 4 pieces of equipment, then the marginal product of labor = 20 and the marginal product of capital = 10. Given this information, is Widgets Inc. optimizing? If not, then what should they do? 2. Currently Moe’s hires 2 employees, has 1 grill (capital), and produces 20 meals per day. If Moe’s decides to double both inputs (employees and grills). If Moe’s experiences decreasing returns to scale how many meals do you expect to produce after inputs double?", "The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $$100$$ workers can produce $$300$$ widgets and $$200$$ whoosits. In two hours, $$60$$ workers can produce $$240$$ widgets and $$300$$ whoosits. In three hours, $$50$$ workers can produce $$150$$ widgets and $$m$$ whoosits. Find $$m$$. (第二十五届AIME2 2007 第4题)", "\\rightarrow & 120 \\ \\mbox{tm} & \\rightarrow & 2 \\ \\mbox{days} \\\\\\\\ 7\\ \\mbox{trucks} & \\rightarrow & x \\ \\mbox{tm} & \\rightarrow & 3 \\ \\mbox{days} \\\\\\\\ & d & & d & \\end{eqnarray}$$ where $$d$$ expresses a direct relation. But it can be the case that the relation between the magnitudes is not direct, but inverse.In these cases, as with the simple rules of three, we will invert those fractions that have an inverse relation with the unknown. It takes $$15$$ days for a team of $$10$$ workers working $$8$$ hours a day to finish an order. How many persons with part time jobs will be needed to make the same work in $$10$$ days? The first thing to do is to make the scheme of the rule of three analyzing the relations between the magnitudes: $$\\begin{eqnarray} & i & & i & \\\\\\\\ 10\\ \\mbox{workers} & \\rightarrow & 8 \\ \\mbox{hours} & \\rightarrow & 15 \\ \\mbox{days} \\\\\\\\ x\\ \\mbox{workers} & \\rightarrow & 4 \\ \\mbox{hours} & \\rightarrow & 10 \\ \\mbox{days} \\\\\\\\ & i & & i & \\end{eqnarray}$$ So, the relation of the unknown with the rest of magnitudes is inverse (i): the more persons, the less hours will have to be worked to finish the order; and less days will require more persons", "widgets wholesale: there’s also rent, employee wages, buying all that music, and so on. Because of overhead, your per-widget average markup can’t be too small: it needs to add up to enough to cover your overhead costs. After that, all the rest of the overhead is profit, and as we’ve seen, the profit margin can be quite small if you’re selling lots of widgets. But, because of overhead, even if you’re selling hojillions of widgets, there’s always going to be some amount below which your per-widget markup cannot go. For instance, if your overhead is, say, $500 a day and you’re selling 2500 widgets a day, then you need there to be, on average, at least a twenty-cent markup per widget to cover overhead. So, if you’re selling the 2500 widgets for$5.40 each, then only $0.20 per widget will be profit, and you’ll make$500 in profit a day. $500 in profit is still not bad, but what if you wanted to be really sure to make$1000 in profit per day? One thing you could try is counting that $1000 per day as part of your overhead. Then, your daily overhead would be$1500 (that is, the actual overhead of $500, plus the profit you want to be sure to make), and you could set your average price at$5.60. As", "to (Figure) except for “widgets” instead of trees. Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 We can use the information from the production function to determine production costs. What we need to know is how many workers are required to produce any quantity of output. If we flip the order of the rows, we “invert” the production function so it shows $L=f\\left(Q\\right)$. Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We’ll eliminate the fractions from the table: Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive ?10 per hour. Multiplying the Workers row by ?10 (and eliminating the blanks) gives us the cost of producing different levels of output. Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour ?10 ?10 ?10 ?10 = Cost ?32.50 ?44.00 ?52.00 ?90.00 This is same cost function with which we began! (shown in (Figure)) Now that we have the basic idea of the cost origins and how they are related to production, let’s drill down", "costs of \\$150 per widget (so it costs \\$50 to produce 1 widget). Let N be the number of widgets produced in a month. a. Find a formula for the manufacturer's total cost C as a function of N. $C=55N+200$ b. The manufacturer sells the widgets for \\$65 each. FInd a formula for the total revenue R as a function of N. $R=58N$ c. Use your answers to parts a and b to find a formula for the profit P of this manufacturer as a function of N. $P=58N-(55N+200)$ or equivalently $3N-200$. d. Use your formula from part c to determine the break-even point for this manufacturer. 66.67 widgets per month It doesn't help me much... I don't know how they got 66.67 and why the math problem I'm having trouble with has two break even points and not just one.", "40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only $$\\frac{1}{4}^{th}$$ of the work was completed, then how many more workers should be recruited so that the work gets completed on time? A. 32 B. 100 C. 200 D. 240 E. 280 Let the daily rate per worker = 1 widget per day, implying that the daily rate for 40 workers = 40 widgets per day. In 20 days, the number of widgets produced by 40 workers = (daily rate for 40 workers)(number of days) = 40*20 = 800 widgets. Since these 800 widgets constitute 1/4 of the total job, we get: $$800 = \\frac{1}{4}j$$ $$j = 3200$$ Remaining work = (total job) - (widgets produced in the first 20 days) = 3200 - 800 = 2400. For 2400 widgets to be produced in the remaining 10 days, the required rate $$= \\frac{work}{time} = \\frac{2400}{10} = 240$$ widgets per day. Since the daily rate must increase from 40 widgets per day to 240 widgets per day, the number of additional workers that must be recruited = 200. _________________ GMAT and GRE Tutor Over 1800 followers [email protected] New York, NY If you find one of my posts", "2,000 thirds, 2,000 thirds, that's how many hours of labor we want. So, evidently in our original problem, where we have this model for our revenue function for our Widgets with $20 per hour of labor, and$2,000 per ton of steel, with a budget of \\$20,000, the allocation that you should make is to buy 10 thirds of a ton of steel and 2,000 thirds hours of labor.", "of hours per month of labor that will be handled by regular employees. To minimize total labor costs, one needs to solve the following optimization problem: $$\\text{minimize}~\\sum_i\\bigl(25\\max(a_i − x,0) + 17.50x\\bigr)$$ • Show how to reformulate this problem as a linear programming problem. • Solve the problem for the specific data [given below]. • Use calculus to find a formula giving the optimal value for $x$. #### Part (a)¶ Let $w_i = \\max(a_i - x)$, the number of extra workers hired in each month. Then the objective function is $$\\zeta = \\sum_{i=1}^n (17.50 x + 25 w_i ) = 17.50 n (x) + \\sum_{i=1}^n 25(w_i)$$ and the problem can be rewritten as \\begin{align} &\\text{minimize} & \\zeta & = 17.50 n (x) + \\sum_{i=1}^n 25(w_i) \\\\ &\\text{subject to} & x + w_1 & \\geq a_1 \\\\ && x+ w_2 & \\geq a_2 \\\\ && & ~~\\vdots \\\\ && x+ w_n & \\geq a_n\\\\ && x, w & \\geq 0 \\end{align} which is basically standard form. #### Part (b)¶ For the sample data, $n=12$, and the optimal solution has $x = 340$. In [321]: c = np.array([17.5*12, # x 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25]) # w_i A_lb = np.hstack([np.ones((1,12)).T, np.eye(12)]) # Sample data (monthly demand in hours) b_lb = np.array([390, 420, 340, 320,", "workers generally. Below, I lay out a laundry list of potential economic explanations for why the portfolio/freelancing approach is not more common. What's interesting to me both academically and as someone working at oDesk is that many of these points are not set-in-stone attributes of the productive process but are instead things that smart features or policies might change. Non-linearity in costs of searching/vetting/bargaining Hiring a freelancer for a small project is like picking out a fancy restaurant; hiring a full time employee is more like buying a house. The effort of searching and vetting (and thus the cost) is related to the stakes of the hire. However, there is no guarantee that those costs scale linearly with the stakes. Suppose it takes nearly as much effort to find a small job as it does to find a large job---then a portfolio approach will generate larger search costs per dollar earned in wages [1]. Non-linearity in job size and productivity If you can make X widgets or Y schwidgets in 1 hour, it doesn't mean you can make X/2 widgets and Y/2 schwidgets in 1 hour. Every job has some fixed set-up costs---getting out the materials, remembering the key details, etc. The larger the costs, the less attractive the small job. On the other hand, productivity eventually wanes", "hardly the norm of workers generally. Below, I lay out a laundry list of potential economic explanations for why the portfolio/freelancing approach is not more common. What's interesting to me both academically and as someone working at oDesk is that many of these points are not set-in-stone attributes of the productive process but are instead things that smart features or policies might change. Non-linearity in costs of searching/vetting/bargaining Hiring a freelancer for a small project is like picking out a fancy restaurant; hiring a full time employee is more like buying a house. The effort of searching and vetting (and thus the cost) is related to the stakes of the hire. However, there is no guarantee that those costs scale linearly with the stakes. Suppose it takes nearly as much effort to find a small job as it does to find a large job---then a portfolio approach will generate larger search costs per dollar earned in wages [1]. Non-linearity in job size and productivity If you can make X widgets or Y schwidgets in 1 hour, it doesn't mean you can make X/2 widgets and Y/2 schwidgets in 1 hour. Every job has some fixed set-up costs---getting out the materials, remembering the key details, etc. The larger the costs, the less attractive the small job. On the other" ]

[ "widgets wholesale: there’s also rent, employee wages, buying all that music, and so on. Because of overhead, your per-widget average markup can’t be too small: it needs to add up to enough to cover your overhead costs. After that, all the rest of the overhead is profit, and as we’ve seen, the profit margin can be quite small if you’re selling lots of widgets. But, because of overhead, even if you’re selling hojillions of widgets, there’s always going to be some amount below which your per-widget markup cannot go. For instance, if your overhead is, say, $500 a day and you’re selling 2500 widgets a day, then you need there to be, on average, at least a twenty-cent markup per widget to cover overhead. So, if you’re selling the 2500 widgets for$5.40 each, then only $0.20 per widget will be profit, and you’ll make$500 in profit a day. $500 in profit is still not bad, but what if you wanted to be really sure to make$1000 in profit per day? One thing you could try is counting that $1000 per day as part of your overhead. Then, your daily overhead would be$1500 (that is, the actual overhead of $500, plus the profit you want to be sure to make), and you could set your average price at$5.60. As", "and widgets sell for $1 each, then how many workers should the firm hire if the market wage is$10/hr and there are no other materials or costs of production? Answer: The key to answering this question is to compare productivity for a given amount of time. With one worker, the firm's productivity is 100 widgets per hour. With two workers, their combined productivity is 133 widgets per hour. With three workers, their combined productivity is 120 widgets per hour. For each hour of operation with two workers, the firm's net profit is $113 from selling 133 widgets at$1 each and paying $20 in wages. With one or three workers, the firm's net profit is only$90 (either $100-$10 or $120-$30). Therefore, profit is maximized by hiring two workers. Hopefully this gives you a taste of how firms use marginal productivity in their decision making with regard to the factors of production. Firms purchase and hire only up to the point where the marginal contribution to revenue of the next resource unit equals the marginal cost. The following are other real-world examples of markets for factors of production. Labor Market – An online application for job listings and available job candidates can serve as a market for employees. Land Market – An auction for industrial or agricultural land is an", "to maximize their profit? 2. A company can produce a maximum of 25 widgets in a day. If they sell x widgets during the day then their profit, in dollars, is given by, $P\\left( x \\right) = 3000 - 40x + 11{x^2} - \\frac{1}{3}{x^3}$ How many widgets should they try to sell in order to maximize their profit? 3. A management company is going to build a new apartment complex. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. will be, $C\\left( x \\right) = 70,000 + \\frac{{2736}}{5}x - \\frac{{211}}{{50}}{x^2} + \\frac{1}{{150}}{x^3}$ The land they have purchased can hold a complex of at most 400 apartments. How many apartments should the complex have in order to minimize the maintenance costs? 4. The production costs of producing x widgets is given by, $C\\left( x \\right) = 2000 + 4x + \\frac{{90,000}}{x}$ If the company can produce at most 200 widgets how many should they produce to minimize the production costs? 5. The production costs, in dollars, per day of producing x widgets is given by, $C\\left( x \\right) = 400 - 3x + 2{x^2} + 0.002{x^3}$ What is the marginal cost when $$x = 20$$ and $$x = 75$$? What do your answers tell you about the production costs? 6. The", "#### Question Between 2000 and 2010, ACME Widgets sold widgets at a continuous rate of $R=R_{0} e^{0.125 t}$ widgets per year, where t is time in years since January 1, 2000. Suppose they were selling widgets at a rate of 1000 per year on January 1, 2000. How many widgets did they sell between 2000 and 2010? How many did they sell if the rate on January 1, 2000 was 1,000,000 widgets per year? Verified #### Step 1 1 of 2 $R(t)=R_0\\,e^{0.125t} \\quad \\& \\quad R_0=1000$ $\\Rightarrow \\quad \\text {Rate of selling: } \\quad \\boxed {R(t)=1000\\,e^{0.125t} }$ The total quantity of widgets sold between $t=0$ (Year 2000.) and $t=10$ (Year 2010.) is given by \\begin{align*} \\int_{0}^{10} 1000\\,e^{0,125t}\\,dt=\\frac {1000} {0.125}\\,e^{0.125t}\\,\\bigg|_{0}^{10}=8000\\,(e^ {1.25 }-1) \\approx \\color{#19804f} \\boxed {\\color {black} 19923 } \\end{align*} If the rate on January 1, 2000. was $10^6$ widgets per year $\\quad \\Rightarrow \\quad R_0=10^6$ $\\Rightarrow \\quad \\text {Rate of selling: } \\quad R(t)=10^6\\,e^{0.125t}$ The total quantity is \\begin{align*} \\int_{0}^{10} 10^6\\,e^{0,125t}\\,dt=\\frac {10^6} {0.125}\\,e^{0.125t}\\,\\bigg|_{0}^{10}=(8\\cdot 10^6)\\cdot(e^ {1.25}-1) \\approx \\color{#4257b2} \\boxed {\\color {black} 19922744 } \\end{align*} ## Create an account to view solutions #### Calculus: Single and Multivariable 6th EditionAndrew M. Gleason, Deborah Hughes-Hallett, William G. McCallum 11,811 solutions #### Calculus: Early Transcendentals 7th EditionJames Stewart 10,070 solutions #### Calculus: Early Transcendentals 8th EditionJames Stewart 11,069 solutions #### Calculus: Early Transcendentals", "costs of \\$150 per widget (so it costs \\$50 to produce 1 widget). Let N be the number of widgets produced in a month. a. Find a formula for the manufacturer's total cost C as a function of N. $C=55N+200$ b. The manufacturer sells the widgets for \\$65 each. FInd a formula for the total revenue R as a function of N. $R=58N$ c. Use your answers to parts a and b to find a formula for the profit P of this manufacturer as a function of N. $P=58N-(55N+200)$ or equivalently $3N-200$. d. Use your formula from part c to determine the break-even point for this manufacturer. 66.67 widgets per month It doesn't help me much... I don't know how they got 66.67 and why the math problem I'm having trouble with has two break even points and not just one.", "40 workers working together have to complete a piece of work in 30 days. If all the workers work at a constant rate and after 20 days, it was found that only $$\\frac{1}{4}^{th}$$ of the work was completed, then how many more workers should be recruited so that the work gets completed on time? A. 32 B. 100 C. 200 D. 240 E. 280 Let the daily rate per worker = 1 widget per day, implying that the daily rate for 40 workers = 40 widgets per day. In 20 days, the number of widgets produced by 40 workers = (daily rate for 40 workers)(number of days) = 40*20 = 800 widgets. Since these 800 widgets constitute 1/4 of the total job, we get: $$800 = \\frac{1}{4}j$$ $$j = 3200$$ Remaining work = (total job) - (widgets produced in the first 20 days) = 3200 - 800 = 2400. For 2400 widgets to be produced in the remaining 10 days, the required rate $$= \\frac{work}{time} = \\frac{2400}{10} = 240$$ widgets per day. Since the daily rate must increase from 40 widgets per day to 240 widgets per day, the number of additional workers that must be recruited = 200. _________________ GMAT and GRE Tutor Over 1800 followers [email protected] New York, NY If you find one of my posts", "4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We’ll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. Try It Now that we have the basic idea of the cost origins and how they are related to production, let’s drill down into the details, by examining average, marginal, fixed, and variable costs. Average and Marginal Costs The cost of producing a firm’s output depends on how much labor and capital the firm uses. A list of the costs involved in producing cars will look very different from the costs involved in producing computer software", "# Weighted average of multiple weighted factors Lets say a factory machine runs as follows: 1. Day 1: $3$ widgets per minute $\\times 1000$ minutes $\\times 1$ kg per widget $= 3000$ kg. 2. Day 2: $5$ widgets per minute $\\times 800$ minutes $\\times 1.5$ kg per widget $= 6000$ kg. 3. Day 3: $8$ widgets per minute $\\times 300$ minutes $\\times 0.5$ kg per widget $= 1200$ kg. So, over the $3$ days the factory machine has run for $2100$ minutes. I am assuming that if I multiply the weighted average number of widgets with the weighted average kg per widget and multiply this by the number of minutes over the three days $(2100)$ this should equal $10,200$ kg $(3000+6000+1200)$. My weighting calculation is below for widgets per minute and kg per widget. $$(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5)+(8\\cdot300\\cdot0.5)/(1000\\cdot1)+(800\\cdot1.5)+(300\\cdot0.5) = 4.833$$ widgets per minute. $$(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5)+(8\\cdot300\\cdot0.5)/(3\\cdot1000\\cdot1)+(5\\cdot800\\cdot1.5) = 1.085$$ kg per widget. However $4.833$ widgets $\\times 1.085$ kg per widget $\\times 2100$ mins $= 11,012$ kg over the three days. Not the correct $10,200$kg. Why is this the case? I have tried weighting by total for the day and this still does not produce the correct answer. I have been using Excel and have noticed that if I keep either \"widgets per minute\" or \"kg per widget\" the same across all three days and", "2. Economics 3. 1 use the following information to answer the next three... # Question: 1 use the following information to answer the next three... ###### Question details 1. Use the following information to answer the next three questions: Widgets Inc. can hire workers at $40 per day and can rent a unit of equipment for$10 per day. Currently Widgets Inc. spends $200 per day on input a- Draw an isocost curve above for Widgets Inc. spending$200. (label the intercepts) b- What is the slope of the isocost curve? c- If Widgets Inc. hires 4 workers and rents 4 pieces of equipment, then the marginal product of labor = 20 and the marginal product of capital = 10. Given this information, is Widgets Inc. optimizing? If not, then what should they do? 2. Currently Moe’s hires 2 employees, has 1 grill (capital), and produces 20 meals per day. If Moe’s decides to double both inputs (employees and grills). If Moe’s experiences decreasing returns to scale how many meals do you expect to produce after inputs double?", "The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $$100$$ workers can produce $$300$$ widgets and $$200$$ whoosits. In two hours, $$60$$ workers can produce $$240$$ widgets and $$300$$ whoosits. In three hours, $$50$$ workers can produce $$150$$ widgets and $$m$$ whoosits. Find $$m$$. (第二十五届AIME2 2007 第4题)", "equation we have $N$N $=$= $60\\log_216-50$60log2​16−50 $=$= $60\\times4-50$60×4−50 $=$= $240-50$240−50 $=$= $190$190 So there will still be $190$190 widgets produced each month. • If the average salary is $\\$500$$500 per month, what is the expected number of widgets produced? By substituting S=500S=500 into the equation we have NN == 60\\log_2\\left(500+16\\right)-5060log2​(500+16)−50 == 60\\log_2516-5060log2​516−50 == 491491 (to nearest whole widget) The warehouse will produce around 490490 widgets if the employees are given \\500$$500 per month. • How many more widgets will be produced if the monthly salary is doubled to $\\$1000$$1000? First we can find the number of widgets produced when the salary is \\1000$$1000, then we can subtract from this the $491$491 widgets expected when the salary is only $\\$500$$500. NN == 60\\log_2\\left(1000+16\\right)-5060log2​(1000+16)−50 == 60\\log_21016-5060log2​1016−50 == 549549 (to nearest whole widget) By doubling the salary, the warehouse will produce and extra 549-491=58549491=58 widgets. #### Practice question A major communications company found that the more they spend on advertising, the higher their revenue. Their sales revenue, in thousands of dollars, is given by R=10+20\\log_4\\left(x+1\\right)R=10+20log4(x+1), where xx represents the amount they spend on advertising (in thousands of dollars). 1. Determine their sales revenue if they spend no money on advertising. 2. Determine their sales revenue if they spend \\14000$$14000 on advertising. Give your answer to the nearest thousand $dollars$dollars. 3. Would", "currently charging$5 for a widget and the average total cost of... Suppose a business is currently charging $5 for a widget and the average total cost of producing a widget is$7. From that information we know that O fixed costs must be really large for the business. O the revenues from selling widgets must be negative o the profit from selling widgets can be negat...", "It is also fairly accurate to state that $$P'(0) = 156\\text{;}$$ that is, at midnight on January 1, 2012, the population of the world was growing by about 156 people per minute (note the units). Twenty days later (or $$28{,}800$$ minutes later) we could reasonably assume the population grew by about $$28{,}800\\cdot156 = 4{,}492{,}800$$ people. #### Example2.2.2.The meaning of the derivative: Manufacturing. The term widget is an economic term for a generic unit of manufacturing output. Suppose a company produces widgets and knows that the market supports a price of $$\\10$$ per widget. Let $$P(n)$$ give the profit, in dollars, earned by manufacturing and selling $$n$$ widgets. The company likely cannot make a (positive) profit making just one widget; the start-up costs will likely exceed $$\\10\\text{.}$$ Mathematically, we would write this as $$P(1) \\lt 0\\text{.}$$ What do $$P(1000) = 500$$ and $$P'(1000)=0.25$$ mean? Approximate $$P(1100)\\text{.}$$ Solution. The equation $$P(1000)=500$$ means that selling $$1000$$ widgets returns a profit of $$\\500\\text{.}$$ We interpret $$P'(1000) = 0.25$$ as meaning that when we are selling $$1000$$ widgets, the profit is increasing at rate of $$\\0.25$$ per widget (the units are “dollars per widget.”) Since we have no other information to use, our best approximation for $$P(1100)$$ is: \\begin{align*} P(1100) \\amp \\approx P(1000) + P'(1000)\\times100\\\\ \\amp= \\$500 + (100\\text{ widgets })\\cdot \\$0.25/\\text{widget}\\\\ \\amp=", "just and fair and all would be good. And this perception is gravely mistaken. To make that clear to you, I offer a little fable. Once upon a time, widgets were made by hand. One person, working for one eight-hour day, could make 100 widgets. Most people were employed making widgets full-time. The wage for making widgets was$1 per widget. Then, an inventor came up with a way to automate the production of widgets. For $100 per day, the same cost to hire a worker to make 100 widgets, the machine could instead make 101 widgets. Because it was 1% more efficient, businesses began adopting the new machine, and now made slightly more widgets than before. But some workers who had previously made widgets were laid off, while others saw their wages fall to only$0.99 per widget. If there were more widgets, but fewer people were getting paid less to make them, where did the extra wealth go? To the inventor, of course, who now owns 10% of all widget production and has billions of dollars. Later, another inventor came up with an even better machine, which could make 102 widgets in a day. And that inventor became a billionare too, while more became unemployed and wages fell to $0.98 per widget. And then there was another inventor,", "$L=g(Q)$ Table 3. Widgets Produced by Workers Widgets (Q) 0.2 0.4 0.8 1 2 3 3.5 3.8 3.95 4 Workers (L) 1 2 3 3.25 4.4 5.2 6 7 8 9 Now focus on the whole number quantities of output. We'll eliminate the fractions (or partial widgets) from the table: Table 4. Number of Widgets Produced Widgets (Q) 1 2 3 4 Workers (L) 3.25 4.4 5.2 9 Suppose widget workers receive $10 per hour. Multiplying the Workers row by$10 (and eliminating the blanks) gives us the cost of producing different levels of output. Table 5. Cost of Producing Widgets Widgets (Q) 1.00 2.00 3.00 4.00 Workers (L) 3.25 4.4 5.2 9 × Wage Rate per hour $10$10 $10$10 = Cost $32.50$44.00 $52.00$90.00 This is same cost function with which we began (shown in Table 1). Figure 2 shows the graph of the cost function. Figure 2. The Total Cost curve for Widgets. This shows cost increasing at an increasing rate as the firm produces more output. #### TRY IT Consider a firm with the following production schedule: Assume that each worker gets paid $15 per hour and that other cost of materials is$2 per item produced. Estimate the cost of producing 45 units. It is ________. Number of Workers Production (units/hr) 1 10 2 30 3 45", "$300$ hours. How many staff hours will the whole project take? Two different methods will be given. Method 1. $15\\%$ of the project takes $300$ hours, therefore $\\phantom{Method 1.}1\\%$ of the project takes $\\frac{300}{15} = 20$ hours (divide both numbers by $15$), so $\\phantom{Method 1.}100\\%$ of the project takes $20 \\times100 = 2000$ hours (multiply both by $100$). Method 2. Write $n$ for the total number of hours needed by staff for the whole project. That $15\\%$ of the total number of hours staff need is $300$ tells us that $\\frac{15}{100}\\times n = 300$. We must solve for $n$: $\\frac{15n}{100} = 300$, so $15n = 100\\times 300$ and therefore $n = \\frac{100\\times 300}{15}= 2000$ hours. 4. ACME Products pays $\\$62.50$for each widget they buy from the importer. They sell each widget for$\\$80.00$. Finding the percentage profit as a proportion of cost price. The profit for each widget is $\\$80.00 - \\$62.50 = \\$17.50$. The profit as a proportion of cost price is$\\frac{17.50}{62.50}$so the percentage profit as a proportion of cost price is$\\frac{17.50}{62.50}\\times100 = 28\\%\\$.", "to try on your own. A gardener charges $20 for each gardening job plus$15 for each hour worked. He charged 80 for a gardening job he did yesterday. a. Write an algebraic equation to represent \\begin{align*}h\\end{align*}, the number of hours that the gardener worked on that80 job. b. Find the number of hours that the gardener worked on that $80 job. Solution Consider part a first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The gardener earned$15 for each hour worked on that job, so you could multiply 15 by \\begin{align*}h\\end{align*}, the number of hours worked, to find how much money the gardener charged for his worktime. \\begin{align*}& \\underline{\\ 20} \\ for \\ each \\ gardening \\ job \\ \\underline{plus} \\ \\underline{\\15 \\ for \\ each \\ hour} \\ worked \\ldots \\underline{charged} \\ \\underline{\\80} \\ for \\ one \\ldots job.\\\\ & \\downarrow \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\downarrow \\qquad \\qquad \\qquad \\downarrow \\qquad \\qquad \\qquad \\qquad \\ \\ \\downarrow \\qquad \\downarrow\\\\ & 20 \\qquad \\qquad \\qquad \\qquad \\qquad \\quad \\ + \\qquad \\qquad \\quad 15h \\qquad \\qquad \\qquad \\qquad = \\quad \\ 80\\end{align*} So, this equation, \\begin{align*}20 + 15h = 80\\end{align*}, represents \\begin{align*}h\\end{align*}, the number of hours the gardener worked on the80 job. Next, consider part", "modeled by the equation $N=60\\log_2\\left(S+16\\right)-50$N=60log2(S+16)50. • How many widgets will be produced if no salary is paid to employees? If no salary is paid, then the value of S will be zero. When we substitute this into the equation we have $N$N $=$= $60\\log_216-50$60log2​16−50 $=$= $60\\times4-50$60×4−50 $=$= $240-50$240−50 $=$= $190$190 So there will still be $190$190 widgets produced each month. • If the average salary is $\\$500$$500 per month, what is the expected number of widgets produced? By substituting S=500S=500 into the equation we have NN == 60\\log_2\\left(500+16\\right)-5060log2​(500+16)−50 == 60\\log_2516-5060log2​516−50 == 491491 (to nearest whole widget) The warehouse will produce around 490490 widgets if the employees are given \\500$$500 per month. • How many more widgets will be produced if the monthly salary is doubled to $\\$1000$$1000? First we can find the number of widgets produced when the salary is \\1000$$1000, then we can subtract from this the $491$491 widgets expected when the salary is only $\\$500$$500. NN == 60\\log_2\\left(1000+16\\right)-5060log2​(1000+16)−50 == 60\\log_21016-5060log2​1016−50 == 549549 (to nearest whole widget) By doubling the salary, the warehouse will produce and extra 549-491=58549491=58 widgets. #### Practice question ##### Question 5 pH is a measure of how acidic or alkaline a substance is, and the pH scale goes from 00 to 1414, 00 being most acidic and 1414 being most alkaline. Water in", "OpenStudy (anonymous): Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $42 to produce 3 widgets,$230 to produce 7 widgets, and \\$645 to produce 12 widgets. Complete the function that represents the cost, c(x), to produce x widgets. c(x)= 2 years ago", "Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 05 Sep 2018, 01:00 Bunuel wrote: If w widgets cost c cents, how many widgets can you get for d dollars? (A) $$\\frac{100dw}{c}$$ (B) $$\\frac{dw}{100c}$$ (C) $$100cdw$$ (D) $$\\frac{dw}{c}$$ (E) $$cdw$$ Unit cost of widgets=$$\\frac{c}{w}$$ cents Or, $$\\frac{c}{100w}$$ dollars are required to purchase 1 widget Or, $1 is required to purchase $$\\frac{100w}{c}$$widgets hence$d is required to purchase $$\\frac{100dw}{c}$$ Ans. (A) _________________ Regards, PKN Rise above the storm, you will find the sunshine Senior Manager Joined: 07 Jul 2012 Posts: 370 Location: India Concentration: Finance, Accounting GPA: 3.5 Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 05 Sep 2018, 01:30 For c cents, w widgets can be bought ......1 cent.., $$\\frac{w}{c,}$$ widgets can be bought .......100d cents...., $$\\frac{w}{c}$$*100d widgets can be bought = $$\\frac{100dw}{c}$$ _________________ Kindly hit kudos if my post helps! Senior Manager Joined: 29 Dec 2017 Posts: 382 Location: United States Concentration: Marketing, Technology GMAT 1: 630 Q44 V33 GMAT 2: 690 Q47 V37 GPA: 3.25 WE: Marketing (Telecommunications) If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] ### Show Tags 15 Sep 2018, 08:40 One of my" ]

[ "of three forms of a three-digit number is always a multiple of $$37$$.", "c\" represents a number: 0.3.", "3 or more than 3 numbers, this equation becomes invalid.", "## Thinking Mathematically (6th Edition) $3$ is a positive number so it is 3 units to the right of $0$. Refer to the attached image in the answer part above for the graph.", "of the three digits is 1 + 3 +4 = 8. 176/8 = 22, with no remainder", "three-digit sequence:", "x = 3 (as 3 is positive) thanks :))", "or a negative number number 2 3 is equal to 3 decay...", "Of the three-digit positive integers whose three digits are all differ 9 25 Apr 2016, 11:28 1 If a,b and c are three positive integers such that at least one 1 25 Feb 2016, 05:24 4 If b/(a-c)=(a+b)/c = a/b for three positive number a,b and c, all diff 6 16 Jan 2015, 05:36 8 Each of the positive integers a, b, and c is a three-digit integer. If 5 17 Sep 2014, 10:51 8 Of the three-digit positive integers that have no digits 9 31 Jan 2007, 06:12 Display posts from previous: Sort by", "numbers. All negative numbers less than $-1000$ can can be expressed as multiple of -1000 + three-digit negative number.", "# Three Symbol The symbol 3 by itself represents a quantity of three. The symbol is also used as a digit which forms part of a number. Format Display Data 3 51", "# Distinct-Digit Positive Numbers How many 3-digit positive numbers are there that comprise of 3 distinct digits? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. 2 digits are distinct if they are not the same. The number 121 is not comprised of 3 distinct digits, as the first and third are the same. ×", "$c_1+c_2 = 1$. Are $c_1$ and $c_2$ both positive?", "they use 3 letters followed by 3 digits? • A three-digit numbers Determine the total number of positive three-digit numbers that contain a digit 6.", "of div1C (Perfect Triples)?", "in $C$ is defined for all positive times $t$.", "c_2+ c_3= 3$ and $c_2+ c_3= 2$", "of remaining three numbers. Verify this", "# Small but distinct Algebra Level 1 What is the smallest positive 3-digit number which comprises of 3 distinct digits? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "the set {1, 2, …, 9}. There are 9 possible choices of digit c 2 such that c 2 ≠ c 1, and there are 8 possible choices of digit c 3 such that c 3 ≠ c 1 and c 3 ≠ c 2. By the product rule it follows that the number of positive integers with the given properties is equal to 9 ⋅ 9 ⋅ 8 = 648." ]

[ "1$), or two more than a multiple of $3$ (in the form $3n + 2$). We can get a multiple of $3$ if we add three multiples of three. Proof: $3a + 3b + 3c = 3(a + b + c)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 1$. Proof: $(3a + 1) + (3b + 1) + (3c + 1) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 2$. Proof: $(3a + 2) + (3b + 2) + (3c + 2) = 3a + 3b + 3c + 6 = 3(a + b + c + 2)$ The above is only if there are $3$ numbers of the same form. What if we don't get $3$ of the same form? Then we must have $2$ of one form, $2$ of another form, $1$ of the last form. $2$ of one form and $1$ of another form does NOT work because the constants ($0$ for $3n$, $1$ and $2$) at the end will never add to $3$. For example: $3a + 3b + (3c + 1) = 3(a + b + c) +", "(3w+1)$ $= 3t + 1 + 3j + 1 + 3w + 1$ $= 3(t+j+w) + 3$ Therefore, it would be a multiple of $3$. $(3x+2) + (3g+2) + (3d+2)$ $= 3x + 2 + 3g + 2 + 3d + 2$ $= 3(x+g+d) + 6$ Therefore, it would be a multiple of $3$. Rin explained why this means that there will always be three numbers that add up to a multiple of three. 1. If there are three or more multiples of $3$ within the five given numbers, we can add those to make a multiple of $3.$ 2. If there are one or two multiples of $3$, other three numbers must be either $3n+1$ or $3n+2.$ If the remaining numbers include: 1. Three or more $3n+1$ s We can add these three to get a multiple of 3. 2. Three or more $3n+1$ s Similarly, we can add these to get a multiple of 3. 3. Otherwise We can pick one each from groups of $3n, 3n+1, 3n+2$ and add together. 3. If there is no multiple of 3, the five numbers can be written as either $3n+1$ or $3n+2.$ For any combination of five numbers, three or more must be in the form of $3n+1$ or $3n+2.$ Adding numbers in the same group would give", "from the options to make $5a+1$ a multiple of 4 is 3. Hence, a = b = 3.", "# Math Help - if c is 3 what is 1. ## if c is 3 what is if c is 3 what is 5(c+4) 2. Originally Posted by andyboy179 if c is 3 what is 5(c+4) $5(c+4) = 5(3+4) = 5(7) = 35$ 3. Originally Posted by Mush $5(c+4) = 5(3+4) = 5(7) = 35$ mush so if c is again 3 would 2c+5 be 11 and thaks for answering the other question 4. Originally Posted by andyboy179 mush so if c is again 3 would 2c+5 be 11 and thaks for answering the other question just plug 3 into anywhere c is, so yeah, 2c+5 would be 11. 2(3)+5 = 6+5 = 11", "has $c + b$ as a multiple of $3.$ 5. Hence, the value of the integer $cba$ is $576.$", "1$), or two more than a multiple of $3$ (in the form $3n + 2$). We can get a multiple of $3$ if we add three multiples of three. Proof: $3a + 3b + 3c = 3(a + b + c)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 1$. Proof: $(3a + 1) + (3b + 1) + (3c + 1) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 2$. Proof: $(3a + 2) + (3b + 2) + (3c + 2) = 3a + 3b + 3c + 6 = 3(a + b + c + 2)$ So if we wanted to find a set of $5$ numbers where we couldn't find $3$ which added up to a multiple of $3$, we would want to avoid five numbers containing $3$ of the same form. If we don't want $3$ of the same form we need $2$ of one form, $2$ another form and $1$ of the last form. This has at least one of each form. If we take one from each form we get a multiple of $3$ Proof: $3a +", "“The three-digit integer $5A4$ is a multiple of six. What is the sum of all the possible values for the digit represented by A?” $0+3+6+9 = 18.$", "+ 0 + 2 = 3$. $3$ is a multiple of $3$, so the sum is divisible by three. So there you have it! This rule only works for multiples of $3$ or $9$, but it makes it very quick and easy to find out whether or not a big number is divisible by $3$ or $9$ without using a calculator. Thank you for all your excellent solutions.", "1$ - this is 1 more than a multiple of $3$ as $0 + 0 + 1 = 1$. How can we add these different constants so that it is $3$ or a multiple of $3$? Take one from each form! Proof: $3a + (3b + 1) + (3c + 2) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ ANSWER: It is possible to get a multiple of $3$ adding $3$ numbers from $5$ random numbers provided that you select $3$ numbers of the same form ($3n$, $3n+ 1$, or $3n + 2$) or if you select $3$ numbers with different forms ($3n$, $3n + 1$ and $3n + 2$). We also received a very colourful solution from Marie and Caitlin from St Leonard's Catholic School, which you can see here. Well done!", "3. 5. Aug 19, 2012 ### HallsofIvy I would recommend an indirect proof. If $p^2$ is NOT a multiple of 3 then it is of the form 3n+1 or 3n+ 2. Now use the fact that $(3k+1)^2= 9k^2+ 6k+ 1= 3(3k^2+ 2k)+ 1$ and that $(3k+ 2)^2= 9k^2+ 12k+ 4= 3(3k^2+ 4k+ 1)+ 1$. 6. Aug 19, 2012 Thanks", "of $$3$$. These are the numbers that can be divided into $$3$$ equal groups. The digit sums for the other two numbers are not multiples of $$3$$. In fact, a number is a multiple of $$3$$ exactly when its digit sum is also a multiple of $$3$$.", "quickly realize that the next consecutive trio, $(1, 0, 6)$, does satisfy this condition, and $1^2 + 0^2 + 6^2 = \\boxed{\\textbf{(D)}\\ 37}$. We come to this quickly after realizing that if the multiple of $55$ was an even multiple, it would have to end in $0$ and thus the $c$ digit would remain unchanged, so it must be an odd multiple, which will carry over the $c$ digit, so $c \\geq{6}$. -Solution by Joeya", "that is a multiple of both $4$ and $5$ is $20$, it follows that $20|a$.", "c_2+ c_3= 3$ and $c_2+ c_3= 2$", "over the property of being a multiple of $$3$$.", "} { 15} + 1$$ $n(n-1)(n+1)$ is clearly a multiple of $3$ for all values of $n$. $n(n-1)(n+1)$ is a multiple of $5$ if $n = 5k, 5k-1, 5k+1$. $3n^2 + 8$ is a multiple of 5 if $n = 5k+2, 5k+3$. Hence, this is an integer. -", "triple, only one can be divisible by 3, not both. • The proof above shows that in any Pythagorean triple $(a,b,c)$, we have $3$ divides $a$ or $3$ divides $b$. This \"or\" as usual includes the possibility of both. For primitive triples, it cannot be both. – André Nicolas Dec 28 '14 at 21:06 \\begin{array}{|c|c|} \\hline n \\pmod 3 & n^2 \\pmod 3 \\\\ \\hline 0 & 0 \\\\ 1 & 1 \\\\ 2 & 1 \\\\ \\hline \\end{array} If neither $a$ nor $b$ is a multiple of $3$, then $a^2 + b^2 \\equiv c^2 \\pmod 3$ becomes $1 + 1 \\equiv c^2 \\pmod 3$, which simplifies to $c^2 \\equiv 2 \\pmod 3$; which has no solution. I'm reading the book too. I came up with a different approach. Of course, it is not as elegant as the answer given, but here goes. $$a = st$$ $$b = (s^2 - t^2)/2$$ Choose values for s and t. If either of s or t is a multiple of 3, then $a = st$ is a multiple of 3. Both s and t cannot be multiples of 3, as that will cause both a and b to be multiples of 3 and the triplet won't be primitive. If neither s nor t is a multiple of 3. $$s = 3x", "of possible ways =$^3{C_3}^3{C_3} = 1$", "(3b + 1) + (3c + 2) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ ANSWER: It is always possible to get a multiple of $3$ adding $3$ numbers from any $5$ random numbers by selecting $3$ numbers of the same form ($3n$, $3n+ 1$, or $3n + 2$) or by selecting $3$ numbers with different forms ($3n$, $3n + 1$ and $3n + 2$), and its always possible to do this. We also received a very colourful solution from Marie and Caitlin from St Leonard's Catholic School, which you can see here. Well done!", "can also say that: $b = 3(a -3 )/4$ $c = 3(b -3 )/4$ $d = 3(c -3 )/4$ $e = 3(d -3 )/4$ Then, as each pile has 3 nuts taken away from it and then is divided into 4 equal piles, we can say that: $a - 3$ is a multiple of 4 $b - 3$ is a multiple of 4 $c - 3$ is a multiple of 4 $d - 3$ is a multiple of 4 Also, as each pile except for $a$ has been created by taking 3/4 of the previous pile minus 3, we can also say that: $b$ is a multiple of 3 and $b - 3$ is a multiple of 4 and 3 $c$ is a multiple of 3 and $c - 3$ is a multiple of 4 and 3 $d$ is a multiple of 3 and $d - 3$ is a multiple of 4 and 3 $e$ is a multiple of 4 and 3, as the castaways were able to divide it exactly into four equal piles. The least number of coconuts the castaways could have ended with will also give you the least number of coconuts they could have started with. The number of coconuts the castaways ended up with, $e$, is a common multiple of 4 and 3," ]

[ "1$), or two more than a multiple of $3$ (in the form $3n + 2$). We can get a multiple of $3$ if we add three multiples of three. Proof: $3a + 3b + 3c = 3(a + b + c)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 1$. Proof: $(3a + 1) + (3b + 1) + (3c + 1) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 2$. Proof: $(3a + 2) + (3b + 2) + (3c + 2) = 3a + 3b + 3c + 6 = 3(a + b + c + 2)$ The above is only if there are $3$ numbers of the same form. What if we don't get $3$ of the same form? Then we must have $2$ of one form, $2$ of another form, $1$ of the last form. $2$ of one form and $1$ of another form does NOT work because the constants ($0$ for $3n$, $1$ and $2$) at the end will never add to $3$. For example: $3a + 3b + (3c + 1) = 3(a + b + c) +", "“The three-digit integer $5A4$ is a multiple of six. What is the sum of all the possible values for the digit represented by A?” $0+3+6+9 = 18.$", "(3w+1)$ $= 3t + 1 + 3j + 1 + 3w + 1$ $= 3(t+j+w) + 3$ Therefore, it would be a multiple of $3$. $(3x+2) + (3g+2) + (3d+2)$ $= 3x + 2 + 3g + 2 + 3d + 2$ $= 3(x+g+d) + 6$ Therefore, it would be a multiple of $3$. Rin explained why this means that there will always be three numbers that add up to a multiple of three. 1. If there are three or more multiples of $3$ within the five given numbers, we can add those to make a multiple of $3.$ 2. If there are one or two multiples of $3$, other three numbers must be either $3n+1$ or $3n+2.$ If the remaining numbers include: 1. Three or more $3n+1$ s We can add these three to get a multiple of 3. 2. Three or more $3n+1$ s Similarly, we can add these to get a multiple of 3. 3. Otherwise We can pick one each from groups of $3n, 3n+1, 3n+2$ and add together. 3. If there is no multiple of 3, the five numbers can be written as either $3n+1$ or $3n+2.$ For any combination of five numbers, three or more must be in the form of $3n+1$ or $3n+2.$ Adding numbers in the same group would give", "# Math Help - if c is 3 what is 1. ## if c is 3 what is if c is 3 what is 5(c+4) 2. Originally Posted by andyboy179 if c is 3 what is 5(c+4) $5(c+4) = 5(3+4) = 5(7) = 35$ 3. Originally Posted by Mush $5(c+4) = 5(3+4) = 5(7) = 35$ mush so if c is again 3 would 2c+5 be 11 and thaks for answering the other question 4. Originally Posted by andyboy179 mush so if c is again 3 would 2c+5 be 11 and thaks for answering the other question just plug 3 into anywhere c is, so yeah, 2c+5 would be 11. 2(3)+5 = 6+5 = 11", "c_2+ c_3= 3$ and $c_2+ c_3= 2$", "of three forms of a three-digit number is always a multiple of $$37$$.", "of possible ways =$^3{C_3}^3{C_3} = 1$", "has $c + b$ as a multiple of $3.$ 5. Hence, the value of the integer $cba$ is $576.$", "+ 0 + 2 = 3$. $3$ is a multiple of $3$, so the sum is divisible by three. So there you have it! This rule only works for multiples of $3$ or $9$, but it makes it very quick and easy to find out whether or not a big number is divisible by $3$ or $9$ without using a calculator. Thank you for all your excellent solutions.", "from the options to make $5a+1$ a multiple of 4 is 3. Hence, a = b = 3.", "1$), or two more than a multiple of $3$ (in the form $3n + 2$). We can get a multiple of $3$ if we add three multiples of three. Proof: $3a + 3b + 3c = 3(a + b + c)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 1$. Proof: $(3a + 1) + (3b + 1) + (3c + 1) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ We can get a multiple of $3$ if we add $3$ numbers in the form $3n + 2$. Proof: $(3a + 2) + (3b + 2) + (3c + 2) = 3a + 3b + 3c + 6 = 3(a + b + c + 2)$ So if we wanted to find a set of $5$ numbers where we couldn't find $3$ which added up to a multiple of $3$, we would want to avoid five numbers containing $3$ of the same form. If we don't want $3$ of the same form we need $2$ of one form, $2$ another form and $1$ of the last form. This has at least one of each form. If we take one from each form we get a multiple of $3$ Proof: $3a +", "= 8$ $(a, b,c) = (8,8+ 1, 8+ 2)$ $= (8,9,10)$ We observe that each triplet consists of one and only one number which is a multiple of 3. Hence, one of any three consecutive positive integers must be divisible by 3. Updated on 10-Oct-2022 13:27:08", "quickly realize that the next consecutive trio, $(1, 0, 6)$, does satisfy this condition, and $1^2 + 0^2 + 6^2 = \\boxed{\\textbf{(D)}\\ 37}$. We come to this quickly after realizing that if the multiple of $55$ was an even multiple, it would have to end in $0$ and thus the $c$ digit would remain unchanged, so it must be an odd multiple, which will carry over the $c$ digit, so $c \\geq{6}$. -Solution by Joeya", "the set {1, 2, …, 9}. There are 9 possible choices of digit c 2 such that c 2 ≠ c 1, and there are 8 possible choices of digit c 3 such that c 3 ≠ c 1 and c 3 ≠ c 2. By the product rule it follows that the number of positive integers with the given properties is equal to 9 ⋅ 9 ⋅ 8 = 648.", "# How do you find the least common multiple of 30 and 3? Least Common Multiple of $30$ and $3$ is $30$. Here, among two numbers $3$ and $30$, as $3$ is a factor of $30$, Least Common Multiple of $30$ and $3$ is $30$.", "1$ - this is 1 more than a multiple of $3$ as $0 + 0 + 1 = 1$. How can we add these different constants so that it is $3$ or a multiple of $3$? Take one from each form! Proof: $3a + (3b + 1) + (3c + 2) = 3a + 3b + 3c + 3 = 3(a + b + c + 1)$ ANSWER: It is possible to get a multiple of $3$ adding $3$ numbers from $5$ random numbers provided that you select $3$ numbers of the same form ($3n$, $3n+ 1$, or $3n + 2$) or if you select $3$ numbers with different forms ($3n$, $3n + 1$ and $3n + 2$). We also received a very colourful solution from Marie and Caitlin from St Leonard's Catholic School, which you can see here. Well done!", "# How many ordered triples (a, b, c) exist if a, b, and c are positive digits and their product is divisible by 20? Edit #1: Remember that a, b, and c are positive digits, so their values are between 1 & 9, inclusive I was thinking casework for this problem, but I got stuck on the last part. I can't figure out whether there are four cases or three. $20 = 2^2 * 5$, so you need two 2's and a 5. Case 1: One five, either a four or an eight, and an odd number that is not five Case 2: Two fives and either a four or an eight Here's where I get confused. I'm not sure whether it should be: Case 3: One five and two even numbers or: Case 3: One five and two even numbers that are different Case 4: One five and two even numbers that are the same Here are the calculations that I have done so far: Case 1: $2* 4 * 3!$, as the five is fixed, you can choose between the 4 & 8, and four choices for the odd number that is not five. The $3!$ accounts for the various permutations of the three that are chosen. Case 2: $2 * 3$. The fives are fixed,", "of $$3$$. These are the numbers that can be divided into $$3$$ equal groups. The digit sums for the other two numbers are not multiples of $$3$$. In fact, a number is a multiple of $$3$$ exactly when its digit sum is also a multiple of $$3$$.", "$-2$ and $-8$; $-3$ and $-7$; $-4$ and $-6$; $-5$ and $-5$, so there are $5$ possibilities for $c$. 3. $3x^2+5x+c$ We now need two integers which add to $5$ and multiply to $3c$, so they must both be positive. Of the pairs which add to $5$ (being $1$ and $4$; $2$ and $3$), only $2\\times3=6$ gives a multiple of $3$, so there is only one possible value for $c$. 4. $10x^2-6x+c$ This time, the integers add to $-6$ and multiply to $10c$; no pairs which add to $-6$ multiply to give a positive multiple of $10$, so there are no possible values for $c$. 3. What are the answers to question 2 if $c$ is only allowed to be a negative integer? 1. $x^2+6x+c$ Now we require two integers which add to $6$ and multiply to the negative integer $c$. There are infinitely many such pairs: $-1$ and $7$; $-2$ and $8$; $-3$ and $9$; …; in general $-n$ and $6+n$ for any positive integer $n$. So there are infinitely many possible values for $c$. 2. $x^2-10x+c$ The same happens here: $-11$ and $1$; $-12$ and $2$, and so on, so again there are infinitely many possible values for $c$. 3. $3x^2+5x+c$ This time, we require two numbers which add to $5$ and multiply to $3c$, a", "a positive integer? A. 70 B. 40 C. 32 D. 28 E. 16 Math Expert Joined: 02 Aug 2009 Posts: 8201 Re: How many ordered triplets (a, b, c) exist such that LCM (a, b) = 1000, [#permalink] Show Tags 15 Oct 2019, 18:49 2 nick1816 wrote: How many ordered triplets (a, b, c) exist such that LCM (a, b) = 1000, LCM(b, c) = 2000, LCM (c,a) = 2000 and HCF (a, b) = k × 125, where k is a positive integer? A. 70 B. 40 C. 32 D. 28 E. 16 Since HCF(a,b) $$= 125*k=5^3*k$$ and LCM(a,b)=$$1000=5^3*2^3$$, at least one of a and b is a multiple of 2^3.. Also as all LCMs are 1000 and 2000, this means a, b and c are multiples of only 2 and 5.. (I) If a is $$2^3$$, b can be $$2^0,2^1,2^2,2^3$$---1*4=4ways (II) If b is $$2^0,2^1,2^2$$, b will be $$2^3$$---3*1=3ways So total ways of (a,b)=4+3=7ways Value of c for each of the 7 pairs of (a,b) As LCM when c is included is 2000 or $$2^45^3$$, while that of (a,b) is 1000, c is surely multiple of $$2^4$$ and it can have any power of c as $$5^3$$ is already included in a and b, so c can be $$2^45^0,2^45^1,2^45^2,2^45^3$$---4 ways So for each pair of (a,b)," ]

[ "length and same breadth. Therefore, those two areas are equal. Hence all squares are not congruent. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two", "areas are equal. Hence all squares are not congruent. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two figures a and B are congruent (", "necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two figures a and B are congruent ( see Fig.1 ), then your two are. All parts equal one figure over the other completely the parallelogram lesson about congruent.. * ( DC ) =", "incredibly important for kids. How many different shapes can you draw that have perimeter of 20 centimeters? Explain using words or a diagram why the perimeter of a rectangle can be found by using a combination of addition and multiplication. 3 Understand and. A book of pentomino lessons, keyed to the Common Core State Standards. McRuffy Pentomino Puzzle Book (Pentominoes sold separately. Each piece is scored on both sides to show one inch segments. •Students will calculate the perimeter and area. A Spatial Reasoning Task Using Pentominoes Posted on November 16, 2015 January 15, 2018 Author Vicki McGinn Categories Math Lesson Observing people of all ages enthusiastically engaging in mathematical tasks is what fuels my passion for the work I do. Which pentominoes tessellate? Make a tessellation drawing for any pentomino. Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Brain Teaser: You know that the 12 pentominoes will always have an area of 60 square units. The perimeter is 600 inches. Area and Perimeter using Pentominoes I began using math stations today. If you are looking for more perimeter and area ideas, I have a HUGE blog post on it HERE! Please Follow Me! Posted by", "In the figure above, the large square is divided into two smaller squa [#permalink] ### Show Tags 30 Sep 2017, 02:43 Since the squares have a perimeter of 8 and 20, the sides of the squares are $$(\\frac{8}{4})2$$ and $$(\\frac{20}{4})5$$. The rectangles share their lengths and breadth with these 2 squares. Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively, and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14 Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E) _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Re: In the figure above, the large square is divided into two smaller squa [#permalink] 30 Sep 2017, 02:43 Display posts from previous: Sort by", "have a perimeter of 12 units, but they are not the same triangle. TRUE. This means that we can obtain one figure from the other through a process of expansion or contraction, possibly followed by translation, rotation or reflection. Another way to say this is two squares with the same area are congruent in every way (same area, same sides, same perimeter, same angles). In other words, if two figures A and B are congruent (see Fig. A BFigures A and Bare congruent andhence they have If two figures are congruent ,equal areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. (ED)*(DG) = the area of the rectangle. So, if two figures X and Y are congruent, they must have equal areas. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. If two triangles have equal areas, then they are congruent. Consider the rectangles shown below. called congruent, if they have the same shape and the same size. If two squares have equal areas, they will also have sides of the same length. Conversely: \"If a rectangle's diagonals are equal, then it is a square\"", "squares are $$(\\frac{8}{4})2$$ and $$(\\frac{20}{4})5$$. The rectangles share their lengths and breadth with these 2 squares. Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively, and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14 Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E) _________________ You've got what it takes, but it will take everything you've got Re: In the figure above, the large square is divided into two smaller squa [#permalink] 30 Sep 2017, 03:43 Display posts from previous: Sort by", "# Two Congruent Circles Geometry Level 2 The above diagram shows 2 congruent circles that are enclosed within a rectangle. Each circle passes through the center of the other circle. Each circle has an area of 10 units. To the nearest whole number, what is the area of the rectangle? Details & Assumptions: 1) Area of a circle$=\\Pi { r }^{ 2 }$. 2) The two circles have the same area. ×", "= |1 + 8| = 9 units The distance of PM = |3 + 5| + |-2 + 5| = |8 – 3| = 5 units Now, The perimeter of the rectangle MNOP = MN + NO + OP + PM = 9 + 5 + 9 + 5 = 28 units Hence, from the above, We can conclude that The perimeter of the given rectangle MNOP is: 28 units Question 5. Jen draws a polygon with vertices E(-2, 3.5), F(3, 3.5), G(3, -1.5), and H(-2, -1.5). Is EFGH a square? Justify your answer. It is given that Jen draws a polygon with vertices E(-2, 3.5), F(3, 3.5), G(3, -1.5), and H(-2, -1.5) Now, Compare the given points with (x, a), and (y, b) Now, We know that, The distance between 2 points = |x + a| + |y + b| Now, The distance of EF = |3 + 2| = 5 units The distance of FG = |3.5 + 1.5| = 5 units The distance of GH = |3 + 2| = 5 units The distance of HE = |3.5 + 1.5| = 5 units Now, We know that, The “Square” has all the same side lengths Hence, from the above, We can conclude that EFGH is a square Question 6. Square ABCD has vertices A(-4.5, 4),", "activity for this game. Six of them have reflection symmetry, and the other 12 are 6 sets of mirror pairs. It is available from Didax, and comes with an eBook version on DVD. Snr Primary • Teacher Pack • Floored by Rectangles • Chicken Coop Challenge • Pandemonium with Pentominoes! Additional Resources. c Logic Puzzles Area and Perimeter Games for That Quiz Unit rate and Price Absolute value. ) and area (1 square inch). Make a 5 by 4 tessellation using four different pentominoes. Plastic pieces are scored in 1\" sections so that visualizing area and perimeter is quick and easy Ages 6-12 Pentominoes, set of 6 (72 pieces) packaged in a plastic bucket. With six pentominoes, the greatest perimeter is 62 in. Available in 6 colors and the large set comes with a storage container. How many different shapes can you draw that have perimeter of 20 centimeters? Explain using words or a diagram why the perimeter of a rectangle can be found by using a combination of addition and multiplication. Find the perimeter of each pentomino above. Fall 2013 Pentominoes Page 3 of 3. Work out the perimeter for each one, what do you notice? Measuring Perimeter of shapes. Use 3 pentomino shapes to form a 3 x 5 rectangle. Perimeter Area Song - Activity: sing", "are (-1,4), and the upper-right coordinates are (3,4). The rectangle has a perimeter of 24 units. Draw the rectangle on the coordinate plane below.... -- 0.016880--", "equal, then they 're exactly the same rectangles: 1., side.! To 108 = the area of 36 units^2, but they are congruent similar!, side '' iii ) if two squares have the same area at. Ii ) if two figures a and B are congruent if they the. Radius, diameter, circumference and surface area and not being congruent is two! And triangles 153 you can superpose one figure over the other such it! And which of the small rectangles need to multiply to 108 used triangle and! Which of them false '' is true for squares, it is not true most... 153 you can superpose one figure over the other completely because they have a constant radius and no sides! End of the parallelogram circles subtend equal angles at their centres and which the. Other words, if two squares have equal areas ( e ) There is no congruence. Equal the building line is said to be square a pair of _____ are.! Sides should equal the building line is said to be similar, their sides have to be (! Rectangles can have the same area then they must have equal areas, areas need be! Careful, let 's compute a/d 153 you can superpose one figure over the other completely be.... Deadlift Grip Width Reddit, Uc Davis Safety", "Day 44 # 44 of 100: Semitricky Semicircles Geometry Level 2 What is the pink area in square units? The figure shows 2 semicircles whose diameters are on the same line; the chord of length 24 units in the larger semicircle is tangent to the smaller semicircle. ×", "all 4 sides the same. So their perimeters are 4x and 4y respectively. What does it mean when there is no flag flying at the White House? Yes, they are similar. And why does a 1 × 1 square have an area of 1 unit?) 4:26 6.9k LIKES. Now, sqrt of y is either a positive sqrt or a negative sqrt. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. Picture below. If two squares have the same area, must they have the same perimeter? 2 rectangles can have the same area with different lengths of sides to form the perimeter, but squares are special. The word for the same in math is congruent. Join Yahoo Answers and get 100 points today. Still have questions? 2+3 produces 5. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 4:26 6.9k LIKES. Hence, if the areas of the two similar triangles are equal, it follows that squares of the corresponding sides are equal. Congruent circles are circles that are equal in terms of radius, diameter, circumference and surface area. She wants two squares that can be placed exactly one over", "# ACT® Math Free Version Difficult ACTMAT-4K1CIE In the figure given below, $(0,0)$ is the center of the circle. The four corners of the quadrilateral lie on the perimeter of the circle. What is the area of the quadrilateral in square units? A 50 B 20 C 25 D 32 E 40", "the five circles is tangent to two other congruent circles and to a smaller inner circle. The perimeter of the figure is composed of 5 line segments of length X and 5 circular arcs of length Y. What is the perimeter of the figure? If 20 red cubes and 7 white cubes, all of equal size, are fitted together to form one large cube, as shown above, what is the greatest fraction of the surface area of the large cube that could be red? If a and b are integers satisfying the equation $a^{2}$+$b^{2}$=145, which of the following could be the value of a+b? Indicate all such values. A restaurant has a total of 16 tables, each of which can seat a maximum of 4 people. If 50 people were sitting at the tables in the restaurant, with no tables empty, what is the greatest possible number of tables that could be occupied by just 1 person? Which of the following set has the greatest number of integers from 1 to 100, inclusive? The number 24 has the property that it is divisible by its units digit, 4. How many of the integers between 10 and 70 are divisible by their respective units digits? #### Quantity A The perimeter of ABCD #### Quantity B The sum of the", "## alainabbyboo22 If the perimeter of square 2 is 200 units, and the perimeter of square 1 is 150 units, what is the perimeter of square 3? A. 62.5 units B. 500 units C. 125 units D. 250 units one year ago one year ago 1. alainabbyboo22 2. alainabbyboo22 I am unsure how to do that. 3. davidkircos Find the side lengths of both. Perimeter/4 and then use the pythagorean theorem. 4. davidkircos Which part are you unsure on? 5. alainabbyboo22 The whole thing. 6. davidkircos Ok so find the side lengths of the Square 1 and 2 first. Side Length = Perimeter /4 Can you do that? 7. davidkircos Then, (Side Length 1)^2+(Side length 2)^2 = (Side Length 3)^2. And once you solve for Side Length 3 you multiply it by 4 to the perimeter. 8. alainabbyboo22 Can this be put into equation form? Like more understandably? I'm just really confused on it right now. :/ 9. davidkircos Sure. Square 1 has a perimeter of 150. That means that each side is one fourth of that because all sides are the same length on a square. So 150/4 = 37.5 Do the same thing for Square 2. 200/4 = 50. |dw:1334176657772:dw| Then you have the triangle: |dw:1334176742025:dw| And when you solve for the hypotenuse, that is the", "to two other congruent circles and to a smaller inner circle. The perimeter of the figure is composed of 5 line segments of length X and 5 circular arcs of length Y. What is the perimeter of the figure? If 20 red cubes and 7 white cubes, all of equal size, are fitted together to form one large cube, as shown above, what is the greatest fraction of the surface area of the large cube that could be red? If a and b are integers satisfying the equation $a^{2}$+$b^{2}$=145, which of the following could be the value of a+b? Indicate all such values. A restaurant has a total of 16 tables, each of which can seat a maximum of 4 people. If 50 people were sitting at the tables in the restaurant, with no tables empty, what is the greatest possible number of tables that could be occupied by just 1 person? Which of the following set has the greatest number of integers from 1 to 100, inclusive? The number 24 has the property that it is divisible by its units digit, 4. How many of the integers between 10 and 70 are divisible by their respective units digits? Quantity A The perimeter of ABCD Quantity B The sum of the lengths of diagonals AC and BD How", "Perimeter of a Rectangle Worksheets, Geometry Worksheets for 4th grade, 5th grade and middle school The term may be used either for the path or its length it can be thought of as the length of the outline of a square. Area A = l × w square units Maharashtra Board Class 7 Maths Chapter 12 Perimeter and Area Practice Set 45 Intext Questions and Activities What is the scale factor of the squares? Perimeters of squares and rectangles Can you find the perimeter of this square? You can earn a trophy if you get at least 9 questions correct and you do this activity online.The diagrams are not drawn to scale. Area of a square = side × side . 2(l + w) = 2(15 + 12) = 2 × 27 = 54 yards. Following quiz provides Multiple Choice Questions (MCQs) related to Perimeter of a Square or a Rectangle. Area And Perimeter Of Rectangles, Triangles And Parallelograms Quiz 19 Questions | By Joeldodd | Last updated: May 21, 2018 | Total Attempts: 529 Questions All questions 5 questions 6 questions 7 questions 8 questions 9 questions 10 questions 11 questions 12 questions 13 questions 14 questions 15 questions 16 questions 17 questions 18 questions 19 questions Each diagonal divides the square into two congruent isosceles right-angled", "4 x $$\\frac{2 m+1}{2}$$ units = 4 x $$\\frac{2 x 6+1}{2}$$ units = 4 x $$\\frac{13}{2}$$ units = 26 units the perimeter of the rectangle is = 2m + 22 units =2 x 9 + 22 = 18 + 22 = 40 units d) If m – 6, the perimeter of the rectangle is greater than the perimeter of the square. Find how many units greater the rectangle’s perimeter is than the square’s perimeter. 40 units Explanation: the perimeters of the two figures if m = 6 Perimeter of a square = 26 units the perimeter of the rectangle is = 40 units Question 26. Math Journal Rita simplified the expression 10w – 5w + 2w in this way: Is Rita’s answer correct? If not, explain why it is incorrect." ]

[ "Utilities) Followers: 11 Kudos [?]: 673 [0], given: 148 Re: The perimeter of square S [#permalink] 25 Aug 2018, 09:54 Carcass wrote: Attachment: square.jpg The perimeter of square S is equal to the perimeter of the rectangle above. Quantity A Quantity B The length of a side of S $$x+3$$ A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. Kudos for R.A.E The perimeter of square is =$$4 * side$$ and the perimeter of rectangle = $$2 *(Length + breadth)$$ Now give, $$4S = 2* (x + x + 6)$$ ( S = side of the square i.e. given) or $$4S = 4x + 12$$ or $$S = x + 3$$ Hence option C _________________ If you found this post useful, please let me know by pressing the Kudos Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Director Joined: 09 Nov 2018 Posts: 508 Followers: 0 Kudos [?]: 27 [0], given: 1 Re: The perimeter of square S [#permalink] 18 Jan 2019, 19:53 the perimeter of rectangle = x+x+x+6+x+6 = 4x+12 side of square= 4x+12 /4= x+3 Re: The perimeter of square S [#permalink] 18 Jan 2019, 19:53 Display posts from previous: Sort by", "the following numbers using diagonal method: (i) 98 $∴ 98^{2}= 9604$ (ii) 273 $∴ 273^{2}= 74529$ (iii) 348 $∴ 348^{2}= 121104$ (iv) 295 $∴ 295^{2}= 87025$ (v) 171 $∴ 171^{2}= 29241$ 3.) Find the squares of the following numbers: We will use visual method as it is the efficient method to solve this problem. (i) We have: 127 = 120 + 7 Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units. Hence, the square of 127 is 16129. (ii) We have: 503 = 500 + 3 Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units. Hence, the square of 503 is 253009. (iii) We have: 451 = 450 + 1 Hence, let us draw a square of having side 451 units. Let us split it into 450 units and 1 units. Hence, the square of 451 is 203401. (iv) We have: 862 = 860 + 2 Hence, let us draw a square having side 862 units. Let us split it into 860 units and 2 units. Hence, the square of 862 is 743044. (v) We have: 265= 260 +5 Hence, let us draw a square having 265 units. Let us split it into 260 units and", "## Basic College Mathematics (9th Edition) A square consists of 4 sides of the same length. The length of a side of a square can be denoted $s$. The area of a square can be defined as $s$$^{2}$. The perimeter of a square can be defined as 4$s$. Each corner of a square is a 90$^{\\circ}$ angle. A rectangle consists of two pairs of sides of the same length. These can be labeled $l$ and $w$. The perimeter of a rectangle can be defined as 2$l$ * 2$w$. The area of a rectangle can be defined as $l$*$w$. Each corner of a rectangle is a 90$^{\\circ}$ angle.", "that breadth of the RECTANGLE is diagonal of the SQUARE =$$d$$ As seen in the diagram the side of the square is $$x$$ Substituting these values in equation 1 gives us $$2d+d=9√2$$ $$3d=9√2$$ $$d=3√2$$ so the diagonal of the square is $$3√2$$ now $$side^2 + side^2 = diagonal ^2$$ {simple pythagorus theorum} $$x^2+x^2= (3√2)^2$$ $$2x^2= 9*2=18$$ $$x^2=\\frac{18}{2} = 9$$ $$x=\\sqrt{9}$$ $$x= 3$$the side of the square is 3 therefore its perimeter is 3*4=12 Attachments Rectangle.png [ 101.26 KiB | Viewed 12103 times ] _________________ Posting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. Intern Joined: 26 Jun 2015 Posts: 37 Location: India Concentration: Entrepreneurship, General Management WE: Engineering (Energy and Utilities) Re: In the figure shown, two identical squares are inscribed in [#permalink] ### Show Tags 02 Jul 2017, 05:35 I solved it in a very easy way. Lets take side of square is x. You can see from figure, two diagonals of squares = length of rectangle. And one diagonal of square = width of rectangle. So, as Length x Width = 36, we can say (2 * root2x)* (root2x) = 36 x", "### Home > INT2 > Chapter 5 > Lesson 5.1.1 > Problem5-6 5-6. For each diagram below, set up an equation and solve for $x$. 1. $\\text{Perimeter}=76\\ \\text{units}$ Use the equation $2x+x+(2x-3)+2x+(x-1)=76$ to solve for $x$. 1. $3x-2=5x-14$ 1. The two angles are supplementary angles.", "# perimeter very easy • Feb 24th 2008, 02:24 AM kri999 perimeter very easy There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long and i have to find the perimeter..how do i do this? • Feb 24th 2008, 02:28 AM Jhevon Quote: Originally Posted by kri999 There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long and i have to find the perimeter..how do i do this? let the side length be $x$. then by Pythagoras' theorem, $x^2 + x^2 = (6 \\sqrt{3})^2 = 108$ solve for $x$. then the perimeter is given by $4x$ • Feb 24th 2008, 02:35 AM kri999 oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem • Feb 24th 2008, 02:38 AM Jhevon Quote: Originally Posted by kri999 oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem why? that would only make your life harder. then you'd have to use sine or cosine or tangent or something if you insist, you could do $\\sin 45 = \\frac x{6 \\sqrt{3}}$ and solve for x from that • Feb 24th 2008, 02:49 AM kri999 Yeah sorry the whole", "## Algebra 1: Common Core (15th Edition) Each side of the hexagon is $4.2$ inches in length. If we divide the perimeter of a regular hexagon by the number of sides in a hexagon, $6$, then we will find the length of each side. length of a side = $25.2 \\div 6$ Divide to solve: length of a side = $4.2$ in.", "of semicircle is 3 units. 1. Given isosceles trapezium ABCD, find the area of rectangle ADFE, If the area of the trapezium is 72. Solution: Area of trapezium=$\\frac{1}{2}$×(AD+BC)×AE $\\frac{1}{2}$ x (8+12)×AE=72 $\\frac{1}{2}$×20×AE=72 10×AE=72 AE=$\\frac{72}{10}$ AE=7.2 Area of rectangle =8×7.2 =57.6 The area of the rectangle is 57.6 sq units. 1. Given a regular hexagon ABCDEF that has a perimeter 144, find the length of diagonal AD. Solution: In a regular hexagon, all sides are equal in length. So, Perimeter=6×side 144=6×BC BC=$\\frac{144}{6}$ BC=24 Also, diagonal=2×side", "$x = 5$, hexagon side length $= 10$ Problem 4 – An Equilateral Triangle and a Rectangle that Share a Common Side Student worksheet solutions • $3x + 9 = 14 + 2x, \\ x = 23$ Problem 5 – A Regular Decagon and 15-gon Student worksheet solutions • Because the side lengths for both polygons all equal $x$, the perimeter of the $15-$gon will be longer by the length $5x$ because it has five more sides. Feb 22, 2012 Aug 19, 2014 Reviews Image Detail Sizes: Medium | Original TI.MAT.ENG.TE.1.Algebra-I.4.2", "# Rectangles Geometry Level 2 Given perimeter of a rectangle is 26 and area of the rectangle is 42. If the angle between diagonals is $$P$$, find $$\\cos P$$. ×", "# Algebra geometry question square? The side of one square is 7 kn longer than the side of another. Find the sides of the squares. The perimeter of the larger square is twice that of the smaller. - Let $x_1$ be the side of one of them and $p_1$ be its perimeter, the we have: $$P_1=4x_1$$ $$P_2=4x_2$$ We also have: $$P_2=2P_1$$ And $$x_2=x_1+7$$ Now you have to solve a system of equations to find $x_1$ and $x_2$. -", "### Home > INT2 > Chapter 6 > Lesson 6.2.2 > Problem6-63 6-63. Calculate the area and perimeter of each polygon below. Show all work. 1. Look at the diagram below. First solve for $x$, then $y$, then $z$. $\\text{P }= 2\\sqrt{10} + 34 \\approx 40.3\\text{ mm, A }= 72 \\text{ sq. mm}$ 1. Start by determining the length of each dashed segment in the diagram below.", "## Elementary Technical Mathematics $x = 1.508 in$ To find the missing length $x$, we just need to subtract all known sides from the perimeter. $P = 6.573 in$ So, $x = 6.573 in - 0.938 in - 0.688 in - 1.313 in - 0.625 in - 1.501 in$ $x = 1.508 in$", "Find the dimensions of the rectangle whose perimeter is $\\quantity{36}{m}$ and which is such that the square of the length of the diagonal is $\\quantity{170}{m^2}$.", "numbers in the brackets) $=$= $100$100 (Perform the multiplication) So we find that one lap of the race track is $100$100 m long. Reflect: First we identified that one lap was equal to the perimeter of the race track, then we applied the formula for the perimeter of a rectangle to find the lap length. #### Worked example 2 A $32$32 cm long piece of wire is bent into the shape of a square. What is the side length of the square? Think: Since the piece of wire was $32$32 cm long, we know that the perimeter of the square will be $32$32 cm. We can try reversing the formula for the perimeter of a square, Perimeter $=$=$4$4$\\times$×Length, to find a solution. Do: We know that $32$32 cm $=4\\times$=4×Length, so we can find the number that multiplies with $4$4 to give $32$32. This can be found by dividing the perimeter by the number of sides, which gives $\\frac{32}{4}=8$324=8. So we find that the side length of the square is $8$8 cm. Reflect: First we identified that the square had a perimeter equal to the length of the wire, then we reversed the formula for the perimeter of a square to find the side length. ### Perimeter of regular polygons Regular polygons are special in that all of their", "in 2:3 so why not make the sides 2 and 3? So we have perimeter of rectangle 10 Since square has same perimeter we have 2.5 for each side of square. Now we can double the perimeter of square (for easier calc later) to 20 so each side is 5. To adjust on R we double those sides to so the new sides for R is 4:6 (which reduces to 2:3 so everything's good) Now we have area of R = 4 * 6 =24 vs area of S = 5 * 5=25 so B _________________ Kudos [?]: 281 [1], given: 52 Manager Joined: 27 Oct 2010 Posts: 178 Kudos [?]: 12 [0], given: 20 Re: If the perimeter of square region S and the perimeter of [#permalink] ### Show Tags 06 Feb 2011, 10:16 4s = 2 (l + b) s = (l + b) / 2 l / b = 2 / 3 l = 2 * b / 3 l * b / s^2 = (2 * b /3) * b / ((l + b)/2)^2 = 8 * b^2/3 / (2 * b / 3 + b)^2 = 24/25 B Kudos [?]: 12 [0], given: 20 Math Forum Moderator Joined: 20 Dec 2010 Posts: 1947 Kudos [?]: 2174 [1], given: 376 Re: If the perimeter", "of the rectangle BCEF are: B (0, 3) C (4, -1), E (2, -3), F (-2, 1), Now, The length of BC = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(4 – 0)² + (3 + 1)²}$$ = 5.65, The length of CE = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(4 – 2)² + (3 – 1)²}$$ = 2.82, The length of EF = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(1 + 3)² + (2 + 2)²}$$ = 5.65, The length of FB = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(0 + 2)² + (3 – 1)²}$$ = 2.82, So, The perimeter of the rectangle BCEF = BC + CE + EF + FB = 5.65 + 2.82 + 5.65 + 2.82 = 16.94 Hence from the above we can conclude that the perimeter of the rectangle BCEF is 16.94. Question 19. Find the perimeter of △ABF. The perimeter is about 12.17 units, Explanation: Question 20. Find the perimeter of quadrilateral ABCD. The perimeter of quadrilateral ABCD is 31.37, Explanation: From the given figure the given vertices of the quadrilateral ABCD are A (-5, 4), B (0, 3), C (4, -1), D (4, -5) Now, The length of AB = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(0 + 5)²", "# Do You Know The Ins And Outs Of This Figure? Geometry Level 2 The above shows a rectilinear figure (all its interior angles are either $90^\\circ$ or $270^\\circ$ only). Calculate the perimeter of this figure. ×", "&= 2(8)+2(6)\\\\P &= 16+12\\\\P &= 28 \\ inches$ Take a minute and copy the formula for finding the perimeter of a rectangle into your notebook. Now let's practice. Example A Find the perimeter of a square with a side length of 7 inches. Solution: 28 inches Example B Find the perimeter of a rectangle with a length of 9 feet and a width of 3 feet. Solution: 24 feet Example C Find the perimeter of a square with a side length of 2 centimeters. Solution: 8 centimeters Now back to Alex and the garden plot. Have you figured out what Alex should do? Here is the original problem once again. Alex is trying to figure out the perimeter of a square plot, a rectangular plot and the perimeter of a plot where the square and the rectangle are next to each other. Let's start with the square plot. $P &= 4s\\\\P &= 4(9) = 36 \\ feet$ The square plot has a perimeter of 36 feet. He will need 36 feet of fencing for the small plot. The rectangular plot has a length of 12 feet and a width of 8 feet. $P &= 2l+2w\\\\P &= 2(12)+2(8)\\\\P &= 24+16\\\\P &= 40 \\ feet$ Alex will need 40 feet of fencing for the rectangular plot. Vocabulary Perimeter the distance", "the following exercises, determine whether you would measure each item using linear, square, or cubic units. 129. amount of water in a fish tank 130. length of dental floss 131. living area of an apartment 132. floor space of a bathroom tile 133. height of a doorway 134. capacity of a truck trailer In the following exercises, find the perimeter and area of each figure. Assume each side of the square is $11$ cm. 135. 136. 137. 138. 139. 140. Use the Properties of Rectangles In the following exercises, find the perimeter and area of each rectangle. 141. The length of a rectangle is $8585$ feet and the width is $4545$ feet. 142. The length of a rectangle is $2626$ inches and the width is $5858$ inches. 143. A rectangular room is $1515$ feet wide by $1414$ feet long. 144. A driveway is in the shape of a rectangle $2020$ feet wide by $3535$ feet long. In the following exercises, solve. 145. Find the length of a rectangle with perimeter $124124$ inches and width $3838$ inches. 146. Find the length of a rectangle with perimeter $20.220.2$ yards and width of $7.87.8$ yards. 147. Find the width of a rectangle with perimeter $9292$ meters and length $1919$ meters. 148. Find the width of a rectangle with perimeter $16.216.2$ meters" ]

[ "## alainabbyboo22 If the perimeter of square 2 is 200 units, and the perimeter of square 1 is 150 units, what is the perimeter of square 3? A. 62.5 units B. 500 units C. 125 units D. 250 units one year ago one year ago 1. alainabbyboo22 2. alainabbyboo22 I am unsure how to do that. 3. davidkircos Find the side lengths of both. Perimeter/4 and then use the pythagorean theorem. 4. davidkircos Which part are you unsure on? 5. alainabbyboo22 The whole thing. 6. davidkircos Ok so find the side lengths of the Square 1 and 2 first. Side Length = Perimeter /4 Can you do that? 7. davidkircos Then, (Side Length 1)^2+(Side length 2)^2 = (Side Length 3)^2. And once you solve for Side Length 3 you multiply it by 4 to the perimeter. 8. alainabbyboo22 Can this be put into equation form? Like more understandably? I'm just really confused on it right now. :/ 9. davidkircos Sure. Square 1 has a perimeter of 150. That means that each side is one fourth of that because all sides are the same length on a square. So 150/4 = 37.5 Do the same thing for Square 2. 200/4 = 50. |dw:1334176657772:dw| Then you have the triangle: |dw:1334176742025:dw| And when you solve for the hypotenuse, that is the", "incredibly important for kids. How many different shapes can you draw that have perimeter of 20 centimeters? Explain using words or a diagram why the perimeter of a rectangle can be found by using a combination of addition and multiplication. 3 Understand and. A book of pentomino lessons, keyed to the Common Core State Standards. McRuffy Pentomino Puzzle Book (Pentominoes sold separately. Each piece is scored on both sides to show one inch segments. •Students will calculate the perimeter and area. A Spatial Reasoning Task Using Pentominoes Posted on November 16, 2015 January 15, 2018 Author Vicki McGinn Categories Math Lesson Observing people of all ages enthusiastically engaging in mathematical tasks is what fuels my passion for the work I do. Which pentominoes tessellate? Make a tessellation drawing for any pentomino. Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Perimeter = _____ square units Brain Teaser: You know that the 12 pentominoes will always have an area of 60 square units. The perimeter is 600 inches. Area and Perimeter using Pentominoes I began using math stations today. If you are looking for more perimeter and area ideas, I have a HUGE blog post on it HERE! Please Follow Me! Posted by", "in order to enclose a 1200 square foot rectangular pig pen. The pig pen is adjacent to the barn so he only needs to form three sides of the rectangular area as shown below. What dimensions should the pen be? 5. A rectangle with perimeter 138 units is divided into 8 congruent rectangles as shown in the diagram below. Find the perimeter and area of one of the 8 congruent rectangles. Perimeter: The distance around a shape. Area: The space inside a shape. 2. (a) $x = \\pm 11$ (b) $x = \\pm 2 \\sqrt{5}$ (c) $x = 4,2$ 3. $4 + 4 + 7 + 7 = 22$ Aug 28, 2013 Jan 21, 2015", "of semicircle is 3 units. 1. Given isosceles trapezium ABCD, find the area of rectangle ADFE, If the area of the trapezium is 72. Solution: Area of trapezium=$\\frac{1}{2}$×(AD+BC)×AE $\\frac{1}{2}$ x (8+12)×AE=72 $\\frac{1}{2}$×20×AE=72 10×AE=72 AE=$\\frac{72}{10}$ AE=7.2 Area of rectangle =8×7.2 =57.6 The area of the rectangle is 57.6 sq units. 1. Given a regular hexagon ABCDEF that has a perimeter 144, find the length of diagonal AD. Solution: In a regular hexagon, all sides are equal in length. So, Perimeter=6×side 144=6×BC BC=$\\frac{144}{6}$ BC=24 Also, diagonal=2×side", "Utilities) Followers: 11 Kudos [?]: 673 [0], given: 148 Re: The perimeter of square S [#permalink] 25 Aug 2018, 09:54 Carcass wrote: Attachment: square.jpg The perimeter of square S is equal to the perimeter of the rectangle above. Quantity A Quantity B The length of a side of S $$x+3$$ A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. Kudos for R.A.E The perimeter of square is =$$4 * side$$ and the perimeter of rectangle = $$2 *(Length + breadth)$$ Now give, $$4S = 2* (x + x + 6)$$ ( S = side of the square i.e. given) or $$4S = 4x + 12$$ or $$S = x + 3$$ Hence option C _________________ If you found this post useful, please let me know by pressing the Kudos Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Director Joined: 09 Nov 2018 Posts: 508 Followers: 0 Kudos [?]: 27 [0], given: 1 Re: The perimeter of square S [#permalink] 18 Jan 2019, 19:53 the perimeter of rectangle = x+x+x+6+x+6 = 4x+12 side of square= 4x+12 /4= x+3 Re: The perimeter of square S [#permalink] 18 Jan 2019, 19:53 Display posts from previous: Sort by", "# perimeter very easy • Feb 24th 2008, 02:24 AM kri999 perimeter very easy There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long and i have to find the perimeter..how do i do this? • Feb 24th 2008, 02:28 AM Jhevon Quote: Originally Posted by kri999 There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long and i have to find the perimeter..how do i do this? let the side length be $x$. then by Pythagoras' theorem, $x^2 + x^2 = (6 \\sqrt{3})^2 = 108$ solve for $x$. then the perimeter is given by $4x$ • Feb 24th 2008, 02:35 AM kri999 oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem • Feb 24th 2008, 02:38 AM Jhevon Quote: Originally Posted by kri999 oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem why? that would only make your life harder. then you'd have to use sine or cosine or tangent or something if you insist, you could do $\\sin 45 = \\frac x{6 \\sqrt{3}}$ and solve for x from that • Feb 24th 2008, 02:49 AM kri999 Yeah sorry the whole", "the following numbers using diagonal method: (i) 98 $∴ 98^{2}= 9604$ (ii) 273 $∴ 273^{2}= 74529$ (iii) 348 $∴ 348^{2}= 121104$ (iv) 295 $∴ 295^{2}= 87025$ (v) 171 $∴ 171^{2}= 29241$ 3.) Find the squares of the following numbers: We will use visual method as it is the efficient method to solve this problem. (i) We have: 127 = 120 + 7 Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units. Hence, the square of 127 is 16129. (ii) We have: 503 = 500 + 3 Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units. Hence, the square of 503 is 253009. (iii) We have: 451 = 450 + 1 Hence, let us draw a square of having side 451 units. Let us split it into 450 units and 1 units. Hence, the square of 451 is 203401. (iv) We have: 862 = 860 + 2 Hence, let us draw a square having side 862 units. Let us split it into 860 units and 2 units. Hence, the square of 862 is 743044. (v) We have: 265= 260 +5 Hence, let us draw a square having 265 units. Let us split it into 260 units and", "problem is forming the triangle. 1M is even, and so are the perimeters of the square and the rectangle; therefore the triangle perimeter must be small and even. We can't use 1 as a side length, because any scalene triangle with one side length = 1 unit will be degenerate. If we use 2 as a side length, in order for the perimeter to be even, the two larger sides must (a) differ by an even amount and (b) adhere to the inequality $2 + b > c$. That is, $2 > c-b$ and $2\\leq c-b$...so 2 is out as well. On to 3, where we take 3 as the smallest side length, and 4 and 5 will work too because they are the next smallest. This removes 12 units. We therefore need to create the smallest rectangle we can where $4|2(d+e)$. Taking $d = 1$ to minimize area of the rectangle, we must have $2|1+e$, where $e\\neq 1,3,5$. The next smallest number is 7, and so we have $f = (1000000-12-16)/4 = 249993$. The combined \"max area\" is therefore $\\frac{1}{2}(3)(4) + (1)(7) + (249,993)^2 = 13 + 62,496,500,049 = 62,496,500,062$. Now let's move on to the minimum area: We're going to maximize the allocation of the perimeter to the triangle, by creating the flattest scalene triangle we", "In the figure above, the large square is divided into two smaller squa [#permalink] ### Show Tags 30 Sep 2017, 02:43 Since the squares have a perimeter of 8 and 20, the sides of the squares are $$(\\frac{8}{4})2$$ and $$(\\frac{20}{4})5$$. The rectangles share their lengths and breadth with these 2 squares. Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively, and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14 Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E) _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Re: In the figure above, the large square is divided into two smaller squa [#permalink] 30 Sep 2017, 02:43 Display posts from previous: Sort by", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of \"2000 AIME I Problems/Problem 4\" ## Problem The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle. $[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]$ ## Solution Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle. The picture shows that \\begin{align*} a_1+a_2 &= a_3\\\\ a_1 + a_3 &= a_4\\\\ a_3 + a_4 &= a_5\\\\ a_4 + a_5 &= a_6\\\\ a_2 + a_3 + a_5 &= a_7\\\\ a_2 + a_7 &= a_8\\\\ a_1 + a_4 + a_6 &= a_9\\\\ a_6 + a_9 &= a_7 + a_8.\\end{align*} With a bit of trial and error and some arithmetic, we can use these equations to find that $5a_1 = 2a_2$; we can guess that $a_1 = 2$. Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started", "+0 # Help -1 296 1 +421 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"1\"? [asy] draw(unitsquare); draw(shift(1,0)*unitsquare); draw(shift(1,-1)*unitsquare); draw(shift(1,1)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); label(\"4\",(0.5,0.5)); label(\"5\",(1.5,0.5)); label(\"6\",(1.5,-0.5)); label(\"3\",(1.5,1.5)); label(\"2\",(2.5,1.5)); label(\"1\",(2.5,2.5)); [/asy] May 12, 2020 #1 +25543 +1 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"$$1$$\"? The product is $$\\mathbf{2\\times 3 \\times 4 \\times 6 = 144}$$ May 13, 2020", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of \"2000 AIME I Problems/Problem 4\" ## Problem The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle. $[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]$ ## Solution Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle. The picture shows that $a_1+a_2= a_3$, $a_1 + a_3 = a_4$, $a_3 + a_4 = a_5$, $a_4 + a_5 = a_6$, $a_2 + a_3 + a_5 = a_7$, $a_2 + a_7 = a_8$, $a_1 + a_4 + a_6 = a_9$, and $a_6 + a_9 = a_7 + a_8$. With a bit of trial and error and some arithmetic, we can use the last equation to find that $5a_1 = 2a_2$; without loss of generality, let $a_1 = 2$. Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$ (relatively prime), and the perimeter is $2(61)+2(69)=\\boxed{398}$.", "Other fractions by whole numbers reciprocal of 1/2 which is 2. in other,! = 216 x for y into the formula for perimeter P = x... ( that is, 0.00925925925.. ) Ask Login ) 144 in^2 B ) 216 in^2 C ) 1,216^2 ). Choices are: 432 units squared 650 units 3/10 divided by 2/4 as a fraction questions like: divided... Need to enter the dividend and divisor values 650 units can use base ten blocks to two! 20 times two is 40 and can use the remainder when P ( )! Entire decimal 31 is 6 with remainder 30 by 6 be enclosed and separated in this way with meters! Is waiting for your help like: what is 216 divided by 1/216 power -4/3 the. Y = 216 x for y into the formula for perimeter P = x... Get the answers you need, now length offense which is 2. in other,... 9 using long division method rectangle will require the smallest total length fence. Division or long division by 5= 7, 35 divided by 1/216 power -4/3 get the answers need... Is 484 m^2 like it can be enclosed by a = xy fixed! The maximum area that can be enclosed by a = xy is at... Division or long division the formula for perimeter P", "of the rectangle BCEF are: B (0, 3) C (4, -1), E (2, -3), F (-2, 1), Now, The length of BC = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(4 – 0)² + (3 + 1)²}$$ = 5.65, The length of CE = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(4 – 2)² + (3 – 1)²}$$ = 2.82, The length of EF = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(1 + 3)² + (2 + 2)²}$$ = 5.65, The length of FB = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(0 + 2)² + (3 – 1)²}$$ = 2.82, So, The perimeter of the rectangle BCEF = BC + CE + EF + FB = 5.65 + 2.82 + 5.65 + 2.82 = 16.94 Hence from the above we can conclude that the perimeter of the rectangle BCEF is 16.94. Question 19. Find the perimeter of △ABF. The perimeter is about 12.17 units, Explanation: Question 20. Find the perimeter of quadrilateral ABCD. The perimeter of quadrilateral ABCD is 31.37, Explanation: From the given figure the given vertices of the quadrilateral ABCD are A (-5, 4), B (0, 3), C (4, -1), D (4, -5) Now, The length of AB = $$\\sqrt{(x2 – x1)² + (y2 – y1)²}$$ = $$\\sqrt{(0 + 5)²", "$$\\frac{21}{6}$$ 3. $$\\frac{28}{12}$$ 4. What do the fraction problems have to do with the previous rectangle decomposition problems? ### 3.3: Finding Equivalent Fractions 1. Accurately draw a rectangle that is 9 units by 4 units. 1. In your rectangle, draw a line segment that decomposes the rectangle into a new rectangle and a square that is as large as possible. Continue until your original rectangle has been entirely decomposed into squares. 2. How many squares of each size are there? 3. What are the side lengths of the last square you drew? 4. Write $$\\frac94$$ as a mixed number. 2. Accurately draw a rectangle that is 27 units by 12 units. 1. In your rectangle, draw a line segment that decomposes the rectangle into a new rectangle and a square that is as large as possible. Continue until your original rectangle has been entirely decomposed into squares. 2. How many squares of each size are there? 3. What are the side lengths of the last square you drew? 4. Write $$\\frac{27}{12}$$ as a mixed number. 5. Compare the diagram you drew for this problem and the one for the previous problem. How are they the same? How are they different? 3. What is the greatest common factor of 9 and 4? What is the greatest common factor of", "squares are $$(\\frac{8}{4})2$$ and $$(\\frac{20}{4})5$$. The rectangles share their lengths and breadth with these 2 squares. Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively, and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14 Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E) _________________ You've got what it takes, but it will take everything you've got Re: In the figure above, the large square is divided into two smaller squa [#permalink] 30 Sep 2017, 03:43 Display posts from previous: Sort by", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. Difference between revisions of \"2000 AIME I Problems/Problem 4\" Problem The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle. $[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]$ Solution Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle. The picture shows that \\begin{align*} a_1+a_2 &= a_3\\\\ a_1 + a_3 &= a_4\\\\ a_3 + a_4 &= a_5\\\\ a_4 + a_5 &= a_6\\\\ a_2 + a_3 + a_5 &= a_7\\\\ a_2 + a_7 &= a_8\\\\ a_1 + a_4 + a_6 &= a_9\\\\ a_6 + a_9 &= a_7 + a_8.\\end{align*} Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$. We can guess that $a_1 = 2$. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need", "# Difference between revisions of \"2010 AMC 10B Problems/Problem 7\" ## Problem A triangle has side lengths $10$, $10$, and $12$. A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle? $\\textbf{(A)}\\ 16 \\qquad \\textbf{(B)}\\ 24 \\qquad \\textbf{(C)}\\ 28 \\qquad \\textbf{(D)}\\ 32 \\qquad \\textbf{(E)}\\ 36$ ## Solution The triangle is isosceles. The height of the triangle is therefore given by $h = \\sqrt{10^2 - ( \\dfrac{12}{2})^2} = \\sqrt{64} = 8$ Now, the area of the triangle is $\\dfrac{bh}{2} = \\dfrac{12*8}{2} = \\dfrac{96}{2} = 48$ We have that the area of the rectangle is the same as the area of the triangle, namely $48$. We also have the width of the rectangle: $4$. The length of the rectangle therefore is: $l = \\dfrac{48}{4} = 12$ The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$ $\\boxed{\\textbf{(D)}\\ 32}$ An alternative way to find the area of the triangle is by using Heron's formula, $A=\\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$. Thus the area equals $\\sqrt{(16)(16-10)(16-10)(16-12)} = \\sqrt{16*6*6*4} = 48.$ 2010 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 •", "the height is one-sixth the perimeter, for a maximum area of one twenty-fourth of the perimeter value (in square units, of course). If you extend that to even more congruent rectangular pens placed end to end it turns out that the width stays the same? If you extend this, as many teachers do, to a shape like this one, it starts to get interesting to the more observant student. The height is now one-eighth of the total , and the width is one-sixth of the perimeter. Even slow students can figure out that the area of the total turns out to be one forty-eighth of the value of the perimeter. Some will even generalize the area of an array of mxn congruent pens to be the perimeter squared divided by [4(m+1)(n+1))].... but it seems to take forever for some one to finally realize that in all these problems, whether there was one rectangle, or one with one side adjacent to a river or barn, or rows of rectangles... there is one invariant... Theorem one for rectangular arrays of congruent rectangular pens \"The amount of fence going right to left is the same as the amount going up and down.\" If you want to maximize a rectangular array with five rows of three pens in each row, we will", "the height is one-sixth the perimeter, for a maximum area of one twenty-fourth of the perimeter value (in square units, of course). If you extend that to even more congruent rectangular pens placed end to end it turns out that the width stays the same? If you extend this, as many teachers do, to a shape like this one, it starts to get interesting to the more observant student. The height is now one-eighth of the total , and the width is one-sixth of the perimeter. Even slow students can figure out that the area of the total turns out to be one forty-eighth of the value of the perimeter. Some will even generalize the area of an array of mxn congruent pens to be the perimeter squared divided by [4(m+1)(n+1))].... but it seems to take forever for some one to finally realize that in all these problems, whether there was one rectangle, or one with one side adjacent to a river or barn, or rows of rectangles... there is one invariant... Theorem one for rectangular arrays of congruent rectangular pens \"The amount of fence going right to left is the same as the amount going up and down.\" If you want to maximize a rectangular array with five rows of three pens in each row, we will" ]

[ "do you notice about the number of bags of lemons, the number of lemons, and the cost? __________________ E. Another way to find the unknown number is to find a factor that generates the equivalent ratio. When both quantities are multiplied by the same number, the result is an equivalent ratio. Find the unknown number using this method. Answer: F. What is the cost of 24 lemons? __________________ Answer: The cost of 24 lemons is$16. Turn and Talk How could you use the relationship between 6 lemons and $4 to find the relationship between 24 lemons and the cost for 24 lemons? Answer: 6 lemons =$4 24 lemons = x x × 6 = 4 × 24 x × 6 = 96 x = 96/6 x = 16 Thus you spend $16 for 24 lemons. 2. Jarrah wants to make a batch of slime. The diagram shows the ratio of hot water to cold water needed to make the slime. He wants to use 18 parts of water in total. How many parts of cold water does he need to make the slime? A. Look at the tape diagram. What is the total parts of water shown? _________________________________ Answer: Hot water has 6 parts = 6 × 2 = 12 parts Cold water has 3 parts = 3", "I mixed up some lemonade in two glasses. The first glass had $200$ml of lemon juice and $300$ml of water. The second glass had $100$ml of lemon juice and $200$ml of water. Which mixture has the stronger tasting lemonade? How do you know? Use the interactivity below to compare different mixtures of lemonade and develop a strategy for deciding which is stronger each time. Full Screen Version This text is usually replaced by the Flash movie. Once you are confident that you can always work out which mixture is stronger, here are some questions to consider: How might you use fractions to help you to work out which mixture is stronger? How might you use ratios?", "quarts of blueberries to make our salad, how can we use a ratio table to find out how many quarts of strawberries were used? We could start with the ratio 1: 3 that was given in the description and then multiply by ten to get 10 and 30. These would be the first values in our table. Then, we would count up by tens in the Blueberries column and count up by 30’s in the Strawberries column. ### Eureka Math Grade 6 Module 1 Lesson 10 Exercise Answer Key Exercise 1. The following tables were made incorrectly. Find the mistakes that were made, create the correct ratio table, and state the ratio that was used to make the correct ratio table. a. b. ### Eureka Math Grade 6 Module 1 Lesson 10 Problem Set Answer Key Question 1. a. Create a ratio table for making lemonade with a lemon juice-to-water ratio of 1: 3. Show how much lemon juice would be needed if you use 36 cups of water to make lemonade. Lemon Juice (cups) Water (cups) 1 3 2 6 3 9 4 12 12 36 12 cups of lemon Juice would be needed if 36 cups of water is used. b. How is the value of the ratio used to create the table? The value of", "• 1 cup of water with $$\\frac14$$ teaspoon of powdered drink mix Students will need three small cups each; they just need a few sips of the mixture in each cup. ### Student Facing • I can use equivalent ratios to describe scaled copies of shapes. • I know that two recipes will taste the same if the ingredients are in equivalent ratios. Building On Building Towards ### Glossary Entries • equivalent ratios Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, $$8:6$$ is equivalent to $$4:3$$, because $$8\\boldcdot\\frac12 = 4$$ and $$6\\boldcdot\\frac12 = 3$$. A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because and are equivalent ratios. cups of water number of lemons 8 6 4 3", "for 1 cup sugar to 6 cups water (after boiling cranberries in it). When you combine the two, you have 2 cups of sugar to 12 cups of liquid in your pink lemonade. Or perhaps you prefer to use only half as much cranberry juice as lemonade, so you’d have 1 1/2 cups sugar to 9 cups liquid. Ratios can be written as fractions, but working with them involves thinking about the situation. You can’t just apply the standard fraction rules blindly, because it may not make sense that way. 1. Kimberly Beezhold says: I will try to find an example. Someone on Marilyn’s blog said you can add both parts of 2 ratios together, when written as a fraction without a common denominator. There was no example. Maybe it’s just confusing what the comment was referencing. I think your example shows this: 1/6 ratio of sugar to liquid for lemonade; 1/6 ratio of sugar to cranberry juice. Put them together and you have 2 cups of sugar to 12 cups of liquid. Does that mean 1/6 + 1/6 = 2/12? Maybe I’ll get it if I write it out. I see now why it can be confusing. Thank you for the example. 2. Kimberly Beezhold says: I think the doubling threw me off,so your example helps. ☺", "24 miles d. 28 miles DOK Level 3) 5. A drink recipe calls for 1 part lemonade, 3 parts orange juice, and 4 parts water. How much lemonade, orange juice, and water are needed to make 64 fluid ounces of the drink using the recipe? Explain how you found your answer. Answer: Sixty-four fluid ounces of the drink will require 8 fluid ounces lemonade, 24 fluid ounces orange juice, and 32 fluid ounces water.My first step was to find the total number of parts: 1 part + 3 parts + 4 parts = 8 parts.Then I divided the total number of fluid ounces by the total number of parts: $\\frac{64\\ fluid\\ ounces}{8\\ parts}=\\frac{8\\ fluid\\ ounces}{1\\ part}$. Next I multiplied the number of parts by $\\frac{8\\ fluid\\ ounces}{1\\ part}$ to find the number of fluid ounces of each ingredient: lemonade - $1\\ part \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=8$ fluid ounces; orange juice - $3\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=24$ fluid ounces; water - $4\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=32$ fluid ounces. I know my answer is reasonable because when I add the number of fluid ounces of each ingredient, I get 64 fluid ounces: (8 fluid ounces lemonade + 24 fluid ounces orange juice + 32 fluid ounces water = 64 fluid ounces drink). (DOK Level 3)", "+0 # Math -1 47 1 +120 For field day, Mrs. Harris is preparing punch. She wants to make 9 gallons of punch. The recipe she found is only for 3 quarts. The recipe calls for 1 1/2 cups of pineapple juice, 1/4 cup of orange juice, and the rest lemon-lime soda. In order to make 9 gallons of punch, how much of each ingredient will Mrs. Harris need? Apr 20, 2021 #1 +496 0 keep in mind that 1 gallon = 4 quarts and 1 quart = 4 cups So the recipe has 12 cups of punch. 1 1/2 cups of it is pineapple, 1/4 cups of it is orange, and 12 - 1 1/2 - 1/4 = 10 1/4 cups of lemon-lime soda. 9 gallons of punch is equivilent to 36 quarts of punch, or 12 times the amount of the recipe. so, she needs: 1 1/2$\\cdot$12=$\\boxed{18 \\ cups}$ of pinaple juice 1/4$\\cdot$ 12 = $\\boxed{3 \\ cups}$ of orange juice 10 1/4 $cdot$ 12 = $\\boxed{123 \\ cups}$ of lemon lime soda. Apr 20, 2021", "of water to pure lemon juice, by volume, in the perfect mix is $$13:5$$.", "Home > CC2 > Chapter Ch5 > Lesson 5.2.4 > Problem5-62 5-62. A lemonade recipe calls for using a ratio of $2$ cups of lemon juice for every $4$ cups of water. Based on the ratio, how much lemon juice and water would be needed to make $6$ cups of lemonade? How much lemon juice is required with every $4$ cups of water to create $6$ cups of lemonade? 3. Angel made $10$ cups of lemonade. She used $3$ cups of lemon juice in her mixture. Did she follow the same recipe? In other words, did she use the same ratio of lemon juice to total liquid? Is $3$ cups of lemon juice for $10$ cups of lemonade the same ratio that you found above? Test to see if Angel followed the same mixture ratio by checking if his ratio is equivalent to the recipe given. She did not; $\\frac{3}{10}\\ne\\frac{2}{6}$ because the two ratios cannot be related by a Giant One.", "24 movie posters, how many band posters should she sell? Explain. The ratio to be maintained is 120 : 100 = $$\\frac{120}{100}$$. After selling 24 of the movie posters, she has sold, $$\\frac{24}{120}$$ × 100 = 20% of them, so she must sell 20% of the other posters too, that is 100 × 20% = 20. She must sell 20 band posters. Question 15. Bob needs to mix 2 cups of liquid lemonade concentrate with 3.5 cups of water to make lemonade. Bob has 6 cups of lemonade concentrate. How much lemonade can he make? The ratio of lemonade concentrate to water is: $$\\frac{\\text { lemonade }}{\\text { water }}=\\frac{2}{3.5}$$ Multiply numerator and denominator by 3 to find the amount of the water he needs to use: $$\\frac{2}{3.5} \\cdot \\frac{3}{3}=\\frac{6}{10.5}$$ He will mix 6 cups of lemonade concentrate and 10.5 cups of water He can make 16.5 cups of lemonade. Question 16. Multistep The ratio of North American butterflies to South American butterflies at a butterfly park is 5:3. The ratio of South American butterflies to European butterflies is 3:2. There are 30 North American butterflies at the butterfly park. a. How many South American butterflies are there? Given ratio: 5 : 3 = $$\\frac{5}{3}$$ The number of North American butterflies is 30 so the equation of proportion", "a recipe for lemonade says, “Mix 2 scoops of lemonade powder with 6 cups of water.”", "### Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have? Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have?... ### How do properties of water influence this relationship How do properties of water influence this relationship... ### Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE...", "+0 # need help 0 78 1 Vonnie likes her lemonade to have a water:lemon juice ratio of 7:2. Lonnie prefers a water: lemon juice ratio of 6:1. How much water would Lonnie have to add to 350 gallons of Vonnie's lemonade in order for it to be to her liking? Jun 27, 2020 In Vonnie's 350 gallons of lemonade, she has $$272\\frac{2}{9}$$ gallons of water $$77 \\frac{7}{9}$$ gallons of lemon juice (in the ratio of 7:2). If we want to add more water to make it the same ratio as Lonnie's lemonade we need to write an equation. $$\\frac{272 \\frac{2}{9}+x}{77 \\frac{7}{9}}=\\frac{6}{1}$$. I am going to leave it up to you to solve the equation.", "numerator: 2/5 = 1/5 X 2 = 0.40 3 Ratio: Ratio is a division finding how many times one entity in to other. And it is a fraction for one in numerator. 4. All are written in fraction form with same value Division and Fractions have applications in daily life where as ratio being an another form of division and fraction does not find place in daily life applications. Therefore ratio is not an independent entity but a name to refer how many times in a division in context. Probably it is good idea not to teach ratio as an independent entity. Therefore ratio is an application of fractions like percentage, rebate, loss and profit. They all use the properties of equivalent fractions. Proportion too is an equivalent fraction. There are several possible fractions one could associate with a given ratio. Say that a recipe for lemonade calls for $$2$$ cups of lemon juice and $$5$$ cups of water. This could be expressed with the ratio $$2:5$$. Here are some different associated fractions: 1. $$2/7$$ of the lemonade is lemon juice. 2. $$5/7$$ of the lemonade is water. 3. $$2/5$$ of a cup of lemon juice is the amount which needs to be added to 1 cup of water. 4. $$5/2$$ of a cup of water is the", "of 4 chairs and 4 tables? Solution: Given that, The cost of a chair =$ 52 The cost of a table = $96 The cost of 4 chairs = 52 x 4 = 208 The cost of 4 tables = 96 x 4 = 384 The total cost of 4 chairs and tables = 208 + 384 = 592 Therefore, the cost of 4 chairs and tables is$592. Question 9: Solution: Given that, The number of cookies found at a friend = 120 ÷ 4 = 30 So, each friend gets 30 cookies. Question 10: Mary requires 4 lemons to make a glass of juice. How many glasses of juice can be prepared with 52 lemons? Solution: Given that, Mary requires 4 lemons to make a glass of juice. The number of lemons = 52 The number of glasses of juice = 52 ÷ 4 = 13 Therefore, 13 glasses of lemon juice can be prepared with 15 lemons. Scroll to Top", "1$$\\frac{11}{12}$$ (C) 9$$\\frac{1}{12}$$ (D) 9$$\\frac{11}{12}$$ Answer: D Explanation: 5$$\\frac{6}{12}$$ + 4$$\\frac{5}{12}$$ = 9$$\\frac{11}{12}$$ is the horse trail Question 13. Multi-Step Students bring 8$$\\frac{7}{8}$$ gallons of lemonade to a picnic. They drink 5$$\\frac{2}{8}$$ gallons with lunch. Then they drink 2$$\\frac{1}{8}$$ gallons with an afternoon snack. How much lemonade is left? (A) 3$$\\frac{5}{8}$$ gallons (B) 6$$\\frac{3}{4}$$gallons (C) 5$$\\frac{3}{4}$$gallons (D) 1$$\\frac{1}{2}$$gallons Answer: D Explanation: 5$$\\frac{2}{8}$$ + 2$$\\frac{1}{8}$$ = 7$$\\frac{3}{8}$$ 8$$\\frac{7}{8}$$ – 7$$\\frac{3}{8}$$ = 1$$\\frac{4}{8}$$ = 1$$\\frac{1}{2}$$gallons lemonade is left TEXAS Test Prep Question 14. Jeff used 4$$\\frac{7}{8}$$ cups of orange juice and 3$$\\frac{1}{8}$$ cups of pineapple juice to make a tropical punch. How much more orange juice than pineapple juice did Jeff use? (A) $$\\frac{3}{4}$$ (B) 1$$\\frac{3}{4}$$ (C) 1$$\\frac{7}{8}$$ (D) 8 cups Answer: B Explanation: 4$$\\frac{7}{8}$$ – 3$$\\frac{1}{8}$$ = 1$$\\frac{6}{8}$$ 1$$\\frac{3}{4}$$ more orange juice than pineapple juice that Jeff use ### Texas Go Math Grade 4 Lesson 5.6 Homework and Practice Answer Key Write the sum as a mixed number with the fractional part less than 1. Question 1. Answer: Explanation: Added the wholes then added the fractions written the sum Question 2. Answer: Explanation: Added the wholes then added the fractions written the sum Question 3. Answer: Explanation: Added the wholes then added the fractions written the sum Question 4. Answer: Explanation: Added the wholes then added the fractions written", "cost of each lemon in the bag? A. $\\2.50$ B. $\\0.60$ C. $\\0.40$ D. $\\0.10$", "mL #### SOLUTION Solution : A 1 L=1000 mL 6 L=6×1000 mL=6000 mL Thus, the total quantity of lemonade to be prepared is 6000 mL. The quantity of lime juice that we have is 3800 mL. Quantity of water required to make the lemonade is given by 6000 mL - 3800 mL = 2200 mL. 4 kg - 400 g = ____ A. 2400 g B. 2600 g C. 3600 g D. 3400 g #### SOLUTION Solution : C Both the quantities 4 kg and 400 g have different units. So let's convert 4 kg to grams. We know that 1 kg = 1000 g (To convert a larger unit (kg) to a smaller unit (g) we multiply by the conversion factor.) Therefore, 4 kg = 4 × 1000 g Now, 4000 g - 400 g = 3600 g", "in the comments on Marilyn’s blog, was ratios, written like fractions, but added the way you said for the test question. So, maybe you can add 1/10 + 1/10 and get 2/20. What do you think? I can’t say I’ve done this activity with the hundred chart. We used it mainly for addition and doubling until we were off the chart. I know your children have read books, so I’m sure you will agree sometimes it’s good to let them have some math time that is just exploring. I also just have math discussions and we used to play games, from Scholastic Professional Books. I also found NCTM to be helpful .Math forum just joined with them. I’m sure you know about these resources. I just wanted to let you know what I’ve actually used personally. 2. Kim Beezhold says: Marilyn’s blog is very interesting.Thank you for sharing it with me. I see ratios being written like fractions, but being added differently. That is very confusing. 1. Kimberly Beezhold says: How would you add 2 ratios together? 1. Adding two ratios is sort of like doubling a recipe. Say you want to make pink lemonade. Your lemonade recipe calls for 1 cup sugar to 6 cups liquid (water and some lemon juice). And your cranberry juice recipe calls", "1,504 Answered Questions for the topic Fractions 1d #### evaluate 17(1-1/2^2)(1-1/3^2)...(1-1/17^2) please also show me the steps by steps solution and the quickest way possible to get the answer,thanks. Fractions Ratios 14d Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.2.5cm to 3cm 16d #### I need help answering this problem Carl the camel drank 15 1/4 quarts of water last week and 21 7/8 quarts of water this week. How many more quarts of water did Carl drink this week? 17d #### word problems! ( elementary math) Kevin ran 6/7 miles and his brother James ran 10 miles. How many miles did the two of them run?​ 17d #### word problems! ( elementary math) Kevin ran 6/7 miles and his brother James ran 10 miles. How many miles did the two of them run?​ 19d #### Basic math fractions! Joseph had 3 gallons of lemonade. He wanted to give each of his 1/8friends of a gallon. How many friends would receive lemonade?​ #### Fractions problem! (5th grade math fractions) Joseph had 3 gallons of lemonade. He wanted to give each of his 1/8friends of a gallon. How many friends would receive lemonade?​ Fractions 05/11/21 #### math / Fractions What fraction of an order of seafood will each of Dwight's" ]

[ "24 miles d. 28 miles DOK Level 3) 5. A drink recipe calls for 1 part lemonade, 3 parts orange juice, and 4 parts water. How much lemonade, orange juice, and water are needed to make 64 fluid ounces of the drink using the recipe? Explain how you found your answer. Answer: Sixty-four fluid ounces of the drink will require 8 fluid ounces lemonade, 24 fluid ounces orange juice, and 32 fluid ounces water.My first step was to find the total number of parts: 1 part + 3 parts + 4 parts = 8 parts.Then I divided the total number of fluid ounces by the total number of parts: $\\frac{64\\ fluid\\ ounces}{8\\ parts}=\\frac{8\\ fluid\\ ounces}{1\\ part}$. Next I multiplied the number of parts by $\\frac{8\\ fluid\\ ounces}{1\\ part}$ to find the number of fluid ounces of each ingredient: lemonade - $1\\ part \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=8$ fluid ounces; orange juice - $3\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=24$ fluid ounces; water - $4\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=32$ fluid ounces. I know my answer is reasonable because when I add the number of fluid ounces of each ingredient, I get 64 fluid ounces: (8 fluid ounces lemonade + 24 fluid ounces orange juice + 32 fluid ounces water = 64 fluid ounces drink). (DOK Level 3)", "do you notice about the number of bags of lemons, the number of lemons, and the cost? __________________ E. Another way to find the unknown number is to find a factor that generates the equivalent ratio. When both quantities are multiplied by the same number, the result is an equivalent ratio. Find the unknown number using this method. Answer: F. What is the cost of 24 lemons? __________________ Answer: The cost of 24 lemons is$16. Turn and Talk How could you use the relationship between 6 lemons and $4 to find the relationship between 24 lemons and the cost for 24 lemons? Answer: 6 lemons =$4 24 lemons = x x × 6 = 4 × 24 x × 6 = 96 x = 96/6 x = 16 Thus you spend $16 for 24 lemons. 2. Jarrah wants to make a batch of slime. The diagram shows the ratio of hot water to cold water needed to make the slime. He wants to use 18 parts of water in total. How many parts of cold water does he need to make the slime? A. Look at the tape diagram. What is the total parts of water shown? _________________________________ Answer: Hot water has 6 parts = 6 × 2 = 12 parts Cold water has 3 parts = 3", "### Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have? Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have?... ### How do properties of water influence this relationship How do properties of water influence this relationship... ### Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE...", "as follows: And…How much would two quarters of a pound cost? Here, we divide the single unit into four parts and take two parts. If we think about it, we can see that two quarters is equivalent to a half. So, two quarters of a pound would cost$2 as well. It can also be expressed as such: ##### Fraction Problems nº 2: Mark bought three quarters of a pound of lemons. How much did he pay? He bought three quarters of a pound, so we divide the unit into four parts and take three of them. Each pound is $4, so we divide 4 / 4 = 1 and multiply by 3. 1 x 3 = 3. He paid$3. When he got home, Mark squeezed the lemons and got a quarter-gallon of lemon juice. After, he added a half-gallon of water to make lemonade. How much lemonade did he make? Let’s draw it out: – He squeezed out a quarter-gallon of lemon juice – He added a half-gallon of water In order to add a quarter to a half, we have to find the common denominator. We should already know that two quarters are equal to one half! So, we add one quarter to two quarters.. And we get three quarters. That means that Mark made three quarters", "+0 # Math -1 47 1 +120 For field day, Mrs. Harris is preparing punch. She wants to make 9 gallons of punch. The recipe she found is only for 3 quarts. The recipe calls for 1 1/2 cups of pineapple juice, 1/4 cup of orange juice, and the rest lemon-lime soda. In order to make 9 gallons of punch, how much of each ingredient will Mrs. Harris need? Apr 20, 2021 #1 +496 0 keep in mind that 1 gallon = 4 quarts and 1 quart = 4 cups So the recipe has 12 cups of punch. 1 1/2 cups of it is pineapple, 1/4 cups of it is orange, and 12 - 1 1/2 - 1/4 = 10 1/4 cups of lemon-lime soda. 9 gallons of punch is equivilent to 36 quarts of punch, or 12 times the amount of the recipe. so, she needs: 1 1/2$\\cdot$12=$\\boxed{18 \\ cups}$ of pinaple juice 1/4$\\cdot$ 12 = $\\boxed{3 \\ cups}$ of orange juice 10 1/4 $cdot$ 12 = $\\boxed{123 \\ cups}$ of lemon lime soda. Apr 20, 2021", "+0 # need help 0 78 1 Vonnie likes her lemonade to have a water:lemon juice ratio of 7:2. Lonnie prefers a water: lemon juice ratio of 6:1. How much water would Lonnie have to add to 350 gallons of Vonnie's lemonade in order for it to be to her liking? Jun 27, 2020 In Vonnie's 350 gallons of lemonade, she has $$272\\frac{2}{9}$$ gallons of water $$77 \\frac{7}{9}$$ gallons of lemon juice (in the ratio of 7:2). If we want to add more water to make it the same ratio as Lonnie's lemonade we need to write an equation. $$\\frac{272 \\frac{2}{9}+x}{77 \\frac{7}{9}}=\\frac{6}{1}$$. I am going to leave it up to you to solve the equation.", "would result in the correct answer? Explain. Explanation: 3$$\\frac{1}{2}$$ is for one side for 2 sides it is 3$$\\frac{1}{2}$$ + 3$$\\frac{1}{2}$$ = 7 7 feet of wood Ezra needs for both sides 7 feet to inches 1 feet = 12 inches 7 x 12 =84 inches Question 2. Leanne needs 100 cups of lemonade for guests at a picnic. Each package of lemonade mix makes 1 gallon and costs $3.00. How much will it cost Leanne to make enough lemonade? A. First find the number of gallons of lemonade Leanne needs. Convert the number of cups needed to a number of pints, then to a number of quarts, and then to a number of gallons. Write an equation to find how many pints of lemonade Leanne needs to make. Answer: 1 pint = 2 cups so, We divide the volume value by 2 100 ÷ 2 = 50 pints 1 quart = 2 pints so, we have to divide the volume value by 2 50 ÷ 2 = 25 quarts 1 gallon = 4 quarts so, we have to divide the volume value by 4 25 ÷ 4 = 6 gallon and 1 quart. Write an equation to find how many quarts of lemonade Leanne needs to make. Answer: Let the unknown quarts of lemonade be n n", "pounds of lemons he uses,and x 2 is the number of labor-hours spent squeezing them. (a) Does Earl have constant returns to scale,decreasing returns to scale, or increasing returns to scale? Decreasing. (b) Where w 1 is the cost of a pound of lemons and w 2 is the wage rate for lemon-squeezers,the cheapest way for Earl to produce lemonade is to use w 1 =w 2 hours of labor per pound of lemons,(Hint,Set the slope of his isoquant equal to the slope of his isocost line.) (c) If he is going to produce y units in the cheapest way possible, then the number of pounds of lemons he will use is x 1 (w 1;w 2;y)= w 1=2 2 y 3=2 =w 1=2 1 and the number of hours of labor that he will use is x 2 (w 1;w 2;y)= w 1=2 1 y 3=2 =w 1=2 2 ,(Hint,Use the production func- tion and the equation you found in the last part of the answer to solve for the input quantities.) (d) The cost to Earl of producing y units at factor prices w 1 and w 2 is c(w 1;w 2;y)=w 1 x 1 (w 1;w 2;y)+w 2 x 2 (w 1;w 2;y)= 2w 1=2 1 w 1=2 2 y 3=2 . 20.5 (0) The", "= lemon juice. then since we know there is $3$ cups of lemon juice, do the math. $3\\cdot2\\cdot4=6\\cdot4=24$ ~ bsu1 ~savannahsolver", "quarts of blueberries to make our salad, how can we use a ratio table to find out how many quarts of strawberries were used? We could start with the ratio 1: 3 that was given in the description and then multiply by ten to get 10 and 30. These would be the first values in our table. Then, we would count up by tens in the Blueberries column and count up by 30’s in the Strawberries column. ### Eureka Math Grade 6 Module 1 Lesson 10 Exercise Answer Key Exercise 1. The following tables were made incorrectly. Find the mistakes that were made, create the correct ratio table, and state the ratio that was used to make the correct ratio table. a. b. ### Eureka Math Grade 6 Module 1 Lesson 10 Problem Set Answer Key Question 1. a. Create a ratio table for making lemonade with a lemon juice-to-water ratio of 1: 3. Show how much lemon juice would be needed if you use 36 cups of water to make lemonade. Lemon Juice (cups) Water (cups) 1 3 2 6 3 9 4 12 12 36 12 cups of lemon Juice would be needed if 36 cups of water is used. b. How is the value of the ratio used to create the table? The value of", "## anonymous one year ago Due to Lara’s success, she is increasing the price of her signature honey lemonade from 0.20 cents a cup to 0.25 cents. What is the percent increase? • This Question is Open 1. anonymous $\\frac{0.25\\, -0.2}{0.2}=\\frac{0.05}{0.2}=\\frac{5}{20}=\\frac{1}{4}=0.25=25 \\%$", "Difference between revisions of \"2020 AMC 8 Problems/Problem 1\" Problem Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\\textbf{(A) }6 \\qquad \\textbf{(B) }8 \\qquad \\textbf{(C) }12 \\qquad \\textbf{(D) }18 \\qquad \\textbf{(E) }24$ Solution 1 (Direct) We have $\\text{water} : \\text{sugar} : \\text{lemon juice} = 4\\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \\cdot 8 = \\boxed{\\textbf{(E) }24}$ cups. Solution 2 (Stepwise) Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\\cdot2=6$ cups of sugar. Since the amount of water is $4$ times the amount of sugar, he uses $6\\cdot4=\\boxed{\\textbf{(E) }24}$ cups of water. ~MRENTHUSIASM Solution 3 (Combination of Solutions 1 and 2) The ratio is $\\text{Water}:\\text{Sugar}:\\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\\cdot4=\\boxed{\\textbf{(E) }24}$ cups of water. Thanks to MRENTHUSIASM for the inspiration! EarthSaver 15:12, 11 June 2021 (EDT) ~savannahsolver Video Solution by The Learning Royal ~The Learning Royal", "24 movie posters, how many band posters should she sell? Explain. The ratio to be maintained is 120 : 100 = $$\\frac{120}{100}$$. After selling 24 of the movie posters, she has sold, $$\\frac{24}{120}$$ × 100 = 20% of them, so she must sell 20% of the other posters too, that is 100 × 20% = 20. She must sell 20 band posters. Question 15. Bob needs to mix 2 cups of liquid lemonade concentrate with 3.5 cups of water to make lemonade. Bob has 6 cups of lemonade concentrate. How much lemonade can he make? The ratio of lemonade concentrate to water is: $$\\frac{\\text { lemonade }}{\\text { water }}=\\frac{2}{3.5}$$ Multiply numerator and denominator by 3 to find the amount of the water he needs to use: $$\\frac{2}{3.5} \\cdot \\frac{3}{3}=\\frac{6}{10.5}$$ He will mix 6 cups of lemonade concentrate and 10.5 cups of water He can make 16.5 cups of lemonade. Question 16. Multistep The ratio of North American butterflies to South American butterflies at a butterfly park is 5:3. The ratio of South American butterflies to European butterflies is 3:2. There are 30 North American butterflies at the butterfly park. a. How many South American butterflies are there? Given ratio: 5 : 3 = $$\\frac{5}{3}$$ The number of North American butterflies is 30 so the equation of proportion", "if you own 10 Lemonade Stands: $cost_{next} = 4 \\times (1.07)^{10} = 7.87$ Meanwhile, the production rate is: $production_{total} = (1.67 \\times 10) \\times 1 = 16.7 / sec$ $^*$It’s technically 3.738, since you get the first one for free and it’s actually the second one that costs 4. Here are the actual values for AdVenture Capitalist (thank you AdCap wiki!): So early on you’re earning well beyond the cost of a new generator every second. Let’s look at that in graph form. Note that when you own 25 and 50 Lemonade Stands, the rate gets a (stacking) x2 multiplier. And we can see a slightly bigger picture if we use a log scale y-axis. Early on, your production rate is high and you can keep buying, and the multipliers bump you back up whenever it gets close. But eventually the costs become overwhelming, and the time you would have to wait to be able to afford the next generator gets prohibitively long. Mathematically, exponential growth (anything of the form $n^x$, for $n > 1$) will eventually catch and far exceed any polynomial growth ($x^k$) no matter how big $k$ is or how small $n$ is. (Thank you pickten for noticing I wrote these backwards at first!) This can take a very long time in some cases, but", "I mixed up some lemonade in two glasses. The first glass had $200$ml of lemon juice and $300$ml of water. The second glass had $100$ml of lemon juice and $200$ml of water. Which mixture has the stronger tasting lemonade? How do you know? Use the interactivity below to compare different mixtures of lemonade and develop a strategy for deciding which is stronger each time. Full Screen Version This text is usually replaced by the Flash movie. Once you are confident that you can always work out which mixture is stronger, here are some questions to consider: How might you use fractions to help you to work out which mixture is stronger? How might you use ratios?", "1$$\\frac{11}{12}$$ (C) 9$$\\frac{1}{12}$$ (D) 9$$\\frac{11}{12}$$ Answer: D Explanation: 5$$\\frac{6}{12}$$ + 4$$\\frac{5}{12}$$ = 9$$\\frac{11}{12}$$ is the horse trail Question 13. Multi-Step Students bring 8$$\\frac{7}{8}$$ gallons of lemonade to a picnic. They drink 5$$\\frac{2}{8}$$ gallons with lunch. Then they drink 2$$\\frac{1}{8}$$ gallons with an afternoon snack. How much lemonade is left? (A) 3$$\\frac{5}{8}$$ gallons (B) 6$$\\frac{3}{4}$$gallons (C) 5$$\\frac{3}{4}$$gallons (D) 1$$\\frac{1}{2}$$gallons Answer: D Explanation: 5$$\\frac{2}{8}$$ + 2$$\\frac{1}{8}$$ = 7$$\\frac{3}{8}$$ 8$$\\frac{7}{8}$$ – 7$$\\frac{3}{8}$$ = 1$$\\frac{4}{8}$$ = 1$$\\frac{1}{2}$$gallons lemonade is left TEXAS Test Prep Question 14. Jeff used 4$$\\frac{7}{8}$$ cups of orange juice and 3$$\\frac{1}{8}$$ cups of pineapple juice to make a tropical punch. How much more orange juice than pineapple juice did Jeff use? (A) $$\\frac{3}{4}$$ (B) 1$$\\frac{3}{4}$$ (C) 1$$\\frac{7}{8}$$ (D) 8 cups Answer: B Explanation: 4$$\\frac{7}{8}$$ – 3$$\\frac{1}{8}$$ = 1$$\\frac{6}{8}$$ 1$$\\frac{3}{4}$$ more orange juice than pineapple juice that Jeff use ### Texas Go Math Grade 4 Lesson 5.6 Homework and Practice Answer Key Write the sum as a mixed number with the fractional part less than 1. Question 1. Answer: Explanation: Added the wholes then added the fractions written the sum Question 2. Answer: Explanation: Added the wholes then added the fractions written the sum Question 3. Answer: Explanation: Added the wholes then added the fractions written the sum Question 4. Answer: Explanation: Added the wholes then added the fractions written", "(a) If lemons cost $1 per pound,the wage rate is$1 per hour,and the price of lemonade is p,Earl¡¯s marginal cost function is MC(y)= 3y 1=2 and his supply function is S(p)= p 2 =9,If lemons cost $4 per pound and the wage rate is$9 per hour,his supply function will be S(p)= p 2 =324. NAME 277 (b) In general,Earl¡¯s marginal cost depends on the price of lemons and the wage rate,At prices w 1 for lemons and w 2 for labor,his mar- ginal cost when he is producing y units of lemonade is MC(w 1;w 2;y)= 3w 1=2 1 w 1=2 2 y 1=2 ,The amount that Earl will supply depends on the three variables,p,w 1 ,w 2 ,As a function of these three variables,Earl¡¯s supply is S(p;w 1;w 2 )= p 2 =9w 1 w 2 . Calculus 22.8 (0) As you may recall from the chapter on cost functions,Irma¡¯s handicrafts has the production function f(x 1;x 2 )=(minfx 1;2x 2 g) 1=2 , where x 1 is the amount of plastic used,x 2 is the amount of labor used, and f(x 1;x 2 ) is the number of lawn ornaments produced,Let w 1 be the price per unit of plastic and w 2 be the wage per unit of labor. (a) Irma¡¯s cost function is c(w", "• 1 cup of water with $$\\frac14$$ teaspoon of powdered drink mix Students will need three small cups each; they just need a few sips of the mixture in each cup. ### Student Facing • I can use equivalent ratios to describe scaled copies of shapes. • I know that two recipes will taste the same if the ingredients are in equivalent ratios. Building On Building Towards ### Glossary Entries • equivalent ratios Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, $$8:6$$ is equivalent to $$4:3$$, because $$8\\boldcdot\\frac12 = 4$$ and $$6\\boldcdot\\frac12 = 3$$. A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because and are equivalent ratios. cups of water number of lemons 8 6 4 3", "pt. E. 100 pt. _________________ Math Expert Joined: 02 Sep 2009 Posts: 47983 ### Show Tags 16 Sep 2014, 00:35 1 2 Official Solution: To make lemonade, a recipe requires 4 lemons, 30 ounces of sugar, and 2 pints of water. From an average lemon one can squeeze out 0.4 cups of 5% citric acid. Marta does not have lemons, but wants to make lemonade from 4 cups of 12% citric acid. If she has sugar and water in abundance, how much water will Marta use? A. 2 pt. B. 6 pt. C. 12 pt. D. 15 pt. E. 100 pt. Standard recipe requires: 4 lemons and 2 pints of water. 4 lemons contain $$4*0.4*\\frac{5}{100}=\\frac{8}{100}$$ cups of citric acid; Now, Marta has $$4*\\frac{12}{100}=\\frac{48}{100}$$ cups of citric acid, so 6 times as many as the standard recipe requires ($$\\frac{48}{8}=6$$), which means that she'll need 6 times as much water as the standard recipe requires. Hence she'll need $$6*2=12$$ pints of water. _________________ Retired Moderator Joined: 29 Apr 2015 Posts: 862 Location: Switzerland Concentration: Economics, Finance Schools: LBS MIF '19 WE: Asset Management (Investment Banking) ### Show Tags 15 Jun 2015, 12:09 Bunuel wrote: Official Solution: To make lemonade, a recipe requires 4 lemons, 30 ounces of sugar, and 2 pints of water. From an average lemon one can squeeze", "mL #### SOLUTION Solution : A 1 L=1000 mL 6 L=6×1000 mL=6000 mL Thus, the total quantity of lemonade to be prepared is 6000 mL. The quantity of lime juice that we have is 3800 mL. Quantity of water required to make the lemonade is given by 6000 mL - 3800 mL = 2200 mL. 4 kg - 400 g = ____ A. 2400 g B. 2600 g C. 3600 g D. 3400 g #### SOLUTION Solution : C Both the quantities 4 kg and 400 g have different units. So let's convert 4 kg to grams. We know that 1 kg = 1000 g (To convert a larger unit (kg) to a smaller unit (g) we multiply by the conversion factor.) Therefore, 4 kg = 4 × 1000 g Now, 4000 g - 400 g = 3600 g" ]

[ "do you notice about the number of bags of lemons, the number of lemons, and the cost? __________________ E. Another way to find the unknown number is to find a factor that generates the equivalent ratio. When both quantities are multiplied by the same number, the result is an equivalent ratio. Find the unknown number using this method. Answer: F. What is the cost of 24 lemons? __________________ Answer: The cost of 24 lemons is$16. Turn and Talk How could you use the relationship between 6 lemons and $4 to find the relationship between 24 lemons and the cost for 24 lemons? Answer: 6 lemons =$4 24 lemons = x x × 6 = 4 × 24 x × 6 = 96 x = 96/6 x = 16 Thus you spend $16 for 24 lemons. 2. Jarrah wants to make a batch of slime. The diagram shows the ratio of hot water to cold water needed to make the slime. He wants to use 18 parts of water in total. How many parts of cold water does he need to make the slime? A. Look at the tape diagram. What is the total parts of water shown? _________________________________ Answer: Hot water has 6 parts = 6 × 2 = 12 parts Cold water has 3 parts = 3", "24 miles d. 28 miles DOK Level 3) 5. A drink recipe calls for 1 part lemonade, 3 parts orange juice, and 4 parts water. How much lemonade, orange juice, and water are needed to make 64 fluid ounces of the drink using the recipe? Explain how you found your answer. Answer: Sixty-four fluid ounces of the drink will require 8 fluid ounces lemonade, 24 fluid ounces orange juice, and 32 fluid ounces water.My first step was to find the total number of parts: 1 part + 3 parts + 4 parts = 8 parts.Then I divided the total number of fluid ounces by the total number of parts: $\\frac{64\\ fluid\\ ounces}{8\\ parts}=\\frac{8\\ fluid\\ ounces}{1\\ part}$. Next I multiplied the number of parts by $\\frac{8\\ fluid\\ ounces}{1\\ part}$ to find the number of fluid ounces of each ingredient: lemonade - $1\\ part \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=8$ fluid ounces; orange juice - $3\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=24$ fluid ounces; water - $4\\ parts \\times \\frac{8\\ fluid\\ ounces}{1\\ part}=32$ fluid ounces. I know my answer is reasonable because when I add the number of fluid ounces of each ingredient, I get 64 fluid ounces: (8 fluid ounces lemonade + 24 fluid ounces orange juice + 32 fluid ounces water = 64 fluid ounces drink). (DOK Level 3)", "+0 # need help 0 78 1 Vonnie likes her lemonade to have a water:lemon juice ratio of 7:2. Lonnie prefers a water: lemon juice ratio of 6:1. How much water would Lonnie have to add to 350 gallons of Vonnie's lemonade in order for it to be to her liking? Jun 27, 2020 In Vonnie's 350 gallons of lemonade, she has $$272\\frac{2}{9}$$ gallons of water $$77 \\frac{7}{9}$$ gallons of lemon juice (in the ratio of 7:2). If we want to add more water to make it the same ratio as Lonnie's lemonade we need to write an equation. $$\\frac{272 \\frac{2}{9}+x}{77 \\frac{7}{9}}=\\frac{6}{1}$$. I am going to leave it up to you to solve the equation.", "24 movie posters, how many band posters should she sell? Explain. The ratio to be maintained is 120 : 100 = $$\\frac{120}{100}$$. After selling 24 of the movie posters, she has sold, $$\\frac{24}{120}$$ × 100 = 20% of them, so she must sell 20% of the other posters too, that is 100 × 20% = 20. She must sell 20 band posters. Question 15. Bob needs to mix 2 cups of liquid lemonade concentrate with 3.5 cups of water to make lemonade. Bob has 6 cups of lemonade concentrate. How much lemonade can he make? The ratio of lemonade concentrate to water is: $$\\frac{\\text { lemonade }}{\\text { water }}=\\frac{2}{3.5}$$ Multiply numerator and denominator by 3 to find the amount of the water he needs to use: $$\\frac{2}{3.5} \\cdot \\frac{3}{3}=\\frac{6}{10.5}$$ He will mix 6 cups of lemonade concentrate and 10.5 cups of water He can make 16.5 cups of lemonade. Question 16. Multistep The ratio of North American butterflies to South American butterflies at a butterfly park is 5:3. The ratio of South American butterflies to European butterflies is 3:2. There are 30 North American butterflies at the butterfly park. a. How many South American butterflies are there? Given ratio: 5 : 3 = $$\\frac{5}{3}$$ The number of North American butterflies is 30 so the equation of proportion", "as follows: And…How much would two quarters of a pound cost? Here, we divide the single unit into four parts and take two parts. If we think about it, we can see that two quarters is equivalent to a half. So, two quarters of a pound would cost$2 as well. It can also be expressed as such: ##### Fraction Problems nº 2: Mark bought three quarters of a pound of lemons. How much did he pay? He bought three quarters of a pound, so we divide the unit into four parts and take three of them. Each pound is $4, so we divide 4 / 4 = 1 and multiply by 3. 1 x 3 = 3. He paid$3. When he got home, Mark squeezed the lemons and got a quarter-gallon of lemon juice. After, he added a half-gallon of water to make lemonade. How much lemonade did he make? Let’s draw it out: – He squeezed out a quarter-gallon of lemon juice – He added a half-gallon of water In order to add a quarter to a half, we have to find the common denominator. We should already know that two quarters are equal to one half! So, we add one quarter to two quarters.. And we get three quarters. That means that Mark made three quarters", "would result in the correct answer? Explain. Explanation: 3$$\\frac{1}{2}$$ is for one side for 2 sides it is 3$$\\frac{1}{2}$$ + 3$$\\frac{1}{2}$$ = 7 7 feet of wood Ezra needs for both sides 7 feet to inches 1 feet = 12 inches 7 x 12 =84 inches Question 2. Leanne needs 100 cups of lemonade for guests at a picnic. Each package of lemonade mix makes 1 gallon and costs $3.00. How much will it cost Leanne to make enough lemonade? A. First find the number of gallons of lemonade Leanne needs. Convert the number of cups needed to a number of pints, then to a number of quarts, and then to a number of gallons. Write an equation to find how many pints of lemonade Leanne needs to make. Answer: 1 pint = 2 cups so, We divide the volume value by 2 100 ÷ 2 = 50 pints 1 quart = 2 pints so, we have to divide the volume value by 2 50 ÷ 2 = 25 quarts 1 gallon = 4 quarts so, we have to divide the volume value by 4 25 ÷ 4 = 6 gallon and 1 quart. Write an equation to find how many quarts of lemonade Leanne needs to make. Answer: Let the unknown quarts of lemonade be n n", "quarts of blueberries to make our salad, how can we use a ratio table to find out how many quarts of strawberries were used? We could start with the ratio 1: 3 that was given in the description and then multiply by ten to get 10 and 30. These would be the first values in our table. Then, we would count up by tens in the Blueberries column and count up by 30’s in the Strawberries column. ### Eureka Math Grade 6 Module 1 Lesson 10 Exercise Answer Key Exercise 1. The following tables were made incorrectly. Find the mistakes that were made, create the correct ratio table, and state the ratio that was used to make the correct ratio table. a. b. ### Eureka Math Grade 6 Module 1 Lesson 10 Problem Set Answer Key Question 1. a. Create a ratio table for making lemonade with a lemon juice-to-water ratio of 1: 3. Show how much lemon juice would be needed if you use 36 cups of water to make lemonade. Lemon Juice (cups) Water (cups) 1 3 2 6 3 9 4 12 12 36 12 cups of lemon Juice would be needed if 36 cups of water is used. b. How is the value of the ratio used to create the table? The value of", "• 1 cup of water with $$\\frac14$$ teaspoon of powdered drink mix Students will need three small cups each; they just need a few sips of the mixture in each cup. ### Student Facing • I can use equivalent ratios to describe scaled copies of shapes. • I know that two recipes will taste the same if the ingredients are in equivalent ratios. Building On Building Towards ### Glossary Entries • equivalent ratios Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, $$8:6$$ is equivalent to $$4:3$$, because $$8\\boldcdot\\frac12 = 4$$ and $$6\\boldcdot\\frac12 = 3$$. A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because and are equivalent ratios. cups of water number of lemons 8 6 4 3", "### Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have? Suski is making lemonade to bring to the beach. She has two containers. One holds one gallon and the other holds 6 quarts. If she fills both containers, how many cups of lemonade will she have?... ### How do properties of water influence this relationship How do properties of water influence this relationship... ### Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE Find The Final cost after discount 8. The original cost was $48 with a 15% discount. The Final cost is$_ PLEASEEE HELPPPP PLEASEE PLEASEE HELPP NOWW PLEASEE...", "+0 # Math -1 47 1 +120 For field day, Mrs. Harris is preparing punch. She wants to make 9 gallons of punch. The recipe she found is only for 3 quarts. The recipe calls for 1 1/2 cups of pineapple juice, 1/4 cup of orange juice, and the rest lemon-lime soda. In order to make 9 gallons of punch, how much of each ingredient will Mrs. Harris need? Apr 20, 2021 #1 +496 0 keep in mind that 1 gallon = 4 quarts and 1 quart = 4 cups So the recipe has 12 cups of punch. 1 1/2 cups of it is pineapple, 1/4 cups of it is orange, and 12 - 1 1/2 - 1/4 = 10 1/4 cups of lemon-lime soda. 9 gallons of punch is equivilent to 36 quarts of punch, or 12 times the amount of the recipe. so, she needs: 1 1/2$\\cdot$12=$\\boxed{18 \\ cups}$ of pinaple juice 1/4$\\cdot$ 12 = $\\boxed{3 \\ cups}$ of orange juice 10 1/4 $cdot$ 12 = $\\boxed{123 \\ cups}$ of lemon lime soda. Apr 20, 2021", "1$$\\frac{11}{12}$$ (C) 9$$\\frac{1}{12}$$ (D) 9$$\\frac{11}{12}$$ Answer: D Explanation: 5$$\\frac{6}{12}$$ + 4$$\\frac{5}{12}$$ = 9$$\\frac{11}{12}$$ is the horse trail Question 13. Multi-Step Students bring 8$$\\frac{7}{8}$$ gallons of lemonade to a picnic. They drink 5$$\\frac{2}{8}$$ gallons with lunch. Then they drink 2$$\\frac{1}{8}$$ gallons with an afternoon snack. How much lemonade is left? (A) 3$$\\frac{5}{8}$$ gallons (B) 6$$\\frac{3}{4}$$gallons (C) 5$$\\frac{3}{4}$$gallons (D) 1$$\\frac{1}{2}$$gallons Answer: D Explanation: 5$$\\frac{2}{8}$$ + 2$$\\frac{1}{8}$$ = 7$$\\frac{3}{8}$$ 8$$\\frac{7}{8}$$ – 7$$\\frac{3}{8}$$ = 1$$\\frac{4}{8}$$ = 1$$\\frac{1}{2}$$gallons lemonade is left TEXAS Test Prep Question 14. Jeff used 4$$\\frac{7}{8}$$ cups of orange juice and 3$$\\frac{1}{8}$$ cups of pineapple juice to make a tropical punch. How much more orange juice than pineapple juice did Jeff use? (A) $$\\frac{3}{4}$$ (B) 1$$\\frac{3}{4}$$ (C) 1$$\\frac{7}{8}$$ (D) 8 cups Answer: B Explanation: 4$$\\frac{7}{8}$$ – 3$$\\frac{1}{8}$$ = 1$$\\frac{6}{8}$$ 1$$\\frac{3}{4}$$ more orange juice than pineapple juice that Jeff use ### Texas Go Math Grade 4 Lesson 5.6 Homework and Practice Answer Key Write the sum as a mixed number with the fractional part less than 1. Question 1. Answer: Explanation: Added the wholes then added the fractions written the sum Question 2. Answer: Explanation: Added the wholes then added the fractions written the sum Question 3. Answer: Explanation: Added the wholes then added the fractions written the sum Question 4. Answer: Explanation: Added the wholes then added the fractions written", "I mixed up some lemonade in two glasses. The first glass had $200$ml of lemon juice and $300$ml of water. The second glass had $100$ml of lemon juice and $200$ml of water. Which mixture has the stronger tasting lemonade? How do you know? Use the interactivity below to compare different mixtures of lemonade and develop a strategy for deciding which is stronger each time. Full Screen Version This text is usually replaced by the Flash movie. Once you are confident that you can always work out which mixture is stronger, here are some questions to consider: How might you use fractions to help you to work out which mixture is stronger? How might you use ratios?", "pt. E. 100 pt. _________________ Math Expert Joined: 02 Sep 2009 Posts: 47983 ### Show Tags 16 Sep 2014, 00:35 1 2 Official Solution: To make lemonade, a recipe requires 4 lemons, 30 ounces of sugar, and 2 pints of water. From an average lemon one can squeeze out 0.4 cups of 5% citric acid. Marta does not have lemons, but wants to make lemonade from 4 cups of 12% citric acid. If she has sugar and water in abundance, how much water will Marta use? A. 2 pt. B. 6 pt. C. 12 pt. D. 15 pt. E. 100 pt. Standard recipe requires: 4 lemons and 2 pints of water. 4 lemons contain $$4*0.4*\\frac{5}{100}=\\frac{8}{100}$$ cups of citric acid; Now, Marta has $$4*\\frac{12}{100}=\\frac{48}{100}$$ cups of citric acid, so 6 times as many as the standard recipe requires ($$\\frac{48}{8}=6$$), which means that she'll need 6 times as much water as the standard recipe requires. Hence she'll need $$6*2=12$$ pints of water. _________________ Retired Moderator Joined: 29 Apr 2015 Posts: 862 Location: Switzerland Concentration: Economics, Finance Schools: LBS MIF '19 WE: Asset Management (Investment Banking) ### Show Tags 15 Jun 2015, 12:09 Bunuel wrote: Official Solution: To make lemonade, a recipe requires 4 lemons, 30 ounces of sugar, and 2 pints of water. From an average lemon one can squeeze", "Home > CC2 > Chapter Ch5 > Lesson 5.2.4 > Problem5-62 5-62. A lemonade recipe calls for using a ratio of $2$ cups of lemon juice for every $4$ cups of water. Based on the ratio, how much lemon juice and water would be needed to make $6$ cups of lemonade? How much lemon juice is required with every $4$ cups of water to create $6$ cups of lemonade? 3. Angel made $10$ cups of lemonade. She used $3$ cups of lemon juice in her mixture. Did she follow the same recipe? In other words, did she use the same ratio of lemon juice to total liquid? Is $3$ cups of lemon juice for $10$ cups of lemonade the same ratio that you found above? Test to see if Angel followed the same mixture ratio by checking if his ratio is equivalent to the recipe given. She did not; $\\frac{3}{10}\\ne\\frac{2}{6}$ because the two ratios cannot be related by a Giant One.", "out 0.4 cups of 5% citric acid. Marta does not have lemons, but wants to make lemonade from 4 cups of 12% citric acid. If she has sugar and water in abundance, how much water will Marta use? A. 2 pt. B. 6 pt. C. 12 pt. D. 15 pt. E. 100 pt. Standard recipe requires: 4 lemons and 2 pints of water. 4 lemons contain $$4*0.4*\\frac{5}{100}=\\frac{8}{100}$$ cups of citric acid; Now, Marta has $$4*\\frac{12}{100}=\\frac{48}{100}$$ cups of citric acid, so 6 times as many as the standard recipe requires ($$\\frac{48}{8}=6$$), which means that she'll need 6 times as much water as the standard recipe requires. Hence she'll need $$6*2=12$$ pints of water. Can we start a discussion here and revive this task here a bit with the following questions? Since this is a mixture problem, how do we account for the different percentages of citric acid? I do not see this in the answer, it's not clear for me. The recipe requires 4 lemons i.e. 4*0.4 cups * 5%? Does this mean, the recipe requires 8 of what... (what? what's the measure after doing the calculations ) _________________ Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS! PS Please send me PM if I do not respond to your question within 24 hours. Intern Joined: 09", "# Difference between revisions of \"2020 AMC 8 Problems/Problem 1\" ## Problem Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\\textbf{(A) }6 \\qquad \\textbf{(B) }8 \\qquad \\textbf{(C) }12 \\qquad \\textbf{(D) }18 \\qquad \\textbf{(E) }24$ ## Solution 1 Luka will need $3\\cdot 2=6$ cups of sugar, and thus $6\\cdot 4=24$ cups of water. The answer is $\\boxed{\\textbf{(E) } 4}$. ## Solution 2 We have that $\\text{lemonade} : \\text{water} : \\text{lemon juice} = 4\\cdot 2 : 2 : 1 = 8 : 2 : 1$, so Luka needs $3 \\cdot 8 = \\boxed{\\textbf{(E) }24}$ cups.", "Difference between revisions of \"2020 AMC 8 Problems/Problem 1\" Problem Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\\textbf{(A) }6 \\qquad \\textbf{(B) }8 \\qquad \\textbf{(C) }12 \\qquad \\textbf{(D) }18 \\qquad \\textbf{(E) }24$ Solution 1 (Direct) We have $\\text{water} : \\text{sugar} : \\text{lemon juice} = 4\\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \\cdot 8 = \\boxed{\\textbf{(E) }24}$ cups. Solution 2 (Stepwise) Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\\cdot2=6$ cups of sugar. Since the amount of water is $4$ times the amount of sugar, he uses $6\\cdot4=\\boxed{\\textbf{(E) }24}$ cups of water. ~MRENTHUSIASM Solution 3 (Combination of Solutions 1 and 2) The ratio is $\\text{Water}:\\text{Sugar}:\\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\\cdot4=\\boxed{\\textbf{(E) }24}$ cups of water. Thanks to MRENTHUSIASM for the inspiration! EarthSaver 15:12, 11 June 2021 (EDT) ~savannahsolver Video Solution by The Learning Royal ~The Learning Royal", "one lemonade cup =$8$$\\frac{3}{4}$$ ÷ $$$\\frac{1}{4}$$ = 35. C. What number could you multiply the numerator and denominator by to write an equivalent fraction with whole numbers? What is the equivalent fraction? Answer: Rule for Multiplication of Fractions: When multiplying fractions, simply multiply the numerators together and then multiply the denominators together. Simplify the result. This works whether the denominators are the same or not. For example: If you multiply the fractions 3/2 and 4/3 together, you get 12/6 Explanation: In math, equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole. Rule for Multiplication of Fractions: When multiplying fractions, simply multiply the numerators together and then multiply the denominators together. Simplify the result. This works whether the denominators are the same or not. If you multiply the fractions$$$\\frac{1}{4}$$ D. Write and solve the new division problem shown by the equivalent fraction. Answer: Explanation: 216 ÷ 8 = 27. E. How can you use this quotient to solve the original division problem? Answer: Quotient × Divisor = Dividend => 27 × 8 = 216. Explanation: Dividend ÷ Divisor = Quotient. 216 ÷ 8 = 27. => Quotient × Divisor = Dividend => 27 × 8 = 216. F. How many cups of lemonade did Marisol sell?", "mL #### SOLUTION Solution : A 1 L=1000 mL 6 L=6×1000 mL=6000 mL Thus, the total quantity of lemonade to be prepared is 6000 mL. The quantity of lime juice that we have is 3800 mL. Quantity of water required to make the lemonade is given by 6000 mL - 3800 mL = 2200 mL. 4 kg - 400 g = ____ A. 2400 g B. 2600 g C. 3600 g D. 3400 g #### SOLUTION Solution : C Both the quantities 4 kg and 400 g have different units. So let's convert 4 kg to grams. We know that 1 kg = 1000 g (To convert a larger unit (kg) to a smaller unit (g) we multiply by the conversion factor.) Therefore, 4 kg = 4 × 1000 g Now, 4000 g - 400 g = 3600 g", "way to see this is by observing that $4\\frac58$ is less than 5 and $3\\frac78$ is less than 4, so $4\\frac58 + 3\\frac78$ is less than 9. Since there are less than 9 gallons of ingredients altogether they will certainly all fit in a 10-gallon container. 2. To see how much total punch is made we need to add the amount of lemon lime soda to the amount of fruit juice. The picture below represents $4\\frac58 + 3\\frac78$. We can write the mixed numbers as a sum of a whole number and a fraction. $$4\\frac58 + 3\\frac78 = (4 + \\frac{5}{8}) + (3 + \\frac{7}{8})$$ Since addition is commutative and associative, we can add the numbers in any order we wish. Let's add the whole numbers together and the fractions together. $$4+\\frac58+3+\\frac78 = 4+3+\\frac58+\\frac78$$ $$4+3+\\frac58+\\frac78 = 7 + \\frac{5+7}{8} = 7 + \\frac{12}{8}$$ Next we can re-write $\\frac{12}{8}$ as a mixed number... $$7 + \\frac{12}{8} = 7 + \\frac{8+4}{8} = 7 + \\frac88 + \\frac48 = 7+1+\\frac48$$ and add the whole numbers once again. $$7+1+\\frac48 = 8+\\frac48$$ Since $\\frac48=\\frac12$, we can write the sum as $8\\frac12$. So we see that this recipe makes $8\\frac12$ gallons of punch." ]

[ "# multiples of five Level pending how many four digit numbers can be formed using 0,1,2,3,4 which are multiples of 5 ×", "1 views Given Jigits $2,2,3,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed? 1. $50$ 2. $51$ 3. $52$ 4. $54$", "• mathmath333 How many even natural numbers divisible by 5 can be formed with digits 0,1,2,3,4,5,6 (if repetition is not allowed) Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. Question: A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is (a) 216 (b) 600 (c) 240 (d) 3125 Solution: (a) 216 A number is divisible by 3 when the sum of the digits of the number is divisible by 3. Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3. 1+2+3+4+5 = 15 0+1+2+4+5 = 12 Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers that can be formed = 5! Similarly, using the digits $0,1,2,4$ and 5, the number that can be formed $=5 !-4 !\\{$ since the first digit cannot be 0$\\}$ $\\therefore$ Total numbers that are possible $=5 !+5 !-4 !=240-24=216$", "Courses Courses for Kids Free study material Free LIVE classes More # Find the number of four digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of those will be even? Last updated date: 26th Mar 2023 Total views: 306.6k Views today: 7.83k 2 $\\times$ 4 $\\times$ 3 $\\times$ 2 = 48", "# Permutations\\Combinations from digits Using digits 1,2,3,4,5,7 only, how many numbers could be made that are between 2500 and 5000, if a digit is not repeated? (If it had been 2000-5000, that would have been easy!!!) The answer at my level should only use $^nP_{r}$, $^nC_{r}$ and\\or factorial notation. -", ". How many two-digit numbers can be formed from the numbers 1,2,3,4 is repetirion is allowed", "Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make? It Figures Stage: 2 Challenge Level: Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?", "Question How many 5 digit numbers divisible by 5 can be formed using 0,1,2,3,4,5 if repeatition not allowed. a)\" \"220 b)\" \"240 c)\" \"None Solution Since repeatition not allowed so we can use a number only once.For number to be divisible by 5, its unit place must have either 0 or 5. Case 1 Lets fix 0 at unit place the we can arrange rest digits in the form of 5xx4xx3xx2 = 120 Case 2 Now if we will 5 at unit place, we can arrange rest numbers as 4xx4xx3xx2 = 96 Hence total numbers = 120 + 96 = 216", "be formed. There are $3^4 = 81$ four-digit numbers that can be formed. There are $3^5 = 243$ five-digit numbers than can be formed. There are $3^6 = 729$ six-digit numbers than can be formed. There are $3^7 = 2187$ seven-digit numbers than can be formed. Therefore: . $n = 7$ I'll let someone else show you a more sophisticated method.", "# A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5. 15 views A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5. A. 1/4 B. 1/8 C. 3/4 D. 5/8 by (121k points) N/a", "if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #2: $$\\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$ 4.8 * 11111 =$$\\color{red}{53332.8}$$", "numbers. However, there are a lot more than 600 even numbers between 1000 and 2000!", "# High five Discrete Mathematics Level pending Suppose a 5-digit number is created from 5 distinct integers chosen among {1,2,3,4,5,6,12}. How many such 5-digit numbers include both 1 and 2 but neither begin nor end with 1? ×", "1, 2, 2, 2 Number of integers formed = 7! 3! 4! = 35 Total number of integers formed = 77.", "# Combinatorics exercise (numbers, digits...) • September 22nd 2011, 07:00 AM Evaldas Combinatorics exercise (numbers, digits...) There are four cards. In each there's one of these numbers written: 1, 2, 3, 4 (every used once). Three digit numbers are made with the cards, putting the cars variously. How many three digit numbers can be made up, that can be divided by three? My first question: is there a formula that'll help me to calculate this, or do I need to write all the possible three digit numbers and choose ones that can be divided by 3? • September 22nd 2011, 07:43 AM emakarov Re: Combinatorics exercise (numbers, digits...) I don't know a formula that would give the final answer. However, I recommend first finding out which (or, rather, how many) subsets of three cards are allowed and then using the standard formula for the number of permutations of each subset. • September 22nd 2011, 07:53 AM Evaldas Re: Combinatorics exercise (numbers, digits...) 4x3x2x1=24? • September 22nd 2011, 08:01 AM emakarov Re: Combinatorics exercise (numbers, digits...) I have a different answer. Do you want to use the suggestion I provided above? • September 22nd 2011, 08:11 AM Evaldas Re: Combinatorics exercise (numbers, digits...) Quote: Originally Posted by emakarov I have a different answer. Do you want to use the", "If the number of elements increase by 3, it increases the number of combinations of the second class of these elements 5 times. How many are the elements? • Phone numbers How many 7-digit telephone numbers can be compiled from the digits 0,1,2,..,8,9 that no digit is repeated? • Math logic There are 20 children in the group, each two children have a different name. Alena and John are among them. How many ways can we choose 8 children to be among the selected A) was John B) was John and Alena C) at least one was Alena, John D) maximum one wa • Possible combinations - word How many ways can the letters F, A, I, R be arranged? • Bridge cards How many bridge hands are possible containing 4 spades,6 diamonds, 1 club, and 2 hearts?", "first case, we do not use 0; the five-digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5P5 ways. (ii) And, in the second case, we do not use 3; the five-digit number can be formed (from the digit 0, 1, 2, 4, 5) in 6 ! −5! × 2 = 480 ways. The total number of such 5 digit number = 5P5 + (5P54P4) = 120 + 96 = 216. ### Practise Question 1. In how many ways can a committee of 5 be made out of 6 men and 4 women containing at least one woman? (A) 246 (B) 222 (C) 186 (D) None of these 2. How many numbers greater than 10 lakhs can be formed from 2, 3,0, 3,4,2,3? (A) 420 (B) 360 (C) 400 (D) 300 ### Conclusion Permutation and combination are explained elaborately in this article, along with the difference between them. We have discussed both the topics here with their formulas, examples and solved questions. With this students will be able when to use permutation and when to use combination formula in a question. Students can also work on Permutations and Combinations of different questions to enhance their knowledge and understanding of this chapter. See More ## JEE Main Important Dates View All Dates JEE Main", "# Questionbank: Permutation and Probability You are here: Home CAT Questionbank CAT Quant Permutation and Probability Question 11 The best combinatorics questions are the ones that involve a bit of Number Systems! ## Counting 5 Digit Numbers Q.11: How many numbers of up to 5 digits can be created using the digits 1, 2, 3 and 5 each at least once such that they are a multiple of 15? 1. 24 2. 18 3. 15 4. 12 • Correct Answer Choice D. 12 ## Explanatory Answer Click to watch video solution Click to view the explanation as a slide show ## Detailed Solution For a number to be a multiple of 15, it has to be a multiple of 3 and of 5. So, the last digit has to be 5 and the sum of digits should be a multiple of 3. We can have either 4–digit or 5–digit numbers. If we have a 4–digit number, sum of the digits will be 1 + 2 + 3 + 5 = 11. No 4–digit number formed with digits 1, 2, 3, 5 exactly once can be a multiple of 3. So, there is no possible 4–digit number. Now, in any 5 digit number, we will have 1, 2, 3, 5 once and one of these 4 digits repeating once.", "# Permutation Problems #2 • October 23rd 2009, 02:03 PM saberteeth Permutation Problems #2 (Hi), How do you solve these? 1) A number of 4 different digits is formed using the digits 1,2,3,4,5,6,7. How many numbers can be formed that are greater than 3400? 2) How many numbers of different digits lying between 100 and 1000 can be formed with the digits 2,3,4,0,8,9? • October 23rd 2009, 03:08 PM Soroban Hello, saberteeth! We have to \"talk\" our way through these . . . Quote: 1) A number of 4 different digits is formed using the digits {1,2,3,4,5,6,7}. How many numbers can be formed that are greater than 3400? There are two cases to be considered: . . [1] The first digit is 3. . . [2] The first digit is 4 or more. [1] The 1st digit is 3: . $3\\:\\_\\:\\_\\:\\_$ . . .The 2nd digit must be {4,5,6,7} . . . 4 choices. . . .The 3rd digit has 5 choices. . . .The 4th digit has 4 choices. There are: . $4\\cdot5\\cdot4 \\:=\\:{\\color{blue}80}$ numbers greater than 3400 which begin with 3 [2] The 1st digit is 4 or more. . . .The 1st digit must be {4,5,6,7} . . . 4 choices. . . .The 2nd digit has 6 choices. . . .The 3rd digit has" ]

[ "have two choices for the units digit. For each such choice, you can use any of the other four digits as the tens digit. Therefore, there are $4 \\cdot 2 = 8$ two-digit even numbers that can be formed using only the digits $3, 4, 5, 6, 7$.", "the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \\cdot 10 \\cdot 5 = 450$. $|A_1 \\cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \\cap A_2| = 4 \\cdot 10 = 40$. $|A_1 \\cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \\cap A_3| = 4 \\cdot 5 = 20$. $|A_2 \\cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \\cap A_3| = 9 \\cdot 5 = 45$. $|A_1 \\cap A_2 \\cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \\cap A_2 \\cap A_3| = 4$. Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is \\begin{align*} |A_1 \\cup A_2 \\cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2|", "# Overpowering tens unit Discrete Mathematics Level 1 How many 2 digit numbers are there, such that the units digit is strictly smaller than the tens digit? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "$3$. The only one-digit positive integer that is divisible by $6$ is $6$ itself, so the number must have at least two digits. Two-digit numbers: If the units digit is $2$, the tens digit must have remainder $1$ when divided by $3$. Hence, the tens digit must be $1$ or $4$. If the units digit is $4$, the tens digit have remainder $2$ when divided by $3$. Hence, the tens digit must be $2$ or $5$. Therefore, there are four two-digit numbers divisible by $6$ that can be formed using the digits $1, 2, 3, 4, 5$ without repetition. They are $12$, $24$, $42$, $54$. Three-digit numbers: If the units digit is $2$, the sum of the hundreds digit and tens digit must have remainder $1$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be at least $1 + 3 = 4$ and at most $5 + 4 = 9$, the only possibilities are that the sum of the hundreds digit and tens digit is $4$ or $7$. Since digits cannot be repeated, the only way to obtain $4$ is to use the digits $1$ and $3$ in either order, and the only way to obtain $7$ is to use the digits $3$ and $4$ in either order. Hence, there are", "picking the 3rd digit, it's less straightforward. You have 10 choices for the 3rd, and 9 choices for the 2nd. But you run into problems when picking the 1st digit, as your number of choices varies depending on whether 0 has already been picked. If it has, you have 8 choices. If it has not been picked, you have 7 choices, since it cannot be digit 2, digit 3, or 0. The main restriction that we need to consider is that the hundreds digit cannot be zero. By choosing the hundreds digit first, we handle this restriction immediately, after which our choices depend only on which digits have already been selected. The restriction that the hundreds digit may not be zero means we have nine choices for the hundreds digit. Since zero may be used in the tens place, we have nine choices for the tens digit (any digit except the hundreds digit). Once the hundreds and tens digits have been selected, we are left with eight choices for the units digits (any digit except the hundreds and tens digits). Therefore, there are $9 \\cdot 9 \\cdot 8 = 648$ three-digit positive integers with distinct digits. If we do not start with the hundreds place, the number of choices we have depends on whether or not the number", "Posts: 51223 Re: The three-digit positive integer x has the hundreds, [#permalink] ### Show Tags 09 Dec 2014, 08:03 1 1 JoostGrijsen wrote: Bunuel wrote: mohnish104 wrote: The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y? A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$. I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$; $$\\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12$$; $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$. Similar questions to practice: k-and-l-are-each-four-digit-positive-integers-with-thousands-91004.html for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html the-function-f-is-defined-for-each-positive-three-digit-100847.html for-a-three-digit-number-xyz-where-x-y-and-z-are-the-59284.html Hope it helps. _________________ Director Status: Professional GMAT Tutor Affiliations: AB, cum laude, Harvard University (Class of '02) Joined: 10 Jul 2015 Posts: 671 Location: United States (CA) Age: 39 GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: Q168 V169", "For units place, We have $$4$$ choices i.e $$2,3,5,7$$ For hundreds place, We have $$8$$ choices $$[10(0 \\quad to \\quad 9)-1$$(Hundreds place cannot be $$0$$)-$$1$$(Digit already used for units place)] For Tens place, We have $$8$$ choices $$[10(0 \\quad to \\quad 9)-1$$(Digit already used for Hundreds place)-$$1$$(Digit already used for units place)] Total Number of 3 digits such that units place of a the number is prime number and having distinct digits:$$8*8*4=256$$ _________________ Good, good Let the kudos flow through you Re: How many 3-digit numbers can be formed so that the units place of a th &nbs [#permalink] 30 Oct 2018, 10:23 Display posts from previous: Sort by # How many 3-digit numbers can be formed so that the units place of a th new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.", "+0 # Number Theory 0 208 1 N = 1991*1993*1995*1997*1999*2001*2003. What is the sum of the hundreds, tens and units digits of N? Jul 4, 2021 #1 +497 0 Since we need to compute the last 3 digits, we can take the whole expression mod 1000: $$1991\\cdot1993\\cdot1995\\cdot1997\\cdot1999\\cdot2001\\cdot2003 (\\text{mod 1000})\\\\ =-9\\cdot-7\\cdot-5\\cdot-3\\cdot-1\\cdot1\\cdot3 (\\text{mod 1000})\\\\ =-2835 (\\text{mod 1000})\\\\ =165(\\text{mod 1000})$$ The sum of the hundreds, tens, and units digit is $$1+6+5=\\boxed{12}$$ Jul 4, 2021", "when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 3 + 4 = 8$ and at most $3 + 4 + 5 = 12$, the sum of the thousands digit, hundreds digit, and tens digit must be $10$. Since digits cannot be repeated, the only way to obtain a sum of $10$ is to use the digits $1$, $4$, and $5$ in some order. There are $3! = 6$ such orders. Hence, there are six four-digit numbers divisible by $6$ with units digit $2$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1452$, $1542$, $4152$, $4512$, $5142$, and $5412$. If the units digit is $4$, the remainder of the sum of the thousands digit, hundreds digit, and tens digit must be $2$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 2 + 3 = 6$ and at most $2 + 3 + 5 = 10$, the sum of the thousands digit, hundreds digit, and tens digit must be $8$. Since digits cannot be repeated, the only way to obtain a sum of $8$ is to use the digits $1$, $2$, and $5$ in", "# A number is composed of 4 thousands and 21 tens. What's the number? Jan 6, 2018 4210 #### Explanation: A thousand =1000 Since there are 4 thousands, we multiply it by 4 to get $4000$ $4 \\cdot 1000 = 4000$ tens = 10 Since there are 21 tens, we multiply to get the total amount so: $21 \\cdot 10 = 210$ $4000 + 210 = 4210$", "contains a zero. We consider cases: 1. There is a zero in the units place: This leaves us with nine choices for the tens place (any digit except zero) and eight choices for the hundreds place (any digit except zero and the digit in the tens place). Hence, there are $1 \\cdot 9 \\cdot 8 = 72$ such numbers. 2. There is a zero in the tens place: This leaves us with nine choices for the units digit (any digit except zero) and eight choices for the hundreds place (any digit except zero and the digit in the units place). Hence, there are $9 \\cdot 1 \\cdot 8 = 72$ such numbers. 3. There is not a zero in either the units place or tens place: This leaves with nine choices for the units place, eight choices for the tens place, and seven choices for the hundreds place. Hence, there are $9 \\cdot 8 \\cdot 7 = 504$ such numbers. Since these cases are mutually exclusive and exhaustive, there are $72 + 72 + 504 = 648$ three-digit positive integers with distinct digits, as we found above without doing tedious casework.", "# Definition:Dozen ## Definition A dozen is another name for a set whose cardinality is $12$. ## Linguistic Note The word dozen derives from the French douzaine, which derives from the word douze, the French for $12$. The word is falling into disuse, but lives on in the rhetorical flourish where dozens is used to mean an unspecified large number.", "The hundreds' digit of 5^n is 6. $$5^3 = 125$$ $$5^4 = 625$$ $$5^5 = 3125$$ $$5^6 = 15625$$ So, for every even power of 5, hundred’s digit = 6 —> Possible values of n = {4, 6, 8, 10, 12, . . . } —> Insufficient Combining (1) & (2), Possible values of n = {4, 14, 24, . . . . } —> Insufficient Option E Posted from my mobile device Director Joined: 07 Mar 2019 Posts: 902 Location: India GMAT 1: 580 Q43 V27 WE: Sales (Energy and Utilities) Re: If n is a positive integer greater than 2, what is the value of n ? [#permalink] ### Show Tags 17 Jan 2020, 10:22 1 If n is a positive integer greater than 2, what is the value of n ? n > 2, n = ? (1) The tens' digit of $$11^n$$ is 4. $$11^4 = 14641$$ for n = 4 $$11^14 = ... 41$$ for n = 14 (Pattern of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 0 at tens' place is repeated for powers 1, 2, 3, 4, 5, 6, 7, 8, 9 & 0 respectively and 11 to 20 and further ..) INSUFFICIENT. (2) The hundreds' digit of $$5^n$$ is 6. $$5^4 = 625$$ for n = 4", "three digit number, the tens digit is half the hundreds digit. The ones digit is $1$ less than the tens digit. If the sum of the digits is $15$, what is the number. Solution Let $x$ be hundreds digit. The tens digit is half of it, so it’s $\\frac{1}{2}x$. Then, the ones digit is $1$ less than the tens digit or $\\frac{1}{2}x - 1$. Since the sum of the three digits is $15$, we can set up the equation $x + \\frac{1}{2}x + (\\frac{1}{2}x - 1) = 15$. Multiplying both sides by $2$, we have $2x + x + (x - 2) = 30$. That gives us $4x - 2 = 30$, $4x = 32$ and $x = 8$. Therefore, the tens digit is $\\frac{1}{2}(8) = 4$ and the ones digits is $4-1 = 3$. So, the number is $843$. The tens digit $4$ is half of the hundreds digit $8$. The ones digit is $3$, one less than the tens digit. PROBLEM 12 The sum of two numbers is $61$. Three times the smaller is $13$ less than the larger. Solution Let $x$ be the smaller number and $61-x$ be the larger number. Three times the smaller, $3x$ is $13$ less than the larger. This means that if we add $13$ to $3x$, it will be equal", "thousand place is given as: $$4!(1+2+4+5) \\cdot 10^4$$ Similarly, suppose we fix some digit in the thousand place, then we have $$3 \\cdot 3!$$ way of arranging the rest of numbers (remember we can't keep zero in first place). Hence the sum of no.s in thousand place is given as: $$3 \\cdot 3! ( 1 + 2 + 4 +5) \\cdot 10^3$$ And, similar results for hundreds , tens and ones place. Now, to get the whole number, we just need to add up the parts of it's expanded form. Hence, $$\\text{number} = 4!(1+2+4+5) \\cdot 10^4 + 3 \\cdot 3! (1+2+4+5) \\cdot 10^3 + 3 \\cdot 3! (1+2+4+5) \\cdot 10^2 + 3 \\cdot 3! (1+2+4+5) \\cdot 10^1 + 3 \\cdot 3! (1+2+4+5) \\cdot 10^0 = 3119976$$", "different ways to solve a division problem. Example 1 In this example, the first non 0 digit after the decimal point is 3 so the decimal becomes 3. 1 rad = 1/2pi tr = approx. The windows calculator can do literally, thousands of. It can estimate the nth root of a large number in any number base up to base 2047. In case of ties among the five-digit numbers, use additional columns to the. 30 Million: 0 WED 05/06:$1. Cooper's computer took 39 days of continuous calculation to verify the prime status of the number, which has over 17 million digits and was discovered January 25. 1 one tenth (1 in the tenths place) 1 one (1 in the ones or units place) 10 ten (1 in the tens place) 100 one hundred (1 in the hundreds. If you're over the nisab threshold, give a minimum of 2. There are 16 ounces in a pound. Cities by ZIP Code™ For more rapid delivery, please use the recommended or recognized city names whenever possible for this ZIP Code ™. Please check your connection and try running the trinket again. The national average has not been this cheap since December 2016. The Power of Compounding - $1 Million Now or Penny Doubling for a Month by Andy Saturday Monday 29", "• Place value (referring to the unit value of each digit in a given number) • Subtraction • Units of ones, tens, hundreds, thousands (referring to place value; 10 ones is the same as 1 unit of ten) • Renaming, changing (instead of carrying or borrowing, e.g., a group of 10 ones is renamed a ten when the ones are bundled and moved from the ones to the tens place; if using $1 bills, they may be changed for a$10 bill when there are enough) ## Materials List The following materials list will be used for the entire four weeks: Materials List.", "positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \\cup A_2 \\cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is $$|A_1 \\cup A_2 \\cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A \\cap A_3| - |A_2 \\cap A_3| + |A_1 \\cap A_2 \\cap A_3|$$ $|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \\cdot 10 \\cdot 10 = 400$. $|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \\cdot 5 \\cdot 10 = 450$. $|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose", "# Identifying Place Values The system that we use to represent numbers is wonderfully efficient and simple. Large numbers can be represented with very few symbols in a neat and organized way. The method rests on two concepts, base and place value, which are briefly discussed in the next few paragraphs. The concepts are introduced using base three for simplicity. Then, the concepts are extended to our base ten number system. The concept of base is best represented by grouping. Imagine a factory where objects are being packaged for shipment around the world. Workers are instructed to ‘bundle things up’ every time they get a group of three. Here are the names for the packaging that they use: • a set of three objects is called a packet • a set of three packets (which is $\\,3\\cdot 3 = 9\\,$ objects) is called a box • a set of three boxes (which is $\\,3\\cdot 9 = 27\\,$ objects) is called a carton Notice that a packet holds $\\,3^1 = 3\\,$ objects; a box holds $\\,3^2 = 9\\,$ objects; and a carton holds $\\,3^3 = 27\\,$ objects. Suppose a shipment of $\\,46\\,$ objects is to go out to Lenox, Massachusetts. How would these $\\,46\\,$ objects get bundled? First, bundle up one carton, leaving $\\,46 - 27 = 19\\,$ objects", "IDENTIFYING PLACE VALUES The system that we use to represent numbers is wonderfully efficient and simple. Large numbers can be represented with very few symbols in a neat and organized way. The method rests on two concepts, base and place value, which are briefly discussed in the next few paragraphs. The concepts are introduced using base three for simplicity. Then, the concepts are extended to our base ten number system. The concept of base is best represented by grouping. Imagine a factory where objects are being packaged for shipment around the world. Workers are instructed to “bundle things up” every time they get a group of three. Here are the names for the packaging that they use: • a set of three objects is called a packet • a set of three packets (which is $\\,3\\cdot 3 = 9\\,$ objects) is called a box • a set of three boxes (which is $\\,3\\cdot 9 = 27\\,$ objects) is called a carton Notice that a packet holds $\\,3^1 = 3\\,$ objects; a box holds $\\,3^2 = 9\\,$ objects; and a carton holds $\\,3^3 = 27\\,$ objects. Suppose a shipment of $\\,46\\,$ objects is to go out to Lenox, Massachusetts. How would these $\\,46\\,$ objects get bundled? First, bundle up one carton, leaving $\\,46 - 27 = 19\\,$ objects remaining." ]

[ "have two choices for the units digit. For each such choice, you can use any of the other four digits as the tens digit. Therefore, there are $4 \\cdot 2 = 8$ two-digit even numbers that can be formed using only the digits $3, 4, 5, 6, 7$.", "# Overpowering tens unit Discrete Mathematics Level 1 How many 2 digit numbers are there, such that the units digit is strictly smaller than the tens digit? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "| {4, 4, 1} | {4, 4, 2} | {4, 4, 3} | {4, 4, 4} | {4, 4, 5} | {4, 5, 1} | {4, 5, 2} | {4, 5, 3} | {4, 5, 4} | {4, 5, 5} |. These are ALL the permutations with repeats under 500. If you count the bold ones you should get 40. Guest Apr 25, 2018 edited by Guest Apr 25, 2018 edited by Guest Apr 25, 2018 edited by Guest Apr 25, 2018 #3 +101733 +2 Using the digits 1, 2, 3, 4, 5, how many even three-digit numbers less than 500 can be formed if each digit can be used more than once? The hundreds digit can be 1,2,3 or 4 that is 4 choice (less than 500) the tens has 5 choices. The units digit has to be even so that is 2 choices so there are 4*5*2= 40 possible numbers Apr 25, 2018", "$3$. The only one-digit positive integer that is divisible by $6$ is $6$ itself, so the number must have at least two digits. Two-digit numbers: If the units digit is $2$, the tens digit must have remainder $1$ when divided by $3$. Hence, the tens digit must be $1$ or $4$. If the units digit is $4$, the tens digit have remainder $2$ when divided by $3$. Hence, the tens digit must be $2$ or $5$. Therefore, there are four two-digit numbers divisible by $6$ that can be formed using the digits $1, 2, 3, 4, 5$ without repetition. They are $12$, $24$, $42$, $54$. Three-digit numbers: If the units digit is $2$, the sum of the hundreds digit and tens digit must have remainder $1$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be at least $1 + 3 = 4$ and at most $5 + 4 = 9$, the only possibilities are that the sum of the hundreds digit and tens digit is $4$ or $7$. Since digits cannot be repeated, the only way to obtain $4$ is to use the digits $1$ and $3$ in either order, and the only way to obtain $7$ is to use the digits $3$ and $4$ in either order. Hence, there are", "formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any number? pls help on the problem above, thanks. $N=5 \\cdot \\frac{8!}{(8-2)!}=5 \\cdot 8 \\cdot 7 =280$", "# Permutation Problems #2 • October 23rd 2009, 02:03 PM saberteeth Permutation Problems #2 (Hi), How do you solve these? 1) A number of 4 different digits is formed using the digits 1,2,3,4,5,6,7. How many numbers can be formed that are greater than 3400? 2) How many numbers of different digits lying between 100 and 1000 can be formed with the digits 2,3,4,0,8,9? • October 23rd 2009, 03:08 PM Soroban Hello, saberteeth! We have to \"talk\" our way through these . . . Quote: 1) A number of 4 different digits is formed using the digits {1,2,3,4,5,6,7}. How many numbers can be formed that are greater than 3400? There are two cases to be considered: . . [1] The first digit is 3. . . [2] The first digit is 4 or more. [1] The 1st digit is 3: . $3\\:\\_\\:\\_\\:\\_$ . . .The 2nd digit must be {4,5,6,7} . . . 4 choices. . . .The 3rd digit has 5 choices. . . .The 4th digit has 4 choices. There are: . $4\\cdot5\\cdot4 \\:=\\:{\\color{blue}80}$ numbers greater than 3400 which begin with 3 [2] The 1st digit is 4 or more. . . .The 1st digit must be {4,5,6,7} . . . 4 choices. . . .The 2nd digit has 6 choices. . . .The 3rd digit has", "picking the 3rd digit, it's less straightforward. You have 10 choices for the 3rd, and 9 choices for the 2nd. But you run into problems when picking the 1st digit, as your number of choices varies depending on whether 0 has already been picked. If it has, you have 8 choices. If it has not been picked, you have 7 choices, since it cannot be digit 2, digit 3, or 0. The main restriction that we need to consider is that the hundreds digit cannot be zero. By choosing the hundreds digit first, we handle this restriction immediately, after which our choices depend only on which digits have already been selected. The restriction that the hundreds digit may not be zero means we have nine choices for the hundreds digit. Since zero may be used in the tens place, we have nine choices for the tens digit (any digit except the hundreds digit). Once the hundreds and tens digits have been selected, we are left with eight choices for the units digits (any digit except the hundreds and tens digits). Therefore, there are $9 \\cdot 9 \\cdot 8 = 648$ three-digit positive integers with distinct digits. If we do not start with the hundreds place, the number of choices we have depends on whether or not the number", "# multiples of five Level pending how many four digit numbers can be formed using 0,1,2,3,4 which are multiples of 5 ×", "Question 64 # A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is Solution The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once. The different possibilities include : Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are : $$\\frac{4!}{3!}$$ = 4. Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are : $$\\frac{4!}{2!\\cdot2!}=\\ 6$$ Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are : $$\\frac{4!}{3!}$$ = 4 Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ A total of 12 + 12 + 12 + 4 + 6 + 4 = 50", "Posts: 51223 Re: The three-digit positive integer x has the hundreds, [#permalink] ### Show Tags 09 Dec 2014, 08:03 1 1 JoostGrijsen wrote: Bunuel wrote: mohnish104 wrote: The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y? A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$. I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$; $$\\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12$$; $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$. Similar questions to practice: k-and-l-are-each-four-digit-positive-integers-with-thousands-91004.html for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html the-function-f-is-defined-for-each-positive-three-digit-100847.html for-a-three-digit-number-xyz-where-x-y-and-z-are-the-59284.html Hope it helps. _________________ Director Status: Professional GMAT Tutor Affiliations: AB, cum laude, Harvard University (Class of '02) Joined: 10 Jul 2015 Posts: 671 Location: United States (CA) Age: 39 GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: Q168 V169", "+0 # Number Theory 0 208 1 N = 1991*1993*1995*1997*1999*2001*2003. What is the sum of the hundreds, tens and units digits of N? Jul 4, 2021 #1 +497 0 Since we need to compute the last 3 digits, we can take the whole expression mod 1000: $$1991\\cdot1993\\cdot1995\\cdot1997\\cdot1999\\cdot2001\\cdot2003 (\\text{mod 1000})\\\\ =-9\\cdot-7\\cdot-5\\cdot-3\\cdot-1\\cdot1\\cdot3 (\\text{mod 1000})\\\\ =-2835 (\\text{mod 1000})\\\\ =165(\\text{mod 1000})$$ The sum of the hundreds, tens, and units digit is $$1+6+5=\\boxed{12}$$ Jul 4, 2021", "the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \\cdot 10 \\cdot 5 = 450$. $|A_1 \\cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \\cap A_2| = 4 \\cdot 10 = 40$. $|A_1 \\cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \\cap A_3| = 4 \\cdot 5 = 20$. $|A_2 \\cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \\cap A_3| = 9 \\cdot 5 = 45$. $|A_1 \\cap A_2 \\cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \\cap A_2 \\cap A_3| = 4$. Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is \\begin{align*} |A_1 \\cup A_2 \\cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2|", "For units place, We have $$4$$ choices i.e $$2,3,5,7$$ For hundreds place, We have $$8$$ choices $$[10(0 \\quad to \\quad 9)-1$$(Hundreds place cannot be $$0$$)-$$1$$(Digit already used for units place)] For Tens place, We have $$8$$ choices $$[10(0 \\quad to \\quad 9)-1$$(Digit already used for Hundreds place)-$$1$$(Digit already used for units place)] Total Number of 3 digits such that units place of a the number is prime number and having distinct digits:$$8*8*4=256$$ _________________ Good, good Let the kudos flow through you Re: How many 3-digit numbers can be formed so that the units place of a th &nbs [#permalink] 30 Oct 2018, 10:23 Display posts from previous: Sort by # How many 3-digit numbers can be formed so that the units place of a th new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.", "four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $2$. They are $132$, $312$, $342$, and $432$. If the units digit is $4$, then the sum of the hundreds digit and tens digit must have remainder $2$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be at least $1 + 2 = 3$ and at most $3 + 5 = 8$, the sum of the hundreds digit and tens digit must be $5$ or $8$. Since digits cannot be repeated, the only way to obtain $5$ is to use the digits $2$ and $3$ in either order, and the only way to obtain $8$ is to use the digits $3$ and $5$ in either order. Hence, there are also four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $4$. They are $234$, $324$, $354$, $534$. Therefore, there are a total of eight three-digit numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. Four-digit numbers: If the units digit is $2$, then the sum of the thousands digit, hundreds digit, and tens digit must have remainder $1$", "three digit number, the tens digit is half the hundreds digit. The ones digit is $1$ less than the tens digit. If the sum of the digits is $15$, what is the number. Solution Let $x$ be hundreds digit. The tens digit is half of it, so it’s $\\frac{1}{2}x$. Then, the ones digit is $1$ less than the tens digit or $\\frac{1}{2}x - 1$. Since the sum of the three digits is $15$, we can set up the equation $x + \\frac{1}{2}x + (\\frac{1}{2}x - 1) = 15$. Multiplying both sides by $2$, we have $2x + x + (x - 2) = 30$. That gives us $4x - 2 = 30$, $4x = 32$ and $x = 8$. Therefore, the tens digit is $\\frac{1}{2}(8) = 4$ and the ones digits is $4-1 = 3$. So, the number is $843$. The tens digit $4$ is half of the hundreds digit $8$. The ones digit is $3$, one less than the tens digit. PROBLEM 12 The sum of two numbers is $61$. Three times the smaller is $13$ less than the larger. Solution Let $x$ be the smaller number and $61-x$ be the larger number. Three times the smaller, $3x$ is $13$ less than the larger. This means that if we add $13$ to $3x$, it will be equal", "# How many numbers can be formed? [closed] How many numbers can be formed from 1, 2, 3, 4, 5, ( without repetition), when the digit at the unit's place must be greater than that in the ten's place? ## closed as off-topic by jvdhooft, Théophile, M. Winter, Leucippus, ShaileshJun 19 '18 at 0:31 This question appears to be off-topic. The users who voted to close gave this specific reason: • \"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.\" – jvdhooft, Théophile, M. Winter, Leucippus, Shailesh If this question can be reworded to fit the rules in the help center, please edit the question. • Please have a look at how to ask a good question. – Théophile Jun 18 '18 at 22:25 • You can explicitly count this yourself on paper. If there are too many, you can try with $1,2,3,4$ or $1,2,3$ and see if you notice a pattern... – Jair Taylor Jun 18 '18 at 22:35 total methods without any restriction $= 5\\cdot4\\cdot3\\cdot2\\cdot1 = 120$ Ex$:$ $12$ & $21$ only one", "when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 3 + 4 = 8$ and at most $3 + 4 + 5 = 12$, the sum of the thousands digit, hundreds digit, and tens digit must be $10$. Since digits cannot be repeated, the only way to obtain a sum of $10$ is to use the digits $1$, $4$, and $5$ in some order. There are $3! = 6$ such orders. Hence, there are six four-digit numbers divisible by $6$ with units digit $2$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1452$, $1542$, $4152$, $4512$, $5142$, and $5412$. If the units digit is $4$, the remainder of the sum of the thousands digit, hundreds digit, and tens digit must be $2$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 2 + 3 = 6$ and at most $2 + 3 + 5 = 10$, the sum of the thousands digit, hundreds digit, and tens digit must be $8$. Since digits cannot be repeated, the only way to obtain a sum of $8$ is to use the digits $1$, $2$, and $5$ in", "# Unit 5 Family Materials Numbers to 1,000 ### Numbers to 1,000 In this unit, students extend their understanding of the base-ten system to include numbers to 1,000. ### Section A: The Value of Three Digits In this section, the unit of a hundred is introduced. Students begin by looking at the large square base-ten block, and its corresponding base-ten drawing, to visualize 100, and to establish that 1 hundred equals 10 tens, which equals 100 ones. After students develop an understanding of a hundred as a unit, students learn that the digits in three-digit numbers represent amounts of hundreds, tens, and ones. Students read and write three-digit numbers in different forms, including using base-ten numerals, number names, and expanded form. Students write expressions and equations based on the base-ten blocks and base-ten drawings that they see. They recognize that the value of the digits in a three-digit number is revealed when using the fewest number of blocks to represent the number. For example, the picture shows 2 hundreds, 11 tens, and 12 ones. However, students recognize that they will need to exchange 10 of the ones for a ten and 10 of the tens for a hundred to find the value of their number. After doing so, they recognize that they have 3 hundreds, 2 tens, and", "them are not divisible by $$5$$; with no digits being repeated? 10. How many even numbers of $$5$$ digits can be formed with the digits $$1, 2, 3, 4, 5$$? 11. How many numbers less than $$1000$$ and divisible by $$5$$ can be formed in which no digit repeats. 12. How many numbers between $$100$$ and $$999$$ can be formed with the digits $$0, 4, 5, 6, 7, 8$$? How many of them are odd? 13. Find the number of even numbers that can be formed with the digits $$0, 1, 2, 3, 4$$; with no digits being repeated. 14. Find the number of numbers of six digits with the digits $$1, 2, 3, 4, 5, 6$$ in which $$5$$ always occupy in the tens place with no digits being repeated. 15. A number of four different digits is formed by using the digits $$1, 2, 3, 4, 5, 6, 7$$. (i) Find how many such numbers can be formed? (ii) How many of them are greater than 3400? 16. Find the number of numbers of $$4$$ digits formed with the digits $$1, 2, 3, 4, 5$$ in which $$3$$ occurs in the thousand’s place and $$5$$ occurs in the unit’s place. 17. Find the number of numbers of $$4$$ digits formed with the digits $$0, 1, 2,", "to ensure you get more than 400: first digit must be either 5, 6, 7, or 9. Once you use up that number, it doesn't matter what the other digits are, you will get a number greater than 400. So there are 5 possibilities for the second digit (any of the original six except for the one used in the first digit), and four for the third digit (any of the original six, except for the two that we used in the first and second digit). – Arturo Magidin Apr 16 '12 at 16:43 I'm assuming each digit chosen from $\\{ 2,3,5,6,7,9\\}$ must be distinct. For the first digit, you must choose a number $\\ge$ 4, so you have four choices. After that, any number will do, as it only needs to be $\\ge 0$. So you have $6-1=5$ choices, and similarly for the last digit you have $5-1=4$ choices, for a total of $4 \\times 5 \\times 4=80$ choices. The total number of choices is $6 \\times 5 \\times 4 = \\frac{6!}{3!}=120$. Therefore the probability of choosing a number greater than 400 is $\\frac{80}{120}=\\frac{2}{3}$." ]